\(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [105]

Optimal result

Integrand size = 36, antiderivative size = 211 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 A+151 i B}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(13 A+83 i B) \sqrt {a+i a \tan (c+d x)}}{30 a^3 d} \]

[Out]

1/8*(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+1/60*(41*A+151*I*B)/a^2/d/
(a+I*a*tan(d*x+c))^(1/2)+1/30*(13*A+83*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+1/5*(I*A-B)*tan(d*x+c)^3/d/(a+I*a*t
an(d*x+c))^(5/2)+1/30*(7*A+17*I*B)*tan(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3676, 3673, 3607, 3561, 212} \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(13 A+83 i B) \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {41 A+151 i B}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(-B+i A) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[In]

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((I*A - B)*Tan[c + d
*x]^3)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((7*A + (17*I)*B)*Tan[c + d*x]^2)/(30*a*d*(a + I*a*Tan[c + d*x])^(
3/2)) + (41*A + (151*I)*B)/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((13*A + (83*I)*B)*Sqrt[a + I*a*Tan[c + d*x
]])/(30*a^3*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3607

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a*f*m)), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^2(c+d x) \left (3 a (i A-B)+\frac {1}{2} a (A+11 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan (c+d x) \left (-a^2 (7 A+17 i B)+\frac {1}{4} a^2 (13 i A-83 B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(13 A+83 i B) \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {\int \frac {-\frac {1}{4} a^2 (13 i A-83 B)-a^2 (7 A+17 i B) \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 A+151 i B}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(13 A+83 i B) \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 A+151 i B}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(13 A+83 i B) \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {(A-i B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d} \\ & = \frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(7 A+17 i B) \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 A+151 i B}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(13 A+83 i B) \sqrt {a+i a \tan (c+d x)}}{30 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.88 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.97 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-240 a^{5/2} (A+5 i B) \tan ^2(c+d x)+240 a^{5/2} B \tan ^3(c+d x)+i \left (72 a^{5/2} (3 i A-13 B)+20 a^{5/2} (19 A+77 i B) (-i+\tan (c+d x))+30 \sqrt {a} (i A+B) (a+i a \tan (c+d x))^2-15 \sqrt {2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (a+i a \tan (c+d x))^{5/2}\right )}{120 a^{5/2} d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(-240*a^(5/2)*(A + (5*I)*B)*Tan[c + d*x]^2 + 240*a^(5/2)*B*Tan[c + d*x]^3 + I*(72*a^(5/2)*((3*I)*A - 13*B) + 2
0*a^(5/2)*(19*A + (77*I)*B)*(-I + Tan[c + d*x]) + 30*Sqrt[a]*(I*A + B)*(a + I*a*Tan[c + d*x])^2 - 15*Sqrt[2]*(
I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*(a + I*a*Tan[c + d*x])^(5/2)))/(120*a^(5/2)*d*(
a + I*a*Tan[c + d*x])^(5/2))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 i B \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a \left (17 i B +7 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{2} \left (7 i B +5 A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3} \left (i B +A \right )}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}}{a^{3} d}\)	\(142\)
default	\(\frac {2 i B \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a \left (17 i B +7 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{2} \left (7 i B +5 A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3} \left (i B +A \right )}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}}{a^{3} d}\)	\(142\)
parts	\(\frac {2 A \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}+\frac {7}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 a}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{2}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{2}}+\frac {2 i B \left (\sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{3}}\)	\(206\)

[In]

int(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/d/a^3*(I*B*(a+I*a*tan(d*x+c))^(1/2)+1/8*a*(7*A+17*I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/12*a^2*(5*A+7*I*B)/(a+I*a*
tan(d*x+c))^(3/2)+1/10*a^3*(A+I*B)/(a+I*a*tan(d*x+c))^(5/2)+1/16*a^(1/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*ta
n(d*x+c))^(1/2)*2^(1/2)/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. 392 vs. \(2 (162) = 324\).

Time = 0.26 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.86 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (83 \, A + 463 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (32 \, A + 97 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (8 \, A + 13 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)
*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2
)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)
/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(
2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/
(I*A + B)) - sqrt(2)*((83*A + 463*I*B)*e^(6*I*d*x + 6*I*c) + 2*(32*A + 97*I*B)*e^(4*I*d*x + 4*I*c) - 2*(8*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)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))*tan(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 480 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (7 \, A + 17 i \, B\right )} a^{2} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (5 \, A + 7 i \, B\right )} a^{3} + 12 \, {\left (A + i \, B\right )} a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{4} d} \]

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/240*(15*sqrt(2)*(A - I*B)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sq
rt(I*a*tan(d*x + c) + a))) - 480*I*sqrt(I*a*tan(d*x + c) + a)*B*a - 4*(15*(I*a*tan(d*x + c) + a)^2*(7*A + 17*I
*B)*a^2 - 10*(I*a*tan(d*x + c) + a)*(5*A + 7*I*B)*a^3 + 12*(A + I*B)*a^4)/(I*a*tan(d*x + c) + a)^(5/2))/(a^4*d
)

Giac [F]

\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*tan(d*x + c)^3/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 7.68 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {B\,1{}\mathrm {i}}{5\,d}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,17{}\mathrm {i}}{4\,a^2\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,d}+\frac {\frac {A\,a^2}{5}+\frac {7\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {5\,A\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}+\frac {\sqrt {2}\,A\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]

[In]

int((tan(c + d*x)^3*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

((B*1i)/(5*d) - (B*(a + a*tan(c + d*x)*1i)*7i)/(6*a*d) + (B*(a + a*tan(c + d*x)*1i)^2*17i)/(4*a^2*d))/(a + a*t
an(c + d*x)*1i)^(5/2) + (B*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/(a^3*d) + ((A*a^2)/5 + (7*A*(a + a*tan(c + d*x)*1
i)^2)/4 - (5*A*a*(a + a*tan(c + d*x)*1i))/6)/(a^2*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (2^(1/2)*B*atan((2^(1/2)*
(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d) + (2^(1/2)*A*atanh((2^(1/2)*(a + a*tan(c +
d*x)*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)