Integrand size = 28, antiderivative size = 119 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {(2-2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \] Output:
(-2+2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)) ^(1/2))/d-2*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/3*(a+I*a*tan (d*x+c))^(3/2)/d/tan(d*x+c)^(3/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(276\) vs. \(2(119)=238\).
Time = 2.45 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.32 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a \left (-3 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))-\frac {3 i \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (i a \tan (c+d x))^{3/2} (-i+\tan (c+d x))}{\sqrt {a}}+\sqrt {1+i \tan (c+d x)} \left (\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (i a \tan (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{a}+a \left (-1-5 i \tan (c+d x)+4 \tan ^2(c+d x)\right )\right )\right )}{3 d \sqrt {1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(5/2),x]
Output:
(2*a*(-3*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Tan[c + d*x]^ (3/2)*(-I + Tan[c + d*x]) - ((3*I)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]] *(I*a*Tan[c + d*x])^(3/2)*(-I + Tan[c + d*x]))/Sqrt[a] + Sqrt[1 + I*Tan[c + d*x]]*((3*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a* Tan[c + d*x]]]*(I*a*Tan[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/a + a* (-1 - (5*I)*Tan[c + d*x] + 4*Tan[c + d*x]^2))))/(3*d*Sqrt[1 + I*Tan[c + d* x]]*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.55 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4031, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4031 |
| \(\displaystyle i \int \frac {(i \tan (c+d x) a+a)^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \int \frac {(i \tan (c+d x) a+a)^{3/2}}{\tan (c+d x)^{3/2}}dx-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4028 |
| \(\displaystyle i \left (2 i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (2 i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle i \left (\frac {4 a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle i \left (\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(5/2),x]
Output:
(-2*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2)) + I*(((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. 367 vs. \(2 (97 ) = 194\).
Time = 1.58 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.09
| method | result | size |
| derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (3 i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-3 \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+12 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+16 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(368\) |
| default | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (3 i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-3 \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+12 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+16 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(368\) |
Input:
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/6/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(3/2)*(3*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)^2-3*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)^2+12*I*(-I*a)^(1/2)*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 *tan(d*x+c)^2+16*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)+4*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c )))^(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. 395 vs. \(2 (89) = 178\).
Time = 0.12 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.32 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \, \sqrt {2} {\left (5 \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{6 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")
Output:
1/6*(4*sqrt(2)*(5*a*e^(5*I*d*x + 5*I*c) + 2*a*e^(3*I*d*x + 3*I*c) - 3*a*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)) - 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2* I*d*x + 2*I*c) + d)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) + sqrt(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e ^(-I*d*x - I*c)/a) + 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) - sqrt(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a ))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+I*a*tan(d*x+c))**(3/2)/tan(d*x+c)**(5/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(3/2)/tan(c + d*x)**(5/2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1008 vs. \(2 (89) = 178\).
Time = 0.24 (sec) , antiderivative size = 1008, normalized size of antiderivative = 8.47 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")
Output:
1/3*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((3*I + 3)*a*cos(3*d*x + 3*c) - (I + 1)*a*cos(d*x + c) + (3*I - 3)*a* sin(3*d*x + 3*c) - (I - 1)*a*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c ), -cos(2*d*x + 2*c) + 1)) + ((3*I - 3)*a*cos(3*d*x + 3*c) - (I - 1)*a*cos (d*x + c) - (3*I + 3)*a*sin(3*d*x + 3*c) + (I + 1)*a*sin(d*x + c))*sin(3/2 *arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*(2*(-(I + 1)*a*cos(2*d*x + 2*c)^2 - (I + 1)*a*sin(2*d*x + 2*c)^2 + (2*I + 2)*a*cos(2 *d*x + 2*c) - (I + 1)*a)*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*cos(2*d*x + 2*c)^2 - (I - 1)* a*sin(2*d*x + 2*c)^2 + (2*I - 2)*a*cos(2*d*x + 2*c) - (I - 1)*a)*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*arctan2(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 + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)...
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),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. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(5/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(5/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a \left (2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}+3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{2} d i \right )}{3 \tan \left (d x +c \right )^{2} d} \] Input:
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(5/2),x)
Output:
(2*sqrt(a)*a*(2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) + 3*int((sqrt(tan(c + d*x)) *sqrt(tan(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**2*d*i))/(3*tan (c + d*x)**2*d)