3.500 \(\int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=270 \[ -\frac {4 a^3 \left (4 c^2-5 c d-3 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^3 (4 c-5 d) (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 (2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 d^2 f (c+d)}+\frac {2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{d f (c+d) \sqrt {c+d \sin (e+f x)}} \]

[Out]

2*(c-d)*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d/(c+d)/f/(c+d*sin(f*x+e))^(1/2)-4/3*a^3*(2*c-d)*cos(f*x+e)*(c+d*sin(f
*x+e))^(1/2)/d^2/(c+d)/f+4/3*a^3*(4*c^2-5*c*d-3*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*
f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/d^3/(c+d)/f/((c+d*sin
(f*x+e))/(c+d))^(1/2)-4/3*a^3*(4*c-5*d)*(c-d)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*El
lipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/d^3/f/(c+d*sin(f*x+e
))^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2762, 2968, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 a^3 \left (4 c^2-5 c d-3 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {4 a^3 (2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 d^2 f (c+d)}+\frac {4 a^3 (4 c-5 d) (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f \sqrt {c+d \sin (e+f x)}}+\frac {2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{d f (c+d) \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{3/2}} \, dx &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(2 a) \int \frac {(a+a \sin (e+f x)) (a (c-2 d)-a (2 c-d) \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}} \, dx}{d (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(2 a) \int \frac {a^2 (c-2 d)+\left (a^2 (c-2 d)-a^2 (2 c-d)\right ) \sin (e+f x)-a^2 (2 c-d) \sin ^2(e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{d (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 (2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 d^2 (c+d) f}-\frac {(4 a) \int \frac {\frac {1}{2} a^2 (c-5 d) d+\frac {1}{2} a^2 \left (4 c^2-5 c d-3 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d^2 (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 (2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 d^2 (c+d) f}+\frac {\left (2 a^3 (4 c-5 d) (c-d)\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d^3}-\frac {\left (2 a^3 \left (4 c^2-5 c d-3 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{3 d^3 (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 (2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 d^2 (c+d) f}-\frac {\left (2 a^3 \left (4 c^2-5 c d-3 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{3 d^3 (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^3 (4 c-5 d) (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{3 d^3 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{d (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 (2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 d^2 (c+d) f}-\frac {4 a^3 \left (4 c^2-5 c d-3 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 d^3 (c+d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^3 (4 c-5 d) (c-d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 d^3 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 1.45, size = 234, normalized size = 0.87 \[ -\frac {2 a^3 (\sin (e+f x)+1)^3 \left (d \cos (e+f x) \left (4 c^2+d (c+d) \sin (e+f x)-5 c d+3 d^2\right )+2 \left (4 c^3-5 c^2 d-4 c d^2+5 d^3\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-2 \left (4 c^3-c^2 d-8 c d^2-3 d^3\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )}{3 d^3 f (c+d) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6 \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*(1 + Sin[e + f*x])^3*(-2*(4*c^3 - c^2*d - 8*c*d^2 - 3*d^3)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c +
d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + 2*(4*c^3 - 5*c^2*d - 4*c*d^2 + 5*d^3)*EllipticF[(-2*e + Pi - 2*f*x)/
4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + d*Cos[e + f*x]*(4*c^2 - 5*c*d + 3*d^2 + d*(c + d)*Sin[e
+ f*x])))/(3*d^3*(c + d)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*Sqrt[c + d*Sin[e + f*x]])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c)/(
d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.41, size = 1031, normalized size = 3.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(8*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*Ellipti
cF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^3*d-16*c^2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+
e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1
/2))*d^2-8*c*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*El
lipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^3+16*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e
)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/
2))*d^4-8*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*Ellip
ticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4+10*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1
)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))
*c^3*d+14*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*Ellip
ticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^2-10*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+
e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1
/2))*c*d^3-6*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*El
lipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^4-c*d^3*sin(f*x+e)^3-d^4*sin(f*x+e)^3-4*c^2*d^2*
sin(f*x+e)^2+5*c*d^3*sin(f*x+e)^2-3*d^4*sin(f*x+e)^2+c*d^3*sin(f*x+e)+d^4*sin(f*x+e)+4*c^2*d^2-5*c*d^3+3*d^4)*
a^3/d^4/(c+d)/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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