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

Optimal. Leaf size=92 \[ -\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0838814, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3527, 3478, 3480, 206} \[ -\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (2*a*Sqrt[a + I*a*Tan[c + d*x]]
)/d + (2*(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 \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d}-i \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d}-(2 i a) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{2 a \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{\left (4 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 \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 a \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.872124, size = 148, normalized size = 1.61 \[ \frac{\sqrt{2} a e^{-\frac{1}{2} i (2 c+3 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\cos \left (\frac{d x}{2}\right )+i \sin \left (\frac{d x}{2}\right )\right ) \left (\sqrt{1+e^{2 i (c+d x)}} (4+i \tan (c+d x)) \sec (c+d x)-6 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos[(d*x)/2]
+ I*Sin[(d*x)/2])*(-6*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]*(4 + I*Tan[c + d*
x])))/(3*d*E^((I/2)*(2*c + 3*d*x)))

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Maple [A]  time = 0.016, size = 70, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{2}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+2\,a\sqrt{a+ia\tan \left ( dx+c \right ) }-2\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.17242, size = 756, normalized size = 8.22 \begin{align*} \frac{2 \, \sqrt{2}{\left (5 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, 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{2}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{a^{3}}{d^{2}}} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + \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 )}}{a}\right ) + 3 \, \sqrt{2}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{a^{3}}{d^{2}}} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} - \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 )}}{a}\right )}{3 \,{\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(tan(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*(2*sqrt(2)*(5*a*e^(2*I*d*x + 2*I*c) + 3*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 3*sqrt(2)*(
d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log((sqrt(2)*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c) + 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) + 3*sqrt(2)*(d*e^
(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(-(sqrt(2)*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c) - 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*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}} \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out