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

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

[Out]

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

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Rubi [A]  time = 0.12496, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3543, 3478, 3480, 206} \[ -\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 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 ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\int (a+i a \tan (c+d x))^{5/2} \, dx\\ &=-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-(2 a) \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\left (4 a^2\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}+\frac{\left (8 i 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 i \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 i a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}\\ \end{align*}

Mathematica [A]  time = 2.56208, size = 170, normalized size = 1.31 \[ \frac{a^2 e^{-i (c+2 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\sin (d x)-i \cos (d x)) \left (\sqrt{1+e^{2 i (c+d x)}} \sec ^3(c+d x) (122 \cos (2 (c+d x))+7 i \tan (c+d x)+19 i \sin (3 (c+d x)) \sec (c+d x)+86)-336 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{42 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2*(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))]*((-I)*Cos[d*x] + Si
n[d*x])*(-336*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^3*(86 + 122*Cos[2*(c + d*x
)] + (19*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (7*I)*Tan[c + d*x])))/(42*Sqrt[2]*d*E^(I*(c + 2*d*x)))

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Maple [A]  time = 0.018, size = 96, normalized size = 0.7 \begin{align*}{\frac{-2\,i}{ad} \left ({\frac{1}{7} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+2\,{a}^{3}\sqrt{a+ia\tan \left ( dx+c \right ) }-2\,{a}^{7/2}\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)^2*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31254, size = 1121, normalized size = 8.62 \begin{align*} \frac{\sqrt{2}{\left (-320 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 616 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 560 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 168 i \, 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 )} - 84 \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 i \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, \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 )}}{4 \, a^{2}}\right ) + 84 \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (-4 i \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, \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 )}}{4 \, a^{2}}\right )}{42 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/42*(sqrt(2)*(-320*I*a^2*e^(6*I*d*x + 6*I*c) - 616*I*a^2*e^(4*I*d*x + 4*I*c) - 560*I*a^2*e^(2*I*d*x + 2*I*c)
- 168*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 84*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I
*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*I*sqrt(2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x
+ 2*I*c) + 4*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) + 84*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*
d*x + 2*I*c) + d)*log(1/4*(-4*I*sqrt(2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c) + 4*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^(6*I*d*x + 6*I*c
) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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)**2*(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)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out