3.698 \(\int \frac{(c+d \sin (e+f x))^4}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=235 \[ \frac{d^2 \left (-3 a^2 d^2+12 a b c d+b^2 \left (-\left (17 c^2+2 d^2\right )\right )\right ) \cos (e+f x)}{3 b^3 f}+\frac{d x \left (8 a^2 b c d^2-2 a^3 d^3-a b^2 d \left (12 c^2+d^2\right )+4 b^3 c \left (2 c^2+d^2\right )\right )}{2 b^4}+\frac{2 (b c-a d)^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 f \sqrt{a^2-b^2}}-\frac{d^3 (8 b c-3 a d) \sin (e+f x) \cos (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f} \]

[Out]

(d*(8*a^2*b*c*d^2 - 2*a^3*d^3 + 4*b^3*c*(2*c^2 + d^2) - a*b^2*d*(12*c^2 + d^2))*x)/(2*b^4) + (2*(b*c - a*d)^4*
ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]*f) + (d^2*(12*a*b*c*d - 3*a^2*d^2 - b^2
*(17*c^2 + 2*d^2))*Cos[e + f*x])/(3*b^3*f) - (d^3*(8*b*c - 3*a*d)*Cos[e + f*x]*Sin[e + f*x])/(6*b^2*f) - (d^2*
Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*b*f)

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Rubi [A]  time = 0.647816, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2793, 3033, 3023, 2735, 2660, 618, 204} \[ \frac{d^2 \left (-3 a^2 d^2+12 a b c d+b^2 \left (-\left (17 c^2+2 d^2\right )\right )\right ) \cos (e+f x)}{3 b^3 f}+\frac{d x \left (8 a^2 b c d^2-2 a^3 d^3-a b^2 d \left (12 c^2+d^2\right )+4 b^3 c \left (2 c^2+d^2\right )\right )}{2 b^4}+\frac{2 (b c-a d)^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 f \sqrt{a^2-b^2}}-\frac{d^3 (8 b c-3 a d) \sin (e+f x) \cos (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(8*a^2*b*c*d^2 - 2*a^3*d^3 + 4*b^3*c*(2*c^2 + d^2) - a*b^2*d*(12*c^2 + d^2))*x)/(2*b^4) + (2*(b*c - a*d)^4*
ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]*f) + (d^2*(12*a*b*c*d - 3*a^2*d^2 - b^2
*(17*c^2 + 2*d^2))*Cos[e + f*x])/(3*b^3*f) - (d^3*(8*b*c - 3*a*d)*Cos[e + f*x]*Sin[e + f*x])/(6*b^2*f) - (d^2*
Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*b*f)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (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*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^4}{a+b \sin (e+f x)} \, dx &=-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}+\frac{\int \frac{(c+d \sin (e+f x)) \left (3 b c^3+2 a d^3+d \left (9 b c^2-a c d+2 b d^2\right ) \sin (e+f x)+d^2 (8 b c-3 a d) \sin ^2(e+f x)\right )}{a+b \sin (e+f x)} \, dx}{3 b}\\ &=-\frac{d^3 (8 b c-3 a d) \cos (e+f x) \sin (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}+\frac{\int \frac{3 \left (2 b^2 c^4+4 a b c d^3-a^2 d^4\right )-b d \left (a d \left (2 c^2-d^2\right )-12 b c \left (2 c^2+d^2\right )\right ) \sin (e+f x)-2 d^2 \left (12 a b c d-3 a^2 d^2-b^2 \left (17 c^2+2 d^2\right )\right ) \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx}{6 b^2}\\ &=\frac{d^2 \left (12 a b c d-3 a^2 d^2-b^2 \left (17 c^2+2 d^2\right )\right ) \cos (e+f x)}{3 b^3 f}-\frac{d^3 (8 b c-3 a d) \cos (e+f x) \sin (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}+\frac{\int \frac{3 b \left (2 b^2 c^4+4 a b c d^3-a^2 d^4\right )+3 d \left (8 a^2 b c d^2-2 a^3 d^3+4 b^3 c \left (2 c^2+d^2\right )-a b^2 d \left (12 c^2+d^2\right )\right ) \sin (e+f x)}{a+b \sin (e+f x)} \, dx}{6 b^3}\\ &=\frac{d \left (8 a^2 b c d^2-2 a^3 d^3+4 b^3 c \left (2 c^2+d^2\right )-a b^2 d \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{d^2 \left (12 a b c d-3 a^2 d^2-b^2 \left (17 c^2+2 d^2\right )\right ) \cos (e+f x)}{3 b^3 f}-\frac{d^3 (8 b c-3 a d) \cos (e+f x) \sin (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}+\frac{(b c-a d)^4 \int \frac{1}{a+b \sin (e+f x)} \, dx}{b^4}\\ &=\frac{d \left (8 a^2 b c d^2-2 a^3 d^3+4 b^3 c \left (2 c^2+d^2\right )-a b^2 d \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{d^2 \left (12 a b c d-3 a^2 d^2-b^2 \left (17 c^2+2 d^2\right )\right ) \cos (e+f x)}{3 b^3 f}-\frac{d^3 (8 b c-3 a d) \cos (e+f x) \sin (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}+\frac{\left (2 (b c-a d)^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{b^4 f}\\ &=\frac{d \left (8 a^2 b c d^2-2 a^3 d^3+4 b^3 c \left (2 c^2+d^2\right )-a b^2 d \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{d^2 \left (12 a b c d-3 a^2 d^2-b^2 \left (17 c^2+2 d^2\right )\right ) \cos (e+f x)}{3 b^3 f}-\frac{d^3 (8 b c-3 a d) \cos (e+f x) \sin (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}-\frac{\left (4 (b c-a d)^4\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (e+f x)\right )\right )}{b^4 f}\\ &=\frac{d \left (8 a^2 b c d^2-2 a^3 d^3+4 b^3 c \left (2 c^2+d^2\right )-a b^2 d \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{2 (b c-a d)^4 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 \sqrt{a^2-b^2} f}+\frac{d^2 \left (12 a b c d-3 a^2 d^2-b^2 \left (17 c^2+2 d^2\right )\right ) \cos (e+f x)}{3 b^3 f}-\frac{d^3 (8 b c-3 a d) \cos (e+f x) \sin (e+f x)}{6 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))^2}{3 b f}\\ \end{align*}

