3.8.40 \(\int \frac {(a+b \sin (e+f x))^3}{\sqrt {c+d \sin (e+f x)}} \, dx\) [740]

Optimal. Leaf size=302 \[ \frac {8 b^2 (b c-3 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d^2 f}-\frac {2 b^2 \cos (e+f x) (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{5 d f}-\frac {2 b \left (30 a b c d-45 a^2 d^2-b^2 \left (8 c^2+9 d^2\right )\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 \left (45 a^2 b c d^2-15 a^3 d^3-15 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+7 c d^2\right )\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^3 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

8/15*b^2*(-3*a*d+b*c)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d^2/f-2/5*b^2*cos(f*x+e)*(a+b*sin(f*x+e))*(c+d*sin(f*x
+e))^(1/2)/d/f+2/15*b*(30*a*b*c*d-45*a^2*d^2-b^2*(8*c^2+9*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/f/((c+
d*sin(f*x+e))/(c+d))^(1/2)+2/15*(45*a^2*b*c*d^2-15*a^3*d^3-15*a*b^2*d*(2*c^2+d^2)+b^3*(8*c^3+7*c*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 = 0.31, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2872, 3102, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {2 b \left (-45 a^2 d^2+30 a b c d-\left (b^2 \left (8 c^2+9 d^2\right )\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 )}{15 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 \left (-15 a^3 d^3+45 a^2 b c d^2-15 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+7 c d^2\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 )}{15 d^3 f \sqrt {c+d \sin (e+f x)}}+\frac {8 b^2 (b c-3 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d^2 f}-\frac {2 b^2 \cos (e+f x) (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{5 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^2*(b*c - 3*a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(15*d^2*f) - (2*b^2*Cos[e + f*x]*(a + b*Sin[e + f*
x])*Sqrt[c + d*Sin[e + f*x]])/(5*d*f) - (2*b*(30*a*b*c*d - 45*a^2*d^2 - b^2*(8*c^2 + 9*d^2))*EllipticE[(e - Pi
/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(15*d^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*(45*a
^2*b*c*d^2 - 15*a^3*d^3 - 15*a*b^2*d*(2*c^2 + d^2) + b^3*(8*c^3 + 7*c*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d
)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^3*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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 2872

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 - 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 3102

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+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)) \sqrt {c+d \sin (e+f x)}}{5 d f}+\frac {2 \int \frac {\frac {1}{2} \left (2 b^3 c+5 a^3 d+a b^2 d\right )-\frac {1}{2} b \left (2 a b c-15 a^2 d-3 b^2 d\right ) \sin (e+f x)-2 b^2 (b c-3 a d) \sin ^2(e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{5 d}\\ &=\frac {8 b^2 (b c-3 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d^2 f}-\frac {2 b^2 \cos (e+f x) (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{5 d f}+\frac {4 \int \frac {\frac {1}{4} d \left (2 b^3 c+15 a^3 d+15 a b^2 d\right )-\frac {1}{4} b \left (30 a b c d-45 a^2 d^2-b^2 \left (8 c^2+9 d^2\right )\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^2}\\ &=\frac {8 b^2 (b c-3 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d^2 f}-\frac {2 b^2 \cos (e+f x) (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{5 d f}-\frac {\left (b \left (30 a b c d-45 a^2 d^2-b^2 \left (8 c^2+9 d^2\right )\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 d^3}-\frac {\left (45 a^2 b c d^2-15 a^3 d^3-15 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+7 c d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^3}\\ &=\frac {8 b^2 (b c-3 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d^2 f}-\frac {2 b^2 \cos (e+f x) (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{5 d f}-\frac {\left (b \left (30 a b c d-45 a^2 d^2-b^2 \left (8 c^2+9 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}{15 d^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {\left (\left (45 a^2 b c d^2-15 a^3 d^3-15 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+7 c 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}{15 d^3 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {8 b^2 (b c-3 a d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 d^2 f}-\frac {2 b^2 \cos (e+f x) (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}{5 d f}-\frac {2 b \left (30 a b c d-45 a^2 d^2-b^2 \left (8 c^2+9 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)}}{15 d^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 \left (45 a^2 b c d^2-15 a^3 d^3-15 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+7 c 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}}}{15 d^3 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 1.32, size = 219, normalized size = 0.73 \begin {gather*} \frac {-2 \left (d^2 \left (2 b^3 c+15 a^3 d+15 a b^2 d\right ) F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )+b \left (-30 a b c d+45 a^2 d^2+b^2 \left (8 c^2+9 d^2\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-2 b^2 d \cos (e+f x) (c+d \sin (e+f x)) (-4 b c+15 a d+3 b d \sin (e+f x))}{15 d^3 f \sqrt {c+d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d^2*(2*b^3*c + 15*a^3*d + 15*a*b^2*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] + b*(-30*a*b*c*d +
45*a^2*d^2 + b^2*(8*c^2 + 9*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)] - 2*b^2*d*Cos[e + f*x]*(c + d*Sin[e + f*
x])*(-4*b*c + 15*a*d + 3*b*d*Sin[e + f*x]))/(15*d^3*f*Sqrt[c + d*Sin[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1084\) vs. \(2(348)=696\).
time = 18.39, size = 1085, normalized size = 3.59

