\(\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx\) [260]

Optimal result

Integrand size = 28, antiderivative size = 163 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 a^4 d}-\frac {2 e^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

-2/77*e^3*sin(d*x+c)/a^4/d/(e*sec(d*x+c))^(1/2)-2/77*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/a^4/d+4/11*I*e^2*(e*sec(d*x+c))^(1/2)/a
/d/(a+I*a*tan(d*x+c))^3-4/77*I*e^4/d/(e*sec(d*x+c))^(3/2)/(a^4+I*a^4*tan(d*x+c))

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3854, 3856, 2720} \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=-\frac {4 i e^4}{77 d \left (a^4+i a^4 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}-\frac {2 e^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 a^4 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3} \]

[In]

Int[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*a^4*d) - (2*e^3*Sin[c + d*x])/(
77*a^4*d*Sqrt[e*Sec[c + d*x]]) + (((4*I)/11)*e^2*Sqrt[e*Sec[c + d*x]])/(a*d*(a + I*a*Tan[c + d*x])^3) - (((4*I
)/77)*e^4)/(d*(e*Sec[c + d*x])^(3/2)*(a^4 + I*a^4*Tan[c + d*x]))

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3581

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*n))), x] - Dist[d^2*((m - 2)/(b^2*(m + 2*n)
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps \begin{align*} \text {integral}& = \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {e^2 \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{11 a^2} \\ & = \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (3 e^4\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{77 a^4} \\ & = -\frac {2 e^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {e^2 \int \sqrt {e \sec (c+d x)} \, dx}{77 a^4} \\ & = -\frac {2 e^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {\left (e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 a^4} \\ & = -\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 a^4 d}-\frac {2 e^3 \sin (c+d x)}{77 a^4 d \sqrt {e \sec (c+d x)}}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{11 a d (a+i a \tan (c+d x))^3}-\frac {4 i e^4}{77 d (e \sec (c+d x))^{3/2} \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.88 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^2(c+d x) (e \sec (c+d x))^{5/2} (\cos (c+d x)+i \sin (c+d x)) \left (37 i \cos (c+d x)+11 i \cos (3 (c+d x))+3 \sin (c+d x)-4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+3 \sin (3 (c+d x))\right )}{154 a^4 d (-i+\tan (c+d x))^4} \]

[In]

Integrate[(e*Sec[c + d*x])^(5/2)/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^2*(e*Sec[c + d*x])^(5/2)*(Cos[c + d*x] + I*Sin[c + d*x])*((37*I)*Cos[c + d*x] + (11*I)*Cos[3*(c
+ d*x)] + 3*Sin[c + d*x] - 4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x
)]) + 3*Sin[3*(c + d*x)]))/(154*a^4*d*(-I + Tan[c + d*x])^4)

Maple [A] (verified)

Time = 6.85 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {2 i \left (\left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (56 \left (\cos ^{4}\left (d x +c \right )\right )-16 \left (\cos ^{2}\left (d x +c \right )\right )-1\right )+\left (\cos ^{4}\left (d x +c \right )\right ) \left (-56 \left (\cos ^{2}\left (d x +c \right )\right )+44\right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e^{2}}{77 a^{4} d}\)	\(143\)

[In]

int((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-2/77*I/a^4/d*((cos(d*x+c)+1)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x
+c)+cot(d*x+c)),I)+I*sin(d*x+c)*cos(d*x+c)*(56*cos(d*x+c)^4-16*cos(d*x+c)^2-1)+cos(d*x+c)^4*(-56*cos(d*x+c)^2+
44))*(e*sec(d*x+c))^(1/2)*e^2

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.77 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (4 i \, \sqrt {2} e^{\frac {5}{2}} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (4 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 17 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 20 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{154 \, a^{4} d} \]

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/154*(4*I*sqrt(2)*e^(5/2)*e^(6*I*d*x + 6*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(4*I*e^2*
e^(6*I*d*x + 6*I*c) + 17*I*e^2*e^(4*I*d*x + 4*I*c) + 20*I*e^2*e^(2*I*d*x + 2*I*c) + 7*I*e^2)*sqrt(e/(e^(2*I*d*
x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-6*I*d*x - 6*I*c)/(a^4*d)

Sympy [F]

\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate((e*sec(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral((e*sec(c + d*x))**(5/2)/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x)
+ 1), x)/a**4

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(5/2)/(I*a*tan(d*x + c) + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^4} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]

[In]

int((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

int((e/cos(c + d*x))^(5/2)/(a + a*tan(c + d*x)*1i)^4, x)