\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) [195]

Optimal result

Integrand size = 38, antiderivative size = 286 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(707 i A-317 B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}} \]

[Out]

(1/8+1/8*I)*(I*A+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+1/60*(707*I*A-3
17*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(1/2)+1/20*(89*A+39*I*B)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*
x+c)^(3/2)-1/60*(361*A+151*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(3/2)+1/5*(A+I*B)/d/tan(d*x+c)^(3/2)
/(a+I*a*tan(d*x+c))^(5/2)+1/30*(21*A+11*I*B)/a/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3677, 3679, 12, 3625, 211} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(-317 B+707 i A) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((1/8 + I/8)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) +
(A + I*B)/(5*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (21*A + (11*I)*B)/(30*a*d*Tan[c + d*x]^(3/2
)*(a + I*a*Tan[c + d*x])^(3/2)) + (89*A + (39*I)*B)/(20*a^2*d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) -
((361*A + (151*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Tan[c + d*x]^(3/2)) + (((707*I)*A - 317*B)*Sqrt[a
+ I*a*Tan[c + d*x]])/(60*a^3*d*Sqrt[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-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] && NeQ[c^2 + d^2, 0]

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[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
] /; FreeQ[{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]

Rule 3679

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] - Dist[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] /; FreeQ[{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]

Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a (13 A+3 i B)-4 a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {3}{4} a^2 (47 A+17 i B)-\frac {3}{2} a^2 (21 i A-11 B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{8} a^3 (361 A+151 i B)-\frac {3}{2} a^3 (89 i A-39 B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{15 a^6} \\ & = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{16} a^4 (707 i A-317 B)-\frac {3}{8} a^4 (361 A+151 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{45 a^7} \\ & = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(707 i A-317 B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {45 a^5 (A-i B) \sqrt {a+i a \tan (c+d x)}}{32 \sqrt {\tan (c+d x)}} \, dx}{45 a^8} \\ & = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(707 i A-317 B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}-\frac {(A-i B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3} \\ & = \frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(707 i A-317 B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {(i A+B) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d} \\ & = \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (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 )}{a^{5/2} d}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(707 i A-317 B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.77 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \sec ^2(c+d x) \left (\frac {15 \sqrt {2} (i A+B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \tan ^2(c+d x)}{\sqrt {i a \tan (c+d x)}}+\frac {\sec ^2(c+d x) (-174 i A+84 B+(747 i A-317 B) \cos (2 (c+d x))+(-493 i A+233 B) \cos (4 (c+d x))-780 A \sin (2 (c+d x))-340 i B \sin (2 (c+d x))+490 A \sin (4 (c+d x))+230 i B \sin (4 (c+d x)))}{\sqrt {a+i a \tan (c+d x)}}\right )}{120 a^2 d \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))^2} \]

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((-1/120*I)*Sec[c + d*x]^2*((15*Sqrt[2]*(I*A + B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c
+ d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Tan[c + d*x]^2)/Sqrt[I*a*Tan[c + d*x]] + (Sec[c + d*x]^2*((-1
74*I)*A + 84*B + ((747*I)*A - 317*B)*Cos[2*(c + d*x)] + ((-493*I)*A + 233*B)*Cos[4*(c + d*x)] - 780*A*Sin[2*(c
+ d*x)] - (340*I)*B*Sin[2*(c + d*x)] + 490*A*Sin[4*(c + d*x)] + (230*I)*B*Sin[4*(c + d*x)]))/Sqrt[a + I*a*Tan
[c + d*x]]))/(a^2*d*Tan[c + d*x]^(3/2)*(-I + Tan[c + d*x])^2)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1228 vs. \(2 (232 ) = 464\).

Time = 0.17 (sec) , antiderivative size = 1229, normalized size of antiderivative = 4.30

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1229\)
default	\(\text {Expression too large to display}\)	\(1229\)
parts	\(\text {Expression too large to display}\)	\(1273\)

[In]

int((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)*(15*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)^6-12260*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c))
)^(1/2)*tan(d*x+c)^3+15*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)^6+15*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+60*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)^3+60*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)^5-2940*I*B*
(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-60*I*B*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*t
an(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)^5-90*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)^4-1268*B
*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^5+2828*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(
d*x+c)))^(1/2)*tan(d*x+c)^5-90*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)^4-60*A*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))*2^(1/2)*a*tan(d*x+c)^3+9868*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan
(d*x+c)))^(1/2)*tan(d*x+c)^4+4468*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4+15*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*ta
n(d*x+c)^2+5660*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+640*I*A*(-I*a)^(1/2)*tan(d*x
+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-6020*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)
^2-480*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-160*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2))/a^3/tan(d*x+c)^(3/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (218) = 436\).

Time = 0.29 (sec) , antiderivative size = 588, normalized size of antiderivative = 2.06 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{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 (9 i \, d x + 9 i \, c\right )} - 2 \, 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 i \, \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 (9 i \, d x + 9 i \, c\right )} - 2 \, 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 i \, \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 (983 \, A + 463 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} - 2 \, {\left (272 \, A + 97 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, {\left (393 \, A + 163 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (381 \, A + 191 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (18 \, A + 13 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, 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 (9 i \, d x + 9 i \, c\right )} - 2 \, 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 )}} \]

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(15*sqrt(1/2)*(a^3*d*e^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqr
t((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*log((2*I*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^(9*I*d*x + 9*I*c) - 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*I*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)*((983*A + 463*I*B)*e^(10*I*d*x + 10*I*c) - 2*(272*A + 97*I*B)*e^(8*I*d*x + 8*I*c) - 3*(39
3*A + 163*I*B)*e^(6*I*d*x + 6*I*c) + (381*A + 191*I*B)*e^(4*I*d*x + 4*I*c) + 2*(18*A + 13*I*B)*e^(2*I*d*x + 2*
I*c) + 3*A + 3*I*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^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(5/2)), x)