Integrand size = 38, antiderivative size = 298 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {3 (-1)^{3/4} a^{5/2} (120 i A+121 B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}+\frac {(4+4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {a^2 (104 i A+107 B) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d} \] Output:
3/64*(-1)^(3/4)*a^(5/2)*(120*I*A+121*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+ c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+(4+4*I)*a^(5/2)*(I*A+B)*arctanh((1+I) *a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+1/64*a^2*(152*A-149* I*B)*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d+1/96*a^2*(104*I*A+107*B)* tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d-1/24*a^2*(8*A-11*I*B)*tan(d*x+ c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2)/d+1/4*I*a*B*tan(d*x+c)^(5/2)*(a+I*a*tan( d*x+c))^(3/2)/d
Time = 6.78 (sec) , antiderivative size = 563, normalized size of antiderivative = 1.89 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{4 d}+\frac {\frac {a (8 A-5 i B) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}{6 d}+\frac {-\frac {i a^4 (40 i A+43 B) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} (-1)^{3/4} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right )+\frac {5}{4} \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}+\frac {1}{2} i \sqrt {1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x)\right )}{4 d \sqrt {1+i \tan (c+d x)}}+\frac {a \left (-\frac {1}{4} i a^3 (8 A-5 i B)-\frac {1}{4} a^3 (40 i A+43 B)\right ) \left (-\frac {4 i \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)}}+\frac {4 i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}+\frac {i \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{\sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}\right )}{d}}{3 a}}{4 a} \] Input:
Integrate[Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d *x]),x]
Output:
(B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2))/(4*d) + ((a*(8*A - (5* I)*B)*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2))/(6*d) + (((-1/4*I)* a^4*((40*I)*A + 43*B)*Sqrt[a + I*a*Tan[c + d*x]]*((-3*(-1)^(3/4)*ArcSinh[( -1)^(1/4)*Sqrt[Tan[c + d*x]]])/4 + (5*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/4 + (I/2)*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(3/2)))/(d*Sqrt[1 + I*Tan[c + d*x]]) + (a*((-1/4*I)*a^3*(8*A - (5*I)*B) - (a^3*((40*I)*A + 43*B))/4)*(((-4*I)*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt [a + I*a*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]])/Sqrt[I*a*Tan[c + d*x]] + ((4*I )*a^(3/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]] *Sqrt[Tan[c + d*x]])/(Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + I*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + (I*Sqrt[a]*ArcSinh[Sqrt [I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) /(Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Tan[c + d*x]])))/d)/(3*a))/(4*a)
Time = 2.07 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.07, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 4077, 27, 3042, 4077, 27, 3042, 4080, 27, 3042, 4080, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^{3/2} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 4077 |
\(\displaystyle \frac {1}{4} \int \frac {1}{2} \tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^{3/2} (a (8 A-5 i B)+a (8 i A+11 B) \tan (c+d x))dx+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^{3/2} (a (8 A-5 i B)+a (8 i A+11 B) \tan (c+d x))dx+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \int \tan (c+d x)^{3/2} (i \tan (c+d x) a+a)^{3/2} (a (8 A-5 i B)+a (8 i A+11 B) \tan (c+d x))dx+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 4077 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \tan ^{\frac {3}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((88 A-85 i B) a^2+(104 i A+107 B) \tan (c+d x) a^2\right )dx-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((88 A-85 i B) a^2+(104 i A+107 B) \tan (c+d x) a^2\right )dx-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \tan (c+d x)^{3/2} \sqrt {i \tan (c+d x) a+a} \left ((88 A-85 i B) a^2+(104 i A+107 B) \tan (c+d x) a^2\right )dx-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 4080 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {\int -\frac {3}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (a^3 (104 i A+107 B)-a^3 (152 A-149 i B) \tan (c+d x)\right )dx}{2 a}+\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (a^3 (104 i A+107 B)-a^3 (152 A-149 i B) \tan (c+d x)\right )dx}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} \left (a^3 (104 i A+107 B)-a^3 (152 A-149 i B) \tan (c+d x)\right )dx}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 4080 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((152 A-149 i B) a^4+3 (120 i A+121 B) \tan (c+d x) a^4\right )}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((152 A-149 i B) a^4+3 (120 i A+121 B) \tan (c+d x) a^4\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((152 A-149 i B) a^4+3 (120 i A+121 B) \tan (c+d x) a^4\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 4084 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {512 a^4 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-3 a^3 (120 A-121 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {512 a^4 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-3 a^3 (120 A-121 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {-3 a^3 (120 A-121 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {1024 i a^6 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\frac {(512-512 i) a^{9/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-3 a^3 (120 A-121 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\frac {(512-512 i) a^{9/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {3 a^5 (120 A-121 i B) \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\frac {(512-512 i) a^{9/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {6 a^5 (120 A-121 i B) \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {a^2 (107 B+104 i A) \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\frac {6 \sqrt [4]{-1} a^{9/2} (120 A-121 i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(512-512 i) a^{9/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a}-\frac {a^3 (152 A-149 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 (8 A-11 i B) \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )+\frac {i a B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{4 d}\) |
Input:
Int[Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x ]
Output:
((I/4)*a*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2))/d + (-1/3*(a^2 *(8*A - (11*I)*B)*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d + ((a^2 *((104*I)*A + 107*B)*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(2*d) - (3*(((6*(-1)^(1/4)*a^(9/2)*(120*A - (121*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a ]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((512 - 512*I)*a^(9 /2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Ta n[c + d*x]]])/d)/(2*a) - (a^3*(152*A - (149*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[ a + I*a*Tan[c + d*x]])/d))/(4*a))/6)/8
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan [e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && !LtQ[n, -1]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 741 vs. \(2 (240 ) = 480\).
