3.3.15 \(\int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx\) [215]

Optimal. Leaf size=170 \[ -\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}+\frac {2 a^3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^3 e (e \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e} \]

[Out]

-10/21*a^3*(e*cos(d*x+c))^(7/2)/d/e+2/3*a^3*e*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/d-2/11*a*(e*cos(d*x+c))^(7/2)*(a
+a*sin(d*x+c))^2/d/e-10/33*(e*cos(d*x+c))^(7/2)*(a^3+a^3*sin(d*x+c))/d/e+2*a^3*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))*(e*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2757, 2748, 2715, 2721, 2719} \begin {gather*} \frac {2 a^3 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}-\frac {10 \left (a^3 \sin (c+d x)+a^3\right ) (e \cos (c+d x))^{7/2}}{33 d e}+\frac {2 a^3 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{3 d}-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{7/2}}{11 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^3,x]

[Out]

(-10*a^3*(e*Cos[c + d*x])^(7/2))/(21*d*e) + (2*a^3*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt
[Cos[c + d*x]]) + (2*a^3*e*(e*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) - (2*a*(e*Cos[c + d*x])^(7/2)*(a + a*Sin
[c + d*x])^2)/(11*d*e) - (10*(e*Cos[c + d*x])^(7/2)*(a^3 + a^3*Sin[c + d*x]))/(33*d*e)

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 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 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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2757

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^3 \, dx &=-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}+\frac {1}{11} (15 a) \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e}+\frac {1}{3} \left (5 a^2\right ) \int (e \cos (c+d x))^{5/2} (a+a \sin (c+d x)) \, dx\\ &=-\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e}+\frac {1}{3} \left (5 a^3\right ) \int (e \cos (c+d x))^{5/2} \, dx\\ &=-\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}+\frac {2 a^3 e (e \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e}+\left (a^3 e^2\right ) \int \sqrt {e \cos (c+d x)} \, dx\\ &=-\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}+\frac {2 a^3 e (e \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e}+\frac {\left (a^3 e^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)}}\\ &=-\frac {10 a^3 (e \cos (c+d x))^{7/2}}{21 d e}+\frac {2 a^3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 a^3 e (e \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {2 a (e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2}{11 d e}-\frac {10 (e \cos (c+d x))^{7/2} \left (a^3+a^3 \sin (c+d x)\right )}{33 d e}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.06, size = 66, normalized size = 0.39 \begin {gather*} -\frac {32\ 2^{3/4} a^3 (e \cos (c+d x))^{7/2} \, _2F_1\left (-\frac {15}{4},\frac {7}{4};\frac {11}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{7 d e (1+\sin (c+d x))^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^3,x]

[Out]

(-32*2^(3/4)*a^3*(e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[-15/4, 7/4, 11/4, (1 - Sin[c + d*x])/2])/(7*d*e*(1 +
Sin[c + d*x])^(7/4))

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Maple [A]
time = 2.44, size = 264, normalized size = 1.55

method	result	size
default	\(\frac {2 a^{3} e^{3} \left (1344 \left (\sin ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2464 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4032 \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4928 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2928 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3080 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+864 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+616 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1908 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+231 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+804 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-111 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{231 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(264\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

2/231/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*e^3*(1344*sin(1/2*d*x+1/2*c)^13-2464*sin(1/2*
d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-4032*sin(1/2*d*x+1/2*c)^11+4928*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+2928*
sin(1/2*d*x+1/2*c)^9-3080*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+864*sin(1/2*d*x+1/2*c)^7+616*sin(1/2*d*x+1/2
*c)^4*cos(1/2*d*x+1/2*c)-1908*sin(1/2*d*x+1/2*c)^5+231*(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))+804*sin(1/2*d*x+1/2*c)^3-111*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))^(5/2)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

e^(5/2)*integrate((a*sin(d*x + c) + a)^3*cos(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 145, normalized size = 0.85 \begin {gather*} \frac {231 i \, \sqrt {2} a^{3} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 231 i \, \sqrt {2} a^{3} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, a^{3} \cos \left (d x + c\right )^{5} e^{\frac {5}{2}} - 132 \, a^{3} \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} - 77 \, {\left (a^{3} \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} - a^{3} \cos \left (d x + c\right ) e^{\frac {5}{2}}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{231 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/231*(231*I*sqrt(2)*a^3*e^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c))) - 231*I*sqrt(2)*a^3*e^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c))) + 2*(21*a^3*cos(d*x + c)^5*e^(5/2) - 132*a^3*cos(d*x + c)^3*e^(5/2) - 77*(a^3*cos(d*x + c)^3*e^(5/2) - a^
3*cos(d*x + c)*e^(5/2))*sin(d*x + c))*sqrt(cos(d*x + c)))/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))**(5/2)*(a+a*sin(d*x+c))**3,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 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))^(5/2)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3*cos(d*x + c)^(5/2)*e^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))^3,x)

[Out]

int((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))^3, x)

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