3.1079 \(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=153 \[ \frac {2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d \sqrt {a^2-b^2}}-\frac {x \left (6 a^2-b^2\right )}{2 b^4}-\frac {3 a \cos (c+d x)}{b^3 d}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac {3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 153, 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, 3050, 3023, 2735, 2660, 618, 204} \[ \frac {2 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d \sqrt {a^2-b^2}}-\frac {x \left (6 a^2-b^2\right )}{2 b^4}-\frac {3 a \cos (c+d x)}{b^3 d}-\frac {\sin ^2(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac {3 \sin (c+d x) \cos (c+d x)}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((6*a^2 - b^2)*x)/(2*b^4) + (2*a*(3*a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[
a^2 - b^2]*d) - (3*a*Cos[c + d*x])/(b^3*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(2*b^2*d) - (Cos[c + d*x]*Sin[c + d
*x]^2)/(b*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 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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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])))

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 129, normalized size = 0.84 \[ \frac {2 \left (b^2-6 a^2\right ) (c+d x)+\frac {8 a \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {4 a^2 b \cos (c+d x)}{a+b \sin (c+d x)}-8 a b \cos (c+d x)+b^2 \sin (2 (c+d x))}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 568, normalized size = 3.71 \[ \left [-\frac {{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x - {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + {\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + {\left ({\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}, -\frac {{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, a^{5} - 7 \, a^{3} b^{2} + a b^{4}\right )} d x + 2 \, {\left (3 \, a^{4} - 2 \, a^{2} b^{2} + {\left (3 \, a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + {\left ({\left (6 \, a^{4} b - 7 \, a^{2} b^{3} + b^{5}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{4} - a b^{6}\right )} d\right )}}\right ] \]

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))^2,x, algorithm="fricas")

[Out]

[-1/2*((a^2*b^3 - b^5)*cos(d*x + c)^3 + (6*a^5 - 7*a^3*b^2 + a*b^4)*d*x - (3*a^4 - 2*a^2*b^2 + (3*a^3*b - 2*a*
b^3)*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)) + (6*a^4*b - 7*a^2*b^3 + b^5)*cos(d*x + c) + ((6*a^4*b - 7*a^2*b^3 + b^5)*d*x + 3*(a^3*b^2 - a*b^4)*co
s(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d*sin(d*x + c) + (a^3*b^4 - a*b^6)*d), -1/2*((a^2*b^3 - b^5)*cos(d*
x + c)^3 + (6*a^5 - 7*a^3*b^2 + a*b^4)*d*x + 2*(3*a^4 - 2*a^2*b^2 + (3*a^3*b - 2*a*b^3)*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^4*b - 7*a^2*b^3 + b^5)*cos(d*x + c
) + ((6*a^4*b - 7*a^2*b^3 + b^5)*d*x + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d*sin(
d*x + c) + (a^3*b^4 - a*b^6)*d)]

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giac [A]  time = 0.19, size = 211, normalized size = 1.38 \[ -\frac {\frac {{\left (6 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {4 \, {\left (3 \, a^{3} - 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {4 \, {\left (a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{3}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}}}{2 \, d} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.51, size = 353, normalized size = 2.31 \[ -\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 a}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{4}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2 a^{2}}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {6 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{4} \sqrt {a^{2}-b^{2}}}-\frac {4 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{2} \sqrt {a^{2}-b^{2}}} \]

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))^2,x)

[Out]

-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^2*a+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*a-6/d/b^4*arctan
(tan(1/2*d*x+1/2*c))*a^2+1/d/b^2*arctan(tan(1/2*d*x+1/2*c))-2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*
c)*b+a)*a*tan(1/2*d*x+1/2*c)-2/d/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a^2+6/d/b^4*a^3/(a^2-b^
2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-4/d/b^2*a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*ta
n(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))

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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))^2,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?

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mupad [B]  time = 10.40, size = 479, normalized size = 3.13 \[ -\frac {\frac {6\,a^2}{b^3}+\frac {9\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2+b^2\right )}{b^3}+\frac {12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{b^2}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{b^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a^2-b^2\right )}{b^3}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sqrt {b^2-a^2}\right )\,\left (3\,a^3\,\sqrt {b^2-a^2}-2\,a\,b^2\,\sqrt {b^2-a^2}\right )}{b^4\,d\,\left (a^2-b^2\right )}-\frac {a\,\ln \left (b+a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (3\,a^2-2\,b^2\right )}{d\,\left (b^6-a^2\,b^4\right )}+\frac {\mathrm {atan}\left (\frac {24\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{24\,a^3-8\,a\,b^2+\frac {144\,a^5}{b^2}}+\frac {144\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{144\,a^5+24\,a^3\,b^2-8\,a\,b^4}-\frac {8\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\frac {24\,a^3}{b^2}-8\,a+\frac {144\,a^5}{b^4}}\right )\,\left (a^2\,6{}\mathrm {i}-b^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b^4\,d} \]

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))^2,x)

[Out]

(atan((24*a^3*tan(c/2 + (d*x)/2))/(24*a^3 - 8*a*b^2 + (144*a^5)/b^2) + (144*a^5*tan(c/2 + (d*x)/2))/(144*a^5 -
8*a*b^4 + 24*a^3*b^2) - (8*a*tan(c/2 + (d*x)/2))/((24*a^3)/b^2 - 8*a + (144*a^5)/b^4))*(a^2*6i - b^2*1i)*1i)/
(b^4*d) - ((6*a^2)/b^3 + (9*a*tan(c/2 + (d*x)/2))/b^2 + (2*tan(c/2 + (d*x)/2)^4*(3*a^2 + b^2))/b^3 + (12*a*tan
(c/2 + (d*x)/2)^3)/b^2 + (3*a*tan(c/2 + (d*x)/2)^5)/b^2 + (2*tan(c/2 + (d*x)/2)^2*(6*a^2 - b^2))/b^3)/(d*(a +
2*b*tan(c/2 + (d*x)/2) + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^6 + 4*b*ta
n(c/2 + (d*x)/2)^3 + 2*b*tan(c/2 + (d*x)/2)^5)) - (log(b + a*tan(c/2 + (d*x)/2) - (b^2 - a^2)^(1/2))*(3*a^3*(b
^2 - a^2)^(1/2) - 2*a*b^2*(b^2 - a^2)^(1/2)))/(b^4*d*(a^2 - b^2)) - (a*log(b + a*tan(c/2 + (d*x)/2) + (b^2 - a
^2)^(1/2))*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 2*b^2))/(d*(b^6 - a^2*b^4))

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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))**2,x)

[Out]

Timed out

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