3.1086 \(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=180 \[ \frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\left (6 a^4-9 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d \left (a^2-b^2\right )^{3/2}}+\frac {3 a x}{b^4}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {3 \cos (c+d x)}{2 b^3 d} \]

[Out]

3*a*x/b^4-(6*a^4-9*a^2*b^2+2*b^4)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^4/(a^2-b^2)^(3/2)/d+3/2*c
os(d*x+c)/b^3/d-1/2*cos(d*x+c)*sin(d*x+c)^2/b/d/(a+b*sin(d*x+c))^2+1/2*a*(3*a^2-2*b^2)*cos(d*x+c)/b^3/(a^2-b^2
)/d/(a+b*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.56, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3048, 3032, 3023, 2735, 2660, 618, 204} \[ -\frac {\left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d \left (a^2-b^2\right )^{3/2}}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 a x}{b^4}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {3 \cos (c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x)/b^4 - ((6*a^4 - 9*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^
(3/2)*d) + (3*Cos[c + d*x])/(2*b^3*d) - (Cos[c + d*x]*Sin[c + d*x]^2)/(2*b*d*(a + b*Sin[c + d*x])^2) + (a*(3*a
^2 - 2*b^2)*Cos[c + d*x])/(2*b^3*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

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])

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 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 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 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 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 3032

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1
))/(b^2*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b
*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f
*x] + b*C*d*(m + 1)*(a^2 - b^2)*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] && LtQ[m, -1]

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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \frac {\sin ^2(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 ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {\int \frac {\sin (c+d x) \left (-2 \left (a^2-b^2\right )+3 \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 ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {b \left (3 a^4-5 a^2 b^2+2 b^4\right )+3 a \left (a^2-b^2\right )^2 \sin (c+d x)-3 b \left (a^2-b^2\right )^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {3 \cos (c+d x)}{2 b^3 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {b^2 \left (3 a^4-5 a^2 b^2+2 b^4\right )+6 a b \left (a^2-b^2\right )^2 \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac {3 a x}{b^4}+\frac {3 \cos (c+d x)}{2 b^3 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (6 a^4-9 a^2 b^2+2 b^4\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )}\\ &=\frac {3 a x}{b^4}+\frac {3 \cos (c+d x)}{2 b^3 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (6 a^4-9 a^2 b^2+2 b^4\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^4 \left (a^2-b^2\right ) d}\\ &=\frac {3 a x}{b^4}+\frac {3 \cos (c+d x)}{2 b^3 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\left (2 \left (6 a^4-9 a^2 b^2+2 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^4 \left (a^2-b^2\right ) d}\\ &=\frac {3 a x}{b^4}-\frac {\left (6 a^4-9 a^2 b^2+2 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^4 \left (a^2-b^2\right )^{3/2} d}+\frac {3 \cos (c+d x)}{2 b^3 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.22, size = 159, normalized size = 0.88 \[ \frac {\frac {a b \left (5 a^2-4 b^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))}-\frac {a^2 b \cos (c+d x)}{(a+b \sin (c+d x))^2}-\frac {2 \left (6 a^4-9 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+6 a (c+d x)+2 b \cos (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*(c + d*x) - (2*(6*a^4 - 9*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^
(3/2) + 2*b*Cos[c + d*x] - (a^2*b*Cos[c + d*x])/(a + b*Sin[c + d*x])^2 + (a*b*(5*a^2 - 4*b^2)*Cos[c + d*x])/((
a - b)*(a + b)*(a + b*Sin[c + d*x])))/(2*b^4*d)