Mathematica [A]  time = 0.567641, size = 203, normalized size = 0.86 \[ \frac{-6 d (e+f x) \left (-8 a^2 b c d^2+2 a^3 d^3+a b^2 d \left (12 c^2+d^2\right )-4 b^3 c \left (2 c^2+d^2\right )\right )-3 b d^2 \left (4 a^2 d^2-16 a b c d+3 b^2 \left (8 c^2+d^2\right )\right ) \cos (e+f x)+\frac{24 (b c-a d)^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-3 b^2 d^3 (4 b c-a d) \sin (2 (e+f x))+b^3 d^4 \cos (3 (e+f x))}{12 b^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*d*(-8*a^2*b*c*d^2 + 2*a^3*d^3 - 4*b^3*c*(2*c^2 + d^2) + a*b^2*d*(12*c^2 + d^2))*(e + f*x) + (24*(b*c - a*d
)^4*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 3*b*d^2*(-16*a*b*c*d + 4*a^2*d^2 + 3*b
^2*(8*c^2 + d^2))*Cos[e + f*x] + b^3*d^4*Cos[3*(e + f*x)] - 3*b^2*d^3*(4*b*c - a*d)*Sin[2*(e + f*x)])/(12*b^4*
f)

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Maple [B]  time = 0.084, size = 948, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/f*d^4/b/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2-2/f*d^4/b^3/(1+tan(1/2*f*x+1/2*e)^2)^3*a^2+4/f*d^3/
b/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5*c+8/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^
4*a*c-8/f/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^3*c*d^3+16/f*d^3/b^2/
(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2*a*c-12/f*d^2/b/(1+tan(1/2*f*x+1/2*e)^2)^3*c^2+8/f*d/b*arctan(t
an(1/2*f*x+1/2*e))*c^3-1/f*d^4/b^2*arctan(tan(1/2*f*x+1/2*e))*a+4/f*d^3/b*arctan(tan(1/2*f*x+1/2*e))*c-2/f*d^4
/b^4*arctan(tan(1/2*f*x+1/2*e))*a^3+2/f/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2
))*c^4-4/3/f*d^4/b/(1+tan(1/2*f*x+1/2*e)^2)^3+12/f/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)
/(a^2-b^2)^(1/2))*a^2*c^2*d^2-8/f/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a
*c^3*d-2/f*d^4/b^3/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^4*a^2-12/f*d^2/b/(1+tan(1/2*f*x+1/2*e)^2)^3*t
an(1/2*f*x+1/2*e)^4*c^2+8/f*d^3/b^3*arctan(tan(1/2*f*x+1/2*e))*a^2*c-12/f*d^2/b^2*arctan(tan(1/2*f*x+1/2*e))*a
*c^2+2/f/b^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^4*d^4+1/f*d^4/b^2/(1+t
an(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)*a-4/f*d^3/b/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)*c-1/f*d^4/
b^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5*a+8/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)^3*a*c-4/f*d^4/b^3/(
1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2*a^2-24/f*d^2/b/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2*
c^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74416, size = 1594, normalized size = 6.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(2*(a^2*b^3 - b^5)*d^4*cos(f*x + e)^3 + 3*(8*(a^2*b^3 - b^5)*c^3*d - 12*(a^3*b^2 - a*b^4)*c^2*d^2 + 4*(2*
a^4*b - a^2*b^3 - b^5)*c*d^3 - (2*a^5 - a^3*b^2 - a*b^4)*d^4)*f*x - 3*(4*(a^2*b^3 - b^5)*c*d^3 - (a^3*b^2 - a*
