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

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

[Out]

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

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Rubi [A]  time = 0.0641361, 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} \[ \frac{2 i}{5 a d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 \frac{\sec ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{1}{(a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{a d}\\ &=\frac{2 i}{5 a d (a+i a \tan (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.252618, size = 39, normalized size = 1.34 \[ -\frac{2 \sqrt{a+i a \tan (c+d x)}}{5 a^4 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, 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))^(7/2),x)

[Out]

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

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Maxima [A]  time = 0.974485, size = 28, normalized size = 0.97 \begin{align*} \frac{2 i}{5 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} 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))^(7/2),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 = 2.15347, size = 212, normalized size = 7.31 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{20 \, a^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

1/20*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(I*e^(6*I*d*x + 6*I*c) + 3*I*e^(4*I*d*x + 4*I*c) + 3*I*e^(2*I*d
*x + 2*I*c) + I)*e^(-5*I*d*x - 5*I*c)/(a^4*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))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^2/(I*a*tan(d*x + c) + a)^(7/2), x)