method	result	size
default	\(\text {Expression too large to display}\)	\(1085\)

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,method=_RETURNVERBOSE)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(b^3*(-2/5/d*sin(f*x+e)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+8/15/d
^2*c*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+4/15/d*c*(1/d*c-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/d^2*c^2)*(1/d*c-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-si
n(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)*((-1/d*c-1)*Ell
ipticE(((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*(-2/3/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2/3*(1/d*c-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)*Ell
ipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-4/3/d*c*(1/d*c-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)*((-1/d*c-
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))))+6*a^2*b*(1/d*c-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)*((-1/d*c-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*(1/d*c-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)*co
s(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 = 0.00, 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*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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.18, size = 605, normalized size = 2.00 \begin {gather*} -\frac {\sqrt {2} {\left (16 \, b^{3} c^{3} - 60 \, a b^{2} c^{2} d + 6 \, {\left (15 \, a^{2} b + 2 \, b^{3}\right )} c d^{2} - 45 \, {\left (a^{3} + a b^{2}\right )} d^{3}\right )} \sqrt {i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right ) + \sqrt {2} {\left (16 \, b^{3} c^{3} - 60 \, a b^{2} c^{2} d + 6 \, {\left (15 \, a^{2} b + 2 \, b^{3}\right )} c d^{2} - 45 \, {\left (a^{3} + a b^{2}\right )} d^{3}\right )} \sqrt {-i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right ) + 3 \, \sqrt {2} {\left (8 i \, b^{3} c^{2} d - 30 i \, a b^{2} c d^{2} + 9 i \, {\left (5 \, a^{2} b + b^{3}\right )} d^{3}\right )} \sqrt {i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right )\right ) + 3 \, \sqrt {2} {\left (-8 i \, b^{3} c^{2} d + 30 i \, a b^{2} c d^{2} - 9 i \, {\left (5 \, a^{2} b + b^{3}\right )} d^{3}\right )} \sqrt {-i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right )\right ) + 6 \, {\left (3 \, b^{3} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (4 \, b^{3} c d^{2} - 15 \, a b^{2} d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{45 \, d^{4} f} \end {gather*}

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]

-1/45*(sqrt(2)*(16*b^3*c^3 - 60*a*b^2*c^2*d + 6*(15*a^2*b + 2*b^3)*c*d^2 - 45*(a^3 + a*b^2)*d^3)*sqrt(I*d)*wei
erstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(
f*x + e) - 2*I*c)/d) + sqrt(2)*(16*b^3*c^3 - 60*a*b^2*c^2*d + 6*(15*a^2*b + 2*b^3)*c*d^2 - 45*(a^3 + a*b^2)*d^
3)*sqrt(-I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x
+ e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) + 3*sqrt(2)*(8*I*b^3*c^2*d - 30*I*a*b^2*c*d^2 + 9*I*(5*a^2*b + b^3)*d^3
)*sqrt(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/
3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d)
) + 3*sqrt(2)*(-8*I*b^3*c^2*d + 30*I*a*b^2*c*d^2 - 9*I*(5*a^2*b + b^3)*d^3)*sqrt(-I*d)*weierstrassZeta(-4/3*(4
*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I
*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)) + 6*(3*b^3*d^3*cos(f*x + e)*sin
(f*x + e) - (4*b^3*c*d^2 - 15*a*b^2*d^3)*cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(d^4*f)

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

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

[Out]

integrate((b*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 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3}{\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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