\(\int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx\) [227]

Optimal result

Integrand size = 28, antiderivative size = 70 \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \]

[Out]

2*I*e^2/a/d/(e*sec(d*x+c))^(1/2)+2*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1
/2*c),2^(1/2))/a/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 70, 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 = {3582, 3856, 2719} \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}} \]

[In]

Int[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x]),x]

[Out]

((2*I)*e^2)/(a*d*Sqrt[e*Sec[c + d*x]]) + (2*e^2*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[
c + d*x]])

Rule 2719

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

Rule 3582

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d^2*(
d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Dist[d^2*((m - 2)/(a*(m + n - 1
))), 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] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 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 {2 i e^2}{a d \sqrt {e \sec (c+d x)}}+\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{a} \\ & = \frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}+\frac {e^2 \int \sqrt {\cos (c+d x)} \, dx}{a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\frac {2 i e e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}}{a d} \]

[In]

Integrate[(e*Sec[c + d*x])^(3/2)/(a + I*a*Tan[c + d*x]),x]

[Out]

((2*I)*e*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))]*Sqrt[e*Sec[c +
d*x]])/(a*d*E^(I*(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. 779 vs. \(2 (89 ) = 178\).

Time = 4.51 (sec) , antiderivative size = 780, normalized size of antiderivative = 11.14

method	result	size
default	\(\text {Expression too large to display}\)	\(780\)

[In]

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/2*e/a/d*(-e*(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(1/2)*(4*I*(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)^(1/2)*((csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*(csc(d*x+
c)^2*(1-cos(d*x+c))^2-1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)-4*I*(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)^(1/2)*((csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*(csc(d*x+c)^2*(1-cos(d*x+c))^
2-1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-I*ln(2*csc(d*x+c)^2*(1-cos(d*x+c))^2+2*(csc(d*x+c)^4*(1-cos
(d*x+c))^4-1)^(1/2))*(csc(d*x+c)^4*(1-cos(d*x+c))^4-1)^(1/2)*((csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*(csc(d*x+c)^2*
(1-cos(d*x+c))^2-1))^(1/2)+I*(csc(d*x+c)^4*(1-cos(d*x+c))^4-1)^(1/2)*((csc(d*x+c)^2*(1-cos(d*x+c))^2+1)*(csc(d
*x+c)^2*(1-cos(d*x+c))^2-1))^(1/2)*ln(csc(d*x+c)^2*(1-cos(d*x+c))^2+(csc(d*x+c)^4*(1-cos(d*x+c))^4-1)^(1/2))+4
*I*csc(d*x+c)^2*(csc(d*x+c)^4*(1-cos(d*x+c))^4-1)^(1/2)*(1-cos(d*x+c))^2+4*csc(d*x+c)^3*(csc(d*x+c)^4*(1-cos(d
*x+c))^4-1)^(1/2)*(1-cos(d*x+c))^3-4*I*(csc(d*x+c)^4*(1-cos(d*x+c))^4-1)^(1/2)-4*(csc(d*x+c)^4*(1-cos(d*x+c))^
4-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(csc(d*x+c)^4*(1-cos(d*x+c))^4-1)^(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 = 96, normalized size of antiderivative = 1.37 \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} e^{\frac {3}{2}} e^{\left (i \, d x + i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {2} {\left (-i \, e e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a d} \]

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2*(-I*sqrt(2)*e^(3/2)*e^(I*d*x + I*c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + s
qrt(2)*(-I*e*e^(2*I*d*x + 2*I*c) - I*e)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-I*d*x -
I*c)/(a*d)

Sympy [F]

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

[In]

integrate((e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c)),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(3/2)/(I*a*tan(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sec (c+d x))^{3/2}}{a+i a \tan (c+d x)} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i),x)

[Out]

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i), x)