\(\int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [124]

Optimal result

Integrand size = 36, antiderivative size = 102 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \sqrt [4]{-1} a^2 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 i A+3 B)}{3 d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[Out]

4*(-1)^(1/4)*a^2*(A-I*B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-2/3*a^2*(5*I*A+3*B)/d/tan(d*x+c)^(1/2)-2/3*A*(a
^2+I*a^2*tan(d*x+c))/d/tan(d*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3674, 3672, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \sqrt [4]{-1} a^2 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (3 B+5 i A)}{3 d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a^2*((5*I)*A + 3*B))/(3*d*Sqrt[Tan[c
+ d*x]]) - (2*A*(a^2 + I*a^2*Tan[c + d*x]))/(3*d*Tan[c + d*x]^(3/2))

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 3672

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

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 \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+i a \tan (c+d x)) \left (\frac {1}{2} a (5 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^2 (5 i A+3 B)}{3 d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {-3 a^2 (A-i B)-3 a^2 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 (5 i A+3 B)}{3 d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (12 a^4 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{-3 a^2 (A-i B)+3 a^2 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {4 \sqrt [4]{-1} a^2 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 i A+3 B)}{3 d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (-A+(-6 i A-3 B) \tan (c+d x)+(3+3 i) \sqrt {2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {\tan (c+d x)}}{\sqrt {2}}\right ) \tan ^{\frac {3}{2}}(c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(2*a^2*(-A + ((-6*I)*A - 3*B)*Tan[c + d*x] + (3 + 3*I)*Sqrt[2]*(I*A + B)*ArcTanh[((1 + I)*Sqrt[Tan[c + d*x]])/
Sqrt[2]]*Tan[c + d*x]^(3/2)))/(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. 221 vs. \(2 (85 ) = 170\).

Time = 0.04 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.18

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (2 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (2 i B -2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(222\)
default	\(\frac {a^{2} \left (-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (2 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (2 i B -2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(222\)
parts	\(\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {\left (2 i B \,a^{2}-A \,a^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {A \,a^{2} \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}-\frac {B \,a^{2} \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\)	\(413\)

[In]

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

[Out]

1/d*a^2*(-2/3*A/tan(d*x+c)^(3/2)-2*(2*I*A+B)/tan(d*x+c)^(1/2)+1/4*(-2*A+2*I*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)))+1/4*(-2*B-2*I*A)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*ta
n(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. 441 vs. \(2 (82) = 164\).

Time = 0.26 (sec) , antiderivative size = 441, normalized size of antiderivative = 4.32 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {3 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 3 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, {\left ({\left (7 \, A - 3 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (5 \, A - 3 i \, B\right )} a^{2}\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-1/3*(3*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^4/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^2*e^(2*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^4/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^2)) - 3*sqrt(-(I*
A^2 + 2*A*B - I*B^2)*a^4/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^2*e^(2
*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^4/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^2)) - 2*((7*A - 3*I*B)*a^2*e^(4*I
*d*x + 4*I*c) + 2*A*a^2*e^(2*I*d*x + 2*I*c) - (5*A - 3*I*B)*a^2)*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]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (82) = 164\).

Time = 0.39 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.74 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} + \frac {4 \, {\left (3 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.89 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\left (2 i - 2\right ) \, \sqrt {2} {\left (-i \, A a^{2} - B a^{2}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (6 i \, A a^{2} \tan \left (d x + c\right ) + 3 \, B a^{2} \tan \left (d x + c\right ) + A a^{2}\right )}}{3 \, d \tan \left (d x + c\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.38 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.18 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\frac {2\,A\,a^2}{3\,d}+\frac {A\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}-\frac {2\,B\,a^2}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (-A\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,A\,a^2\,\ln \left (-A\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-4\,B\,a^2\,d+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,B\,a^2\,\ln \left (-4\,B\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

[In]

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

[Out]

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