3.1.42 \(\int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx\) [42]

Optimal. Leaf size=154 \[ \frac {2 \left (9 a^2+2 b^2\right ) e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 d \sqrt {\sin (c+d x)}}-\frac {2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac {22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e} \]

[Out]

-2/45*(9*a^2+2*b^2)*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/d+22/63*a*b*(e*sin(d*x+c))^(7/2)/d/e+2/9*b*(a+b*cos(d*x+
c))*(e*sin(d*x+c))^(7/2)/d/e-2/15*(9*a^2+2*b^2)*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d
*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/sin(d*x+c)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2771, 2748, 2715, 2721, 2719} \begin {gather*} \frac {2 e^2 \left (9 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 d \sqrt {\sin (c+d x)}}-\frac {2 e \left (9 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac {22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac {2 b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))}{9 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9*a^2 + 2*b^2)*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(15*d*Sqrt[Sin[c + d*x]]) - (2*(
9*a^2 + 2*b^2)*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(45*d) + (22*a*b*(e*Sin[c + d*x])^(7/2))/(63*d*e) + (2*b
*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(7/2))/(9*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 2771

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[1/(m + p), Int[(g*Cos[e + f*x]
)^p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ
[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ
[m])

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx &=\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac {2}{9} \int \left (\frac {9 a^2}{2}+b^2+\frac {11}{2} a b \cos (c+d x)\right ) (e \sin (c+d x))^{5/2} \, dx\\ &=\frac {22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac {1}{9} \left (9 a^2+2 b^2\right ) \int (e \sin (c+d x))^{5/2} \, dx\\ &=-\frac {2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac {22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac {1}{15} \left (\left (9 a^2+2 b^2\right ) e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx\\ &=-\frac {2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac {22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac {\left (\left (9 a^2+2 b^2\right ) e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{15 \sqrt {\sin (c+d x)}}\\ &=\frac {2 \left (9 a^2+2 b^2\right ) e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 d \sqrt {\sin (c+d x)}}-\frac {2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac {22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}\\ \end {align*}

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Mathematica [A]
time = 0.84, size = 116, normalized size = 0.75 \begin {gather*} -\frac {(e \sin (c+d x))^{5/2} \left (84 \left (9 a^2+2 b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+\left (21 \left (12 a^2+b^2\right ) \cos (c+d x)+5 b (-36 a+36 a \cos (2 (c+d x))+7 b \cos (3 (c+d x)))\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{630 d \sin ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/630*((e*Sin[c + d*x])^(5/2)*(84*(9*a^2 + 2*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + (21*(12*a^2 + b^2)*Co
s[c + d*x] + 5*b*(-36*a + 36*a*Cos[2*(c + d*x)] + 7*b*Cos[3*(c + d*x)]))*Sin[c + d*x]^(3/2)))/(d*Sin[c + d*x]^
(5/2))

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Maple [A]
time = 0.15, size = 332, normalized size = 2.16

method	result	size
default	\(\frac {\frac {4 a b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 e}-\frac {e^{3} \left (10 b^{2} \left (\sin ^{6}\left (d x +c \right )\right )+54 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+12 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-27 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-18 a^{2} \left (\sin ^{4}\left (d x +c \right )\right )-14 \left (\sin ^{4}\left (d x +c \right )\right ) b^{2}+18 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+4 \left (\sin ^{2}\left (d x +c \right )\right ) b^{2}\right )}{45 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(332\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4/7/e*a*b*(e*sin(d*x+c))^(7/2)-1/45*e^3*(10*b^2*sin(d*x+c)^6+54*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*
sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+12*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1
/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-27*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2
)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2-6*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c
)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-18*a^2*sin(d*x+c)^4-14*sin(d*x+c)
^4*b^2+18*sin(d*x+c)^2*a^2+4*sin(d*x+c)^2*b^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/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((a+b*cos(d*x+c))^2*(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

e^(5/2)*integrate((b*cos(d*x + c) + a)^2*sin(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 = 0.15, size = 157, normalized size = 1.02 \begin {gather*} \frac {21 i \, \sqrt {2} \sqrt {-i} {\left (9 \, a^{2} + 2 \, b^{2}\right )} 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 ) - 21 i \, \sqrt {2} \sqrt {i} {\left (9 \, a^{2} + 2 \, b^{2}\right )} 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 (35 \, b^{2} \cos \left (d x + c\right )^{3} e^{\frac {5}{2}} + 90 \, a b \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} - 90 \, a b e^{\frac {5}{2}} + 21 \, {\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right ) e^{\frac {5}{2}}\right )} \sin \left (d x + c\right )^{\frac {3}{2}}}{315 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(21*I*sqrt(2)*sqrt(-I)*(9*a^2 + 2*b^2)*e^(5/2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x +
c) + I*sin(d*x + c))) - 21*I*sqrt(2)*sqrt(I)*(9*a^2 + 2*b^2)*e^(5/2)*weierstrassZeta(4, 0, weierstrassPInvers
e(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(35*b^2*cos(d*x + c)^3*e^(5/2) + 90*a*b*cos(d*x + c)^2*e^(5/2) - 9
0*a*b*e^(5/2) + 21*(3*a^2 - b^2)*cos(d*x + c)*e^(5/2))*sin(d*x + c)^(3/2))/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((a+b*cos(d*x+c))**2*(e*sin(d*x+c))**(5/2),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((a+b*cos(d*x+c))^2*(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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