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

Optimal. Leaf size=98 \[ -\frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a^2 (-B+i A) \sqrt{\tan (c+d x)}}{d}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{\tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.213756, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3593, 3592, 3533, 205} \[ -\frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a^2 (-B+i A) \sqrt{\tan (c+d x)}}{d}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3593

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] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3592

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

Mathematica [A]  time = 3.24912, size = 85, normalized size = 0.87 \[ -\frac{2 a^2 \left (-2 (A-i B) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+A+B \tan (c+d x)\right )}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(A - 2*(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]] + B*Tan[c + d*x]))/(d*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.016, size = 484, normalized size = 4.9 \begin{align*} -2\,{\frac{{a}^{2}B\sqrt{\tan \left ( dx+c \right ) }}{d}}-2\,{\frac{{a}^{2}A}{d\sqrt{\tan \left ( dx+c \right ) }}}+{\frac{iA{a}^{2}\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{iA{a}^{2}\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{{\frac{i}{2}}{a}^{2}A\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{{a}^{2}B\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{{a}^{2}B\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{{a}^{2}B\sqrt{2}}{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}}{a}^{2}B\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{iB{a}^{2}\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{iB{a}^{2}\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{a}^{2}A\sqrt{2}}{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{{a}^{2}A\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{a}^{2}A\sqrt{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))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.83462, size = 230, normalized size = 2.35 \begin{align*} -\frac{4 \, B a^{2} \sqrt{\tan \left (d x + c\right )} -{\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 \, A a^{2}}{\sqrt{\tan \left (d x + c\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.76831, size = 1010, normalized size = 10.31 \begin{align*} \frac{\sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) +{\left ({\left (-8 i \, A - 8 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-8 i \, A + 8 \, 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}}}{4 \,{\left (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))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/4*(sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((4*(A - I*B)*a^2*e^(2*I*d*x
+ 2*I*c) + sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + 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)/((2*I*A + 2*B)*a^2)) - sqrt((16*I*A^2 + 32*A*B - 16*I
*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B
- 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + 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)/((2*I*A + 2*B)*a^2)) + ((-8*I*A - 8*B)*a^2*e^(2*I*d*x + 2*I*c) + (-8*I*A + 8*B)*a^2)*sqr
t((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(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^{2} \left (\int \frac{A}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int - A \sqrt{\tan{\left (c + d x \right )}}\, dx + \int \frac{B}{\sqrt{\tan{\left (c + d x \right )}}}\, dx + \int - B \tan ^{\frac{3}{2}}{\left (c + d x \right )}\, dx + \int \frac{2 i A}{\sqrt{\tan{\left (c + d x \right )}}}\, dx + \int 2 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))**2*(A+B*tan(d*x+c))/tan(d*x+c)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.26082, size = 95, normalized size = 0.97 \begin{align*} -\frac{2 \, B a^{2} \sqrt{\tan \left (d x + c\right )}}{d} + \frac{\left (i + 1\right ) \, \sqrt{2}{\left (8 i \, A a^{2} + 8 \, 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 )}{4 \, d} - \frac{2 \, A a^{2}}{d \sqrt{\tan \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*B*a^2*sqrt(tan(d*x + c))/d + (1/4*I + 1/4)*sqrt(2)*(8*I*A*a^2 + 8*B*a^2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt
(tan(d*x + c)))/d - 2*A*a^2/(d*sqrt(tan(d*x + c)))