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

Optimal. Leaf size=72 \[ \frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.041887, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3478, 3480, 206} \[ \frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 0.440727, size = 79, normalized size = 1.1 \[ \frac{2 i a e^{-i (c+d x)} \left (e^{i (c+d x)}-\sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*a*(E^(I*(c + d*x)) - Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*x]]
)/(d*E^(I*(c + d*x)))

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Maple [A]  time = 0.012, size = 54, normalized size = 0.8 \begin{align*}{\frac{2\,ia}{d} \left ( \sqrt{a+ia\tan \left ( dx+c \right ) }-\sqrt{a}\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),x)

[Out]

2*I/d*a*((a+I*a*tan(d*x+c))^(1/2)-a^(1/2)*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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.16074, size = 645, normalized size = 8.96 \begin{align*} \frac{4 i \, \sqrt{2} a \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 2 \, \sqrt{2} \sqrt{-\frac{a^{3}}{d^{2}}} d \log \left (\frac{{\left (2 i \, \sqrt{2} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right ) - 2 \, \sqrt{2} \sqrt{-\frac{a^{3}}{d^{2}}} d \log \left (\frac{{\left (-2 i \, \sqrt{2} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*tan(c + d*x) + a)**(3/2), 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),x, algorithm="giac")

[Out]

Timed out