\(\int \sin ^4(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [138]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 325 \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\left (a^2+11 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b f}-\frac {2 (4 a+3 b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f}-\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{35 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-1/35*(a^2+11*a*b+8*b^2)*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f-2/35*(4*a+3*b)*cos(f*x+e)*sin(f*x+
e)^3*(a+b*sin(f*x+e)^2)^(1/2)/f-1/7*b*cos(f*x+e)*sin(f*x+e)^5*(a+b*sin(f*x+e)^2)^(1/2)/f-2/35*(a+2*b)*(a^2-4*a
*b-4*b^2)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/b^2/f/(1
+b*sin(f*x+e)^2/a)^(1/2)+1/35*a*(a+b)*(2*a^2-5*a*b-8*b^2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f
*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3267, 488, 596, 538, 437, 435, 432, 430} \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{35 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{35 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\left (a^2+11 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b f}-\frac {b \sin ^5(e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {2 (4 a+3 b) \sin ^3(e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f} \]

[In]

Int[Sin[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-1/35*((a^2 + 11*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(b*f) - (2*(4*a + 3*b)*Cos
[e + f*x]*Sin[e + f*x]^3*Sqrt[a + b*Sin[e + f*x]^2])/(35*f) - (b*Cos[e + f*x]*Sin[e + f*x]^5*Sqrt[a + b*Sin[e
+ f*x]^2])/(7*f) - (2*(a + 2*b)*(a^2 - 4*a*b - 4*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b
/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(35*b^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*(a + b)*(2*a^2 -
5*a*b - 8*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f
*x]^2)/a])/(35*b^2*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

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

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

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

Rule 488

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*e*(m + n*(p + q) + 1))), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 596

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
 x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q +
 1) + 1))), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 3267

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff^(m + 1)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[x^m*((a + b*ff^2*
x^2)^p/Sqrt[1 - ff^2*x^2]), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^{3/2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = -\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^4 \left (-a (7 a+5 b)-2 b (4 a+3 b) x^2\right )}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{7 f} \\ & = -\frac {2 (4 a+3 b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f}-\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^2 \left (-6 a b (4 a+3 b)-3 b \left (a^2+11 a b+8 b^2\right ) x^2\right )}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{35 b f} \\ & = -\frac {\left (a^2+11 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b f}-\frac {2 (4 a+3 b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f}-\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-3 a b \left (a^2+11 a b+8 b^2\right )+6 b (a+2 b) \left (a^2-4 a b-4 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{105 b^2 f} \\ & = -\frac {\left (a^2+11 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b f}-\frac {2 (4 a+3 b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f}-\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}+\frac {\left (a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{35 b^2 f}-\frac {\left (2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{35 b^2 f} \\ & = -\frac {\left (a^2+11 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b f}-\frac {2 (4 a+3 b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f}-\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {\left (2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{35 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{35 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \\ & = -\frac {\left (a^2+11 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b f}-\frac {2 (4 a+3 b) \cos (e+f x) \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 f}-\frac {b \cos (e+f x) \sin ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{7 f}-\frac {2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{35 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{35 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.77 \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {-128 a \left (a^3-2 a^2 b-12 a b^2-8 b^3\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+64 a \left (2 a^3-3 a^2 b-13 a b^2-8 b^3\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\sqrt {2} b \left (-32 a^3-496 a^2 b-684 a b^2-250 b^3+b \left (144 a^2+480 a b+299 b^2\right ) \cos (2 (e+f x))-2 b^2 (26 a+27 b) \cos (4 (e+f x))+5 b^3 \cos (6 (e+f x))\right ) \sin (2 (e+f x))}{2240 b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

Integrate[Sin[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-128*a*(a^3 - 2*a^2*b - 12*a*b^2 - 8*b^3)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] +
 64*a*(2*a^3 - 3*a^2*b - 13*a*b^2 - 8*b^3)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] +
 Sqrt[2]*b*(-32*a^3 - 496*a^2*b - 684*a*b^2 - 250*b^3 + b*(144*a^2 + 480*a*b + 299*b^2)*Cos[2*(e + f*x)] - 2*b
^2*(26*a + 27*b)*Cos[4*(e + f*x)] + 5*b^3*Cos[6*(e + f*x)])*Sin[2*(e + f*x)])/(2240*b^2*f*Sqrt[2*a + b - b*Cos
[2*(e + f*x)]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(601\) vs. \(2(299)=598\).

Time = 4.04 (sec) , antiderivative size = 602, normalized size of antiderivative = 1.85

method result size
default \(\frac {5 b^{4} \left (\sin ^{9}\left (f x +e \right )\right )+13 a \,b^{3} \left (\sin ^{7}\left (f x +e \right )\right )+b^{4} \left (\sin ^{7}\left (f x +e \right )\right )+9 a^{2} b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+4 a \,b^{3} \left (\sin ^{5}\left (f x +e \right )\right )+2 b^{4} \left (\sin ^{5}\left (f x +e \right )\right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{4}-3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3} b -13 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b^{2}-8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{3}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{4}+4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3} b +24 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b^{2}+16 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{3}+a^{3} b \left (\sin ^{3}\left (f x +e \right )\right )+2 a^{2} b^{2} \left (\sin ^{3}\left (f x +e \right )\right )-9 a \,b^{3} \left (\sin ^{3}\left (f x +e \right )\right )-8 b^{4} \left (\sin ^{3}\left (f x +e \right )\right )-a^{3} b \sin \left (f x +e \right )-11 a^{2} b^{2} \sin \left (f x +e \right )-8 a \,b^{3} \sin \left (f x +e \right )}{35 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) \(602\)

[In]

int(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/35*(5*b^4*sin(f*x+e)^9+13*a*b^3*sin(f*x+e)^7+b^4*sin(f*x+e)^7+9*a^2*b^2*sin(f*x+e)^5+4*a*b^3*sin(f*x+e)^5+2*
b^4*sin(f*x+e)^5+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^4-
3*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3*b-13*(cos(f*x+e)^
2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b^2-8*(cos(f*x+e)^2)^(1/2)*((a+
b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b^3-2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)
/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^4+4*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*Ellipti
cE(sin(f*x+e),(-1/a*b)^(1/2))*a^3*b+24*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),
(-1/a*b)^(1/2))*a^2*b^2+16*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/
2))*a*b^3+a^3*b*sin(f*x+e)^3+2*a^2*b^2*sin(f*x+e)^3-9*a*b^3*sin(f*x+e)^3-8*b^4*sin(f*x+e)^3-a^3*b*sin(f*x+e)-1
1*a^2*b^2*sin(f*x+e)-8*a*b^3*sin(f*x+e))/b^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

Fricas [F]

\[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(-(b*cos(f*x + e)^6 - (a + 3*b)*cos(f*x + e)^4 + (2*a + 3*b)*cos(f*x + e)^2 - a - b)*sqrt(-b*cos(f*x +
 e)^2 + a + b), x)

Sympy [F(-1)]

Timed out. \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**4*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)

Giac [F]

\[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

[In]

int(sin(e + f*x)^4*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(sin(e + f*x)^4*(a + b*sin(e + f*x)^2)^(3/2), x)