3.380 \(\int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d}+\frac{8 i \sqrt{a+i a \tan (c+d x)}}{a^4 d}+\frac{8 i}{a^3 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(8*I)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((8*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^4*d) - (((2*I)/3)*(a + I*a*Ta
n[c + d*x])^(3/2))/(a^5*d)

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Rubi [A]  time = 0.0795285, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3487, 43} \[ -\frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d}+\frac{8 i \sqrt{a+i a \tan (c+d x)}}{a^4 d}+\frac{8 i}{a^3 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(8*I)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((8*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^4*d) - (((2*I)/3)*(a + I*a*Ta
n[c + d*x])^(3/2))/(a^5*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{(a-x)^2}{(a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (\frac{4 a^2}{(a+x)^{3/2}}-\frac{4 a}{\sqrt{a+x}}+\sqrt{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=\frac{8 i}{a^3 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i \sqrt{a+i a \tan (c+d x)}}{a^4 d}-\frac{2 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d}\\ \end{align*}

Mathematica [A]  time = 0.295884, size = 61, normalized size = 0.73 \[ \frac{2 i \sec ^2(c+d x) (5 i \sin (2 (c+d x))+11 \cos (2 (c+d x))+12)}{3 a^3 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((2*I)/3)*Sec[c + d*x]^2*(12 + 11*Cos[2*(c + d*x)] + (5*I)*Sin[2*(c + d*x)]))/(a^3*d*Sqrt[a + I*a*Tan[c + d*x
]])

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Maple [A]  time = 0.28, size = 88, normalized size = 1.1 \begin{align*}{\frac{24\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+24\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +22\,i\cos \left ( dx+c \right ) +2\,\sin \left ( dx+c \right ) }{3\,{a}^{4}d\cos \left ( dx+c \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

2/3/d/a^4*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(12*I*cos(d*x+c)^3+12*cos(d*x+c)^2*sin(d*x+c)+11*I*co
s(d*x+c)+sin(d*x+c))/cos(d*x+c)

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Maxima [A]  time = 0.978093, size = 84, normalized size = 1. \begin{align*} \frac{2 i \,{\left (\frac{12}{\sqrt{i \, a \tan \left (d x + c\right ) + a} a^{2}} - \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} - 12 \, \sqrt{i \, a \tan \left (d x + c\right ) + a} a}{a^{4}}\right )}}{3 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/3*I*(12/(sqrt(I*a*tan(d*x + c) + a)*a^2) - ((I*a*tan(d*x + c) + a)^(3/2) - 12*sqrt(I*a*tan(d*x + c) + a)*a)/
a^4)/(a*d)

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Fricas [A]  time = 2.16177, size = 243, normalized size = 2.89 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (32 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 48 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 12 i\right )} e^{\left (i \, d x + i \, c\right )}}{3 \,{\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(32*I*e^(4*I*d*x + 4*I*c) + 48*I*e^(2*I*d*x + 2*I*c) + 12*I)*e^(
I*d*x + I*c)/(a^4*d*e^(4*I*d*x + 4*I*c) + a^4*d*e^(2*I*d*x + 2*I*c))

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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)**6/(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 )^{6}}{{\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)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^6/(I*a*tan(d*x + c) + a)^(7/2), x)