\(\int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \, dx\) [1517]

Optimal result

Integrand size = 33, antiderivative size = 341 \[ \int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \, dx=-\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}+\frac {3 (c+d)^2 \left (6 b c^2-13 a c d-10 b d^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {7}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}}{65 \sqrt {2} d^3 f \sqrt {1+\sin (e+f x)} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}}-\frac {3 (c-d) (c+d)^2 (6 b c-13 a d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}}{65 \sqrt {2} d^3 f \sqrt {1+\sin (e+f x)} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}} \]

[Out]

-3/130*(-13*a*d+6*b*c)*cos(f*x+e)*(c+d*sin(f*x+e))^(7/3)/d^2/f+3/13*b*cos(f*x+e)*sin(f*x+e)*(c+d*sin(f*x+e))^(
7/3)/d/f+3/130*(c+d)^2*(-13*a*c*d+6*b*c^2-10*b*d^2)*AppellF1(1/2,-7/3,1/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*s
in(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^(1/3)/d^3/f/((c+d*sin(f*x+e))/(c+d))^(1/3)*2^(1/2)/(1+sin(f*x+e))^(1/2)
-3/130*(c-d)*(c+d)^2*(-13*a*d+6*b*c)*AppellF1(1/2,-4/3,1/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(
f*x+e)*(c+d*sin(f*x+e))^(1/3)/d^3/f/((c+d*sin(f*x+e))/(c+d))^(1/3)*2^(1/2)/(1+sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3001, 3113, 3102, 2835, 2744, 144, 143} \[ \int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \, dx=\frac {3 (c+d)^2 \left (-13 a c d+6 b c^2-10 b d^2\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {7}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{65 \sqrt {2} d^3 f \sqrt {\sin (e+f x)+1} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}}-\frac {3 (c-d) (c+d)^2 (6 b c-13 a d) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{65 \sqrt {2} d^3 f \sqrt {\sin (e+f x)+1} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}}-\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f} \]

[In]

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

[Out]

(-3*(6*b*c - 13*a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(7/3))/(130*d^2*f) + (3*b*Cos[e + f*x]*Sin[e + f*x]*(c
+ d*Sin[e + f*x])^(7/3))/(13*d*f) + (3*(c + d)^2*(6*b*c^2 - 13*a*c*d - 10*b*d^2)*AppellF1[1/2, 1/2, -7/3, 3/2,
(1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1/3))/(65*Sqrt[2]*d^
3*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^(1/3)) - (3*(c - d)*(c + d)^2*(6*b*c - 13*a*d)*Appel
lF1[1/2, 1/2, -4/3, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x
])^(1/3))/(65*Sqrt[2]*d^3*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^(1/3))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2744

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rule 2835

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(b*
c - a*d)/b, Int[(a + b*Sin[e + f*x])^m, x], x] + Dist[d/b, Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{
a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3001

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

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]

