\(\int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [118]

Optimal result

Integrand size = 34, antiderivative size = 103 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \sqrt [4]{-1} a (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (i A+B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3672, 3610, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \sqrt [4]{-1} a (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(2*(-1)^(1/4)*a*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a*A)/(5*d*Tan[c + d*x]^(5/2)) - (2*a*(
I*A + B))/(3*d*Tan[c + d*x]^(3/2)) + (2*a*(A - I*B))/(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 3610

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 + 1)/(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*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.81 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.55 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \left (-3 A-5 i (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right ) \tan (c+d x)\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[((a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(2*a*(-3*A - (5*I)*(A - I*B)*Hypergeometric2F1[-3/2, 1, -1/2, I*Tan[c + d*x]]*Tan[c + d*x]))/(15*d*Tan[c + d*x
]^(5/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (84 ) = 168\).

Time = 0.04 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.29

method	result	size
derivativedivides	\(\frac {a \left (-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (i B -A \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 \left (i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-i A -B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-i B +A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(236\)
default	\(\frac {a \left (-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (i B -A \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 \left (i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-i A -B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-i B +A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(236\)
parts	\(\frac {\left (i a A +B a \right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {a A \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {i a B \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\)	\(327\)

[In]

int((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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. 486 vs. \(2 (81) = 162\).

Time = 0.28 (sec) , antiderivative size = 486, normalized size of antiderivative = 4.72 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \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}\right ) - 15 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \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}\right ) + 4 \, {\left ({\left (-23 i \, A - 20 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (i \, A + 10 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (11 i \, A + 20 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-13 i \, A - 10 \, B\right )} a\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}}}{30 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/30*(15*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(-I*A^2 - 2*A*
B + I*B^2)*a^2/d^2)*log(2*((A - I*B)*a*e^(2*I*d*x + 2*I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*B
+ I*B^2)*a^2/d^2)*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)) - 15*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(-I*A^2 - 2*
A*B + I*B^2)*a^2/d^2)*log(2*((A - I*B)*a*e^(2*I*d*x + 2*I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A
*B + I*B^2)*a^2/d^2)*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)) + 4*((-23*I*A - 20*B)*a*e^(6*I*d*x + 6*I*c) + (I*A + 10*B)*a*e^(4*I*d*x + 4*I*c) + (11*I*A + 20*B)*a*
e^(2*I*d*x + 2*I*c) + (-13*I*A - 10*B)*a)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(
6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=i a \left (\int \frac {A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {i A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\right )\, dx + \int \left (- \frac {i B}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

I*a*(Integral(A/tan(c + d*x)**(5/2), x) + Integral(B/tan(c + d*x)**(3/2), x) + Integral(-I*A/tan(c + d*x)**(7/
2), x) + Integral(-I*B/tan(c + d*x)**(5/2), 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. 187 vs. \(2 (81) = 162\).

Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.82 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a - \frac {8 \, {\left (15 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{2} + 5 \, {\left (-i \, A - B\right )} a \tan \left (d x + c\right ) - 3 \, A a\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{60 \, d} \]

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/60*(15*(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 - 8*(15*(A - I*B)*a*tan(d*x + c)^2 + 5*(-I*A - B)*a*tan(d*x + c) - 3*A*a)/ta
n(d*x + c)^(5/2))/d

Giac [A] (verification not implemented)

none

Time = 0.80 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.90 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\left (i - 1\right ) \, \sqrt {2} {\left (A a - i \, B a\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} + \frac {2 \, {\left (15 \, A a \tan \left (d x + c\right )^{2} - 15 i \, B a \tan \left (d x + c\right )^{2} - 5 i \, A a \tan \left (d x + c\right ) - 5 \, B a \tan \left (d x + c\right ) - 3 \, A a\right )}}{15 \, d \tan \left (d x + c\right )^{\frac {5}{2}}} \]

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

-(I - 1)*sqrt(2)*(A*a - I*B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d + 2/15*(15*A*a*tan(d*x + c)
^2 - 15*I*B*a*tan(d*x + c)^2 - 5*I*A*a*tan(d*x + c) - 5*B*a*tan(d*x + c) - 3*A*a)/(d*tan(d*x + c)^(5/2))

Mupad [B] (verification not implemented)

Time = 9.97 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.19 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\frac {2\,B\,a}{3\,d}+\frac {B\,a\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d}-\frac {2\,A\,a\,\left (15\,{\left (-1\right )}^{1/4}\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )-15\,{\mathrm {tan}\left (c+d\,x\right )}^2+3+\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}\right )}{15\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \]

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i))/tan(c + d*x)^(7/2),x)

[Out]

- ((2*B*a)/(3*d) + (B*a*tan(c + d*x)*2i)/d)/tan(c + d*x)^(3/2) - (2^(1/2)*B*a*atan(2^(1/2)*tan(c + d*x)^(1/2)*
(1/2 - 1i/2))*(1 + 1i))/d - (2*A*a*(tan(c + d*x)*5i - 15*tan(c + d*x)^2 + 15*(-1)^(1/4)*tan(c + d*x)^(5/2)*ata
nh((-1)^(1/4)*tan(c + d*x)^(1/2)) + 3))/(15*d*tan(c + d*x)^(5/2))