3.1287 \(\int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=148 \[ \frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d}+\frac{a x \left (2 a^2-b^2\right )}{2 b^4}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.458729, antiderivative size = 148, normalized size of antiderivative = 1., 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, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac{\left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^3 d}-\frac{2 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d}+\frac{a x \left (2 a^2-b^2\right )}{2 b^4}-\frac{a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])))

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

Mathematica [A]  time = 0.25919, size = 130, normalized size = 0.88 \[ -\frac{3 b \left (b^2-4 a^2\right ) \cos (c+d x)+24 a^2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-12 a^3 c-12 a^3 d x+3 a b^2 \sin (2 (c+d x))+6 a b^2 c+6 a b^2 d x+b^3 \cos (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-12*a^3*c + 6*a*b^2*c - 12*a^3*d*x + 6*a*b^2*d*x + 24*a^2*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sq
rt[a^2 - b^2]] + 3*b*(-4*a^2 + b^2)*Cos[c + d*x] + b^3*Cos[3*(c + d*x)] + 3*a*b^2*Sin[2*(c + d*x)])/(12*b^4*d)

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Maple [B]  time = 0.082, size = 318, normalized size = 2.2 \begin{align*}{\frac{a}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}{a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+4\,{\frac{{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{a}{d{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{{a}^{2}}{d{b}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-{\frac{2}{3\,bd} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{3}}{d{b}^{4}}}-{\frac{a}{d{b}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{a}^{2}\sqrt{{a}^{2}-{b}^{2}}}{d{b}^{4}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

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)),x)

[Out]

1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c
)^4*a^2-2/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2*tan(1/2*d
*x+1/2*c)^2-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2-2/3
/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3+2/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^3-1/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))-2/
d*a^2*(a^2-b^2)^(1/2)/b^4*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)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60131, size = 736, normalized size = 4.97 \begin{align*} \left [-\frac{2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 3 \, \sqrt{-a^{2} + b^{2}} a^{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 ) - 3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}, -\frac{2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, a^{2} b \cos \left (d x + c\right ) - 6 \, \sqrt{a^{2} - b^{2}} a^{2} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \,{\left (2 \, a^{3} - a b^{2}\right )} d x}{6 \, b^{4} d}\right ] \end{align*}

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)),x, algorithm="fricas")

[Out]

[-1/6*(2*b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)*sin(d*x + c) - 6*a^2*b*cos(d*x + c) - 3*sqrt(-a^2 + b^2)*a^
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)) - 3*(2*a^3 - a*b^2)*d*x)/(b
^4*d), -1/6*(2*b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)*sin(d*x + c) - 6*a^2*b*cos(d*x + c) - 6*sqrt(a^2 - b^
2)*a^2*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 3*(2*a^3 - a*b^2)*d*x)/(b^4*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)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.19669, size = 279, normalized size = 1.89 \begin{align*} \frac{\frac{3 \,{\left (2 \, a^{3} - a b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{12 \,{\left (a^{4} - a^{2} b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \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{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \end{align*}

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)),x, algorithm="giac")

[Out]

1/6*(3*(2*a^3 - a*b^2)*(d*x + c)/b^4 - 12*(a^4 - a^2*b^2)*(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)))/(sqrt(a^2 - b^2)*b^4) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*t
an(1/2*d*x + 1/2*c)^4 - 6*b^2*tan(1/2*d*x + 1/2*c)^4 + 12*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2
*c) + 6*a^2 - 2*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d