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

Optimal. Leaf size=375 \[ \frac{2 b \left (-105 a^2 d^2+42 a b c d+b^2 \left (-\left (8 c^2+25 d^2\right )\right )\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 d^2 f}+\frac{2 b \left (c^2-d^2\right ) \left (-105 a^2 d^2+42 a b c d+b^2 \left (-\left (8 c^2+25 d^2\right )\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d^3 f \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (105 a^2 b c d^2+105 a^3 d^3-21 a b^2 d \left (2 c^2-9 d^2\right )+b^3 \left (8 c^3+19 c d^2\right )\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d^3 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f} \]

[Out]

(2*b*(42*a*b*c*d - 105*a^2*d^2 - b^2*(8*c^2 + 25*d^2))*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(105*d^2*f) + (8
*b^2*(b*c - 4*a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(35*d^2*f) - (2*b^2*Cos[e + f*x]*(a + b*Sin[e + f*
x])*(c + d*Sin[e + f*x])^(3/2))/(7*d*f) + (2*(105*a^2*b*c*d^2 + 105*a^3*d^3 - 21*a*b^2*d*(2*c^2 - 9*d^2) + b^3
*(8*c^3 + 19*c*d^2))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(105*d^3*f*Sqrt[(c
+ d*Sin[e + f*x])/(c + d)]) + (2*b*(c^2 - d^2)*(42*a*b*c*d - 105*a^2*d^2 - b^2*(8*c^2 + 25*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]])

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Rubi [A]  time = 0.677772, antiderivative size = 375, normalized size of antiderivative = 1., 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 = {2793, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 b \left (-105 a^2 d^2+42 a b c d+b^2 \left (-\left (8 c^2+25 d^2\right )\right )\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 d^2 f}+\frac{2 b \left (c^2-d^2\right ) \left (-105 a^2 d^2+42 a b c d+b^2 \left (-\left (8 c^2+25 d^2\right )\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d^3 f \sqrt{c+d \sin (e+f x)}}+\frac{2 \left (105 a^2 b c d^2+105 a^3 d^3-21 a b^2 d \left (2 c^2-9 d^2\right )+b^3 \left (8 c^3+19 c d^2\right )\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{105 d^3 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(42*a*b*c*d - 105*a^2*d^2 - b^2*(8*c^2 + 25*d^2))*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(105*d^2*f) + (8
*b^2*(b*c - 4*a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(35*d^2*f) - (2*b^2*Cos[e + f*x]*(a + b*Sin[e + f*
x])*(c + d*Sin[e + f*x])^(3/2))/(7*d*f) + (2*(105*a^2*b*c*d^2 + 105*a^3*d^3 - 21*a*b^2*d*(2*c^2 - 9*d^2) + b^3
*(8*c^3 + 19*c*d^2))*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(105*d^3*f*Sqrt[(c
+ d*Sin[e + f*x])/(c + d)]) + (2*b*(c^2 - d^2)*(42*a*b*c*d - 105*a^2*d^2 - b^2*(8*c^2 + 25*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 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int (a+b \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)} \, dx &=-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac{2 \int \sqrt{c+d \sin (e+f x)} \left (\frac{1}{2} \left (2 b^3 c+7 a^3 d+3 a b^2 d\right )-\frac{1}{2} b \left (2 a b c-21 a^2 d-5 b^2 d\right ) \sin (e+f x)-2 b^2 (b c-4 a d) \sin ^2(e+f x)\right ) \, dx}{7 d}\\ &=\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac{4 \int \sqrt{c+d \sin (e+f x)} \left (-\frac{1}{4} d \left (2 b^3 c-35 a^3 d-63 a b^2 d\right )-\frac{1}{4} b \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\right ) \sin (e+f x)\right ) \, dx}{35 d^2}\\ &=\frac{2 b \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 d^2 f}+\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac{8 \int \frac{\frac{1}{8} d \left (105 a^3 c d+147 a b^2 c d+105 a^2 b d^2+b^3 \left (2 c^2+25 d^2\right )\right )+\frac{1}{8} \left (105 a^2 b c d^2+105 a^3 d^3-21 a b^2 d \left (2 c^2-9 d^2\right )+b^3 \left (8 c^3+19 c d^2\right )\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{105 d^2}\\ &=\frac{2 b \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 d^2 f}+\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac{\left (b \left (c^2-d^2\right ) \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{105 d^3}+\frac{\left (105 a^2 b c d^2+105 a^3 d^3-21 a b^2 d \left (2 c^2-9 d^2\right )+b^3 \left (8 c^3+19 c d^2\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{105 d^3}\\ &=\frac{2 b \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 d^2 f}+\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac{\left (\left (105 a^2 b c d^2+105 a^3 d^3-21 a b^2 d \left (2 c^2-9 d^2\right )+b^3 \left (8 c^3+19 c d^2\right )\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 (b \left (c^2-d^2\right ) \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\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{2 b \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{105 d^2 f}+\frac{8 b^2 (b c-4 a d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 d^2 f}-\frac{2 b^2 \cos (e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}{7 d f}+\frac{2 \left (105 a^2 b c d^2+105 a^3 d^3-21 a b^2 d \left (2 c^2-9 d^2\right )+b^3 \left (8 c^3+19 c d^2\right )\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{2 b \left (c^2-d^2\right ) \left (42 a b c d-105 a^2 d^2-b^2 \left (8 c^2+25 d^2\right )\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*}

