3.1 \(\int \frac{\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\)

Optimal. Leaf size=136 \[ -\frac{\cos (x) \left (b^2-c (a+2 c)\right )}{c^3}+\frac{b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^4}+\frac{\left (-2 b^2 c (2 a+c)+2 c^2 (a+c)^2+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{b^2-4 a c}}\right )}{c^4 \sqrt{b^2-4 a c}}+\frac{b \cos ^2(x)}{2 c^2}-\frac{\cos ^3(x)}{3 c} \]

[Out]

((b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c))*ArcTanh[(b + 2*c*Cos[x])/Sqrt[b^2 - 4*a*c]])/(c^4*Sqrt[b^2 - 4*a*
c]) - ((b^2 - c*(a + 2*c))*Cos[x])/c^3 + (b*Cos[x]^2)/(2*c^2) - Cos[x]^3/(3*c) + (b*(b^2 - 2*c*(a + c))*Log[a
+ b*Cos[x] + c*Cos[x]^2])/(2*c^4)

________________________________________________________________________________________

Rubi [A]  time = 0.23179, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3259, 1657, 634, 618, 206, 628} \[ -\frac{\cos (x) \left (b^2-c (a+2 c)\right )}{c^3}+\frac{b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^4}+\frac{\left (-2 b^2 c (2 a+c)+2 c^2 (a+c)^2+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{b^2-4 a c}}\right )}{c^4 \sqrt{b^2-4 a c}}+\frac{b \cos ^2(x)}{2 c^2}-\frac{\cos ^3(x)}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^5/(a + b*Cos[x] + c*Cos[x]^2),x]

[Out]

((b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c))*ArcTanh[(b + 2*c*Cos[x])/Sqrt[b^2 - 4*a*c]])/(c^4*Sqrt[b^2 - 4*a*
c]) - ((b^2 - c*(a + 2*c))*Cos[x])/c^3 + (b*Cos[x]^2)/(2*c^2) - Cos[x]^3/(3*c) + (b*(b^2 - 2*c*(a + c))*Log[a
+ b*Cos[x] + c*Cos[x]^2])/(2*c^4)

Rule 3259

Int[((a_.) + (b_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n_.) + (c_.)*(cos[(d_.) + (e_.)*(x_)]*(f_.))^(n2_.))^(p_.)
*sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Module[{g = FreeFactors[Cos[d + e*x], x]}, -Dist[g/e, Subst[Int[(
1 - g^2*x^2)^((m - 1)/2)*(a + b*(f*g*x)^n + c*(f*g*x)^(2*n))^p, x], x, Cos[d + e*x]/g], x]] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] && EqQ[n2, 2*n] && IntegerQ[(m - 1)/2]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^5(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{a+b x+c x^2} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{b^2-c (a+2 c)}{c^3}-\frac{b x}{c^2}+\frac{x^2}{c}-\frac{-a^2 c-c^3+a \left (b^2-2 c^2\right )+b \left (b^2-2 c (a+c)\right ) x}{c^3 \left (a+b x+c x^2\right )}\right ) \, dx,x,\cos (x)\right )\\ &=-\frac{\left (b^2-c (a+2 c)\right ) \cos (x)}{c^3}+\frac{b \cos ^2(x)}{2 c^2}-\frac{\cos ^3(x)}{3 c}+\frac{\operatorname{Subst}\left (\int \frac{-a^2 c-c^3+a \left (b^2-2 c^2\right )+b \left (b^2-2 c (a+c)\right ) x}{a+b x+c x^2} \, dx,x,\cos (x)\right )}{c^3}\\ &=-\frac{\left (b^2-c (a+2 c)\right ) \cos (x)}{c^3}+\frac{b \cos ^2(x)}{2 c^2}-\frac{\cos ^3(x)}{3 c}+\frac{\left (b \left (b^2-2 c (a+c)\right )\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,\cos (x)\right )}{2 c^4}-\frac{\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,\cos (x)\right )}{2 c^4}\\ &=-\frac{\left (b^2-c (a+2 c)\right ) \cos (x)}{c^3}+\frac{b \cos ^2(x)}{2 c^2}-\frac{\cos ^3(x)}{3 c}+\frac{b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^4}+\frac{\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c \cos (x)\right )}{c^4}\\ &=\frac{\left (b^4+2 c^2 (a+c)^2-2 b^2 c (2 a+c)\right ) \tanh ^{-1}\left (\frac{b+2 c \cos (x)}{\sqrt{b^2-4 a c}}\right )}{c^4 \sqrt{b^2-4 a c}}-\frac{\left (b^2-c (a+2 c)\right ) \cos (x)}{c^3}+\frac{b \cos ^2(x)}{2 c^2}-\frac{\cos ^3(x)}{3 c}+\frac{b \left (b^2-2 c (a+c)\right ) \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.534214, size = 239, normalized size = 1.76 \[ \frac{\frac{6 \left (b^3 \sqrt{b^2-4 a c}+2 b^2 c (2 a+c)-2 b c (a+c) \sqrt{b^2-4 a c}-2 c^2 (a+c)^2-b^4\right ) \log \left (\sqrt{b^2-4 a c}-b-2 c \cos (x)\right )}{\sqrt{b^2-4 a c}}+\frac{6 \left (b^3 \sqrt{b^2-4 a c}-2 b^2 c (2 a+c)-2 b c (a+c) \sqrt{b^2-4 a c}+2 c^2 (a+c)^2+b^4\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c \cos (x)\right )}{\sqrt{b^2-4 a c}}+3 c \cos (x) \left (c (4 a+7 c)-4 b^2\right )+3 b c^2 \cos (2 x)+c^3 (-\cos (3 x))}{12 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^5/(a + b*Cos[x] + c*Cos[x]^2),x]

