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

Optimal. Leaf size=123 \[ -\frac{4 i \sec (c+d x)}{a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{4 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.169913, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3491, 3489, 206} \[ -\frac{4 i \sec (c+d x)}{a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{4 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac{2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3491

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

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

Mathematica [A]  time = 0.728286, size = 82, normalized size = 0.67 \[ -\frac{2 \sec (c+d x) \left (\tan (c+d x)-6 i \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )+7 i\right )}{3 a^2 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[c + d*x]*(7*I - (6*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] + Tan[c + d
*x]))/(3*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.285, size = 281, normalized size = 2.3 \begin{align*}{\frac{2}{3\,d{a}^{3} \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) \cos \left ( dx+c \right ) } \left ( 3\,\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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \sqrt{2}+3\,\sqrt{2}\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 ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) +8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-7\,i\cos \left ( dx+c \right ) +8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -i-\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))^(5/2),x)

[Out]

2/3/d/a^3*(3*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*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)*cos(d*x+c)*2^(1/2)+3*2^(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*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+
8*I*cos(d*x+c)^2-7*I*cos(d*x+c)+8*cos(d*x+c)*sin(d*x+c)-I-sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))
^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)

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Maxima [B]  time = 2.17721, size = 1445, normalized size = 11.75 \begin{align*} \text{result too large to display} \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))^(5/2),x, algorithm="maxima")

[Out]

-1/3*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((12*I*sqrt(2)*cos(2*d*x + 2*c)
- 12*sqrt(2)*sin(2*d*x + 2*c) + 16*I*sqrt(2))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 4*(3
*sqrt(2)*cos(2*d*x + 2*c) + 3*I*sqrt(2)*sin(2*d*x + 2*c) + 4*sqrt(2))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) + 1)))*sqrt(a) + (6*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x
+ 2*c) + sqrt(2))*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 6*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt
(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^
2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^
2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)
) - 1) + (-3*I*sqrt(2)*cos(2*d*x + 2*c)^2 - 3*I*sqrt(2)*sin(2*d*x + 2*c)^2 - 6*I*sqrt(2)*cos(2*d*x + 2*c) - 3*
I*sqrt(2))*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*si
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(
2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + (3*I*sqrt(2)*cos(2*d*x
+ 2*c)^2 + 3*I*sqrt(2)*sin(2*d*x + 2*c)^2 + 6*I*sqrt(2)*cos(2*d*x + 2*c) + 3*I*sqrt(2))*log(sqrt(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1))^2 - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1))*sqrt(a))/((a^3*cos(2*d*x + 2*c)^2 + a^3*sin(2*d*x + 2*c
)^2 + 2*a^3*cos(2*d*x + 2*c) + a^3)*d)

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Fricas [B]  time = 2.16131, size = 872, normalized size = 7.09 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-12 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )} e^{\left (i \, d x + i \, c\right )} + \sqrt{2}{\left (6 i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt{\frac{1}{a^{5} d^{2}}} \log \left ({\left (\sqrt{2} a^{3} d \sqrt{\frac{1}{a^{5} 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}{\left (-6 i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} - 6 i \, a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt{\frac{1}{a^{5} d^{2}}} \log \left (-{\left (\sqrt{2} a^{3} d \sqrt{\frac{1}{a^{5} 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 \,{\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )}} \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))^(5/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-12*I*e^(2*I*d*x + 2*I*c) - 16*I)*e^(I*d*x + I*c) + sqrt(2)*(6
*I*a^3*d*e^(3*I*d*x + 3*I*c) + 6*I*a^3*d*e^(I*d*x + I*c))*sqrt(1/(a^5*d^2))*log((sqrt(2)*a^3*d*sqrt(1/(a^5*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)) + sqrt(2)*(-6*I*a^3*d*e^(3*I*d*x + 3*I*c) - 6*I*a^3*d*e^(I*d*x + I*c))*sqrt(1/(a^5*d^2))*log(-(s
qrt(2)*a^3*d*sqrt(1/(a^5*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)))/(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + 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)**5/(a+I*a*tan(d*x+c))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

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