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

Optimal. Leaf size=146 \[ -\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{16 a^3 (5 A-6 i B) \sqrt{\tan (c+d x)}}{15 d}-\frac{2 (5 A-9 i B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d} \]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (16*a^3*(5*A - (6*I)*B)*Sqrt[Tan[c + d
*x]])/(15*d) + (((2*I)/5)*a*B*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^2)/d - (2*(5*A - (9*I)*B)*Sqrt[Tan[c +
d*x]]*(a^3 + I*a^3*Tan[c + d*x]))/(15*d)

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Rubi [A]  time = 0.376554, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3594, 3592, 3533, 205} \[ -\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{16 a^3 (5 A-6 i B) \sqrt{\tan (c+d x)}}{15 d}-\frac{2 (5 A-9 i B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (16*a^3*(5*A - (6*I)*B)*Sqrt[Tan[c + d
*x]])/(15*d) + (((2*I)/5)*a*B*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^2)/d - (2*(5*A - (9*I)*B)*Sqrt[Tan[c +
d*x]]*(a^3 + I*a^3*Tan[c + d*x]))/(15*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))^3 (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx &=\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac{2}{5} \int \frac{(a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (5 A-i B)+\frac{1}{2} a (5 i A+9 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac{2 (5 A-9 i B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac{4}{15} \int \frac{(a+i a \tan (c+d x)) \left (a^2 (5 A-3 i B)+2 a^2 (5 i A+6 B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{16 a^3 (5 A-6 i B) \sqrt{\tan (c+d x)}}{15 d}+\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac{2 (5 A-9 i B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac{4}{15} \int \frac{15 a^3 (A-i B)+15 a^3 (i A+B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{16 a^3 (5 A-6 i B) \sqrt{\tan (c+d x)}}{15 d}+\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac{2 (5 A-9 i B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac{\left (120 a^6 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{15 a^3 (A-i B)-15 a^3 (i A+B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{16 a^3 (5 A-6 i B) \sqrt{\tan (c+d x)}}{15 d}+\frac{2 i a B \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac{2 (5 A-9 i B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}\\ \end{align*}

Mathematica [A]  time = 7.25288, size = 273, normalized size = 1.87 \[ \frac{\cos ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\frac{8 e^{-3 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac{1}{15} (\cos (3 c)-i \sin (3 c)) \sqrt{\tan (c+d x)} \sec ^2(c+d x) (5 (3 B+i A) \sin (2 (c+d x))+9 (5 A-7 i B) \cos (2 (c+d x))+45 A-57 i B)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^4*((8*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(
-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((3*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]) - (Sec[c + d*x]^2*(Cos[3*c] - I*Sin[3*c])*(45*A - (57*I)*B + 9*(5*A - (7*I)*B)*Cos[2*(c + d*
x)] + 5*(I*A + 3*B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/15)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(
d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.016, size = 538, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/d*a^3*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)+2*I/d*a^
3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-2/d*a^3*B*tan(d*x+c)^(3/2)-2*I/d*a^3*B*arctan(1+2^(1/2)*tan(d*x
+c)^(1/2))*2^(1/2)-6*a^3*A*tan(d*x+c)^(1/2)/d-I/d*a^3*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*I/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2/3*I/d*a^3*A*ta
n(d*x+c)^(3/2)+2/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/d*a^3*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)))+2/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1
/2)-2/5*I/d*a^3*B*tan(d*x+c)^(5/2)+8*I/d*a^3*B*tan(d*x+c)^(1/2)+2*I/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)
)*2^(1/2)+1/d*a^3*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
)+2/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+2/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)

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Maxima [A]  time = 2.11047, size = 262, normalized size = 1.79 \begin{align*} -\frac{6 i \, B a^{3} \tan \left (d x + c\right )^{\frac{5}{2}} + 10 \,{\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac{3}{2}} + 2 \,{\left (45 \, A - 60 i \, B\right )} a^{3} \sqrt{\tan \left (d x + c\right )} + 15 \,{\left (\sqrt{2}{\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\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 (2 i + 2\right ) \, A + \left (2 i - 2\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^{3}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

-1/15*(6*I*B*a^3*tan(d*x + c)^(5/2) + 10*(I*A + 3*B)*a^3*tan(d*x + c)^(3/2) + 2*(45*A - 60*I*B)*a^3*sqrt(tan(d
*x + c)) + 15*(sqrt(2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + sqr
t(2)*(-(2*I + 2)*A + (2*I - 2)*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^3)/d

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Fricas [B]  time = 1.8495, size = 1214, normalized size = 8.32 \begin{align*} \frac{15 \, \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 15 \, \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 16 \,{\left ({\left (25 \, A - 39 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \,{\left (15 \, A - 19 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \,{\left (5 \, A - 6 i \, B\right )} a^{3}\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}}}{60 \,{\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))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/60*(15*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*
log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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)/((4*I*A + 4*B)*
a^3)) - 15*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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)/((4*I*A + 4*
B)*a^3)) - 16*((25*A - 39*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 3*(15*A - 19*I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(5*A -
6*I*B)*a^3)*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.38623, size = 171, normalized size = 1.17 \begin{align*} \frac{\left (4 i - 4\right ) \, \sqrt{2}{\left (-i \, A a^{3} - B a^{3}\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 \, B a^{3} d^{4} \tan \left (d x + c\right )^{\frac{5}{2}} + 10 i \, A a^{3} d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} + 30 \, B a^{3} d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} + 90 \, A a^{3} d^{4} \sqrt{\tan \left (d x + c\right )} - 120 i \, B a^{3} d^{4} \sqrt{\tan \left (d x + c\right )}}{15 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="giac")

[Out]

(4*I - 4)*sqrt(2)*(-I*A*a^3 - B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/15*(6*I*B*a^3*d^4
*tan(d*x + c)^(5/2) + 10*I*A*a^3*d^4*tan(d*x + c)^(3/2) + 30*B*a^3*d^4*tan(d*x + c)^(3/2) + 90*A*a^3*d^4*sqrt(
tan(d*x + c)) - 120*I*B*a^3*d^4*sqrt(tan(d*x + c)))/d^5