Integrand size = 38, antiderivative size = 240 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}} \] Output:
(1/8+1/8*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+ c))^(1/2))/a^(5/2)/d+1/5*(A+I*B)/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/ 2)+1/30*(17*A+7*I*B)/a/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2)+1/60*(1 51*A+41*I*B)/a^2/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-1/60*(317*A+6 7*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(1/2)
Time = 2.67 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.90 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^2(c+d x) \left (\frac {15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (-i \cos (2 (c+d x))+\sin (2 (c+d x))) \tan (c+d x)}{\sqrt {i a \tan (c+d x)}}+\frac {2 i (340 i A-80 B+(149 A+19 i B) \tan (c+d x)+\cos (2 (c+d x)) (-460 i A+80 B+(466 A+86 i B) \tan (c+d x)))}{\sqrt {a+i a \tan (c+d x)}}\right )}{120 a^2 d \sqrt {\tan (c+d x)} (-i+\tan (c+d x))^2} \] Input:
Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^ (5/2)),x]
Output:
(Sec[c + d*x]^2*((15*Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d *x]])/Sqrt[a + I*a*Tan[c + d*x]]]*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x) ])*Tan[c + d*x])/Sqrt[I*a*Tan[c + d*x]] + ((2*I)*((340*I)*A - 80*B + (149* A + (19*I)*B)*Tan[c + d*x] + Cos[2*(c + d*x)]*((-460*I)*A + 80*B + (466*A + (86*I)*B)*Tan[c + d*x])))/Sqrt[a + I*a*Tan[c + d*x]]))/(120*a^2*d*Sqrt[T an[c + d*x]]*(-I + Tan[c + d*x])^2)
Time = 1.42 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4081, 27, 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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{3/2} (a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {a (11 A+i B)-6 a (i A-B) \tan (c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (11 A+i B)-6 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (11 A+i B)-6 a (i A-B) \tan (c+d x)}{\tan (c+d x)^{3/2} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (83 A+13 i B)-4 a^2 (17 i A-7 B) \tan (c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (83 A+13 i B)-4 a^2 (17 i A-7 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (83 A+13 i B)-4 a^2 (17 i A-7 B) \tan (c+d x)}{\tan (c+d x)^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (317 A+67 i B)-2 a^3 (151 i A-41 B) \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (317 A+67 i B)-2 a^3 (151 i A-41 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (317 A+67 i B)-2 a^3 (151 i A-41 B) \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{2 a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \int \frac {15 a^4 (i A+B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a^3 (317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{2 a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^3 (317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{2 a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^3 (317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{2 a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {\frac {\frac {-\frac {30 i a^5 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a^3 (317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{2 a^2}+\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {a^2 (151 A+41 i B)}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\frac {(15-15 i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^3 (317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{2 a^2}}{6 a^2}+\frac {a (17 A+7 i B)}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}\) |
Input:
Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) ,x]
Output:
(A + I*B)/(5*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(17* A + (7*I)*B))/(3*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + ((a^ 2*(151*A + (41*I)*B))/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((15 - 15*I)*a^(7/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x] ])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^3*(317*A + (67*I)*B)*Sqrt[a + I*a *Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(2*a^2))/(6*a^2))/(10*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 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[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1157 vs. \(2 (194 ) = 388\).
Time = 0.24 (sec) , antiderivative size = 1158, normalized size of antiderivative = 4.82
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1158\) |
| default | \(\text {Expression too large to display}\) | \(1158\) |
| parts | \(\text {Expression too large to display}\) | \(1188\) |
Input:
int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_R ETURNVERBOSE)
Output:
-1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(15*I*B*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*tan(d*x+c)^5+1268*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I *a)^(1/2)*tan(d*x+c)^4-1060*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I* a)^(1/2)*tan(d*x+c)^2-15*A*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*tan(d*x+ c)^5+908*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3 +15*I*B*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*tan(d*x+c)-4468*I*A*(a*tan( d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+60*B*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*tan(d*x+c)^4-5660*A*(a*tan(d*x+c)*(1+I*tan(d*x+ c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+268*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c) ))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4-60*I*A*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*tan(d*x+c)^2+90*A*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*tan(d*x+c) ^3+60*I*A*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*tan(d*x+c)^4+2940*I*A*(a* tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-60*B*2^(1/2)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (182) = 364\).
Time = 0.13 (sec) , antiderivative size = 529, normalized size of antiderivative = 2.20 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\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}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\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}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left ({\left (-463 i \, A + 83 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-269 i \, A + 19 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 20 \, {\left (-11 i \, A + 4 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (29 i \, A - 19 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\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}}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, al gorithm="fricas")
Output:
1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c) )*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*log((2*sqrt(1/2)*a^3*d*sqrt((I*A ^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I *d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) - 15*sqrt(1/ 2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((I*A^2 + 2 *A*B - I*B^2)/(a^5*d^2))*log(-(2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B ^2)/(a^5*d^2))*e^(I*d*x + I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + sqrt(2)*((-463*I*A + 83*B) *e^(8*I*d*x + 8*I*c) + (-269*I*A + 19*B)*e^(6*I*d*x + 6*I*c) - 20*(-11*I*A + 4*B)*e^(4*I*d*x + 4*I*c) + (29*I*A - 19*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*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)))/(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, al gorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),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 TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Ar gument Ty
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:
int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2) ),x)
Output:
int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2) ), x)
\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-\left (\int \frac {1}{\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}-2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2} i -\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}d x \right ) a -\left (\int \frac {1}{\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) b}{\sqrt {a}\, a^{2}} \] Input:
int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - (int(1/(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3 - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2*i - sqrt(tan (c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)),x)*a + int(1/(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt(tan(c + d*x))*s qrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x))*sqrt(tan(c + d *x)*i + 1)),x)*b))/(sqrt(a)*a**2)