________________________________________________________________________________________

fricas [B]  time = 0.76, size = 919, normalized size = 5.11 \[ \left [\frac {12 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} - 12 \, {\left (a^{7} - a^{5} b^{2} - a^{3} b^{4} + a b^{6}\right )} d x - {\left (6 \, a^{6} - 3 \, a^{4} b^{2} - 7 \, a^{2} b^{4} + 2 \, b^{6} - {\left (6 \, a^{4} b^{2} - 9 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{5} b - 9 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (6 \, a^{6} b - 9 \, a^{4} b^{3} + a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right ) - 2 \, {\left (12 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d x + {\left (9 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + 8 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{6} - 2 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b^{5} - 2 \, a^{3} b^{7} + a b^{9}\right )} d \sin \left (d x + c\right ) - {\left (a^{6} b^{4} - a^{4} b^{6} - a^{2} b^{8} + b^{10}\right )} d\right )}}, \frac {6 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d x \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{7} - a^{5} b^{2} - a^{3} b^{4} + a b^{6}\right )} d x - {\left (6 \, a^{6} - 3 \, a^{4} b^{2} - 7 \, a^{2} b^{4} + 2 \, b^{6} - {\left (6 \, a^{4} b^{2} - 9 \, a^{2} b^{4} + 2 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{5} b - 9 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (6 \, a^{6} b - 9 \, a^{4} b^{3} + a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right ) - {\left (12 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d x + {\left (9 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + 8 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{6} - 2 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b^{5} - 2 \, a^{3} b^{7} + a b^{9}\right )} d \sin \left (d x + c\right ) - {\left (a^{6} b^{4} - a^{4} b^{6} - a^{2} b^{8} + b^{10}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(12*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*x*cos(d*x + c)^2 + 4*(a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 - 12*
(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*x - (6*a^6 - 3*a^4*b^2 - 7*a^2*b^4 + 2*b^6 - (6*a^4*b^2 - 9*a^2*b^4 + 2*b^
6)*cos(d*x + c)^2 + 2*(6*a^5*b - 9*a^3*b^3 + 2*a*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*(6*a^6*b - 9*a^4*b^3 + a^2*b^5 + 2*b^7)*cos(d*x +
c) - 2*(12*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*x + (9*a^5*b^2 - 17*a^3*b^4 + 8*a*b^6)*cos(d*x + c))*sin(d*x + c))/
((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c)^2 - 2*(a^5*b^5 - 2*a^3*b^7 + a*b^9)*d*sin(d*x + c) - (a^6*b^4 - a
^4*b^6 - a^2*b^8 + b^10)*d), 1/2*(6*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*x*cos(d*x + c)^2 + 2*(a^4*b^3 - 2*a^2*b^5
+ b^7)*cos(d*x + c)^3 - 6*(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*x - (6*a^6 - 3*a^4*b^2 - 7*a^2*b^4 + 2*b^6 - (6*
a^4*b^2 - 9*a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(6*a^5*b - 9*a^3*b^3 + 2*a*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))) - (6*a^6*b - 9*a^4*b^3 + a^2*b^5 + 2*b^7)*cos(d*x
+ c) - (12*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*x + (9*a^5*b^2 - 17*a^3*b^4 + 8*a*b^6)*cos(d*x + c))*sin(d*x + c))
/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c)^2 - 2*(a^5*b^5 - 2*a^3*b^7 + a*b^9)*d*sin(d*x + c) - (a^6*b^4 -
a^4*b^6 - a^2*b^8 + b^10)*d)]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 302, normalized size = 1.68 \[ -\frac {\frac {{\left (6 \, a^{4} - 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{4} - 3 \, a^{2} b^{2}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} - \frac {3 \, {\left (d x + c\right )} a}{b^{4}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((6*a^4 - 9*a^2*b^2 + 2*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)/s
qrt(a^2 - b^2)))/((a^2*b^4 - b^6)*sqrt(a^2 - b^2)) - (3*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^3*tan(1/2*d*x + 1
/2*c)^3 + 4*a^4*tan(1/2*d*x + 1/2*c)^2 + 5*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*b^4*tan(1/2*d*x + 1/2*c)^2 + 13*
a^3*b*tan(1/2*d*x + 1/2*c) - 10*a*b^3*tan(1/2*d*x + 1/2*c) + 4*a^4 - 3*a^2*b^2)/((a^2*b^3 - b^5)*(a*tan(1/2*d*
x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) - 3*(d*x + c)*a/b^4 - 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3))/d

