3.7.70 \(\int \frac {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx\) [670]

Optimal. Leaf size=122 \[ \frac {6 \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

6*cos(d*x+c)^(7/2)*(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^2/d
/(e*cos(d*x+c))^(7/2)-6*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(7/2)+4*I*cos(d*x+c)^2/d/(e*cos(d*x+c))^(
7/2)/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]
time = 0.12, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3853, 3856, 2719} \begin {gather*} \frac {6 \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac {6 \sin (c+d x) \cos ^3(c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(6*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2])/(a^2*d*(e*Cos[c + d*x])^(7/2)) - (6*Cos[c + d*x]^3*Sin[c + d*
x])/(a^2*d*(e*Cos[c + d*x])^(7/2)) + ((4*I)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(7/2)*(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 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[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 {1}{(e \cos (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx &=\frac {\int \frac {(e \sec (c+d x))^{7/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \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 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=-\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \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 (e \cos (c+d x))^{7/2} (e \sec (c+d x))^{7/2}}\\ &=-\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 \cos ^{\frac {7}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2 (e \cos (c+d x))^{7/2}}\\ &=\frac {6 \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (e \cos (c+d x))^{7/2}}-\frac {6 \cos ^3(c+d x) \sin (c+d x)}{a^2 d (e \cos (c+d x))^{7/2}}+\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(255\) vs. \(2(122)=244\).
time = 1.06, size = 255, normalized size = 2.09 \begin {gather*} \frac {10 i \cos (c+d x)-2 \sin (c+d x)-6 i F\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right ) (\cos (c+d x)-i \sin (c+d x)) \sqrt {1-i \cos (c+d x)+\sin (c+d x)} \sqrt {-i \cos (c+d x)+\cos (2 (c+d x))+\sin (c+d x)+i \sin (2 (c+d x))}+6 E\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right ) \sqrt {1-i \cos (c+d x)+\sin (c+d x)} (i \cos (c+d x)+\sin (c+d x)) \sqrt {-i \cos (c+d x)+\cos (2 (c+d x))+\sin (c+d x)+i \sin (2 (c+d x))}}{a^2 d e^3 \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

((10*I)*Cos[c + d*x] - 2*Sin[c + d*x] - (6*I)*EllipticF[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1]*(C
os[c + d*x] - I*Sin[c + d*x])*Sqrt[1 - I*Cos[c + d*x] + Sin[c + d*x]]*Sqrt[(-I)*Cos[c + d*x] + Cos[2*(c + d*x)
] + Sin[c + d*x] + I*Sin[2*(c + d*x)]] + 6*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1]*Sqrt[
1 - I*Cos[c + d*x] + Sin[c + d*x]]*(I*Cos[c + d*x] + Sin[c + d*x])*Sqrt[(-I)*Cos[c + d*x] + Cos[2*(c + d*x)] +
Sin[c + d*x] + I*Sin[2*(c + d*x)]])/(a^2*d*e^3*Sqrt[e*Cos[c + d*x]])

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Maple [A]
time = 1.01, size = 135, normalized size = 1.11

method	result	size
default	\(-\frac {2 \left (4 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{e^{3} a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-2/e^3/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(4*I*sin(1/2*d*x+1/2*c)^3+2*sin(1/2*d*x+1/2*
c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))-2*I*sin(1/2*d*x+1/2*c))/d

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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(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

e^(-7/2)*integrate(1/((I*a*tan(d*x + c) + a)^2*cos(d*x + c)^(7/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 105, normalized size = 0.86 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} {\left (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 3 \, {\left (-i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{a^{2} d e^{\frac {7}{2}} + a^{2} d e^{\left (2 i \, d x + 2 i \, c + \frac {7}{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2*(2*sqrt(1/2)*sqrt(e^(2*I*d*x + 2*I*c) + 1)*(-3*I*e^(2*I*d*x + 2*I*c) - 2*I)*e^(-1/2*I*d*x - 1/2*I*c) + 3*(-
I*sqrt(2)*e^(2*I*d*x + 2*I*c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c)))
)/(a^2*d*e^(7/2) + a^2*d*e^(2*I*d*x + 2*I*c + 7/2))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (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(1/((e*cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

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

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