3.3.63 \(\int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx\) [263]

Optimal. Leaf size=180 \[ \frac {78 e^8 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 a^4 d \sqrt {e \cos (c+d x)}}+\frac {78 e^7 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^4 d}+\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

234/35*e^5*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a^4/d+4*e*(e*cos(d*x+c))^(13/2)/a/d/(a+a*sin(d*x+c))^3+52/5*e^3*(e*
cos(d*x+c))^(9/2)/d/(a^4+a^4*sin(d*x+c))+78/7*e^8*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a^4/d/(e*cos(d*x+c))^(1/2)+78/7*e^7*sin(d*x+c)*(e*cos(d*x+c))^(1/2)
/a^4/d

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Rubi [A]
time = 0.13, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {78 e^8 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 a^4 d \sqrt {e \cos (c+d x)}}+\frac {78 e^7 \sin (c+d x) \sqrt {e \cos (c+d x)}}{7 a^4 d}+\frac {234 e^5 \sin (c+d x) (e \cos (c+d x))^{5/2}}{35 a^4 d}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(15/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(78*e^8*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^4*d*Sqrt[e*Cos[c + d*x]]) + (78*e^7*Sqrt[e*Cos[c +
d*x]]*Sin[c + d*x])/(7*a^4*d) + (234*e^5*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(35*a^4*d) + (4*e*(e*Cos[c + d*x
])^(13/2))/(a*d*(a + a*Sin[c + d*x])^3) + (52*e^3*(e*Cos[c + d*x])^(9/2))/(5*d*(a^4 + a^4*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))^{15/2}}{(a+a \sin (c+d x))^4} \, dx &=\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {\left (13 e^2\right ) \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx}{a^2}\\ &=\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\left (117 e^4\right ) \int (e \cos (c+d x))^{7/2} \, dx}{5 a^4}\\ &=\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\left (117 e^6\right ) \int (e \cos (c+d x))^{3/2} \, dx}{7 a^4}\\ &=\frac {78 e^7 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^4 d}+\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\left (39 e^8\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{7 a^4}\\ &=\frac {78 e^7 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^4 d}+\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\left (39 e^8 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 a^4 \sqrt {e \cos (c+d x)}}\\ &=\frac {78 e^8 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 a^4 d \sqrt {e \cos (c+d x)}}+\frac {78 e^7 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^4 d}+\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \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 = 0.30, size = 66, normalized size = 0.37 \begin {gather*} -\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{17/2} \, _2F_1\left (\frac {3}{4},\frac {17}{4};\frac {21}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{17 a^4 d e (1+\sin (c+d x))^{17/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(15/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(-2*2^(1/4)*(e*Cos[c + d*x])^(17/2)*Hypergeometric2F1[3/4, 17/4, 21/4, (1 - Sin[c + d*x])/2])/(17*a^4*d*e*(1 +
Sin[c + d*x])^(17/4))

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Maple [A]
time = 2.56, size = 225, normalized size = 1.25

method	result	size
default	\(-\frac {2 e^{8} \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-224 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-280 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+336 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+195 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+160 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+392 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-252 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-2/35/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^8*(80*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^8-120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-224*sin(1/2*d*x+1/2*c)^7-280*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1
/2*c)+336*sin(1/2*d*x+1/2*c)^5+195*(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))+160*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+392*sin(1/2*d*x+1/2*c)^3-252*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))^(15/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

e^(15/2)*integrate(cos(d*x + c)^(15/2)/(a*sin(d*x + c) + a)^4, x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 110, normalized size = 0.61 \begin {gather*} \frac {-195 i \, \sqrt {2} e^{\frac {15}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 195 i \, \sqrt {2} e^{\frac {15}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (28 \, \cos \left (d x + c\right )^{2} e^{\frac {15}{2}} - 5 \, {\left (\cos \left (d x + c\right )^{2} e^{\frac {15}{2}} - 17 \, e^{\frac {15}{2}}\right )} \sin \left (d x + c\right ) - 280 \, e^{\frac {15}{2}}\right )} \sqrt {\cos \left (d x + c\right )}}{35 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/35*(-195*I*sqrt(2)*e^(15/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 195*I*sqrt(2)*e^(15/
2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(28*cos(d*x + c)^2*e^(15/2) - 5*(cos(d*x + c)
^2*e^(15/2) - 17*e^(15/2))*sin(d*x + c) - 280*e^(15/2))*sqrt(cos(d*x + c)))/(a^4*d)

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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((e*cos(d*x+c))**(15/2)/(a+a*sin(d*x+c))**4,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((e*cos(d*x+c))^(15/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(15/2)*e^(15/2)/(a*sin(d*x + c) + a)^4, 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 )}^{15/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(15/2)/(a + a*sin(c + d*x))^4,x)

[Out]

int((e*cos(c + d*x))^(15/2)/(a + a*sin(c + d*x))^4, x)

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