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

Optimal. Leaf size=390 \[ -\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}-\frac {4 a^3 \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d^2 f}-\frac {4 a^3 \left (c^2-d^2\right ) \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^3 f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^3 \left (4 c^4-27 c^3 d+111 c^2 d^2+579 c d^3+357 d^4\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d^2 f}-\frac {2 \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))^{5/2}}{9 d f} \]

[Out]

-4/315*a^3*(4*c^2-27*c*d+119*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d^2/f+8/63*a^3*(c-5*d)*cos(f*x+e)*(c+d*sin
(f*x+e))^(5/2)/d^2/f-2/9*cos(f*x+e)*(a^3+a^3*sin(f*x+e))*(c+d*sin(f*x+e))^(5/2)/d/f-4/315*a^3*(4*c^3-27*c^2*d+
114*c*d^2+165*d^3)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d^2/f-4/315*a^3*(4*c^4-27*c^3*d+111*c^2*d^2+579*c*d^3+357
*d^4)*(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)+4/315*a^3*(c^2-d^2)*(4*c^3-27*c
^2*d+114*c*d^2+165*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)*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 = 0.78, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}-\frac {4 a^3 \left (-27 c^2 d+4 c^3+114 c d^2+165 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d^2 f}-\frac {4 a^3 \left (c^2-d^2\right ) \left (-27 c^2 d+4 c^3+114 c d^2+165 d^3\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^3 f \sqrt {c+d \sin (e+f x)}}+\frac {4 a^3 \left (111 c^2 d^2-27 c^3 d+4 c^4+579 c d^3+357 d^4\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{315 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 d^2 f}-\frac {2 \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))^{5/2}}{9 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3*(4*c^3 - 27*c^2*d + 114*c*d^2 + 165*d^3)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(315*d^2*f) - (4*a^3*(
4*c^2 - 27*c*d + 119*d^2)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(315*d^2*f) + (8*a^3*(c - 5*d)*Cos[e + f*x]
*(c + d*Sin[e + f*x])^(5/2))/(63*d^2*f) - (2*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x])*(c + d*Sin[e + f*x])^(5/2))
/(9*d*f) + (4*a^3*(4*c^4 - 27*c^3*d + 111*c^2*d^2 + 579*c*d^3 + 357*d^4)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(
c + d)]*Sqrt[c + d*Sin[e + f*x]])/(315*d^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (4*a^3*(c^2 - d^2)*(4*c^3 -
27*c^2*d + 114*c*d^2 + 165*d^3)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d
)])/(315*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 (c+d \sin (e+f x))^{3/2} \, dx &=-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{5/2}}{9 d f}+\frac {2 \int (a+a \sin (e+f x)) \left (a^2 (c+7 d)-2 a^2 (c-5 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2} \, dx}{9 d}\\ &=-\frac {2 \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^{5/2}}{9 d f}+\frac {2 \int (c+d \sin (e+f x))^{3/2} \left (a^3 (c+7 d)+\left (-2 a^3 (c-5 d)+a^3 (c+7 d)\right ) \sin (e+f x)-2 a^3 (c-5 d) \sin ^2(e+f x)\right ) \, dx}{9 d}\\ &=\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 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))^{5/2}}{9 d f}+\frac {4 \int (c+d \sin (e+f x))^{3/2} \left (-\frac {3}{2} a^3 (c-33 d) d+\frac {1}{2} a^3 \left (4 c^2-27 c d+119 d^2\right ) \sin (e+f x)\right ) \, dx}{63 d^2}\\ &=-\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 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))^{5/2}}{9 d f}+\frac {8 \int \sqrt {c+d \sin (e+f x)} \left (-\frac {3}{4} a^3 d \left (c^2-138 c d-119 d^2\right )+\frac {3}{4} a^3 \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \sin (e+f x)\right ) \, dx}{315 d^2}\\ &=-\frac {4 a^3 \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d^2 f}-\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 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))^{5/2}}{9 d f}+\frac {16 \int \frac {\frac {3}{8} a^3 d \left (c^3+387 c^2 d+471 c d^2+165 d^3\right )+\frac {3}{8} a^3 \left (4 c^4-27 c^3 d+111 c^2 d^2+579 c d^3+357 d^4\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{945 d^2}\\ &=-\frac {4 a^3 \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d^2 f}-\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 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))^{5/2}}{9 d f}-\frac {\left (2 a^3 \left (c^2-d^2\right ) \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{315 d^3}+\frac {\left (2 a^3 \left (4 c^4-27 c^3 d+111 c^2 d^2+579 c d^3+357 d^4\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{315 d^3}\\ &=-\frac {4 a^3 \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d^2 f}-\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 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))^{5/2}}{9 d f}+\frac {\left (2 a^3 \left (4 c^4-27 c^3 d+111 c^2 d^2+579 c d^3+357 d^4\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{315 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^3-27 c^2 d+114 c d^2+165 d^3\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}{315 d^3 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {4 a^3 \left (4 c^3-27 c^2 d+114 c d^2+165 d^3\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{315 d^2 f}-\frac {4 a^3 \left (4 c^2-27 c d+119 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{315 d^2 f}+\frac {8 a^3 (c-5 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{63 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))^{5/2}}{9 d f}+\frac {4 a^3 \left (4 c^4-27 c^3 d+111 c^2 d^2+579 c d^3+357 d^4\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)}}{315 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^3-27 c^2 d+114 c d^2+165 d^3\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}}}{315 d^3 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.27, size = 318, normalized size = 0.82 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (d (c+d \sin (e+f x)) \left (2 d \left (5 d (10 c+27 d) \cos (3 (e+f x))-\sin (2 (e+f x)) \left (6 c^2+432 c d-35 d^2 \cos (2 (e+f x))+511 d^2\right )\right )+\left (32 c^3-216 c^2 d-3828 c d^2-2910 d^3\right ) \cos (e+f x)\right )-16 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left (d^2 \left (c^3+387 c^2 d+471 c d^2+165 d^3\right ) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )+\left (4 c^4-27 c^3 d+111 c^2 d^2+579 c d^3+357 d^4\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )\right )\right )}{1260 d^3 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6 \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Sin[e + f*x])^3*(-16*(d^2*(c^3 + 387*c^2*d + 471*c*d^2 + 165*d^3)*EllipticF[(-2*e + Pi - 2*f*x)/4, (
2*d)/(c + d)] + (4*c^4 - 27*c^3*d + 111*c^2*d^2 + 579*c*d^3 + 357*d^4)*((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)] + d
*(c + d*Sin[e + f*x])*((32*c^3 - 216*c^2*d - 3828*c*d^2 - 2910*d^3)*Cos[e + f*x] + 2*d*(5*d*(10*c + 27*d)*Cos[
3*(e + f*x)] - (6*c^2 + 432*c*d + 511*d^2 - 35*d^2*Cos[2*(e + f*x)])*Sin[2*(e + f*x)]))))/(1260*d^3*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 = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{3} d \cos \left (f x + e\right )^{4} + 4 \, a^{3} c + 4 \, a^{3} d - {\left (3 \, a^{3} c + 5 \, a^{3} d\right )} \cos \left (f x + e\right )^{2} + {\left (4 \, a^{3} c + 4 \, a^{3} d - {\left (a^{3} c + 3 \, a^{3} d\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}, x\right ) \]

