3.1080 \(\int \frac{\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=106 \[ -\frac{2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d \sqrt{a^2-b^2}}+\frac{\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}+\frac{2 a x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.152587, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2863, 2735, 2660, 618, 204} \[ -\frac{2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d \sqrt{a^2-b^2}}+\frac{\cos (c+d x) (2 a+b \sin (c+d x))}{b^2 d (a+b \sin (c+d x))}+\frac{2 a x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2863

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

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

Mathematica [A]  time = 0.853078, size = 130, normalized size = 1.23 \[ \frac{\frac{4 a^2 c+4 a^2 d x+4 a b (c+d x) \sin (c+d x)+4 a b \cos (c+d x)+b^2 \sin (2 (c+d x))}{a+b \sin (c+d x)}-\frac{4 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.107, size = 229, normalized size = 2.2 \begin{align*} 2\,{\frac{1}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) a}{d{b}^{3}}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{bd \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{a}{d{b}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-4\,{\frac{{a}^{2}}{d{b}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{1}{bd\sqrt{{a}^{2}-{b}^{2}}}\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)/(a+b*sin(d*x+c))^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63915, size = 1025, normalized size = 9.67 \begin{align*} \left [\frac{4 \,{\left (a^{4} - a^{2} b^{2}\right )} d x +{\left (2 \, a^{3} - a b^{2} +{\left (2 \, a^{2} b - 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 ) + 4 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) + 2 \,{\left (2 \,{\left (a^{3} b - a b^{3}\right )} d x +{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{4} - b^{6}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{3} - a b^{5}\right )} d\right )}}, \frac{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d x +{\left (2 \, a^{3} - a b^{2} +{\left (2 \, a^{2} b - 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 ) + 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) +{\left (2 \,{\left (a^{3} b - a b^{3}\right )} d x +{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{4} - b^{6}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{3} - a b^{5}\right )} 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)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.21664, size = 258, normalized size = 2.43 \begin{align*} \frac{2 \,{\left (\frac{{\left (d x + c\right )} a}{b^{3}} - \frac{{\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 )}{\left (2 \, a^{2} - b^{2}\right )}}{\sqrt{a^{2} - b^{2}} b^{3}} + \frac{b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, 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^{2}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((d*x + c)*a/b^3 - (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)))*(2*a^2 - b^2)/(sqrt(a^2 - b^2)*b^3) + (b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 3*b*tan
(1/2*d*x + 1/2*c) + 2*a)/((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)*b^2))/d