Mathematica [A]  time = 1.41046, size = 306, normalized size = 0.82 \[ \frac{b d (c+d \sin (e+f x)) \left (\left (-420 a^2 d^2-84 a b c d+b^2 \left (16 c^2-115 d^2\right )\right ) \cos (e+f x)+3 b d (5 b d \cos (3 (e+f x))-2 (21 a d+b c) \sin (2 (e+f x)))\right )-4 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left (d^2 \left (105 a^2 b d^2+105 a^3 c d+147 a b^2 c d+b^3 \left (2 c^2+25 d^2\right )\right ) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+\left (105 a^2 b c d^2+105 a^3 d^3+21 a b^2 d \left (9 d^2-2 c^2\right )+b^3 \left (8 c^3+19 c d^2\right )\right ) \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )\right )}{210 d^3 f \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(d^2*(105*a^3*c*d + 147*a*b^2*c*d + 105*a^2*b*d^2 + b^3*(2*c^2 + 25*d^2))*EllipticF[(-2*e + Pi - 2*f*x)/4,
(2*d)/(c + d)] + (105*a^2*b*c*d^2 + 105*a^3*d^3 + 21*a*b^2*d*(-2*c^2 + 9*d^2) + b^3*(8*c^3 + 19*c*d^2))*((c +
d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]))*Sqrt
[(c + d*Sin[e + f*x])/(c + d)] + b*d*(c + d*Sin[e + f*x])*((-84*a*b*c*d - 420*a^2*d^2 + b^2*(16*c^2 - 115*d^2)
)*Cos[e + f*x] + 3*b*d*(5*b*d*Cos[3*(e + f*x)] - 2*(b*c + 21*a*d)*Sin[2*(e + f*x)])))/(210*d^3*f*Sqrt[c + d*Si
n[e + f*x]])

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Maple [B]  time = 4.738, size = 1561, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(b^3*d*(-2/7/d*sin(f*x+e)^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+12
/35*c/d^2*sin(f*x+e)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)-2/3*(5/7+24/35*c^2/d^2)/d*(-(-d*sin(f*x+e)-c)*cos
(f*x+e)^2)^(1/2)+2*(-4/35*c^2/d^2+5/21)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*
((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/
2),((c-d)/(c+d))^(1/2))+2/105*(-48*c^3-44*c*d^2)/d^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/
(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d
*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))
+(3*a*b^2*d+b^3*c)*(-2/5/d*sin(f*x+e)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+8/15*c/d^2*(-(-d*sin(f*x+e)-c)*c
os(f*x+e)^2)^(1/2)+4/15*c/d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e
)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c
+d))^(1/2))+2*(3/5+8/15*c^2/d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(
f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^
(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+(3*a^2*b*d+3*a*b^2*
c)*(-2/3/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2/3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e)
)/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x
+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-4/3*c/d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))
^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f
*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+2*(a^
3*d+3*a^2*b*c)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))
^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d)
)^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))+2*a^3*c*(c/d-1)*((c+d*sin(f*x+e))/(c-d
))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2
)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right ) + a\right )}^{3} \sqrt{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (3 \, a b^{2} \cos \left (f x + e\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{d \sin \left (f x + e\right ) + c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (e + f x \right )}\right )^{3} \sqrt{c + d \sin{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out