3.706 \(\int \frac{(c+d \sin (e+f x))^4}{(a+b \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=306 \[ \frac{d (2 b c-a d) \left (-3 a^2 d^2+2 a b c d+b^2 \left (-\left (c^2-2 d^2\right )\right )\right ) \cos (e+f x)}{b^3 f \left (a^2-b^2\right )}+\frac{d^2 \left (-3 a^2 d^2+4 a b c d+b^2 \left (-\left (2 c^2-d^2\right )\right )\right ) \sin (e+f x) \cos (e+f x)}{2 b^2 f \left (a^2-b^2\right )}-\frac{d^2 x \left (-6 a^2 d^2+16 a b c d+b^2 \left (-\left (12 c^2+d^2\right )\right )\right )}{2 b^4}+\frac{2 \left (3 a^2 d+a b c-4 b^2 d\right ) (b c-a d)^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.940713, antiderivative size = 306, 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 = {2792, 3033, 3023, 2735, 2660, 618, 204} \[ \frac{d (2 b c-a d) \left (-3 a^2 d^2+2 a b c d+b^2 \left (-\left (c^2-2 d^2\right )\right )\right ) \cos (e+f x)}{b^3 f \left (a^2-b^2\right )}+\frac{d^2 \left (-3 a^2 d^2+4 a b c d+b^2 \left (-\left (2 c^2-d^2\right )\right )\right ) \sin (e+f x) \cos (e+f x)}{2 b^2 f \left (a^2-b^2\right )}-\frac{d^2 x \left (-6 a^2 d^2+16 a b c d+b^2 \left (-\left (12 c^2+d^2\right )\right )\right )}{2 b^4}+\frac{2 \left (3 a^2 d+a b c-4 b^2 d\right ) (b c-a d)^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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))^2} \, dx &=\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\int \frac{(c+d \sin (e+f x)) \left (4 b^2 c^2 d+2 a^2 d^3-a b c \left (c^2+5 d^2\right )-d \left (a^2 c d-3 b^2 c d+a b \left (c^2+d^2\right )\right ) \sin (e+f x)+d \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \sin ^2(e+f x)\right )}{a+b \sin (e+f x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{d^2 \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{2 b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\int \frac{8 b^3 c^3 d+8 a^2 b c d^3-3 a^3 d^4-a b^2 \left (2 c^4+12 c^2 d^2-d^4\right )+b d \left (b^2 d \left (12 c^2+d^2\right )-4 a b c \left (c^2+2 d^2\right )-a^2 \left (2 c^2 d-d^3\right )\right ) \sin (e+f x)-2 d (2 b c-a d) \left (b^2 c^2-2 a b c d+3 a^2 d^2-2 b^2 d^2\right ) \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{d (2 b c-a d) \left (2 a b c d-3 a^2 d^2-b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{b^3 \left (a^2-b^2\right ) f}+\frac{d^2 \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{2 b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\int \frac{b \left (8 b^3 c^3 d+8 a^2 b c d^3-3 a^3 d^4-a b^2 \left (2 c^4+12 c^2 d^2-d^4\right )\right )+\left (a^2-b^2\right ) d^2 \left (16 a b c d-6 a^2 d^2-b^2 \left (12 c^2+d^2\right )\right ) \sin (e+f x)}{a+b \sin (e+f x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{d^2 \left (16 a b c d-6 a^2 d^2-b^2 \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{d (2 b c-a d) \left (2 a b c d-3 a^2 d^2-b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{b^3 \left (a^2-b^2\right ) f}+\frac{d^2 \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{2 b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left ((b c-a d)^3 \left (a b c+3 a^2 d-4 b^2 d\right )\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{b^4 \left (a^2-b^2\right )}\\ &=-\frac{d^2 \left (16 a b c d-6 a^2 d^2-b^2 \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{d (2 b c-a d) \left (2 a b c d-3 a^2 d^2-b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{b^3 \left (a^2-b^2\right ) f}+\frac{d^2 \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{2 b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left (2 (b c-a d)^3 \left (a b c+3 a^2 d-4 b^2 d\right )\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 \left (a^2-b^2\right ) f}\\ &=-\frac{d^2 \left (16 a b c d-6 a^2 d^2-b^2 \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{d (2 b c-a d) \left (2 a b c d-3 a^2 d^2-b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{b^3 \left (a^2-b^2\right ) f}+\frac{d^2 \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{2 b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\left (4 (b c-a d)^3 \left (a b c+3 a^2 d-4 b^2 d\right )\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 \left (a^2-b^2\right ) f}\\ &=-\frac{d^2 \left (16 a b c d-6 a^2 d^2-b^2 \left (12 c^2+d^2\right )\right ) x}{2 b^4}+\frac{2 (b c-a d)^3 \left (a b c+3 a^2 d-4 b^2 d\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2} f}+\frac{d (2 b c-a d) \left (2 a b c d-3 a^2 d^2-b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{b^3 \left (a^2-b^2\right ) f}+\frac{d^2 \left (4 a b c d-3 a^2 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{2 b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))^2}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.01492, size = 199, normalized size = 0.65 \[ -\frac{-2 d^2 (e+f x) \left (6 a^2 d^2-16 a b c d+b^2 \left (12 c^2+d^2\right )\right )+\frac{8 (a d-b c)^3 \left (3 a^2 d+a b c-4 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+8 b d^3 (2 b c-a d) \cos (e+f x)-\frac{4 b (b c-a d)^4 \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}+b^2 d^4 \sin (2 (e+f x))}{4 b^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.109, size = 1303, normalized size = 4.3 \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))^2,x)