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))^(3/2),x, algorithm="fricas")

[Out]

integral((a^3*d*cos(f*x + e)^4 + 4*a^3*c + 4*a^3*d - (3*a^3*c + 5*a^3*d)*cos(f*x + e)^2 + (4*a^3*c + 4*a^3*d -
(a^3*c + 3*a^3*d)*cos(f*x + e)^2)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c), x)

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

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 = 1.48, size = 1613, 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))^(3/2),x)

[Out]

-2/315*a^3*(492*((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^4-4*c^4*d^2-54*((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*d-1158*((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^5-8*((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^5*d+60*((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^2-1048*((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^3*d^3-1104*((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^4+1056*((c+d*s
in(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^5+214*((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^4*d^2+1212*
((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^3+27*c^3*d^3+330*c*d^5+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^6-714*((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^6+1044*((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^6-85*c*d^5*sin(f*x+e)^5-53*c^2*d^4*sin(f*x+e)^4-351*c*d^5*sin(f*x+e)^4+c^3*d^3*sin(f*x+e)^3-
243*c^2*d^4*sin(f*x+e)^3-619*c*d^5*sin(f*x+e)^3+4*c^4*d^2*sin(f*x+e)^2-27*c^3*d^3*sin(f*x+e)^2-413*c^2*d^4*sin
(f*x+e)^2+21*c*d^5*sin(f*x+e)^2-c^3*d^3*sin(f*x+e)+243*c^2*d^4*sin(f*x+e)+704*c*d^5*sin(f*x+e)-135*d^6*sin(f*x
+e)^5-203*d^6*sin(f*x+e)^4-195*d^6*sin(f*x+e)^3+238*d^6*sin(f*x+e)^2+330*d^6*sin(f*x+e)-35*d^6*sin(f*x+e)^6+46
6*c^2*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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}\,{d x} \]

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 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int c \sqrt {c + d \sin {\left (e + f x \right )}}\, dx + \int 3 c \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int 3 c \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int c \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )}\, dx + \int d \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int 3 d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int 3 d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )}\, dx + \int d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{4}{\left (e + f x \right )}\, dx\right ) \]

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))**(3/2),x)

[Out]

a**3*(Integral(c*sqrt(c + d*sin(e + f*x)), x) + Integral(3*c*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integ
ral(3*c*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(c*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3, x)
+ Integral(d*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integral(3*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**
2, x) + Integral(3*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3, x) + Integral(d*sqrt(c + d*sin(e + f*x))*sin(e
+ f*x)**4, x))

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