3.2.87 \(\int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx\) [187]

Optimal. Leaf size=90 \[ -\frac {2 a e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}+\frac {2 a e \sqrt {e \sec (c+d x)} \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3567, 3853, 3856, 2719} \begin {gather*} -\frac {2 a e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}+\frac {2 a e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d} \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*a*e^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((2*I)/3)*a*(e*Sec[c + d*x
])^(3/2))/d + (2*a*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d

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 3567

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

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx &=\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}+a \int (e \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}+\frac {2 a e \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}-\left (a e^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}+\frac {2 a e \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}-\frac {\left (a e^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {2 a e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}+\frac {2 a e \sqrt {e \sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.82, size = 102, normalized size = 1.13 \begin {gather*} \frac {2 a e e^{-2 i d x} \sqrt {e \sec (c+d x)} (\cos (c+3 d x)+i \sin (c+3 d x)) \left (-2 i+i \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+\tan (c+d x)\right )}{3 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*a*e*Sqrt[e*Sec[c + d*x]]*(Cos[c + 3*d*x] + I*Sin[c + 3*d*x])*(-2*I + I*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyperg
eometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + Tan[c + d*x]))/(3*d*E^((2*I)*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. 350 vs. \(2 (105 ) = 210\).
time = 0.52, size = 351, normalized size = 3.90

method	result	size
default	\(-\frac {2 a \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-i \sin \left (d x +c \right )+3 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{3 d \sin \left (d x +c \right )^{5}}\)	\(351\)

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]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 113, normalized size = 1.26 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (3 i \, a e^{\left (3 i \, d x + 3 i \, c + \frac {3}{2}\right )} + i \, a e^{\left (i \, d x + 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}} + 3 \, {\left (i \, \sqrt {2} a e^{\frac {3}{2}} + i \, \sqrt {2} a e^{\left (2 i \, d x + 2 i \, c + \frac {3}{2}\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \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]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\, dx\right ) \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*a*(Integral(-I*(e*sec(c + d*x))**(3/2), x) + Integral((e*sec(c + d*x))**(3/2)*tan(c + d*x), x))

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,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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