\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx\) [412]

Optimal result

Integrand size = 30, antiderivative size = 38 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

-2/5*I*(a+I*a*tan(d*x+c))^(5/2)/d/(e*sec(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {3569} \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \]

[In]

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

[Out]

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

Rule 3569

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 9.62 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {2 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}{5 d \sqrt {e \sec \left (d x +c \right )}\, e^{2}}\)	\(57\)
risch	\(-\frac {2 i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{5 e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(74\)

[In]

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

[Out]

2/5*I/d*cos(d*x+c)^2*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(tan(d*x+c)-I)^2/(e*sec(d*x+c))^(1/2)/e^2

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. 80 vs. \(2 (28) = 56\).

Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.11 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (-i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, d e^{3}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

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. 76 vs. \(2 (28) = 56\).

Time = 0.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.00 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {2 i \, a^{\frac {5}{2}} {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {5}{2}}}{5 \, d e^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/5*I*a^(5/2)*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(5/2)/(d*e^(5/
2)*(-sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(5/2))

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.74 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx=-\frac {a^2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-\sin \left (c+d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,1{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{5\,d\,e^3} \]

[In]

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

[Out]

-(a^2*(e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(
cos(c + d*x)*1i - sin(c + d*x) + cos(3*c + 3*d*x)*1i - sin(3*c + 3*d*x)))/(5*d*e^3)