3.1.35 \(\int (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2} \, dx\) [35]

Optimal. Leaf size=100 \[ \frac {2 a e^2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d \sqrt {e \sin (c+d x)}}-\frac {2 a e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}+\frac {2 b (e \sin (c+d x))^{5/2}}{5 d e} \]

[Out]

2/5*b*(e*sin(d*x+c))^(5/2)/d/e-2/3*a*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)*Ellipti
cF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)-2/3*a*e*cos(d*x+c)*(e*sin(d*x+c)
)^(1/2)/d

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Rubi [A]
time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2748, 2715, 2721, 2720} \begin {gather*} \frac {2 a e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d \sqrt {e \sin (c+d x)}}-\frac {2 a e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}+\frac {2 b (e \sin (c+d x))^{5/2}}{5 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.50, size = 80, normalized size = 0.80 \begin {gather*} \frac {2 (e \sin (c+d x))^{3/2} \left (-5 a F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+\sqrt {\sin (c+d x)} \left (-5 a \cos (c+d x)+3 b \sin ^2(c+d x)\right )\right )}{15 d \sin ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*Sin[c + d*x])^(3/2)*(-5*a*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + Sqrt[Sin[c + d*x]]*(-5*a*Cos[c + d*x] +
3*b*Sin[c + d*x]^2)))/(15*d*Sin[c + d*x]^(3/2))

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Maple [A]
time = 0.13, size = 116, normalized size = 1.16

method	result	size
default	\(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 e}-\frac {e^{2} a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/5/e*b*(e*sin(d*x+c))^(5/2)-1/3*e^2*a*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipti
cF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/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))*(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 104, normalized size = 1.04 \begin {gather*} \frac {5 \, \sqrt {2} \sqrt {-i} a e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} \sqrt {i} a e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (3 \, b \cos \left (d x + c\right )^{2} e^{\frac {3}{2}} + 5 \, a \cos \left (d x + c\right ) e^{\frac {3}{2}} - 3 \, b e^{\frac {3}{2}}\right )} \sqrt {\sin \left (d x + c\right )}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(5*sqrt(2)*sqrt(-I)*a*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*sqrt(I
)*a*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(3*b*cos(d*x + c)^2*e^(3/2) + 5*a*cos
(d*x + c)*e^(3/2) - 3*b*e^(3/2))*sqrt(sin(d*x + c)))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (a + b \cos {\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*sin(c + d*x))**(3/2)*(a + b*cos(c + d*x)), x)

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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))*(e*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)*e^(3/2)*sin(d*x + c)^(3/2), 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 )}^{3/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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