Time = 0.32 (sec) , antiderivative size = 742, normalized size of antiderivative = 2.49
method | result | size |
derivativedivides | \(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-96 B \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-128 A \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+272 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+416 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+447 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -894 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+428 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-384 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -456 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +912 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-768 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +384 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -768 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(742\) |
default | \(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-96 B \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-128 A \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+272 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+416 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+447 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -894 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+428 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-384 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -456 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +912 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-768 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +384 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -768 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(742\) |
parts | \(\text {Expression too large to display}\) | \(942\) |
Input:
int(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_R ETURNVERBOSE)
Output:
1/384/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(-96*B*(I*a)^(1/2) *(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-128*A*(I* a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+2 72*I*B*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I* a)^(1/2)+416*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^ (1/2)*tan(d*x+c)+447*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan (d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a-894*I*B*(a*tan( d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+428*B*(I*a)^(1/2)* (-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-384*I*(I*a)^ (1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c))) ^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-456*A*ln(1/2*(2*I*a*tan(d*x+c )+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I* a)^(1/2)*a+912*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^ (1/2)-768*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a+384*(I*a)^(1/2)*2^(1/2)*ln(- (-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan (d*x+c))/(tan(d*x+c)+I))*a-768*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1 +I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*(-I*a)^(1/2))/(a*tan(d *x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1017 vs. \(2 (224) = 448\).
Time = 0.13 (sec) , antiderivative size = 1017, normalized size of antiderivative = 3.41 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, al gorithm="fricas")
Output:
1/384*(768*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log((I*sqr t(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*(( -I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^2)*sqrt(a/(e^(2*I*d*x + 2 *I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))* e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 768*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I* B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2* I*d*x + 2*I*c) + d)*log((-I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2) *d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) + 2*sqrt (2)*(13*(56*A - 65*I*B)*a^2*e^(7*I*d*x + 7*I*c) + 3*(504*A - 425*I*B)*a^2* e^(5*I*d*x + 5*I*c) + (1096*A - 1135*I*B)*a^2*e^(3*I*d*x + 3*I*c) + 3*(104 *A - 107*I*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt( (-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - 9*sqrt(-(14400*I *A^2 + 29040*A*B - 14641*I*B^2)*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4 *I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((-120*I*A - 1 21*B)*a^2*e^(2*I*d*x + 2*I*c) + (-120*I*A - 121*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1 )) + 2*I*sqrt(-(14400*I*A^2 + 29040*A*B - 14641*I*B^2)*a^5/d^2)*d*e^(I*...
Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(tan(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
Output:
Timed out
\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, al gorithm="maxima")
Output:
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^( 3/2), x)
Exception generated. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, al gorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeWarning, need to choose a branch for the root of a polynomial with par
Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] Input:
int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2), x)
Output:
int(tan(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2), x)
\[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4}d x \right ) b -\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}d x \right ) a +2 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}d x \right ) b i +2 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}d x \right ) a i +\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) a \right ) \] Input:
int(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
Output:
sqrt(a)*a**2*( - int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d *x)**4,x)*b - int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x) **3,x)*a + 2*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)* *3,x)*b*i + 2*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x) **2,x)*a*i + int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)* *2,x)*b + int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)* a)