3.1517 \(\int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \, dx\)

Optimal. Leaf size=341 \[ \frac{3 (c+d)^2 \left (-13 a c d+6 b c^2-10 b d^2\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{7}{3};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{65 \sqrt{2} d^3 f \sqrt{\sin (e+f x)+1} \sqrt [3]{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{3 (c-d) (c+d)^2 (6 b c-13 a d) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{65 \sqrt{2} d^3 f \sqrt{\sin (e+f x)+1} \sqrt [3]{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac{3 b \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f} \]

[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))

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Rubi [A]  time = 0.651351, antiderivative size = 341, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2922, 3034, 3023, 2756, 2665, 139, 138} \[ \frac{3 (c+d)^2 \left (-13 a c d+6 b c^2-10 b d^2\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{7}{3};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{65 \sqrt{2} d^3 f \sqrt{\sin (e+f x)+1} \sqrt [3]{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{3 (c-d) (c+d)^2 (6 b c-13 a d) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{65 \sqrt{2} d^3 f \sqrt{\sin (e+f x)+1} \sqrt [3]{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{3 (6 b c-13 a d) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{130 d^2 f}+\frac{3 b \sin (e+f x) \cos (e+f x) (c+d \sin (e+f x))^{7/3}}{13 d f} \]

Antiderivative was successfully verified.

[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 2922

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 3034

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]

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]

Rule 2756

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 2665

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 139

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 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

\begin{align*} \int \cos ^2(e+f x) (a+b \sin (e+f x)) (c+d \sin (e+f x))^{4/3} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) F_1\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) F_1\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]  time = 5.06916, size = 398, normalized size = 1.17 \[ \frac{3 \sec (e+f x) \sqrt [3]{c+d \sin (e+f x)} \left (3 \left (52 a c^3 d+663 a c d^3+84 b c^2 d^2-24 b c^4+160 b d^4\right ) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{-\frac{d (\sin (e+f x)+1)}{c-d}} (c+d \sin (e+f x)) F_1\left (\frac{4}{3};\frac{1}{2},\frac{1}{2};\frac{7}{3};\frac{c+d \sin (e+f x)}{c-d},\frac{c+d \sin (e+f x)}{c+d}\right )+12 \left (d^2-c^2\right ) \left (52 a c^2 d+91 a d^3-24 b c^3+68 b c d^2\right ) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{-\frac{d (\sin (e+f x)+1)}{c-d}} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};\frac{c+d \sin (e+f x)}{c-d},\frac{c+d \sin (e+f x)}{c+d}\right )-4 d^2 \cos ^2(e+f x) \left (-2 d \left (286 a c d+8 b c^2+45 b d^2\right ) \sin (e+f x)+14 d^2 (13 a d+14 b c) \cos (2 (e+f x))-52 a c^2 d+91 a d^3+24 b c^3+128 b c d^2+70 b d^3 \sin (3 (e+f x))\right )\right )}{14560 d^4 f} \]

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 0.264, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b\sin \left ( fx+e \right ) \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) ^{{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b d \cos \left (f x + e\right )^{4} -{\left (b c + a d\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) -{\left (a c + b d\right )} \cos \left (f x + e\right )^{2}\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{1}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)