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

Optimal. Leaf size=78 \[ \frac{2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a (B+i A)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.126943, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3591, 3529, 3533, 205} \[ \frac{2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a (B+i A)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3591

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 3529

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 3533

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 205

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

Rubi steps

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

Mathematica [A]  time = 1.87451, size = 94, normalized size = 1.21 \[ -\frac{2 a \left (-3 i (A-i B) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+A \cot (c+d x)+3 i A+3 B\right )}{3 d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*((3*I)*A + 3*B + A*Cot[c + d*x] - (3*I)*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*
(c + d*x)))]]*Sqrt[I*Tan[c + d*x]]))/(3*d*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.015, size = 474, normalized size = 6.1 \begin{align*} -{\frac{2\,Aa}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,iaA}{d}{\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}}-2\,{\frac{aB}{d\sqrt{\tan \left ( dx+c \right ) }}}+{\frac{{\frac{i}{2}}aB\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{{\frac{i}{4}}aB\sqrt{2}}{d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{{\frac{i}{2}}aB\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{Aa\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{Aa\sqrt{2}}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{Aa\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{\frac{i}{4}}aA\sqrt{2}}{d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{{\frac{i}{2}}aA\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{\frac{i}{2}}aA\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{aB\sqrt{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{aB\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{aB\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.80105, size = 231, normalized size = 2.96 \begin{align*} \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 + \frac{8 \,{\left (3 \,{\left (-i \, A - B\right )} a \tan \left (d x + c\right ) - A a\right )}}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(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 + 8*(3*(-I*A - B)*a*tan(d*x + c) - A*a)/tan(d*x + c)^(3/2))/d

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Fricas [B]  time = 1.90797, size = 1135, normalized size = 14.55 \begin{align*} -\frac{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 )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 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 ) - 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 )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 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 ) - 8 \,{\left ({\left (4 \, A - 3 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (2 \, A - 3 i \, 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}}}{12 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*log(
(2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) + (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((-4*I*A^2 - 8*A*B + 4*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)) - 3*(d*e^(
4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*a*
e^(2*I*d*x + 2*I*c) + (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((-4*I*A^2 - 8*A*B + 4*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)) - 8*((4*A - 3*I*B)*a*e^(
4*I*d*x + 4*I*c) + 2*A*a*e^(2*I*d*x + 2*I*c) - (2*A - 3*I*B)*a)*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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 \frac{i A}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{i B}{\sqrt{\tan{\left (c + d x \right )}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27302, size = 95, normalized size = 1.22 \begin{align*} -\frac{\left (i - 1\right ) \, \sqrt{2}{\left (-i \, A a - 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{6 i \, A a \tan \left (d x + c\right ) + 6 \, B a \tan \left (d x + c\right ) + 2 \, A a}{3 \, d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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