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

Optimal. Leaf size=93 \[ -\frac{a^2 b x}{\left (a^2+b^2\right )^2}+\frac{b x}{2 \left (a^2+b^2\right )}+\frac{a \sin ^2(x)}{2 \left (a^2+b^2\right )}+\frac{b \sin (x) \cos (x)}{2 \left (a^2+b^2\right )}-\frac{a b^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

[Out]

-((a^2*b*x)/(a^2 + b^2)^2) + (b*x)/(2*(a^2 + b^2)) - (a*b^2*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^2 + (b*Cos[x
]*Sin[x])/(2*(a^2 + b^2)) + (a*Sin[x]^2)/(2*(a^2 + b^2))

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Rubi [A]  time = 0.133434, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3109, 2635, 8, 2564, 30, 3098, 3133} \[ -\frac{a^2 b x}{\left (a^2+b^2\right )^2}+\frac{b x}{2 \left (a^2+b^2\right )}+\frac{a \sin ^2(x)}{2 \left (a^2+b^2\right )}+\frac{b \sin (x) \cos (x)}{2 \left (a^2+b^2\right )}-\frac{a b^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*b*x)/(a^2 + b^2)^2) + (b*x)/(2*(a^2 + b^2)) - (a*b^2*Log[a*Cos[x] + b*Sin[x]])/(a^2 + b^2)^2 + (b*Cos[x
]*Sin[x])/(2*(a^2 + b^2)) + (a*Sin[x]^2)/(2*(a^2 + b^2))

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(x) \sin (x)}{a \cos (x)+b \sin (x)} \, dx &=\frac{a \int \cos (x) \sin (x) \, dx}{a^2+b^2}+\frac{b \int \cos ^2(x) \, dx}{a^2+b^2}-\frac{(a b) \int \frac{\cos (x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}\\ &=-\frac{a^2 b x}{\left (a^2+b^2\right )^2}+\frac{b \cos (x) \sin (x)}{2 \left (a^2+b^2\right )}-\frac{\left (a b^2\right ) \int \frac{b \cos (x)-a \sin (x)}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac{a \operatorname{Subst}(\int x \, dx,x,\sin (x))}{a^2+b^2}+\frac{b \int 1 \, dx}{2 \left (a^2+b^2\right )}\\ &=-\frac{a^2 b x}{\left (a^2+b^2\right )^2}+\frac{b x}{2 \left (a^2+b^2\right )}-\frac{a b^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2}+\frac{b \cos (x) \sin (x)}{2 \left (a^2+b^2\right )}+\frac{a \sin ^2(x)}{2 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.303178, size = 82, normalized size = 0.88 \[ \frac{b \left (a^2+b^2\right ) \sin (2 x)-a \left (a^2+b^2\right ) \cos (2 x)+4 i a b^2 \tan ^{-1}(\tan (x))-2 b \left (a b \log \left ((a \cos (x)+b \sin (x))^2\right )+x (a+i b)^2\right )}{4 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*I)*a*b^2*ArcTan[Tan[x]] - a*(a^2 + b^2)*Cos[2*x] - 2*b*((a + I*b)^2*x + a*b*Log[(a*Cos[x] + b*Sin[x])^2])
+ b*(a^2 + b^2)*Sin[2*x])/(4*(a^2 + b^2)^2)

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Maple [A]  time = 0.069, size = 175, normalized size = 1.9 \begin{align*} -{\frac{a{b}^{2}\ln \left ( a+b\tan \left ( x \right ) \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\tan \left ( x \right ){a}^{2}b}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }}+{\frac{\tan \left ( x \right ){b}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }}-{\frac{{a}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }}-{\frac{a{b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }}+{\frac{\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) a{b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( x \right ) \right ){a}^{2}b}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( x \right ) \right ){b}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2*sin(x)/(a*cos(x)+b*sin(x)),x)

[Out]

-a*b^2/(a^2+b^2)^2*ln(a+b*tan(x))+1/2/(a^2+b^2)^2/(tan(x)^2+1)*tan(x)*a^2*b+1/2/(a^2+b^2)^2/(tan(x)^2+1)*tan(x
)*b^3-1/2/(a^2+b^2)^2/(tan(x)^2+1)*a^3-1/2/(a^2+b^2)^2/(tan(x)^2+1)*a*b^2+1/2/(a^2+b^2)^2*ln(tan(x)^2+1)*a*b^2
-1/2/(a^2+b^2)^2*arctan(tan(x))*a^2*b+1/2/(a^2+b^2)^2*arctan(tan(x))*b^3

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Maxima [B]  time = 1.6452, size = 286, normalized size = 3.08 \begin{align*} -\frac{a b^{2} \log \left (-a - \frac{2 \, b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{a b^{2} \log \left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a^{2} b - b^{3}\right )} \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{\frac{b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{2 \, a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac{b \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}}{a^{2} + b^{2} + \frac{2 \,{\left (a^{2} + b^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{{\left (a^{2} + b^{2}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b^2*log(-a - 2*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/(a^4 + 2*a^2*b^2 + b^4) + a*b^2*log(sin(x
)^2/(cos(x) + 1)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) - (a^2*b - b^3)*arctan(sin(x)/(cos(x) + 1))/(a^4 + 2*a^2*b^2 +
b^4) + (b*sin(x)/(cos(x) + 1) + 2*a*sin(x)^2/(cos(x) + 1)^2 - b*sin(x)^3/(cos(x) + 1)^3)/(a^2 + b^2 + 2*(a^2
+ b^2)*sin(x)^2/(cos(x) + 1)^2 + (a^2 + b^2)*sin(x)^4/(cos(x) + 1)^4)

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Fricas [A]  time = 0.51382, size = 223, normalized size = 2.4 \begin{align*} -\frac{a b^{2} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}\right ) +{\left (a^{3} + a b^{2}\right )} \cos \left (x\right )^{2} -{\left (a^{2} b + b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} b - b^{3}\right )} x}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*b^2*log(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2) + (a^3 + a*b^2)*cos(x)^2 - (a^2*b + b^3)*cos
(x)*sin(x) + (a^2*b - b^3)*x)/(a^4 + 2*a^2*b^2 + b^4)

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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(x)**2*sin(x)/(a*cos(x)+b*sin(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.1426, size = 211, normalized size = 2.27 \begin{align*} -\frac{a b^{3} \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac{a b^{2} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{{\left (a^{2} b - b^{3}\right )} x}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{a b^{2} \tan \left (x\right )^{2} - a^{2} b \tan \left (x\right ) - b^{3} \tan \left (x\right ) + a^{3} + 2 \, a b^{2}}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b^3*log(abs(b*tan(x) + a))/(a^4*b + 2*a^2*b^3 + b^5) + 1/2*a*b^2*log(tan(x)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4)
- 1/2*(a^2*b - b^3)*x/(a^4 + 2*a^2*b^2 + b^4) - 1/2*(a*b^2*tan(x)^2 - a^2*b*tan(x) - b^3*tan(x) + a^3 + 2*a*b^
2)/((a^4 + 2*a^2*b^2 + b^4)*(tan(x)^2 + 1))