3.3.37 \(\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx\) [237]

Optimal. Leaf size=115 \[ \frac {6 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

6*e^4*(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^2/d/cos(d*x+c)^(
1/2)/(e*sec(d*x+c))^(1/2)-6*e^3*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/a^2/d+4*I*e^2*(e*sec(d*x+c))^(3/2)/d/(a^2+I*a^
2*tan(d*x+c))

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Rubi [A]
time = 0.07, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3853, 3856, 2719} \begin {gather*} \frac {6 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{a^2 d}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(6*e^4*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*e^3*Sqrt[e*Sec[c + d*x]
]*Sin[c + d*x])/(a^2*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(3/2))/(d*(a^2 + I*a^2*Tan[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 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 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 \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx &=\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (3 e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{a^2}\\ &=-\frac {6 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 e^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{a^2}\\ &=-\frac {6 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 e^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {6 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {4 i e^2 (e \sec (c+d x))^{3/2}}{d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.59, size = 80, normalized size = 0.70 \begin {gather*} \frac {2 i e^3 e^{-i (c+d x)} \left (-1+3 \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 )\right ) \sqrt {e \sec (c+d x)}}{a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(7/2)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((2*I)*e^3*(-1 + 3*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^2*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. 351 vs. \(2 (131 ) = 262\).
time = 0.70, size = 352, normalized size = 3.06

method	result	size
default	\(\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )-1\right )}{a^{2} d \sin \left (d x +c \right )^{5}}\)	\(352\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

2/a^2/d*(1+cos(d*x+c))^2*cos(d*x+c)^3*(-1+cos(d*x+c))^2*(e/cos(d*x+c))^(7/2)*(3*I*(1/(1+cos(d*x+c)))^(1/2)*(co
s(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
)+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)*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)*sin(d*x+c)*EllipticE(I*(-1+cos(d*x+c))/
sin(d*x+c),I)+2*I*cos(d*x+c)*sin(d*x+c)-2*cos(d*x+c)^2+3*cos(d*x+c)-1)/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))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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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 = 0.79 \begin {gather*} -\frac {2 \, {\left (-3 i \, \sqrt {2} e^{\left (i \, d x + i \, c + \frac {7}{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 (-2 i \, e^{\frac {7}{2}} - 3 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {7}{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^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2*(-3*I*sqrt(2)*e^(I*d*x + I*c + 7/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + s
qrt(2)*(-2*I*e^(7/2) - 3*I*e^(2*I*d*x + 2*I*c + 7/2))*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^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

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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))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(e^(7/2)*sec(d*x + c)^(7/2)/(I*a*tan(d*x + c) + a)^2, 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 )}^{7/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

int((e/cos(c + d*x))^(7/2)/(a + a*tan(c + d*x)*1i)^2, x)

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