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

Optimal. Leaf size=101 \[ \frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d}-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d} \]

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

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Rubi [A]  time = 0.105723, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3543, 3478, 3480, 206} \[ \frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d}-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(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 - (((2*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(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))^{3/2} \, dx &=-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d}-\int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac{2 i a \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a 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{2 i (a+i a \tan (c+d x))^{5/2}}{5 a 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}-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d}\\ \end{align*}

Mathematica [A]  time = 1.22321, size = 162, normalized size = 1.6 \[ \frac{a e^{-\frac{1}{2} i (2 c+3 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\sin \left (\frac{d x}{2}\right )-i \cos \left (\frac{d x}{2}\right )\right ) \left (\sqrt{1+e^{2 i (c+d x)}} \sec ^3(c+d x) (2 i \sin (2 (c+d x))+7 \cos (2 (c+d x))+5)-20 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{5 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*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)/2] +
Sin[(d*x)/2])*(-20*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^3*(5 + 7*Cos[2*(c + d
*x)] + (2*I)*Sin[2*(c + d*x)])))/(5*Sqrt[2]*d*E^((I/2)*(2*c + 3*d*x)))

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Maple [A]  time = 0.018, size = 76, normalized size = 0.8 \begin{align*}{\frac{-2\,i}{ad} \left ({\frac{1}{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{a}^{2}\sqrt{a+ia\tan \left ( dx+c \right ) }-{a}^{{\frac{5}{2}}}\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(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.26217, size = 944, normalized size = 9.35 \begin{align*} \frac{\sqrt{2}{\left (-36 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 40 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 20 i \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 10 \, \sqrt{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 )} \sqrt{-\frac{a^{3}}{d^{2}}} \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 ) + 10 \, \sqrt{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 )} \sqrt{-\frac{a^{3}}{d^{2}}} \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 )}{10 \,{\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)^2*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/10*(sqrt(2)*(-36*I*a*e^(4*I*d*x + 4*I*c) - 40*I*a*e^(2*I*d*x + 2*I*c) - 20*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*e^(I*d*x + I*c) - 10*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-a^3/d^2)*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) + 10*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x
+ 2*I*c) + d)*sqrt(-a^3/d^2)*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*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]  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 ^{2}{\left (c + d x \right )}\, dx \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))**(3/2),x)

[Out]

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

[Out]

Timed out