3.3.45 \(\int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx\) [245]

Optimal. Leaf size=112 \[ \frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt {e \cos (c+d x)}}+\frac {10 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

4*e*(e*cos(d*x+c))^(5/2)/d/(a^2+a^2*sin(d*x+c))+10/3*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a^2/d/(e*cos(d*x+c))^(1/2)+10/3*e^3*sin(d*x+c)*(e*cos(d*x+c)
)^(1/2)/a^2/d

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Rubi [A]
time = 0.07, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2759, 2715, 2721, 2720} \begin {gather*} \frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt {e \cos (c+d x)}}+\frac {10 e^3 \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 a^2 d}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2 \sin (c+d x)+a^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*Sqrt[e*Cos[c + d*x]]) + (10*e^3*Sqrt[e*Cos[c +
d*x]]*Sin[c + d*x])/(3*a^2*d) + (4*e*(e*Cos[c + d*x])^(5/2))/(d*(a^2 + a^2*Sin[c + d*x]))

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{7/2}}{(a+a \sin (c+d x))^2} \, dx &=\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (5 e^2\right ) \int (e \cos (c+d x))^{3/2} \, dx}{a^2}\\ &=\frac {10 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (5 e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{3 a^2}\\ &=\frac {10 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (5 e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2 \sqrt {e \cos (c+d x)}}\\ &=\frac {10 e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt {e \cos (c+d x)}}+\frac {10 e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2+a^2 \sin (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 = 0.07, size = 66, normalized size = 0.59 \begin {gather*} -\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{9/2} \, _2F_1\left (\frac {3}{4},\frac {9}{4};\frac {13}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{9 a^2 d e (1+\sin (c+d x))^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*2^(1/4)*(e*Cos[c + d*x])^(9/2)*Hypergeometric2F1[3/4, 9/4, 13/4, (1 - Sin[c + d*x])/2])/(9*a^2*d*e*(1 + Si
n[c + d*x])^(9/4))

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Maple [A]
time = 1.94, size = 155, normalized size = 1.38

method	result	size
default	\(-\frac {2 e^{4} \left (-4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^4*(-4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c
)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*sin(
1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+12*sin(1/2*d*x+1/2*c)^3-6*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((e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

e^(7/2)*integrate(cos(d*x + c)^(7/2)/(a*sin(d*x + c) + a)^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 = 83, normalized size = 0.74 \begin {gather*} \frac {-5 i \, \sqrt {2} e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (e^{\frac {7}{2}} \sin \left (d x + c\right ) - 6 \, e^{\frac {7}{2}}\right )} \sqrt {\cos \left (d x + c\right )}}{3 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-5*I*sqrt(2)*e^(7/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*e^(7/2)*weie
rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(e^(7/2)*sin(d*x + c) - 6*e^(7/2))*sqrt(cos(d*x + 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*cos(d*x+c))**(7/2)/(a+a*sin(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3878 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*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(7/2)*e^(7/2)/(a*sin(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 (e\,\cos \left (c+d\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\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*sin(c + d*x))^2,x)

[Out]

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

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