[Out]

6/f*d^4/b^4*arctan(tan(1/2*f*x+1/2*e))*a^2+12/f*d^2/b^2*arctan(tan(1/2*f*x+1/2*e))*c^2+2/f*b/(tan(1/2*f*x+1/2*
e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*c^4+2/f/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^
2-b^2)^(1/2))*a*c^4+1/f*d^4/b^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3-1/f*d^4/b^2/(1+tan(1/2*f*x+1/2
*e)^2)^2*tan(1/2*f*x+1/2*e)+4/f*d^4/b^3/(1+tan(1/2*f*x+1/2*e)^2)^2*a-8/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)^2*c-
16/f*d^3/b^3*arctan(tan(1/2*f*x+1/2*e))*a*c+2/f/b^3/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2
)*a^4*d^4-6/f/b^4/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^5*d^4+8/f/b^2/(a^
2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^3*d^4-8/f*b/(a^2-b^2)^(3/2)*arctan(1/2
*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*c^3*d-8/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^
2-b^2)*a*c^3*d+24/f/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a*c^2*d^2+4/f*d^4
/b^3/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*a-8/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*
e)^2*c-8/f/b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a^2*tan(1/2*f*x+1/2*e)*c*d^3+12/f/(ta
n(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a*tan(1/2*f*x+1/2*e)*c^2*d^2+2/f/b^2/(tan(1/2*f*x+1/2
*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a^3*tan(1/2*f*x+1/2*e)*d^4-8/f*b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2
*f*x+1/2*e)*b+a)/(a^2-b^2)*tan(1/2*f*x+1/2*e)*c^3*d+2/f*b^2/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/
(a^2-b^2)/a*tan(1/2*f*x+1/2*e)*c^4-8/f/b^2/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a^3*c*d
^3+1/f*d^4/b^2*arctan(tan(1/2*f*x+1/2*e))+12/f/b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a
^2*c^2*d^2+16/f/b^3/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^4*c*d^3-12/f/b^
2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^3*c^2*d^2-24/f/b/(a^2-b^2)^(3/2)*
arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2*c*d^3

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.96677, size = 2984, normalized size = 9.75 \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))^2,x, algorithm="fricas")

[Out]

