\(\int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx\) [238]

Optimal result

Integrand size = 28, antiderivative size = 90 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 a^2 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

-2/3*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(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^2/d+4/3*I*e^2*(e*sec(d*x+c))^(1/2)/d/(a^2+I*a^2*tan(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3581, 3856, 2720} \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 a^2 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[In]

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

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*a^2*d) + (((4*I)/3)*e^2*Sqrt[e*S
ec[c + d*x]])/(d*(a^2 + I*a^2*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 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)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {e^2 \int \sqrt {e \sec (c+d x)} \, dx}{3 a^2} \\ & = \frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \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}{3 a^2} \\ & = -\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)}}{3 a^2 d}+\frac {4 i e^2 \sqrt {e \sec (c+d x)}}{3 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 (e \sec (c+d x))^{5/2} \left (-2 i \cos (c+d x)+\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (c+d x)+i \sin (c+d x))\right ) (\cos (c+d x)+i \sin (c+d x))}{3 a^2 d (-i+\tan (c+d x))^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.51 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {2 e^{2} \left (i \cos \left (d x +c \right ) F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i \left (\cos ^{2}\left (d x +c \right )\right )+2 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}}{3 a^{2} d}\)	\(164\)

[In]

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

[Out]

2/3/a^2/d*e^2*(I*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(
1/2)*cos(d*x+c)+I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^
(1/2)+2*I*cos(d*x+c)^2+2*sin(d*x+c)*cos(d*x+c))*(e*sec(d*x+c))^(1/2)

Fricas [C] (verification not implemented)

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

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08 \[ \int \frac {(e \sec (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} e^{\frac {5}{2}} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{3 \, a^{2} d} \]

[In]

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

[Out]

-2/3*(-I*sqrt(2)*e^(5/2)*e^(2*I*d*x + 2*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-I*e^2*e^(
2*I*d*x + 2*I*c) - 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^(-2*I*d*x - 2*I*c)/(a^2
*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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