3.1288 \(\int \frac{\cos ^2(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=100 \[ \frac{2 a \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{x \left (2 a^2-b^2\right )}{2 b^3}-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.165019, antiderivative size = 100, 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 = {2865, 2735, 2660, 618, 204} \[ \frac{2 a \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{x \left (2 a^2-b^2\right )}{2 b^3}-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2865

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 + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[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)} \, dx &=-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d}+\frac{\int \frac{-a b-\left (2 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^2}\\ &=-\frac{\left (2 a^2-b^2\right ) x}{2 b^3}-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d}+\frac{\left (a \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^3}\\ &=-\frac{\left (2 a^2-b^2\right ) x}{2 b^3}-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d}+\frac{\left (2 a \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^3 d}\\ &=-\frac{\left (2 a^2-b^2\right ) x}{2 b^3}-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d}-\frac{\left (4 a \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^3 d}\\ &=-\frac{\left (2 a^2-b^2\right ) x}{2 b^3}+\frac{2 a \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^3 d}-\frac{\cos (c+d x) (2 a-b \sin (c+d x))}{2 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.23189, size = 104, normalized size = 1.04 \[ \frac{8 a \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-4 a^2 c-4 a^2 d x-4 a b \cos (c+d x)+b^2 \sin (2 (c+d x))+2 b^2 c+2 b^2 d x}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.072, size = 214, normalized size = 2.1 \begin{align*} -{\frac{1}{bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{a}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{3}}}+{\frac{1}{bd}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{a\sqrt{{a}^{2}-{b}^{2}}}{d{b}^{3}}\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)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50194, size = 637, normalized size = 6.37 \begin{align*} \left [\frac{b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left (2 \, a^{2} - b^{2}\right )} d x - 2 \, a b \cos \left (d x + c\right ) + \sqrt{-a^{2} + b^{2}} a \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 \, b^{3} d}, \frac{b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left (2 \, a^{2} - b^{2}\right )} d x - 2 \, a b \cos \left (d x + c\right ) - 2 \, \sqrt{a^{2} - b^{2}} a \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{2 \, b^{3} 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)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.17071, size = 215, normalized size = 2.15 \begin{align*} -\frac{\frac{{\left (2 \, a^{2} - b^{2}\right )}{\left (d x + c\right )}}{b^{3}} - \frac{4 \,{\left (a^{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 )}}{\sqrt{a^{2} - b^{2}} b^{3}} + \frac{2 \,{\left (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} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, 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)),x, algorithm="giac")

[Out]

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