[1/2*((a^4*b^3 - 2*a^2*b^5 + b^7)*d^4*cos(f*x + e)^3 + (12*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c^2*d^2 - 16*(a^6*b -
2*a^4*b^3 + a^2*b^5)*c*d^3 + (6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*d^4)*f*x + (a^2*b^4*c^4 - 4*a*b^5*c^3*d
- 6*(a^4*b^2 - 2*a^2*b^4)*c^2*d^2 + 4*(2*a^5*b - 3*a^3*b^3)*c*d^3 - (3*a^6 - 4*a^4*b^2)*d^4 + (a*b^5*c^4 - 4*
b^6*c^3*d - 6*(a^3*b^3 - 2*a*b^5)*c^2*d^2 + 4*(2*a^4*b^2 - 3*a^2*b^4)*c*d^3 - (3*a^5*b - 4*a^3*b^3)*d^4)*sin(f
*x + e))*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)) +
(2*(a^2*b^5 - b^7)*c^4 - 8*(a^3*b^4 - a*b^6)*c^3*d + 12*(a^4*b^3 - a^2*b^5)*c^2*d^2 - 8*(2*a^5*b^2 - 3*a^3*b^4
+ a*b^6)*c*d^3 + (6*a^6*b - 11*a^4*b^3 + 6*a^2*b^5 - b^7)*d^4)*cos(f*x + e) + ((12*(a^4*b^3 - 2*a^2*b^5 + b^7
)*c^2*d^2 - 16*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c*d^3 + (6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d^4)*f*x - (8*(a
^4*b^3 - 2*a^2*b^5 + b^7)*c*d^3 - 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d^4)*cos(f*x + e))*sin(f*x + e))/((a^4*b^5 -
2*a^2*b^7 + b^9)*f*sin(f*x + e) + (a^5*b^4 - 2*a^3*b^6 + a*b^8)*f), 1/2*((a^4*b^3 - 2*a^2*b^5 + b^7)*d^4*cos(
f*x + e)^3 + (12*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c^2*d^2 - 16*(a^6*b - 2*a^4*b^3 + a^2*b^5)*c*d^3 + (6*a^7 - 11*
a^5*b^2 + 4*a^3*b^4 + a*b^6)*d^4)*f*x - 2*(a^2*b^4*c^4 - 4*a*b^5*c^3*d - 6*(a^4*b^2 - 2*a^2*b^4)*c^2*d^2 + 4*(
2*a^5*b - 3*a^3*b^3)*c*d^3 - (3*a^6 - 4*a^4*b^2)*d^4 + (a*b^5*c^4 - 4*b^6*c^3*d - 6*(a^3*b^3 - 2*a*b^5)*c^2*d^
2 + 4*(2*a^4*b^2 - 3*a^2*b^4)*c*d^3 - (3*a^5*b - 4*a^3*b^3)*d^4)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(-(a*sin(
f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + (2*(a^2*b^5 - b^7)*c^4 - 8*(a^3*b^4 - a*b^6)*c^3*d + 12*(a^4*b
^3 - a^2*b^5)*c^2*d^2 - 8*(2*a^5*b^2 - 3*a^3*b^4 + a*b^6)*c*d^3 + (6*a^6*b - 11*a^4*b^3 + 6*a^2*b^5 - b^7)*d^4
)*cos(f*x + e) + ((12*(a^4*b^3 - 2*a^2*b^5 + b^7)*c^2*d^2 - 16*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*c*d^3 + (6*a^6*b
- 11*a^4*b^3 + 4*a^2*b^5 + b^7)*d^4)*f*x - (8*(a^4*b^3 - 2*a^2*b^5 + b^7)*c*d^3 - 3*(a^5*b^2 - 2*a^3*b^4 + a*b
^6)*d^4)*cos(f*x + e))*sin(f*x + e))/((a^4*b^5 - 2*a^2*b^7 + b^9)*f*sin(f*x + e) + (a^5*b^4 - 2*a^3*b^6 + a*b^
8)*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.52387, size = 698, normalized size = 2.28 \begin{align*} \frac{\frac{4 \,{\left (a b^{4} c^{4} - 4 \, b^{5} c^{3} d - 6 \, a^{3} b^{2} c^{2} d^{2} + 12 \, a b^{4} c^{2} d^{2} + 8 \, a^{4} b c d^{3} - 12 \, a^{2} b^{3} c d^{3} - 3 \, a^{5} d^{4} + 4 \, a^{3} b^{2} 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 )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{4 \,{\left (b^{5} c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 4 \, a b^{4} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, a^{2} b^{3} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 4 \, a^{3} b^{2} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{4} b d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a b^{4} c^{4} - 4 \, a^{2} b^{3} c^{3} d + 6 \, a^{3} b^{2} c^{2} d^{2} - 4 \, a^{4} b c d^{3} + a^{5} d^{4}\right )}}{{\left (a^{3} b^{3} - a b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a\right )}} + \frac{{\left (12 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} + 6 \, a^{2} d^{4} + b^{2} d^{4}\right )}{\left (f x + e\right )}}{b^{4}} + \frac{2 \,{\left (b d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 8 \, b c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 4 \, a d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - b d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 8 \, b c d^{3} + 4 \, a d^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, 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))^2,x, algorithm="giac")

[Out]

1/2*(4*(a*b^4*c^4 - 4*b^5*c^3*d - 6*a^3*b^2*c^2*d^2 + 12*a*b^4*c^2*d^2 + 8*a^4*b*c*d^3 - 12*a^2*b^3*c*d^3 - 3*
a^5*d^4 + 4*a^3*b^2*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a
^2 - b^2)))/((a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 4*(b^5*c^4*tan(1/2*f*x + 1/2*e) - 4*a*b^4*c^3*d*tan(1/2*f*x +
1/2*e) + 6*a^2*b^3*c^2*d^2*tan(1/2*f*x + 1/2*e) - 4*a^3*b^2*c*d^3*tan(1/2*f*x + 1/2*e) + a^4*b*d^4*tan(1/2*f*x
+ 1/2*e) + a*b^4*c^4 - 4*a^2*b^3*c^3*d + 6*a^3*b^2*c^2*d^2 - 4*a^4*b*c*d^3 + a^5*d^4)/((a^3*b^3 - a*b^5)*(a*t
an(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)) + (12*b^2*c^2*d^2 - 16*a*b*c*d^3 + 6*a^2*d^4 + b^2*d^4)
*(f*x + e)/b^4 + 2*(b*d^4*tan(1/2*f*x + 1/2*e)^3 - 8*b*c*d^3*tan(1/2*f*x + 1/2*e)^2 + 4*a*d^4*tan(1/2*f*x + 1/
2*e)^2 - b*d^4*tan(1/2*f*x + 1/2*e) - 8*b*c*d^3 + 4*a*d^4)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*b^3))/f