∫ 1____________ (a+a sin(e+fx))2(c+d sin(e+fx))3 dx

3.469 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=294 \[ \frac{d^2 \left (12 c^2+16 c d+7 d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^4 (c+d)^2 \sqrt{c^2-d^2}}-\frac{d \left (-16 c^2 d+2 c^3-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 f (c-d)^4 (c+d)^2 (c+d \sin (e+f x))}-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))^2} \]

[Out]

(d^2*(12*c^2 + 16*c*d + 7*d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^4*(c + d)^2*Sqrt

[c^2 - d^2]*f) - (d*(2*c^2 - 16*c*d - 21*d^2)*Cos[e + f*x])/(6*a^2*(c - d)^3*(c + d)*f*(c + d*Sin[e + f*x])^2)

- ((c - 8*d)*Cos[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - Cos[e + f*x]/(3*(c

- d)*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2) - (d*(2*c^3 - 16*c^2*d - 59*c*d^2 - 32*d^3)*Cos[e + f*x

])/(6*a^2*(c - d)^4*(c + d)^2*f*(c + d*Sin[e + f*x]))

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Rubi [A] time = 0.615488, antiderivative size = 294, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2766, 2978, 2754, 12, 2660, 618, 204} \[ \frac{d^2 \left (12 c^2+16 c d+7 d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^4 (c+d)^2 \sqrt{c^2-d^2}}-\frac{d \left (-16 c^2 d+2 c^3-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 f (c-d)^4 (c+d)^2 (c+d \sin (e+f x))}-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))^2} \]

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^3),x]

[Out]

(d^2*(12*c^2 + 16*c*d + 7*d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^4*(c + d)^2*Sqrt

[c^2 - d^2]*f) - (d*(2*c^2 - 16*c*d - 21*d^2)*Cos[e + f*x])/(6*a^2*(c - d)^3*(c + d)*f*(c + d*Sin[e + f*x])^2)

- ((c - 8*d)*Cos[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x])^2) - Cos[e + f*x]/(3*(c

- d)*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2) - (d*(2*c^3 - 16*c^2*d - 59*c*d^2 - 32*d^3)*Cos[e + f*x

])/(6*a^2*(c - d)^4*(c + d)^2*f*(c + d*Sin[e + f*x]))

Rule 2766

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

p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis

t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*

(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,

0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ

[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*

x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +

1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*

(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2

- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,

0])

Rule 2754

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

b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2

)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /

; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^3} \, dx &=-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{\int \frac{-a (c-5 d)-3 a d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx}{3 a^2 (c-d)}\\ &=-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}+\frac{\int \frac{21 a^2 d^2+2 a^2 (c-8 d) d \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{3 a^4 (c-d)^2}\\ &=-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{\int \frac{-2 a^2 d^2 (19 c+16 d)-a^2 d \left (2 c (c-8 d)-21 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{6 a^4 (c-d)^3 (c+d)}\\ &=-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{d \left (2 c^3-16 c^2 d-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 (c-d)^4 (c+d)^2 f (c+d \sin (e+f x))}+\frac{\int \frac{3 a^2 d^2 \left (12 c^2+16 c d+7 d^2\right )}{c+d \sin (e+f x)} \, dx}{6 a^4 (c-d)^4 (c+d)^2}\\ &=-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{d \left (2 c^3-16 c^2 d-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 (c-d)^4 (c+d)^2 f (c+d \sin (e+f x))}+\frac{\left (d^2 \left (12 c^2+16 c d+7 d^2\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{2 a^2 (c-d)^4 (c+d)^2}\\ &=-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{d \left (2 c^3-16 c^2 d-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 (c-d)^4 (c+d)^2 f (c+d \sin (e+f x))}+\frac{\left (d^2 \left (12 c^2+16 c d+7 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{a^2 (c-d)^4 (c+d)^2 f}\\ &=-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{d \left (2 c^3-16 c^2 d-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 (c-d)^4 (c+d)^2 f (c+d \sin (e+f x))}-\frac{\left (2 d^2 \left (12 c^2+16 c d+7 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac{1}{2} (e+f x)\right )\right )}{a^2 (c-d)^4 (c+d)^2 f}\\ &=\frac{d^2 \left (12 c^2+16 c d+7 d^2\right ) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{a^2 (c-d)^4 (c+d)^2 \sqrt{c^2-d^2} f}-\frac{d \left (2 c^2-16 c d-21 d^2\right ) \cos (e+f x)}{6 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^2}-\frac{(c-8 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^2}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac{d \left (2 c^3-16 c^2 d-59 c d^2-32 d^3\right ) \cos (e+f x)}{6 a^2 (c-d)^4 (c+d)^2 f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A] time = 1.18258, size = 338, normalized size = 1.15 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\frac{6 d^2 \left (12 c^2+16 c d+7 d^2\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d)^2 \sqrt{c^2-d^2}}+\frac{3 d^3 (7 c+4 d) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{(c+d)^2 (c+d \sin (e+f x))}+\frac{3 d^3 (c-d) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{(c+d) (c+d \sin (e+f x))^2}+4 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )+4 (c-10 d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-2 (c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{6 a^2 f (c-d)^4 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^3),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(4*(c - d)*Sin[(e + f*x)/2] - 2*(c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*

