3.1085 \(\int \frac{\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=266 \[ -\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 d \left (a^2-b^2\right )}+\frac{a \left (-19 a^2 b^2+12 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \left (a^2-b^2\right )^{3/2}}-\frac{\left (4 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{\left (6 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac{x \left (12 a^2-b^2\right )}{2 b^5}-\frac{\sin ^3(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

-((12*a^2 - b^2)*x)/(2*b^5) + (a*(12*a^4 - 19*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]
])/(b^5*(a^2 - b^2)^(3/2)*d) - (a*(12*a^2 - 11*b^2)*Cos[c + d*x])/(2*b^4*(a^2 - b^2)*d) + ((6*a^2 - 5*b^2)*Cos
[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2)*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(2*b*d*(a + b*Sin[c + d*x])^2) -
((4*a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(2*b^2*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.877846, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2889, 3048, 3049, 3023, 2735, 2660, 618, 204} \[ -\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 d \left (a^2-b^2\right )}+\frac{a \left (-19 a^2 b^2+12 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^5 d \left (a^2-b^2\right )^{3/2}}-\frac{\left (4 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{\left (6 a^2-5 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac{x \left (12 a^2-b^2\right )}{2 b^5}-\frac{\sin ^3(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((12*a^2 - b^2)*x)/(2*b^5) + (a*(12*a^4 - 19*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]
])/(b^5*(a^2 - b^2)^(3/2)*d) - (a*(12*a^2 - 11*b^2)*Cos[c + d*x])/(2*b^4*(a^2 - b^2)*d) + ((6*a^2 - 5*b^2)*Cos
[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2)*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(2*b*d*(a + b*Sin[c + d*x])^2) -
((4*a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x]^2)/(2*b^2*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 2889

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

Rule 3048

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

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

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{\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \frac{\sin ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^3} \, dx\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\int \frac{\sin ^2(c+d x) \left (-3 \left (a^2-b^2\right )+4 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{\sin (c+d x) \left (2 \left (4 a^4-7 a^2 b^2+3 b^4\right )-a b \left (a^2-b^2\right ) \sin (c+d x)-2 \left (6 a^2-5 b^2\right ) \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (6 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{-2 a \left (6 a^4-11 a^2 b^2+5 b^4\right )+2 b \left (2 a^4-3 a^2 b^2+b^4\right ) \sin (c+d x)+2 a \left (12 a^2-11 b^2\right ) \left (a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{4 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}+\frac{\left (6 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{-2 a b \left (6 a^4-11 a^2 b^2+5 b^4\right )-2 \left (a^2-b^2\right )^2 \left (12 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{4 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (12 a^2-b^2\right ) x}{2 b^5}-\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}+\frac{\left (6 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (a \left (12 a^4-19 a^2 b^2+6 b^4\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 b^5 \left (a^2-b^2\right )}\\ &=-\frac{\left (12 a^2-b^2\right ) x}{2 b^5}-\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}+\frac{\left (6 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (a \left (12 a^4-19 a^2 b^2+6 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^5 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (12 a^2-b^2\right ) x}{2 b^5}-\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}+\frac{\left (6 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\left (2 a \left (12 a^4-19 a^2 b^2+6 b^4\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} (c+d x)\right )\right )}{b^5 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (12 a^2-b^2\right ) x}{2 b^5}+\frac{a \left (12 a^4-19 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^5 \left (a^2-b^2\right )^{3/2} d}-\frac{a \left (12 a^2-11 b^2\right ) \cos (c+d x)}{2 b^4 \left (a^2-b^2\right ) d}+\frac{\left (6 a^2-5 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (4 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.70776, size = 288, normalized size = 1.08 \[ \frac{\frac{4 a \left (-19 a^2 b^2+12 a^4+6 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{-2 b^2 \left (-13 a^2 b^2+12 a^4+b^4\right ) (c+d x) \sin ^2(c+d x)+a^2 \left (-\left (2 \left (-13 a^2 b^2+12 a^4+b^4\right ) (c+d x)+\left (18 a^2 b^2-17 b^4\right ) \sin (2 (c+d x))\right )\right )-4 a b \left (-13 a^2 b^2+12 a^4+b^4\right ) (c+d x) \sin (c+d x)+\cos (c+d x) \left (-8 a b^3 \left (a^2-b^2\right ) \sin ^2(c+d x)+2 b^4 \left (a^2-b^2\right ) \sin ^3(c+d x)+22 a^3 b^3-24 a^5 b\right )}{(a+b \sin (c+d x))^2}}{4 b^5 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*(12*a^4 - 19*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (-4*a*
b*(12*a^4 - 13*a^2*b^2 + b^4)*(c + d*x)*Sin[c + d*x] - 2*b^2*(12*a^4 - 13*a^2*b^2 + b^4)*(c + d*x)*Sin[c + d*x
]^2 + Cos[c + d*x]*(-24*a^5*b + 22*a^3*b^3 - 8*a*b^3*(a^2 - b^2)*Sin[c + d*x]^2 + 2*b^4*(a^2 - b^2)*Sin[c + d*
x]^3) - a^2*(2*(12*a^4 - 13*a^2*b^2 + b^4)*(c + d*x) + (18*a^2*b^2 - 17*b^4)*Sin[2*(c + d*x)]))/(a + b*Sin[c +
d*x])^2)/(4*(a - b)*b^5*(a + b)*d)

