3.389 \(\int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=121 \[ \frac{6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{3 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((-3*I)*Sqrt[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(a^(7/2)*d) - ((2*I)*Sec
[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((6*I)*Sec[c + d*x])/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2))

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Rubi [A]  time = 0.21665, antiderivative size = 121, 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 = {3501, 3502, 3489, 206} \[ \frac{6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{3 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-3*I)*Sqrt[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(a^(7/2)*d) - ((2*I)*Sec
[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((6*I)*Sec[c + d*x])/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2))

Rule 3501

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d^2*
(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(d^2*(m - 2))/(a*(m + n -
1)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
+ b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*a)/(b*f), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, 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 \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}}+\frac{6 \int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx}{a}\\ &=-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}}+\frac{12 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{12 \int \frac{\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{a^2}\\ &=-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}}+\frac{6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{3 \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{a^3}\\ &=-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}}+\frac{6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{1}{2-a x^2} \, dx,x,\frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^3 d}\\ &=-\frac{3 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac{2 i \sec ^3(c+d x)}{a d (a+i a \tan (c+d x))^{5/2}}+\frac{6 i \sec (c+d x)}{a^2 d (a+i a \tan (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.766839, size = 126, normalized size = 1.04 \[ \frac{16 e^{5 i (c+d x)} \left (-3 e^{2 i (c+d x)}+3 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-1\right )}{a^3 d \left (1+e^{2 i (c+d x)}\right )^4 (\tan (c+d x)-i)^3 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(16*E^((5*I)*(c + d*x))*(-1 - 3*E^((2*I)*(c + d*x)) + 3*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcT
anh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(a^3*d*(1 + E^((2*I)*(c + d*x)))^4*(-I + Tan[c + d*x])^3*Sqrt[a + I*a*Tan
[c + d*x]])

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Maple [B]  time = 0.279, size = 318, normalized size = 2.6 \begin{align*} -{\frac{1}{2\,{a}^{4}d} \left ( 3\,i\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}\cos \left ( dx+c \right ) +3\,i\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}+3\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) -8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -4\,\sin \left ( dx+c \right ) \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)^5/(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

-1/2/d/a^4*(3*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)
/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)+3*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*
2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+3*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2))*2^(1/2)*sin(d*x+c)-8*I*cos(d*x+c)^3-8*cos(d*x+c)^2*sin(d*x+c)-4*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c
))/cos(d*x+c))^(1/2)

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

[Out]

Timed out

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Fricas [B]  time = 2.11668, size = 756, normalized size = 6.25 \begin{align*} \frac{{\left (-3 i \, \sqrt{2} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left ({\left (\sqrt{2} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 i \, \sqrt{2} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-{\left (\sqrt{2} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{2 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/2*(-3*I*sqrt(2)*a^4*d*sqrt(1/(a^7*d^2))*e^(3*I*d*x + 3*I*c)*log((sqrt(2)*a^4*d*sqrt(1/(a^7*d^2))*e^(I*d*x +
I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c))
+ 3*I*sqrt(2)*a^4*d*sqrt(1/(a^7*d^2))*e^(3*I*d*x + 3*I*c)*log(-(sqrt(2)*a^4*d*sqrt(1/(a^7*d^2))*e^(I*d*x + I*c
) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + s
qrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(6*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(I*d*x + I*c))*e^(-3*I*d*x - 3*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)**5/(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 )^{5}}{{\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)^5/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/(I*a*tan(d*x + c) + a)^(7/2), x)