x)/2]) + 4*(c - 10*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (6*d^2*(12*c^2 + 16*c*d + 7*d

^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/((c + d)^2*Sqrt[

c^2 - d^2]) + (3*(c - d)*d^3*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/((c + d)*(c + d*Sin[e + f*x

])^2) + (3*d^3*(7*c + 4*d)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/((c + d)^2*(c + d*Sin[e + f*x

]))))/(6*a^2*(c - d)^4*f*(1 + Sin[e + f*x])^2)

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Maple [B] time = 0.127, size = 1313, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^3,x)

[Out]

9/f/a^2*d^4/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3

+4/f/a^2*d^5/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3-

2/f/a^2*d^6/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3

+8/f/a^2*d^3/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*c^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e

)^2+4/f/a^2*d^4/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2*c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*

e)^2+15/f/a^2*d^5/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*

e)^2+8/f/a^2*d^6/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2

*e)^2-2/f/a^2*d^7/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/c^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+

1/2*e)^2+23/f/a^2*d^4/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c*tan(1/2*f*

x+1/2*e)+12/f/a^2*d^5/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+

1/2*e)-2/f/a^2*d^6/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1

/2*e)+8/f/a^2*d^3/(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c^2+4/f/a^2*d^4/

(c-d)^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)*c-1/f/a^2*d^5/(c-d)^4/(c*tan(1/2*f

*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)^2/(c^2+2*c*d+d^2)+12/f/a^2*d^2/(c-d)^4/(c^2+2*c*d+d^2)/(c^2-d^2)^(1/2)*a

rctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*c^2+16/f/a^2*d^3/(c-d)^4/(c^2+2*c*d+d^2)/(c^2-d^2)^(1/

2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*c+7/f/a^2*d^4/(c-d)^4/(c^2+2*c*d+d^2)/(c^2-d^2)^(1

/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))-2/f/a^2/(c-d)^4/(tan(1/2*f*x+1/2*e)+1)*c+8/f/a^2/

(c-d)^4/(tan(1/2*f*x+1/2*e)+1)*d-4/3/f/a^2/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^3+2/f/a^2/(c-d)^3/(tan(1/2*f*x+1/2*e

)+1)^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(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B] time = 2.81252, size = 7625, normalized size = 25.94 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/12*(4*c^7 - 4*c^6*d - 12*c^5*d^2 + 12*c^4*d^3 + 12*c^3*d^4 - 12*c^2*d^5 - 4*c*d^6 + 4*d^7 - 2*(2*c^5*d^2 -

16*c^4*d^3 - 61*c^3*d^4 - 16*c^2*d^5 + 59*c*d^6 + 32*d^7)*cos(f*x + e)^4 - 2*(4*c^6*d - 28*c^5*d^2 - 118*c^4*

d^3 - 106*c^3*d^4 + 71*c^2*d^5 + 134*c*d^6 + 43*d^7)*cos(f*x + e)^3 + 2*(2*c^7 - 12*c^6*d - 36*c^5*d^2 - 54*c^