Rule 3113

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*d*(C*(m + 2) + A*(
m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C,
m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \left (1-\sin ^2(e+f x)\right ) \, dx \\ & = \frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}+\frac {3 \int (c+d \sin (e+f x))^{4/3} \left (\frac {1}{3} (-3 b c+13 a d)+b d \sin (e+f x)+\frac {1}{3} (6 b c-13 a d) \sin ^2(e+f x)\right ) \, dx}{13 d} \\ & = -\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}+\frac {9 \int (c+d \sin (e+f x))^{4/3} \left (\frac {1}{3} d (4 b c+13 a d)-\frac {1}{3} \left (6 b c^2-13 a c d-10 b d^2\right ) \sin (e+f x)\right ) \, dx}{130 d^2} \\ & = -\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}+\frac {\left (3 (6 b c-13 a d) \left (c^2-d^2\right )\right ) \int (c+d \sin (e+f x))^{4/3} \, dx}{130 d^3}-\frac {\left (3 \left (6 b c^2-13 a c d-10 b d^2\right )\right ) \int (c+d \sin (e+f x))^{7/3} \, dx}{130 d^3} \\ & = -\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}+\frac {\left (3 (6 b c-13 a d) \left (c^2-d^2\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^{4/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{130 d^3 f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}-\frac {\left (3 \left (6 b c^2-13 a c d-10 b d^2\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^{7/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{130 d^3 f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}-\frac {\left (3 (-c-d) (6 b c-13 a d) \left (c^2-d^2\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {\left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^{4/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{130 d^3 f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)} \sqrt [3]{-\frac {c+d \sin (e+f x)}{-c-d}}}-\frac {\left (3 (-c-d)^2 \left (6 b c^2-13 a c d-10 b d^2\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {\left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^{7/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{130 d^3 f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)} \sqrt [3]{-\frac {c+d \sin (e+f x)}{-c-d}}} \\ & = -\frac {3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac {3 b \cos (e+f x) \sin (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f}+\frac {3 (c+d)^2 \left (6 b c^2-13 a c d-10 b d^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {7}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}}{65 \sqrt {2} d^3 f \sqrt {1+\sin (e+f x)} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}}-\frac {3 (c-d) (c+d)^2 (6 b c-13 a d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}}{65 \sqrt {2} d^3 f \sqrt {1+\sin (e+f x)} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 4.28 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.17 \[ \int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \, dx=\frac {3 \sec (e+f x) \sqrt [3]{c+d \sin (e+f x)} \left (12 \left (-c^2+d^2\right ) \left (-24 b c^3+52 a c^2 d+68 b c d^2+91 a d^3\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right ) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {-\frac {d (1+\sin (e+f x))}{c-d}}+3 \left (-24 b c^4+52 a c^3 d+84 b c^2 d^2+663 a c d^3+160 b d^4\right ) \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right ) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {-\frac {d (1+\sin (e+f x))}{c-d}} (c+d \sin (e+f x))-4 d^2 \cos ^2(e+f x) \left (24 b c^3-52 a c^2 d+128 b c d^2+91 a d^3+14 d^2 (14 b c+13 a d) \cos (2 (e+f x))-2 d \left (8 b c^2+286 a c d+45 b d^2\right ) \sin (e+f x)+70 b d^3 \sin (3 (e+f x))\right )\right )}{14560 d^4 f} \]

[In]

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

[Out]

(3*Sec[e + f*x]*(c + d*Sin[e + f*x])^(1/3)*(12*(-c^2 + d^2)*(-24*b*c^3 + 52*a*c^2*d + 68*b*c*d^2 + 91*a*d^3)*A
ppellF1[1/3, 1/2, 1/2, 4/3, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e + f*x])/(c + d)]*Sqrt[-((d*(-1 + Sin[e
+ f*x]))/(c + d))]*Sqrt[-((d*(1 + Sin[e + f*x]))/(c - d))] + 3*(-24*b*c^4 + 52*a*c^3*d + 84*b*c^2*d^2 + 663*a*
c*d^3 + 160*b*d^4)*AppellF1[4/3, 1/2, 1/2, 7/3, (c + d*Sin[e + f*x])/(c - d), (c + d*Sin[e + f*x])/(c + d)]*Sq
rt[-((d*(-1 + Sin[e + f*x]))/(c + d))]*Sqrt[-((d*(1 + Sin[e + f*x]))/(c - d))]*(c + d*Sin[e + f*x]) - 4*d^2*Co
s[e + f*x]^2*(24*b*c^3 - 52*a*c^2*d + 128*b*c*d^2 + 91*a*d^3 + 14*d^2*(14*b*c + 13*a*d)*Cos[2*(e + f*x)] - 2*d
*(8*b*c^2 + 286*a*c*d + 45*b*d^2)*Sin[e + f*x] + 70*b*d^3*Sin[3*(e + f*x)])))/(14560*d^4*f)

Maple [F]

\[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{\frac {4}{3}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(b*d*cos(f*x + e)^4 - (b*c + a*d)*cos(f*x + e)^2*sin(f*x + e) - (a*c + b*d)*cos(f*x + e)^2)*(d*sin(f
*x + e) + c)^(1/3), x)

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*sin(e + f*x))*(c + d*sin(e + f*x))**(4/3)*cos(e + f*x)**2, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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