3.101 \(\int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=119 \[ \frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{4 \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.10184, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3527, 3478, 3480, 206} \[ \frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{4 \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 a (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

Mathematica [A]  time = 1.1283, size = 154, normalized size = 1.29 \[ \frac{a^2 e^{-i (c+2 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x)) \left (\sqrt{1+e^{2 i (c+d x)}} \sec ^3(c+d x) (11 i \sin (2 (c+d x))+41 \cos (2 (c+d x))+35)-120 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{15 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(a^2*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] + I*Sin[d
*x])*(-120*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^3*(35 + 41*Cos[2*(c + d*x)] +
(11*I)*Sin[2*(c + d*x)])))/(15*Sqrt[2]*d*E^(I*(c + 2*d*x)))

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Maple [A]  time = 0.016, size = 89, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{2}{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(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.28733, size = 929, normalized size = 7.81 \begin{align*} \frac{2 \,{\left (2 \, \sqrt{2}{\left (26 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 35 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 15 \, \sqrt{2} \sqrt{\frac{a^{5}}{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 (\sqrt{2} \sqrt{\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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^{2}}\right ) + 15 \, \sqrt{2} \sqrt{\frac{a^{5}}{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 (\sqrt{2} \sqrt{\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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^{2}}\right )\right )}}{15 \,{\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(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/15*(2*sqrt(2)*(26*a^2*e^(4*I*d*x + 4*I*c) + 35*a^2*e^(2*I*d*x + 2*I*c) + 15*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*e^(I*d*x + I*c) - 15*sqrt(2)*sqrt(a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log((s
qrt(2)*sqrt(a^5/d^2)*d*e^(2*I*d*x + 2*I*c) + sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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) + 15*sqrt(2)*sqrt(a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^
(2*I*d*x + 2*I*c) + d)*log(-(sqrt(2)*sqrt(a^5/d^2)*d*e^(2*I*d*x + 2*I*c) - sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) +
a^2)*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))/(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(tan(d*x+c)*(a+I*a*tan(d*x+c))**(5/2),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(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out