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Maple [B]  time = 0.149, size = 845, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^2*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*a-12/d/b^5*arcta
n(tan(1/2*d*x+1/2*c))*a^2+1/d/b^3*arctan(tan(1/2*d*x+1/2*c))-5/d*a^4/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x
+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+4/d*a^2/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a
^2-b^2)*tan(1/2*d*x+1/2*c)^3-6/d*a^5/b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2
*d*x+1/2*c)^2-7/d*a^3/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2+1
0/d*a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-19/d*a^4/b^3/(tan(1/2
*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)+16/d*a^2/b/(tan(1/2*d*x+1/2*c)^2*a+2*
tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)-6/d*a^5/b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c
)*b+a)^2/(a^2-b^2)+5/d*a^3/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)+12/d*a^5/b^5/(a^2
-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-19/d*a^3/b^3/(a^2-b^2)^(3/2)*arctan(1/2*(
2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+6/d*a/b/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(
a^2-b^2)^(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(cos(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.1118, size = 2309, normalized size = 8.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*(12*a^6*b^2 - 25*a^4*b^4 + 14*a^2*b^6 - b^8)*d*x*cos(d*x + c)^2 + 8*(a^5*b^3 - 2*a^3*b^5 + a*b^7)*cos
(d*x + c)^3 - 2*(12*a^8 - 13*a^6*b^2 - 11*a^4*b^4 + 13*a^2*b^6 - b^8)*d*x + (12*a^7 - 7*a^5*b^2 - 13*a^3*b^4 +
6*a*b^6 - (12*a^5*b^2 - 19*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + 2*(12*a^6*b - 19*a^4*b^3 + 6*a^2*b^5)*sin(d*x
+ c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)
*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(
12*a^7*b - 19*a^5*b^3 + 3*a^3*b^5 + 4*a*b^7)*cos(d*x + c) - 2*((a^4*b^4 - 2*a^2*b^6 + b^8)*cos(d*x + c)^3 + 2*
(12*a^7*b - 25*a^5*b^3 + 14*a^3*b^5 - a*b^7)*d*x + (18*a^6*b^2 - 36*a^4*b^4 + 19*a^2*b^6 - b^8)*cos(d*x + c))*
sin(d*x + c))/((a^4*b^7 - 2*a^2*b^9 + b^11)*d*cos(d*x + c)^2 - 2*(a^5*b^6 - 2*a^3*b^8 + a*b^10)*d*sin(d*x + c)
- (a^6*b^5 - a^4*b^7 - a^2*b^9 + b^11)*d), -1/2*((12*a^6*b^2 - 25*a^4*b^4 + 14*a^2*b^6 - b^8)*d*x*cos(d*x + c
)^2 + 4*(a^5*b^3 - 2*a^3*b^5 + a*b^7)*cos(d*x + c)^3 - (12*a^8 - 13*a^6*b^2 - 11*a^4*b^4 + 13*a^2*b^6 - b^8)*d
*x - (12*a^7 - 7*a^5*b^2 - 13*a^3*b^4 + 6*a*b^6 - (12*a^5*b^2 - 19*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + 2*(12*a
^6*b - 19*a^4*b^3 + 6*a^2*b^5)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos
(d*x + c))) - (12*a^7*b - 19*a^5*b^3 + 3*a^3*b^5 + 4*a*b^7)*cos(d*x + c) - ((a^4*b^4 - 2*a^2*b^6 + b^8)*cos(d*
x + c)^3 + 2*(12*a^7*b - 25*a^5*b^3 + 14*a^3*b^5 - a*b^7)*d*x + (18*a^6*b^2 - 36*a^4*b^4 + 19*a^2*b^6 - b^8)*c
os(d*x + c))*sin(d*x + c))/((a^4*b^7 - 2*a^2*b^9 + b^11)*d*cos(d*x + c)^2 - 2*(a^5*b^6 - 2*a^3*b^8 + a*b^10)*d
*sin(d*x + c) - (a^6*b^5 - a^4*b^7 - a^2*b^9 + b^11)*d)]

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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(d*x+c)**2*sin(d*x+c)**3/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.36636, size = 722, normalized size = 2.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(12*a^5 - 19*a^3*b^2 + 6*a*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*
c) + b)/sqrt(a^2 - b^2)))/((a^2*b^5 - b^7)*sqrt(a^2 - b^2)) - 2*(6*a^4*b*tan(1/2*d*x + 1/2*c)^7 - 5*a^2*b^3*ta
n(1/2*d*x + 1/2*c)^7 + 12*a^5*tan(1/2*d*x + 1/2*c)^6 + 5*a^3*b^2*tan(1/2*d*x + 1/2*c)^6 - 14*a*b^4*tan(1/2*d*x
+ 1/2*c)^6 + 54*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 45*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 4*b^5*tan(1/2*d*x + 1/2*c)
^5 + 36*a^5*tan(1/2*d*x + 1/2*c)^4 + 15*a^3*b^2*tan(1/2*d*x + 1/2*c)^4 - 44*a*b^4*tan(1/2*d*x + 1/2*c)^4 + 90*
a^4*b*tan(1/2*d*x + 1/2*c)^3 - 87*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*b^5*tan(1/2*d*x + 1/2*c)^3 + 36*a^5*tan(1
/2*d*x + 1/2*c)^2 - a^3*b^2*tan(1/2*d*x + 1/2*c)^2 - 30*a*b^4*tan(1/2*d*x + 1/2*c)^2 + 42*a^4*b*tan(1/2*d*x +
1/2*c) - 39*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*a^5 - 11*a^3*b^2)/((a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^4 + 2
*b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) - (12*a^2 - b^2)*(d*
x + c)/b^5)/d