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

Optimal. Leaf size=70 \[ \frac {2 i e^2}{a d \sqrt {e \sec (c+d x)}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \]

[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)

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Rubi [A]
time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3582, 3856, 2719} \begin {gather*} \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)}} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.38, size = 74, normalized size = 1.06 \begin {gather*} \frac {2 i e e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\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} \end {gather*}

Antiderivative was successfully verified.

[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)))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (89 ) = 178\).
time = 4.61, size = 347, normalized size = 4.96

method	result	size
default	\(-\frac {2 \cos \left (d x +c \right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-i \cos \left (d x +c \right ) \sin \left (d x +c \right )+\cos ^{2}\left (d x +c \right )-\cos \left (d x +c \right )\right )}{a d \sin \left (d x +c \right )^{5}}\)	\(347\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a/d*cos(d*x+c)*(e/cos(d*x+c))^(3/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*(I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-I*(1/(1+cos(d*x+c)))
^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+I*(1/
(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-I
*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),
I)-I*cos(d*x+c)*sin(d*x+c)+cos(d*x+c)^2-cos(d*x+c))/sin(d*x+c)^5

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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.

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 91, normalized size = 1.30 \begin {gather*} -\frac {2 \, {\left (-i \, \sqrt {2} e^{\left (i \, d x + i \, c + \frac {3}{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \frac {\sqrt {2} {\left (-i \, e^{\frac {3}{2}} - i \, e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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