\(\int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [373]

Optimal result

Integrand size = 26, antiderivative size = 123 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {4 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}-\frac {4 i \sec (c+d x)}{a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[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)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3572, 3570, 212} \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {4 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {4 i \sec (c+d x)}{a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3570

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 3572

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*} \text {integral}& = -\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) \text {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} \text {arctanh}\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] (verified)

Time = 1.48 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 \sec (c+d x) \left (7 i-6 i \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )+\tan (c+d x)\right )}{3 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[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]])

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (102 ) = 204\).

Time = 9.84 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.18

method	result	size
default	\(\frac {2 \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )+i\right )^{5} \left (6 \sqrt {2}\, \arctan \left (\frac {\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right ) \sqrt {2}}{2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right ) {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )}^{\frac {3}{2}}-7 i \left (\csc ^{3}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{3}+9 i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+9 \left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-7\right )}{3 d {\left (-\frac {a \left (2 i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )}^{\frac {5}{2}} {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )}^{4}}\)	\(268\)

[In]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/d*(-csc(d*x+c)+cot(d*x+c)+I)^5*(6*2^(1/2)*arctan(1/2*(I*(csc(d*x+c)-cot(d*x+c))-1)*2^(1/2)/(csc(d*x+c)^2*(
1-cos(d*x+c))^2-1)^(1/2))*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(3/2)-7*I*csc(d*x+c)^3*(1-cos(d*x+c))^3+9*I*(csc(d
*x+c)-cot(d*x+c))+9*csc(d*x+c)^2*(1-cos(d*x+c))^2-7)/(-a*(2*I*(csc(d*x+c)-cot(d*x+c))-csc(d*x+c)^2*(1-cos(d*x+
c))^2+1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(5/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^4

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (96) = 192\).

Time = 0.25 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.20 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {2} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-\frac {16 \, {\left ({\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) + 3 \, \sqrt {2} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-\frac {16 \, {\left ({\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i\right )}\right )}}{3 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

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

[Out]

-2/3*(3*sqrt(2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(1/(a^5*d^2))*log(-16*((I*a^2*d*e^(2*I*d*x + 2*I*c
) + I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) + 3*sqrt(2)*(-
I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(1/(a^5*d^2))*log(-16*((-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqr
t(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) + 2*sqrt(2)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*(3*I*e^(2*I*d*x + 2*I*c) + 4*I))/(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1074 vs. \(2 (96) = 192\).

Time = 0.48 (sec) , antiderivative size = 1074, normalized size of antiderivative = 8.73 \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/3*(4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((-3*I*sqrt(2)*cos(2*d*x + 2*c
) + 3*sqrt(2)*sin(2*d*x + 2*c) - 4*I*sqrt(2))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (3*sq
rt(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) - 3*(2*(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*arct
an2(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) - 2*(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) - (I*sqrt(2)*cos(2*d*x + 2*c)^2 + I*sqrt(2)*sin(2*d*x + 2*c)^2 + 2*I*sqrt(2)*cos(2*d*x + 2*c) + 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*ar
ctan2(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) - (-I*sqrt(2)*cos(2*d*x + 2*c)^2
- I*sqrt(2)*sin(2*d*x + 2*c)^2 - 2*I*sqrt(2)*cos(2*d*x + 2*c) - 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*c
os(2*d*x + 2*c) + a^3)*d)

Giac [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[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)