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

Optimal. Leaf size=325 \[ -\frac{\left (a^2+11 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{35 b f}+\frac{a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{35 b^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{35 b^2 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{b \sin ^5(e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{7 f}-\frac{2 (4 a+3 b) \sin ^3(e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{35 f} \]

[Out]

-((a^2 + 11*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(35*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])

________________________________________________________________________________________

Rubi [A]  time = 0.482311, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3188, 477, 582, 524, 426, 424, 421, 419} \[ -\frac{\left (a^2+11 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{35 b f}+\frac{a (a+b) \left (2 a^2-5 a b-8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{35 b^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 b) \left (a^2-4 a b-4 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{35 b^2 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{b \sin ^5(e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{7 f}-\frac{2 (4 a+3 b) \sin ^3(e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{35 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 + 11*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(35*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 3188

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]

Rule 477

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 582

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 524

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 426

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

Rule 424

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

Rule 421

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

Rule 419

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

Rubi steps

\begin{align*} \int \sin ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\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)} F\left (\sin ^{-1}(\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]  time = 2.74001, size = 249, normalized size = 0.77 \[ \frac{\sqrt{2} b \sin (2 (e+f x)) \left (b \left (144 a^2+480 a b+299 b^2\right ) \cos (2 (e+f x))-496 a^2 b-32 a^3-2 b^2 (26 a+27 b) \cos (4 (e+f x))-684 a b^2+5 b^3 \cos (6 (e+f x))-250 b^3\right )+64 a \left (-3 a^2 b+2 a^3-13 a b^2-8 b^3\right ) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-128 a \left (-2 a^2 b+a^3-12 a b^2-8 b^3\right ) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{2240 b^2 f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[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]  time = 1.324, size = 602, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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-sin(f*x+e)*a^3*b-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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b \cos \left (f x + e\right )^{6} -{\left (a + 3 \, b\right )} \cos \left (f x + e\right )^{4} +{\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)