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

Optimal. Leaf size=107 \[ -\frac{2 \sqrt{2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (B+i A) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 B (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0999415, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3527, 3478, 3480, 206} \[ -\frac{2 \sqrt{2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (B+i A) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 B (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 2.54725, size = 190, normalized size = 1.78 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{2}{3} (\cos (c)-i \sin (c)) \sqrt{\sec (c+d x)} (\sin (d x)+i \cos (d x)) (3 A+B \tan (c+d x)-4 i B)-\frac{2 i \sqrt{2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}\right )}{d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*(((-2*I)*Sqrt[2]*(A - I*B)*ArcSinh[E^(I*(c + d*x))])/((E^(I
*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) + (2*Sqrt[Sec[c + d*x]]*(Cos[c]
- I*Sin[c])*(I*Cos[d*x] + Sin[d*x])*(3*A - (4*I)*B + B*Tan[c + d*x]))/3))/(d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x
] + B*Sin[c + d*x]))

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Maple [A]  time = 0.018, size = 99, normalized size = 0.9 \begin{align*}{\frac{2\,i}{d} \left ( -{\frac{i}{3}}B \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}-iBa\sqrt{a+ia\tan \left ( dx+c \right ) }+A\sqrt{a+ia\tan \left ( dx+c \right ) }a-{a}^{{\frac{3}{2}}} \left ( A-iB \right ) \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.70316, size = 1017, normalized size = 9.5 \begin{align*} \frac{\sqrt{2}{\left ({\left (12 i \, A + 20 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (12 i \, A + 12 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 3 \, \sqrt{-\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (\sqrt{2}{\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + \sqrt{-\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) + 3 \, \sqrt{-\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (\sqrt{2}{\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (2 i \, A + 2 \, B\right )} a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - \sqrt{-\frac{{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{6 \,{\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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(sqrt(2)*((12*I*A + 20*B)*a*e^(2*I*d*x + 2*I*c) + (12*I*A + 12*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(
I*d*x + I*c) - 3*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A +
2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + sqrt(-(8*A^2
- 16*I*A*B - 8*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 3*sqrt(-(8*A^2
- 16*I*A*B - 8*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (
2*I*A + 2*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a^3/d^2)*
d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)))/(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*} \int \left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}} \left (A + B \tan{\left (c + d x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out