Integrand size = 28, antiderivative size = 156 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \] Output:
(-4-4*I)*a^(5/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)) ^(1/2))/d+4*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/3*I*a*(a+I*a *tan(d*x+c))^(3/2)/d/tan(d*x+c)^(3/2)-2/5*(a+I*a*tan(d*x+c))^(5/2)/d/tan(d *x+c)^(5/2)
Time = 3.65 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.95 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {a^2 \left (75 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (1+i \tan (c+d x)) \tan ^2(c+d x) \sqrt {i a \tan (c+d x)}-75 (-1)^{3/4} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {5}{2}}(c+d x) (-i+\tan (c+d x))+2 \sqrt {1+i \tan (c+d x)} \left (-30 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \tan ^2(c+d x) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}+a \left (-3-14 i \tan (c+d x)+49 \tan ^2(c+d x)+38 i \tan ^3(c+d x)\right )\right )\right )}{15 d \sqrt {1+i \tan (c+d x)} \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]
Output:
(a^2*(75*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*(1 + I*Tan[c + d* x])*Tan[c + d*x]^2*Sqrt[I*a*Tan[c + d*x]] - 75*(-1)^(3/4)*a*ArcSinh[(-1)^( 1/4)*Sqrt[Tan[c + d*x]]]*Tan[c + d*x]^(5/2)*(-I + Tan[c + d*x]) + 2*Sqrt[1 + I*Tan[c + d*x]]*(-30*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/S qrt[a + I*a*Tan[c + d*x]]]*Tan[c + d*x]^2*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + a*(-3 - (14*I)*Tan[c + d*x] + 49*Tan[c + d*x]^2 + (38* I)*Tan[c + d*x]^3))))/(15*d*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(5/2)*Sq rt[a + I*a*Tan[c + d*x]])
Time = 0.74 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4031, 3042, 4028, 3042, 4028, 3042, 4027, 218}
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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan (c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4031 |
| \(\displaystyle i \int \frac {(i \tan (c+d x) a+a)^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \int \frac {(i \tan (c+d x) a+a)^{5/2}}{\tan (c+d x)^{5/2}}dx-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4028 |
| \(\displaystyle i \left (2 i a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (2 i a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{\tan (c+d x)^{3/2}}dx-\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4028 |
| \(\displaystyle i \left (2 i a \left (2 i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (2 i a \left (2 i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle i \left (2 i a \left (\frac {4 a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle i \left (2 i a \left (\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]
Output:
(-2*(a + I*a*Tan[c + d*x])^(5/2))/(5*d*Tan[c + d*x]^(5/2)) + I*((-2*a*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2)) + (2*I)*a*(((2 + 2*I)*a ^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d *x]]])/d - (2*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])))
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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - b*d)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0 ] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Ta n[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Simp[a/(a*c - b*d) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d , e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^ 2, 0] && EqQ[m + n + 1, 0] && !LtQ[m, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (127 ) = 254\).
Time = 1.57 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.65
| method | result | size |
| derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (15 i \sqrt {2}\, \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+15 \sqrt {2}\, \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+60 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{3}-22 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+76 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}-6 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{15 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(414\) |
| default | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (15 i \sqrt {2}\, \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+15 \sqrt {2}\, \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+60 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{3}-22 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+76 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}-6 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{15 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(414\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(15*I*2^(1/2)*(I*a)^(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*tan(d*x+c)^3+15*2^(1/2)*(I*a)^(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*tan(d*x+c)^3+60*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*t an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c) ^3-22*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a )^(1/2)+76*(-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)^2-6*(-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)^(5/2)/(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. 466 vs. \(2 (118) = 236\).
Time = 0.08 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.99 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {8 \, \sqrt {2} {\left (-26 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 9 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 20 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 15 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 15 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{30 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")
Output:
-1/30*(8*sqrt(2)*(-26*I*a^2*e^(7*I*d*x + 7*I*c) + 9*I*a^2*e^(5*I*d*x + 5*I *c) + 20*I*a^2*e^(3*I*d*x + 3*I*c) - 15*I*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)) - 15*sqrt(32*I*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(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + 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)) + I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) + 15*sqrt(32*I*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(1/4*(4*sqrt( 2)*(a^2*e^(2*I*d*x + 2*I*c) + 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)) - I*sqrt(32*I*a^5/ d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^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)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)/tan(d*x+c)**(7/2),x)
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1284 vs. \(2 (118) = 236\).
Time = 0.29 (sec) , antiderivative size = 1284, normalized size of antiderivative = 8.23 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")
Output:
2/15*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((30*I - 30)*a^2*cos(3*d*x + 3*c) - (31*I - 31)*a^2*cos(d*x + c) - (30 *I + 30)*a^2*sin(3*d*x + 3*c) + (31*I + 31)*a^2*sin(d*x + c))*cos(3/2*arct an2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(30*I + 30)*a^2*cos(3*d* x + 3*c) + (31*I + 31)*a^2*cos(d*x + c) - (30*I - 30)*a^2*sin(3*d*x + 3*c) + (31*I - 31)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), -cos(2* d*x + 2*c) + 1)))*sqrt(a) + 15*(2*(-(I - 1)*a^2*cos(2*d*x + 2*c)^2 - (I - 1)*a^2*sin(2*d*x + 2*c)^2 + (2*I - 2)*a^2*cos(2*d*x + 2*c) - (I - 1)*a^2)* arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1) ^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^ (1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + ((I + 1)*a^2*cos(2*d*x + 2*c)^2 + (I + 1)*a^2*sin(2*d*x + 2*c)^2 - (2*I + 2)*a^2*cos(2*d*x + 2*c) + (I + 1)*a^2)*log(cos(d*x + c)^2 + sin(d* x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2* c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin (1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arct an2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)* sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x...
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="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 TypeDone
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(7/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(7/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a^{2} \left (-22 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-11 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -3 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}-30 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{3} d \right )}{15 \tan \left (d x +c \right )^{3} d} \] Input:
int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x)
Output:
(2*sqrt(a)*a**2*( - 22*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 11*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 3*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) - 30*int((sqrt(tan(c + d*x) )*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**3*d))/(15*tan (c + d*x)**3*d)