[Out]

(3*c*(-4*b^2 + c*(4*a + 7*c))*Cos[x] + 3*b*c^2*Cos[2*x] - c^3*Cos[3*x] + (6*(-b^4 - 2*c^2*(a + c)^2 + 2*b^2*c*
(2*a + c) + b^3*Sqrt[b^2 - 4*a*c] - 2*b*c*(a + c)*Sqrt[b^2 - 4*a*c])*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*Cos[x]])
/Sqrt[b^2 - 4*a*c] + (6*(b^4 + 2*c^2*(a + c)^2 - 2*b^2*c*(2*a + c) + b^3*Sqrt[b^2 - 4*a*c] - 2*b*c*(a + c)*Sqr
t[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*Cos[x]])/Sqrt[b^2 - 4*a*c])/(12*c^4)

________________________________________________________________________________________

Maple [B]  time = 0.027, size = 344, normalized size = 2.5 \begin{align*} -{\frac{ \left ( \cos \left ( x \right ) \right ) ^{3}}{3\,c}}+{\frac{b \left ( \cos \left ( x \right ) \right ) ^{2}}{2\,{c}^{2}}}+{\frac{\cos \left ( x \right ) a}{{c}^{2}}}-{\frac{\cos \left ( x \right ){b}^{2}}{{c}^{3}}}+2\,{\frac{\cos \left ( x \right ) }{c}}-{\frac{\ln \left ( a+b\cos \left ( x \right ) +c \left ( \cos \left ( x \right ) \right ) ^{2} \right ) ab}{{c}^{3}}}+{\frac{\ln \left ( a+b\cos \left ( x \right ) +c \left ( \cos \left ( x \right ) \right ) ^{2} \right ){b}^{3}}{2\,{c}^{4}}}-{\frac{b\ln \left ( a+b\cos \left ( x \right ) +c \left ( \cos \left ( x \right ) \right ) ^{2} \right ) }{{c}^{2}}}-2\,{\frac{{a}^{2}}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{b+2\,c\cos \left ( x \right ) }{\sqrt{4\,ac-{b}^{2}}}} \right ) }+4\,{\frac{a{b}^{2}}{{c}^{3}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{b+2\,c\cos \left ( x \right ) }{\sqrt{4\,ac-{b}^{2}}}} \right ) }-4\,{\frac{a}{c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{b+2\,c\cos \left ( x \right ) }{\sqrt{4\,ac-{b}^{2}}}} \right ) }-2\,{\frac{1}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{b+2\,c\cos \left ( x \right ) }{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{4}}{{c}^{4}}\arctan \left ({(b+2\,c\cos \left ( x \right ) ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+2\,{\frac{{b}^{2}}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{b+2\,c\cos \left ( x \right ) }{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x)

[Out]

