3.107 \(\int (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=130 \[ \frac{8 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{8 i \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

((-8*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((8*I)*a^3*Sqrt[a + I*a*Tan
[c + d*x]])/d + (((4*I)/3)*a^2*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*I)/5)*a*(a + I*a*Tan[c + d*x])^(5/2))/d

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Rubi [A]  time = 0.0815251, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3478, 3480, 206} \[ \frac{8 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{8 i \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-8*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((8*I)*a^3*Sqrt[a + I*a*Tan
[c + d*x]])/d + (((4*I)/3)*a^2*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*I)/5)*a*(a + I*a*Tan[c + d*x])^(5/2))/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))^{7/2} \, dx &=\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}+(2 a) \int (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac{4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (4 a^2\right ) \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac{8 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (8 a^3\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{8 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{\left (16 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac{8 i \sqrt{2} a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{8 i a^3 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.62234, size = 166, normalized size = 1.28 \[ \frac{i \sqrt{2} a^3 e^{-\frac{1}{2} i (2 c+5 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\cos \left (\frac{3 d x}{2}\right )+i \sin \left (\frac{3 d x}{2}\right )\right ) \left (\sqrt{1+e^{2 i (c+d x)}} \sec ^3(c+d x) (8 i \sin (2 (c+d x))+38 \cos (2 (c+d x))+35)-120 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((I/15)*Sqrt[2]*a^3*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos
[(3*d*x)/2] + I*Sin[(3*d*x)/2])*(-120*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^3*
(35 + 38*Cos[2*(c + d*x)] + (8*I)*Sin[2*(c + d*x)])))/(d*E^((I/2)*(2*c + 5*d*x)))

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Maple [A]  time = 0.013, size = 92, normalized size = 0.7 \begin{align*}{\frac{2\,ia}{d} \left ({\frac{1}{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,a}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+4\,{a}^{2}\sqrt{a+ia\tan \left ( dx+c \right ) }-4\,{a}^{5/2}\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((a+I*a*tan(d*x+c))^(7/2),x)

[Out]

2*I/d*a*(1/5*(a+I*a*tan(d*x+c))^(5/2)+2/3*(a+I*a*tan(d*x+c))^(3/2)*a+4*a^2*(a+I*a*tan(d*x+c))^(1/2)-4*a^(5/2)*
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))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.08696, size = 973, normalized size = 7.48 \begin{align*} \frac{\sqrt{2}{\left (368 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 560 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 240 i \, a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 120 \, \sqrt{2} \sqrt{-\frac{a^{7}}{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 i \, \sqrt{2} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\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 )}}{8 \, a^{3}}\right ) - 120 \, \sqrt{2} \sqrt{-\frac{a^{7}}{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 i \, \sqrt{2} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\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 )}}{8 \, a^{3}}\right )}{30 \,{\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))^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*(368*I*a^3*e^(4*I*d*x + 4*I*c) + 560*I*a^3*e^(2*I*d*x + 2*I*c) + 240*I*a^3)*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*e^(I*d*x + I*c) + 120*sqrt(2)*sqrt(-a^7/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) +
d)*log(1/8*(8*I*sqrt(2)*sqrt(-a^7/d^2)*d*e^(2*I*d*x + 2*I*c) + 8*sqrt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*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) - 120*sqrt(2)*sqrt(-a^7/d^2)*(d*e^(4*I
*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/8*(-8*I*sqrt(2)*sqrt(-a^7/d^2)*d*e^(2*I*d*x + 2*I*c) + 8*sq
rt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*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))/(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))**(7/2),x)

[Out]

Timed out

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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))^(7/2),x, algorithm="giac")

[Out]

Timed out