b^4)*d^4)*cos(f*x + e)*sin(f*x + e) - 3*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4
)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2 + 2*(a*cos(f*x + e)*sin(
f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) - 6*(6*(a^
2*b^3 - b^5)*c^2*d^2 - 4*(a^3*b^2 - a*b^4)*c*d^3 + (a^4*b - b^5)*d^4)*cos(f*x + e))/((a^2*b^4 - b^6)*f), 1/6*(
2*(a^2*b^3 - b^5)*d^4*cos(f*x + e)^3 + 3*(8*(a^2*b^3 - b^5)*c^3*d - 12*(a^3*b^2 - a*b^4)*c^2*d^2 + 4*(2*a^4*b
- a^2*b^3 - b^5)*c*d^3 - (2*a^5 - a^3*b^2 - a*b^4)*d^4)*f*x - 3*(4*(a^2*b^3 - b^5)*c*d^3 - (a^3*b^2 - a*b^4)*d
^4)*cos(f*x + e)*sin(f*x + e) - 6*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt
(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) - 6*(6*(a^2*b^3 - b^5)*c^2*d^2 - 4*(a
^3*b^2 - a*b^4)*c*d^3 + (a^4*b - b^5)*d^4)*cos(f*x + e))/((a^2*b^4 - b^6)*f)]

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

[Out]

Timed out

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Giac [B]  time = 1.37152, size = 628, normalized size = 2.67 \begin{align*} \frac{\frac{3 \,{\left (8 \, b^{3} c^{3} d - 12 \, a b^{2} c^{2} d^{2} + 8 \, a^{2} b c d^{3} + 4 \, b^{3} c d^{3} - 2 \, a^{3} d^{4} - a b^{2} d^{4}\right )}{\left (f x + e\right )}}{b^{4}} + \frac{12 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} b^{4}} + \frac{2 \,{\left (12 \, b^{2} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 3 \, a b d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 36 \, b^{2} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 24 \, a b c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 6 \, a^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 72 \, b^{2} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 48 \, a b c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, a^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, b^{2} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, b^{2} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, a b d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 36 \, b^{2} c^{2} d^{2} + 24 \, a b c d^{3} - 6 \, a^{2} d^{4} - 4 \, b^{2} d^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(8*b^3*c^3*d - 12*a*b^2*c^2*d^2 + 8*a^2*b*c*d^3 + 4*b^3*c*d^3 - 2*a^3*d^4 - a*b^2*d^4)*(f*x + e)/b^4 +
12*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*s
gn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^4) + 2*(12*b^2*c*d^3*tan(1/2*
f*x + 1/2*e)^5 - 3*a*b*d^4*tan(1/2*f*x + 1/2*e)^5 - 36*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^4 + 24*a*b*c*d^3*tan(1
/2*f*x + 1/2*e)^4 - 6*a^2*d^4*tan(1/2*f*x + 1/2*e)^4 - 72*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 + 48*a*b*c*d^3*ta
n(1/2*f*x + 1/2*e)^2 - 12*a^2*d^4*tan(1/2*f*x + 1/2*e)^2 - 12*b^2*d^4*tan(1/2*f*x + 1/2*e)^2 - 12*b^2*c*d^3*ta
n(1/2*f*x + 1/2*e) + 3*a*b*d^4*tan(1/2*f*x + 1/2*e) - 36*b^2*c^2*d^2 + 24*a*b*c*d^3 - 6*a^2*d^4 - 4*b^2*d^4)/(
(tan(1/2*f*x + 1/2*e)^2 + 1)^3*b^3))/f