3.498 \(\int (a+a \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=318 \[ -\frac {4 a^3 \left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 d^2 f}-\frac {4 a^3 \left (c^2-d^2\right ) \left (4 c^2-21 c d+65 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^3 \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))^{3/2}}{7 d f} \]

[Out]

8/35*a^3*(c-4*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d^2/f-2/7*cos(f*x+e)*(a^3+a^3*sin(f*x+e))*(c+d*sin(f*x+e))^
(3/2)/d/f-4/105*a^3*(4*c^2-21*c*d+65*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d^2/f-4/105*a^3*(4*c^3-21*c^2*d+62
*c*d^2+147*d^3)*(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/f/((c+d*sin(f*x+e))/(c+d))^(1/2)+4/105*a^3*(c^2-d^2)*(
4*c^2-21*c*d+65*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)*EllipticF(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 = 0.58, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2763, 2968, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 a^3 \left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 d^2 f}-\frac {4 a^3 \left (c^2-d^2\right ) \left (4 c^2-21 c d+65 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^3 \left (-21 c^2 d+4 c^3+62 c d^2+147 d^3\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))^{3/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3*(4*c^2 - 21*c*d + 65*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(105*d^2*f) + (8*a^3*(c - 4*d)*Cos[e
+ f*x]*(c + d*Sin[e + f*x])^(3/2))/(35*d^2*f) - (2*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x])*(c + d*Sin[e + f*x])^
(3/2))/(7*d*f) + (4*a^3*(4*c^3 - 21*c^2*d + 62*c*d^2 + 147*d^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*S
qrt[c + d*Sin[e + f*x]])/(105*d^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (4*a^3*(c^2 - d^2)*(4*c^2 - 21*c*d +
65*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(105*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 2753

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

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, 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 (a+a \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)} \, dx &=-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac {2 \int (a+a \sin (e+f x)) \left (a^2 (c+5 d)-2 a^2 (c-4 d) \sin (e+f x)\right ) \sqrt {c+d \sin (e+f x)} \, dx}{7 d}\\ &=-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac {2 \int \sqrt {c+d \sin (e+f x)} \left (a^3 (c+5 d)+\left (-2 a^3 (c-4 d)+a^3 (c+5 d)\right ) \sin (e+f x)-2 a^3 (c-4 d) \sin ^2(e+f x)\right ) \, dx}{7 d}\\ &=\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac {4 \int \sqrt {c+d \sin (e+f x)} \left (-\frac {1}{2} a^3 (c-49 d) d+\frac {1}{2} a^3 \left (4 c^2-21 c d+65 d^2\right ) \sin (e+f x)\right ) \, dx}{35 d^2}\\ &=-\frac {4 a^3 \left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 d^2 f}+\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac {8 \int \frac {\frac {1}{4} a^3 d \left (c^2+126 c d+65 d^2\right )+\frac {1}{4} a^3 \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{105 d^2}\\ &=-\frac {4 a^3 \left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 d^2 f}+\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}-\frac {\left (2 a^3 \left (c^2-d^2\right ) \left (4 c^2-21 c d+65 d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{105 d^3}+\frac {\left (2 a^3 \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{105 d^3}\\ &=-\frac {4 a^3 \left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 d^2 f}+\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac {\left (2 a^3 \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{105 d^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {\left (2 a^3 \left (c^2-d^2\right ) \left (4 c^2-21 c d+65 d^2\right ) \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}{105 d^3 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {4 a^3 \left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 d^2 f}+\frac {8 a^3 (c-4 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac {4 a^3 \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\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)}}{105 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {4 a^3 \left (c^2-d^2\right ) \left (4 c^2-21 c d+65 d^2\right ) 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}}}{105 d^3 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.72, size = 266, normalized size = 0.84 \[ -\frac {a^3 \left (-2 d \cos (e+f x) \left (16 c^3+d \left (4 c^2-336 c d-565 d^2\right ) \sin (e+f x)-84 c^2 d+18 d^2 (2 c+7 d) \cos (2 (e+f x))-556 c d^2+15 d^3 \sin (3 (e+f x))-126 d^3\right )-16 \left (4 c^4-21 c^3 d+61 c^2 d^2+21 c d^3-65 d^4\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )+16 \left (4 c^4-17 c^3 d+41 c^2 d^2+209 c d^3+147 d^4\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )}{420 d^3 f \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/420*(a^3*(16*(4*c^4 - 17*c^3*d + 41*c^2*d^2 + 209*c*d^3 + 147*d^4)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(
c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - 16*(4*c^4 - 21*c^3*d + 61*c^2*d^2 + 21*c*d^3 - 65*d^4)*EllipticF[
(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - 2*d*Cos[e + f*x]*(16*c^3 - 84*c^2*d
- 556*c*d^2 - 126*d^3 + 18*d^2*(2*c + 7*d)*Cos[2*(e + f*x)] + d*(4*c^2 - 336*c*d - 565*d^2)*Sin[e + f*x] + 15
*d^3*Sin[3*(e + f*x)])))/(d^3*f*Sqrt[c + d*Sin[e + f*x]])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}, x\right ) \]

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))^(1/2),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),
x)

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giac [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))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.38, size = 1316, normalized size = 4.14 \[ \text {result too large to display} \]

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))^(1/2),x)

[Out]

2/105*a^3*(42*((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)*E
llipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4*d-130*c*d^4-21*c^2*d^3+4*c^3*d^2-8*((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)*EllipticE(((c+d*sin(f*x+e
))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^5+294*((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))*d^5-424*((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^5+15*d^5*sin(f*x+e)^5+63*d^5*sin(f*x+e)^4+115*d^5*sin(f*x+e)^3-63*d^5*s
in(f*x+e)^2-130*d^5*sin(f*x+e)-4*c^3*d^2*sin(f*x+e)^2+21*c^2*d^3*sin(f*x+e)^2+112*c*d^4*sin(f*x+e)^2+c^2*d^3*s
in(f*x+e)-84*c*d^4*sin(f*x+e)+18*c*d^4*sin(f*x+e)^4-c^2*d^3*sin(f*x+e)^3+84*c*d^4*sin(f*x+e)^3-116*((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^2+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)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4*d-48*((c+d*si
n(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))*c^3*d^2+416*((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))*c^2*d^3+48*(
(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))*c*d^4-336*((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^2*d^3
+124*((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^4)/d^4/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{3} \sqrt {d \sin \left (f x + e\right ) + c}\,{d x} \]

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))^(1/2),x, algorithm="maxima")

[Out]

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

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

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))^(1/2),x)

[Out]

int((a + a*sin(e + f*x))^3*(c + d*sin(e + f*x))^(1/2), x)

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

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))**(1/2),x)

[Out]

a**3*(Integral(3*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integral(3*sqrt(c + d*sin(e + f*x))*sin(e + f*x)*
*2, x) + Integral(sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3, x) + Integral(sqrt(c + d*sin(e + f*x)), x))

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