3.1087 \(\int \frac{\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=167 \[ \frac{a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d \left (a^2-b^2\right )^{3/2}}-\frac{a \cos ^3(c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{\cos (c+d x) \left (2 \left (a^2-b^2\right )+a b \sin (c+d x)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.27907, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2864, 2863, 2735, 2660, 618, 204} \[ \frac{a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d \left (a^2-b^2\right )^{3/2}}-\frac{a \cos ^3(c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{\cos (c+d x) \left (2 \left (a^2-b^2\right )+a b \sin (c+d x)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2864

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

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

Mathematica [A]  time = 1.89661, size = 289, normalized size = 1.73 \[ \frac{\frac{\frac{2 a \left (-20 a^2 b^2+8 a^4+15 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{3 b \left (-7 a^2 b^2+4 a^4+2 b^4\right ) \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac{a b \left (4 a^2-3 b^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))^2}-8 (c+d x)}{b^3}-\frac{\frac{6 a b \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{\cos (c+d x) \left (b \left (a^2+2 b^2\right ) \sin (c+d x)+a \left (2 a^2+b^2\right )\right )}{(a+b \sin (c+d x))^2}}{(a-b)^2 (a+b)^2}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*(c + d*x) + (2*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b
^2)^(5/2) + (a*b*(4*a^2 - 3*b^2)*Cos[c + d*x])/((a - b)*(a + b)*(a + b*Sin[c + d*x])^2) - (3*b*(4*a^4 - 7*a^2*
b^2 + 2*b^4)*Cos[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])))/b^3 - ((6*a*b*ArcTan[(b + a*Tan[(c + d*
x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (Cos[c + d*x]*(a*(2*a^2 + b^2) + b*(a^2 + 2*b^2)*Sin[c + d*x]))/(a
+ b*Sin[c + d*x])^2)/((a - b)^2*(a + b)^2))/(8*d)

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Maple [B]  time = 0.132, size = 576, normalized size = 3.5 \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)/(a+b*sin(d*x+c))^3,x)

[Out]

-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))-1/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)*ta
n(1/2*d*x+1/2*c)^3-2/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
)^2-3/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+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)^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-7/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)+4/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/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)+1/
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)+2/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))-3/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)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.68554, size = 1702, normalized size = 10.19 \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)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.28848, size = 346, normalized size = 2.07 \begin{align*} \frac{\frac{{\left (2 \, a^{3} - 3 \, a 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 )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} - \frac{a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 7 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a^{4} - a^{2} b^{2}}{{\left (a^{3} b^{2} - a b^{4}\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{d x + c}{b^{3}}}{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))^3,x, algorithm="giac")

[Out]

((2*a^3 - 3*a*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)))/((a^2*b^3 - b^5)*sqrt(a^2 - b^2)) - (a^3*b*tan(1/2*d*x + 1/2*c)^3 + 2*a^4*tan(1/2*d*x + 1/2*c)^2 + 3*a^2
*b^2*tan(1/2*d*x + 1/2*c)^2 - 2*b^4*tan(1/2*d*x + 1/2*c)^2 + 7*a^3*b*tan(1/2*d*x + 1/2*c) - 4*a*b^3*tan(1/2*d*
x + 1/2*c) + 2*a^4 - a^2*b^2)/((a^3*b^2 - a*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2)
- (d*x + c)/b^3)/d