4*d^3 - 39*c^3*d^4 + 39*c^2*d^5 + 73*c*d^6 + 27*d^7)*cos(f*x + e)^2 + 3*(24*c^4*d^2 + 80*c^3*d^3 + 102*c^2*d^4

+ 60*c*d^5 + 14*d^6 + (12*c^2*d^4 + 16*c*d^5 + 7*d^6)*cos(f*x + e)^4 - (24*c^3*d^3 + 44*c^2*d^4 + 30*c*d^5 +

7*d^6)*cos(f*x + e)^3 - (12*c^4*d^2 + 64*c^3*d^3 + 107*c^2*d^4 + 76*c*d^5 + 21*d^6)*cos(f*x + e)^2 + (12*c^4*d

^2 + 40*c^3*d^3 + 51*c^2*d^4 + 30*c*d^5 + 7*d^6)*cos(f*x + e) + (24*c^4*d^2 + 80*c^3*d^3 + 102*c^2*d^4 + 60*c*

d^5 + 14*d^6 - (12*c^2*d^4 + 16*c*d^5 + 7*d^6)*cos(f*x + e)^3 - 2*(12*c^3*d^3 + 28*c^2*d^4 + 23*c*d^5 + 7*d^6)

*cos(f*x + e)^2 + (12*c^4*d^2 + 40*c^3*d^3 + 51*c^2*d^4 + 30*c*d^5 + 7*d^6)*cos(f*x + e))*sin(f*x + e))*sqrt(-

c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e)

+ d*cos(f*x + e))*sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 4*(2*c^7 - 5*c^6

*d - 36*c^5*d^2 - 75*c^4*d^3 - 39*c^3*d^4 + 60*c^2*d^5 + 73*c*d^6 + 20*d^7)*cos(f*x + e) - 2*(2*c^7 - 2*c^6*d

- 6*c^5*d^2 + 6*c^4*d^3 + 6*c^3*d^4 - 6*c^2*d^5 - 2*c*d^6 + 2*d^7 + (2*c^5*d^2 - 16*c^4*d^3 - 61*c^3*d^4 - 16*

c^2*d^5 + 59*c*d^6 + 32*d^7)*cos(f*x + e)^3 - (4*c^6*d - 30*c^5*d^2 - 102*c^4*d^3 - 45*c^3*d^4 + 87*c^2*d^5 +

75*c*d^6 + 11*d^7)*cos(f*x + e)^2 - 2*(c^7 - 4*c^6*d - 33*c^5*d^2 - 78*c^4*d^3 - 42*c^3*d^4 + 63*c^2*d^5 + 74*

c*d^6 + 19*d^7)*cos(f*x + e))*sin(f*x + e))/((a^2*c^8*d^2 - 2*a^2*c^7*d^3 - 2*a^2*c^6*d^4 + 6*a^2*c^5*d^5 - 6*

a^2*c^3*d^7 + 2*a^2*c^2*d^8 + 2*a^2*c*d^9 - a^2*d^10)*f*cos(f*x + e)^4 - (2*a^2*c^9*d - 3*a^2*c^8*d^2 - 6*a^2*

c^7*d^3 + 10*a^2*c^6*d^4 + 6*a^2*c^5*d^5 - 12*a^2*c^4*d^6 - 2*a^2*c^3*d^7 + 6*a^2*c^2*d^8 - a^2*d^10)*f*cos(f*

x + e)^3 - (a^2*c^10 + 2*a^2*c^9*d - 7*a^2*c^8*d^2 - 8*a^2*c^7*d^3 + 18*a^2*c^6*d^4 + 12*a^2*c^5*d^5 - 22*a^2*

c^4*d^6 - 8*a^2*c^3*d^7 + 13*a^2*c^2*d^8 + 2*a^2*c*d^9 - 3*a^2*d^10)*f*cos(f*x + e)^2 + (a^2*c^10 - 5*a^2*c^8*

d^2 + 10*a^2*c^6*d^4 - 10*a^2*c^4*d^6 + 5*a^2*c^2*d^8 - a^2*d^10)*f*cos(f*x + e) + 2*(a^2*c^10 - 5*a^2*c^8*d^2

