\(\int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [131]

Optimal result

Integrand size = 36, antiderivative size = 136 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^3 A \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {\tan (c+d x)}} \]

[Out]

8*(-1)^(1/4)*a^3*(A-I*B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-16/3*a^3*A*tan(d*x+c)^(1/2)/d-2/3*a*A*(a+I*a*ta
n(d*x+c))^2/d/tan(d*x+c)^(3/2)-2/3*(7*I*A+3*B)*(a^3+I*a^3*tan(d*x+c))/d/tan(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3674, 3673, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 (3 B+7 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {\tan (c+d x)}}-\frac {16 a^3 A \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (16*a^3*A*Sqrt[Tan[c + d*x]])/(3*d) - (
2*a*A*(a + I*a*Tan[c + d*x])^2)/(3*d*Tan[c + d*x]^(3/2)) - (2*((7*I)*A + 3*B)*(a^3 + I*a^3*Tan[c + d*x]))/(3*d
*Sqrt[Tan[c + d*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 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 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 3674

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+i a \tan (c+d x))^2 \left (\frac {1}{2} a (7 i A+3 B)+\frac {1}{2} a (A+3 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx \\ & = -\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {\tan (c+d x)}}+\frac {4}{3} \int \frac {(a+i a \tan (c+d x)) \left (-a^2 (5 A-3 i B)+2 i a^2 A \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {16 a^3 A \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {\tan (c+d x)}}+\frac {4}{3} \int \frac {-3 a^3 (A-i B)-3 a^3 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {16 a^3 A \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {\tan (c+d x)}}+\frac {\left (24 a^6 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{-3 a^3 (A-i B)+3 a^3 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {8 \sqrt [4]{-1} a^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^3 A \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.00 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (-A+(-9 i A-3 B) \tan (c+d x)+12 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x)-3 i B \tan ^2(c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(2*a^3*(-A + ((-9*I)*A - 3*B)*Tan[c + d*x] + 12*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Tan
[c + d*x]^(3/2) - (3*I)*B*Tan[c + d*x]^2))/(3*d*Tan[c + d*x]^(3/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (114 ) = 228\).

Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.72

method	result	size
derivativedivides	\(\frac {a^{3} \left (-2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (3 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (4 i B -4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(234\)
default	\(\frac {a^{3} \left (-2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (3 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (4 i B -4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(234\)
parts	\(\frac {\left (-i A \,a^{3}-3 B \,a^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {\left (3 i B \,a^{3}-3 A \,a^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {A \,a^{3} \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}-\frac {i B \,a^{3} \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(529\)

[In]

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

[Out]

1/d*a^3*(-2*I*B*tan(d*x+c)^(1/2)-2/3*A/tan(d*x+c)^(3/2)-2*(3*I*A+B)/tan(d*x+c)^(1/2)+1/4*(4*I*B-4*A)*2^(1/2)*(
ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x
+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-4*I*A-4*B)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan
(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan
(d*x+c)^(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. 438 vs. \(2 (110) = 220\).

Time = 0.26 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.22 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (3 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 3 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 2 \, {\left ({\left (5 \, A - 3 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (A + 3 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, A a^{3}\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-2/3*(3*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((
A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt
((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((-I*A - B)*a^3)) - 3*sqrt(-(I*
A^2 + 2*A*B - I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^3*e^(2
*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((-I*e^(2*I*d*x
+ 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((-I*A - B)*a^3)) - 2*((5*A - 3*I*B)*a^3*e^(4*I
*d*x + 4*I*c) + (A + 3*I*B)*a^3*e^(2*I*d*x + 2*I*c) - 4*A*a^3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.40 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 i \, B a^{3} \sqrt {\tan \left (d x + c\right )} + 3 \, {\left (2 \, \sqrt {2} {\left (\left (i + 1\right ) \, A - \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i + 1\right ) \, A - \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} - \frac {2 \, {\left (3 \, {\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{3 \, d} \]

[In]

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

[Out]

-1/3*(6*I*B*a^3*sqrt(tan(d*x + c)) + 3*(2*sqrt(2)*((I + 1)*A - (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt
(tan(d*x + c)))) + 2*sqrt(2)*((I + 1)*A - (I - 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - s
qrt(2)*((I - 1)*A + (I + 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*((I - 1)*A + (I +
1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 - 2*(3*(-3*I*A - B)*a^3*tan(d*x + c) - A*a^3)/t
an(d*x + c)^(3/2))/d

Giac [A] (verification not implemented)

none

Time = 1.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 i \, B a^{3} \sqrt {\tan \left (d x + c\right )}}{d} - \frac {\left (4 i - 4\right ) \, \sqrt {2} {\left (-i \, A a^{3} - B a^{3}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} + \frac {2 \, {\left (-9 i \, A a^{3} \tan \left (d x + c\right ) - 3 \, B a^{3} \tan \left (d x + c\right ) - A a^{3}\right )}}{3 \, d \tan \left (d x + c\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-2*I*B*a^3*sqrt(tan(d*x + c))/d - (4*I - 4)*sqrt(2)*(-I*A*a^3 - B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(
d*x + c)))/d + 2/3*(-9*I*A*a^3*tan(d*x + c) - 3*B*a^3*tan(d*x + c) - A*a^3)/(d*tan(d*x + c)^(3/2))

Mupad [B] (verification not implemented)

Time = 8.66 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.76 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\frac {2\,A\,a^3}{3\,d}+\frac {A\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,6{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}-\frac {2\,B\,a^3}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}-\frac {B\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (-A\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,A\,a^3\,\ln \left (-A\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (-8\,B\,a^3\,d+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,B\,a^3\,\ln \left (-8\,B\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

[In]

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

[Out]

(2^(1/2)*A*a^3*log(- A*a^3*d*8i - 2^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - (2*B*a^3)/(d*tan(
c + d*x)^(1/2)) - (B*a^3*tan(c + d*x)^(1/2)*2i)/d - ((2*A*a^3)/(3*d) + (A*a^3*tan(c + d*x)*6i)/d)/tan(c + d*x)
^(3/2) - ((-16i)^(1/2)*A*a^3*log(2*(-16i)^(1/2)*A*a^3*d*tan(c + d*x)^(1/2) - A*a^3*d*8i))/d + (2^(1/2)*B*a^3*l
og(- 8*B*a^3*d - 2^(1/2)*B*a^3*d*tan(c + d*x)^(1/2)*(4 + 4i))*(2 + 2i))/d - (16i^(1/2)*B*a^3*log(2*16i^(1/2)*B
*a^3*d*tan(c + d*x)^(1/2) - 8*B*a^3*d))/d