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

Optimal. Leaf size=29 \[ -\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

[Out]

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

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Rubi [A]  time = 0.060965, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3487, 32} \[ -\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a d}\\ &=-\frac{2 i (a+i a \tan (c+d x))^{5/2}}{5 a d}\\ \end{align*}

Mathematica [B]  time = 0.265127, size = 69, normalized size = 2.38 \[ \frac{2 a \sec ^2(c+d x) (\cos (d x)-i \sin (d x)) \sqrt{a+i a \tan (c+d x)} (\sin (2 c+3 d x)-i \cos (2 c+3 d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sec[c + d*x]^2*(Cos[d*x] - I*Sin[d*x])*((-I)*Cos[2*c + 3*d*x] + Sin[2*c + 3*d*x])*Sqrt[a + I*a*Tan[c + d*
x]])/(5*d)

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Maple [A]  time = 0.03, size = 24, normalized size = 0.8 \begin{align*}{\frac{-{\frac{2\,i}{5}}}{ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*I*(a+I*a*tan(d*x+c))^(5/2)/a/d

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Maxima [A]  time = 1.10683, size = 28, normalized size = 0.97 \begin{align*} -\frac{2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{5 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*I*(I*a*tan(d*x + c) + a)^(5/2)/(a*d)

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Fricas [B]  time = 2.34836, size = 170, normalized size = 5.86 \begin{align*} -\frac{8 i \, \sqrt{2} a \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (5 i \, d x + 5 i \, c\right )}}{5 \,{\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(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-8/5*I*sqrt(2)*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(5*I*d*x + 5*I*c)/(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(sec(d*x+c)**2*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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