+ 10*a^2*c^6*d^4 - 10*a^2*c^4*d^6 + 5*a^2*c^2*d^8 - a^2*d^10)*f - ((a^2*c^8*d^2 - 2*a^2*c^7*d^3 - 2*a^2*c^6*d

^4 + 6*a^2*c^5*d^5 - 6*a^2*c^3*d^7 + 2*a^2*c^2*d^8 + 2*a^2*c*d^9 - a^2*d^10)*f*cos(f*x + e)^3 + 2*(a^2*c^9*d -

a^2*c^8*d^2 - 4*a^2*c^7*d^3 + 4*a^2*c^6*d^4 + 6*a^2*c^5*d^5 - 6*a^2*c^4*d^6 - 4*a^2*c^3*d^7 + 4*a^2*c^2*d^8 +

a^2*c*d^9 - a^2*d^10)*f*cos(f*x + e)^2 - (a^2*c^10 - 5*a^2*c^8*d^2 + 10*a^2*c^6*d^4 - 10*a^2*c^4*d^6 + 5*a^2*

c^2*d^8 - a^2*d^10)*f*cos(f*x + e) - 2*(a^2*c^10 - 5*a^2*c^8*d^2 + 10*a^2*c^6*d^4 - 10*a^2*c^4*d^6 + 5*a^2*c^2

*d^8 - a^2*d^10)*f)*sin(f*x + e)), -1/6*(2*c^7 - 2*c^6*d - 6*c^5*d^2 + 6*c^4*d^3 + 6*c^3*d^4 - 6*c^2*d^5 - 2*c

*d^6 + 2*d^7 - (2*c^5*d^2 - 16*c^4*d^3 - 61*c^3*d^4 - 16*c^2*d^5 + 59*c*d^6 + 32*d^7)*cos(f*x + e)^4 - (4*c^6*

d - 28*c^5*d^2 - 118*c^4*d^3 - 106*c^3*d^4 + 71*c^2*d^5 + 134*c*d^6 + 43*d^7)*cos(f*x + e)^3 + (2*c^7 - 12*c^6

*d - 36*c^5*d^2 - 54*c^4*d^3 - 39*c^3*d^4 + 39*c^2*d^5 + 73*c*d^6 + 27*d^7)*cos(f*x + e)^2 + 3*(24*c^4*d^2 + 8

0*c^3*d^3 + 102*c^2*d^4 + 60*c*d^5 + 14*d^6 + (12*c^2*d^4 + 16*c*d^5 + 7*d^6)*cos(f*x + e)^4 - (24*c^3*d^3 + 4

4*c^2*d^4 + 30*c*d^5 + 7*d^6)*cos(f*x + e)^3 - (12*c^4*d^2 + 64*c^3*d^3 + 107*c^2*d^4 + 76*c*d^5 + 21*d^6)*cos

(f*x + e)^2 + (12*c^4*d^2 + 40*c^3*d^3 + 51*c^2*d^4 + 30*c*d^5 + 7*d^6)*cos(f*x + e) + (24*c^4*d^2 + 80*c^3*d^

3 + 102*c^2*d^4 + 60*c*d^5 + 14*d^6 - (12*c^2*d^4 + 16*c*d^5 + 7*d^6)*cos(f*x + e)^3 - 2*(12*c^3*d^3 + 28*c^2*

d^4 + 23*c*d^5 + 7*d^6)*cos(f*x + e)^2 + (12*c^4*d^2 + 40*c^3*d^3 + 51*c^2*d^4 + 30*c*d^5 + 7*d^6)*cos(f*x + e

))*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + 2*(2*c^7 - 5*c

^6*d - 36*c^5*d^2 - 75*c^4*d^3 - 39*c^3*d^4 + 60*c^2*d^5 + 73*c*d^6 + 20*d^7)*cos(f*x + e) - (2*c^7 - 2*c^6*d