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

Optimal. Leaf size=104 \[ -\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (3 A-5 i B) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 i B \sqrt{\tan (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.224437, antiderivative size = 104, 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 = {3594, 3592, 3533, 205} \[ -\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (3 A-5 i B) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 i B \sqrt{\tan (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f
*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, 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))}{\sqrt{\tan (c+d x)}} \, dx &=\frac{2 i B \sqrt{\tan (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{2}{3} \int \frac{(a+i a \tan (c+d x)) \left (\frac{1}{2} a (3 A-i B)+\frac{1}{2} a (3 i A+5 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (3 A-5 i B) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 i B \sqrt{\tan (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\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 (3 A-5 i B) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 i B \sqrt{\tan (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{\left (12 a^4 (A-i B)^2\right ) \operatorname{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) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (3 A-5 i B) \sqrt{\tan (c+d x)}}{3 d}+\frac{2 i B \sqrt{\tan (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 3.3122, size = 110, normalized size = 1.06 \[ -\frac{2 a^2 \sqrt{\tan (c+d x)} \left (\sqrt{i \tan (c+d x)} (3 A+B \tan (c+d x)-6 i B)-6 (A-i B) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{3 d \sqrt{i \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sqrt[Tan[c + d*x]]*(-6*(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*A - (6*I)*B + B*Tan[c + d*x])))/(3*d*Sqrt[I*Tan[c + d*x]])

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Maple [B]  time = 0.014, size = 500, normalized size = 4.8 \begin{align*} -{\frac{2\,{a}^{2}B}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-2\,{\frac{{a}^{2}A\sqrt{\tan \left ( dx+c \right ) }}{d}}+{\frac{4\,i{a}^{2}B}{d}\sqrt{\tan \left ( dx+c \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{{\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{{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 ) }+{\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{{\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{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{{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{{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 ) } \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)^(1/2),x)

[Out]

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

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Maxima [B]  time = 2.07537, size = 238, normalized size = 2.29 \begin{align*} -\frac{4 \, B a^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 4 \,{\left (3 \, A - 6 i \, B\right )} a^{2} \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^{2}}{6 \, 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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.77262, size = 1034, normalized size = 9.94 \begin{align*} \frac{3 \, \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 (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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 3 \, \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 (-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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \,{\left ({\left (3 \, A - 7 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (3 \, A - 5 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}}}{12 \,{\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)^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*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)*(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)/((2*I*A + 2*B)*a^2)) - 3*sqrt((-16*I*A^2 - 3
2*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)*(-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)/((2*I*A + 2*B)*a^2)) - 8*((3*A - 7*I*B)*a^2*e^(2*I*d*x + 2*I*c) + (3*A
- 5*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^(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}{\sqrt{\tan{\left (c + d x \right )}}}\, dx + \int - A \tan ^{\frac{3}{2}}{\left (c + d x \right )}\, dx + \int B \sqrt{\tan{\left (c + d x \right )}}\, dx + \int - B \tan ^{\frac{5}{2}}{\left (c + d x \right )}\, dx + \int 2 i A \sqrt{\tan{\left (c + d x \right )}}\, dx + \int 2 i B \tan ^{\frac{3}{2}}{\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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.36067, size = 124, normalized size = 1.19 \begin{align*} \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 (B a^{2} d^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 3 \, A a^{2} d^{2} \sqrt{\tan \left (d x + c\right )} - 6 i \, B a^{2} d^{2} \sqrt{\tan \left (d x + c\right )}\right )}}{3 \, d^{3}} \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)^(1/2),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*(B*a^2*d^2*tan(
d*x + c)^(3/2) + 3*A*a^2*d^2*sqrt(tan(d*x + c)) - 6*I*B*a^2*d^2*sqrt(tan(d*x + c)))/d^3