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

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

[Out]

4*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d-4*I*sec(d*x+c)/a^2/d/(a
+I*a*tan(d*x+c))^(1/2)-2/3*I*sec(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.17, antiderivative size = 123, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.87, 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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fricas [B]  time = 0.52, size = 267, normalized size = 2.17 \[ \frac {\sqrt {2} {\left (6 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (\frac {{\left (2 \, {\left (8 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + 16 i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) + \sqrt {2} {\left (-6 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (\frac {{\left (2 \, {\left (-8 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + 16 i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-12 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{3 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

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)*(6*I*a^3*d*e^(2*I*d*x + 2*I*c) + 6*I*a^3*d)*sqrt(1/(a^5*d^2))*log((2*(8*I*a^2*d*e^(2*I*d*x + 2*I*
c) + 8*I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + 16*I)*e^(-I*d*x - I*c)/(a^2*d)) + sqrt(2
)*(-6*I*a^3*d*e^(2*I*d*x + 2*I*c) - 6*I*a^3*d)*sqrt(1/(a^5*d^2))*log((2*(-8*I*a^2*d*e^(2*I*d*x + 2*I*c) - 8*I*
a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + 16*I)*e^(-I*d*x - I*c)/(a^2*d)) + 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))/(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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)

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maple [B]  time = 1.21, size = 281, normalized size = 2.28 \[ \frac {2 \left (3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}+3 \sqrt {2}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+8 i \left (\cos ^{2}\left (d x +c \right )\right )+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )-7 i \cos \left (d x +c \right )-\sin \left (d x +c \right )-i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{3 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) a^{3}} \]

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

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maxima [B]  time = 1.21, size = 1070, normalized size = 8.70 \[ \text {result too large to display} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

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]

Integral(sec(c + d*x)**5/(I*a*(tan(c + d*x) - I))**(5/2), x)

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