________________________________________________________________________________________

maple [B]  time = 0.54, size = 711, normalized size = 3.95 \[ \frac {2}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {6 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}+\frac {3 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {6 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {13 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {10 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {4 a^{4}}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {3 a^{2}}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {6 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{4}}{d \,b^{4} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}+\frac {9 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{d \,b^{2} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}-\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)+6/d/b^4*a*arctan(tan(1/2*d*x+1/2*c))+3/d/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^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3-2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2
*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+4/d/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)^2*a^4+5/d/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)^2*a^
2-6/d*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)^2+13/d/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^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)-10/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(
1/2*d*x+1/2*c)*b+a)^2*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)+4/d/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^4/(a^2-b^2)-3/d/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/(a^2-b^2)-6/d/b^4/(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))*a^4+9/d/b^2/(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))*a^2-2/d/(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))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 15.79, size = 3031, normalized size = 16.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((6*a^4 - 5*a^2*b^2)/(b^3*(a^2 - b^2)) + (3*tan(c/2 + (d*x)/2)^4*(2*a^4 - 2*b^4 + a^2*b^2))/(b^3*(a^2 - b^2))
+ (2*tan(c/2 + (d*x)/2)^2*(6*a^4 - 7*b^4 + 3*a^2*b^2))/(b^3*(a^2 - b^2)) - (3*tan(c/2 + (d*x)/2)*(6*a*b^2 - 7*
a^3))/(b^2*(a^2 - b^2)) - (tan(c/2 + (d*x)/2)^5*(2*a*b^2 - 3*a^3))/(b^2*(a^2 - b^2)) - (4*tan(c/2 + (d*x)/2)^3
*(5*a*b^2 - 6*a^3))/(b^2*(a^2 - b^2)))/(d*(tan(c/2 + (d*x)/2)^2*(3*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(3*a^2
+ 4*b^2) + a^2*tan(c/2 + (d*x)/2)^6 + a^2 + 8*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*tan(c/2 + (d*x)/2)^5 + 4*a*b*ta
n(c/2 + (d*x)/2))) + (6*a*atan((192*a^2*b^7*tan(c/2 + (d*x)/2))/((192*a^2*b^19)/(b^12 - 2*a^2*b^10 + a^4*b^8)
- (384*a^4*b^17)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (48*a^6*b^15)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (288*a^8*b^13)/
(b^12 - 2*a^2*b^10 + a^4*b^8) - (144*a^10*b^11)/(b^12 - 2*a^2*b^10 + a^4*b^8)) - (144*a^6*b^3*tan(c/2 + (d*x)/
2))/((192*a^2*b^19)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (384*a^4*b^17)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (48*a^6*b^1
5)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (288*a^8*b^13)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (144*a^10*b^11)/(b^12 - 2*a^
2*b^10 + a^4*b^8))))/(b^4*d) + (atan((((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*(36*a^4*
b^7 - 72*a^6*b^5 + 36*a^8*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(4*a*b^11 - 108*a^3*b^9
+ 285*a^5*b^7 - 252*a^7*b^5 + 72*a^9*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^
4 + b^4 - (9*a^2*b^2)/2)*((8*tan(c/2 + (d*x)/2)*(8*a*b^14 - 44*a^3*b^12 + 60*a^5*b^10 - 24*a^7*b^8))/(b^13 - 2
*a^2*b^11 + a^4*b^9) - (8*(8*a^2*b^12 - 14*a^4*b^10 + 6*a^6*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + ((-(a + b)^3
*(a - b)^3)^(1/2)*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*
x)/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(3*a^4 + b^4 - (9*a
^2*b^2)/2))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))*1i)/(b^10 - 3
*a^2*b^8 + 3*a^4*b^6 - a^6*b^4) + ((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*(36*a^4*b^7
- 72*a^6*b^5 + 36*a^8*b^3))/(b^12 - 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(4*a*b^11 - 108*a^3*b^9 + 28
5*a^5*b^7 - 252*a^7*b^5 + 72*a^9*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 +
b^4 - (9*a^2*b^2)/2)*((8*(8*a^2*b^12 - 14*a^4*b^10 + 6*a^6*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 +
(d*x)/2)*(8*a*b^14 - 44*a^3*b^12 + 60*a^5*b^10 - 24*a^7*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^3*(a
- b)^3)^(1/2)*((8*(4*a^2*b^15 - 8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2
)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(3*a^4 + b^4 - (9*a^2*b
^2)/2))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))*1i)/(b^10 - 3*a^2
*b^8 + 3*a^4*b^6 - a^6*b^4))/((16*(54*a^8 - 12*a^2*b^6 + 72*a^4*b^4 - 117*a^6*b^2))/(b^12 - 2*a^2*b^10 + a^4*b
^8) + (16*tan(c/2 + (d*x)/2)*(216*a^9 - 72*a^3*b^6 + 396*a^5*b^4 - 540*a^7*b^2))/(b^13 - 2*a^2*b^11 + a^4*b^9)
- ((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*(36*a^4*b^7 - 72*a^6*b^5 + 36*a^8*b^3))/(b^
12 - 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(4*a*b^11 - 108*a^3*b^9 + 285*a^5*b^7 - 252*a^7*b^5 + 72*a^
9*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*tan(c/
2 + (d*x)/2)*(8*a*b^14 - 44*a^3*b^12 + 60*a^5*b^10 - 24*a^7*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) - (8*(8*a^2*b^
12 - 14*a^4*b^10 + 6*a^6*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) + ((-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a^2*b^15 -
8*a^4*b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(12*a*b^17 - 32*a^3*b^15 + 28
*a^5*b^13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(3*a^4 + b^4 - (9*a^2*b^2)/2))/(b^10 - 3*a^2*b^8 + 3*a
^4*b^6 - a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4) + ((-(
a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*(36*a^4*b^7 - 72*a^6*b^5 + 36*a^8*b^3))/(b^12 - 2*
a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(4*a*b^11 - 108*a^3*b^9 + 285*a^5*b^7 - 252*a^7*b^5 + 72*a^9*b^3))
/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*(8*a^2*b^12 -
14*a^4*b^10 + 6*a^6*b^8))/(b^12 - 2*a^2*b^10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(8*a*b^14 - 44*a^3*b^12 + 60*
a^5*b^10 - 24*a^7*b^8))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^3*(a - b)^3)^(1/2)*((8*(4*a^2*b^15 - 8*a^4*
b^13 + 4*a^6*b^11))/(b^12 - 2*a^2*b^10 + a^4*b^8) + (8*tan(c/2 + (d*x)/2)*(12*a*b^17 - 32*a^3*b^15 + 28*a^5*b^
13 - 8*a^7*b^11))/(b^13 - 2*a^2*b^11 + a^4*b^9))*(3*a^4 + b^4 - (9*a^2*b^2)/2))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6
- a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))/(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4)))*(-(a + b)^3
*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*2i)/(d*(b^10 - 3*a^2*b^8 + 3*a^4*b^6 - a^6*b^4))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________