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

________________________________________________________________________________________

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(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.59324, size = 1125, normalized size = 8.27 \begin{align*} \left [-\frac{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \cos \left (x\right )^{3} - 3 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} \cos \left (x\right )^{2} - 3 \,{\left (b^{4} - 4 \, a b^{2} c + 4 \, a c^{3} + 2 \, c^{4} + 2 \,{\left (a^{2} - b^{2}\right )} c^{2}\right )} \sqrt{b^{2} - 4 \, a c} \log \left (-\frac{2 \, c^{2} \cos \left (x\right )^{2} + 2 \, b c \cos \left (x\right ) + b^{2} - 2 \, a c + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c \cos \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a}\right ) + 6 \,{\left (b^{4} c - 5 \, a b^{2} c^{2} + 8 \, a c^{4} + 2 \,{\left (2 \, a^{2} - b^{2}\right )} c^{3}\right )} \cos \left (x\right ) - 3 \,{\left (b^{5} - 6 \, a b^{3} c + 8 \, a b c^{3} + 2 \,{\left (4 \, a^{2} b - b^{3}\right )} c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{6 \,{\left (b^{2} c^{4} - 4 \, a c^{5}\right )}}, -\frac{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \cos \left (x\right )^{3} - 3 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} \cos \left (x\right )^{2} - 6 \,{\left (b^{4} - 4 \, a b^{2} c + 4 \, a c^{3} + 2 \, c^{4} + 2 \,{\left (a^{2} - b^{2}\right )} c^{2}\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c \cos \left (x\right ) + b\right )}}{b^{2} - 4 \, a c}\right ) + 6 \,{\left (b^{4} c - 5 \, a b^{2} c^{2} + 8 \, a c^{4} + 2 \,{\left (2 \, a^{2} - b^{2}\right )} c^{3}\right )} \cos \left (x\right ) - 3 \,{\left (b^{5} - 6 \, a b^{3} c + 8 \, a b c^{3} + 2 \,{\left (4 \, a^{2} b - b^{3}\right )} c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{6 \,{\left (b^{2} c^{4} - 4 \, a c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x, algorithm="fricas")

[Out]

[-1/6*(2*(b^2*c^3 - 4*a*c^4)*cos(x)^3 - 3*(b^3*c^2 - 4*a*b*c^3)*cos(x)^2 - 3*(b^4 - 4*a*b^2*c + 4*a*c^3 + 2*c^
4 + 2*(a^2 - b^2)*c^2)*sqrt(b^2 - 4*a*c)*log(-(2*c^2*cos(x)^2 + 2*b*c*cos(x) + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)
*(2*c*cos(x) + b))/(c*cos(x)^2 + b*cos(x) + a)) + 6*(b^4*c - 5*a*b^2*c^2 + 8*a*c^4 + 2*(2*a^2 - b^2)*c^3)*cos(
x) - 3*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3)*c^2)*log(c*cos(x)^2 + b*cos(x) + a))/(b^2*c^4 - 4*a*c^
5), -1/6*(2*(b^2*c^3 - 4*a*c^4)*cos(x)^3 - 3*(b^3*c^2 - 4*a*b*c^3)*cos(x)^2 - 6*(b^4 - 4*a*b^2*c + 4*a*c^3 + 2
*c^4 + 2*(a^2 - b^2)*c^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*cos(x) + b)/(b^2 - 4*a*c)) + 6*(b
^4*c - 5*a*b^2*c^2 + 8*a*c^4 + 2*(2*a^2 - b^2)*c^3)*cos(x) - 3*(b^5 - 6*a*b^3*c + 8*a*b*c^3 + 2*(4*a^2*b - b^3
)*c^2)*log(c*cos(x)^2 + b*cos(x) + a))/(b^2*c^4 - 4*a*c^5)]

________________________________________________________________________________________

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(sin(x)**5/(a+b*cos(x)+c*cos(x)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.14366, size = 207, normalized size = 1.52 \begin{align*} -\frac{2 \, c^{2} \cos \left (x\right )^{3} - 3 \, b c \cos \left (x\right )^{2} + 6 \, b^{2} \cos \left (x\right ) - 6 \, a c \cos \left (x\right ) - 12 \, c^{2} \cos \left (x\right )}{6 \, c^{3}} + \frac{{\left (b^{3} - 2 \, a b c - 2 \, b c^{2}\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, c^{4}} - \frac{{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2} - 2 \, b^{2} c^{2} + 4 \, a c^{3} + 2 \, c^{4}\right )} \arctan \left (\frac{2 \, c \cos \left (x\right ) + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^5/(a+b*cos(x)+c*cos(x)^2),x, algorithm="giac")

[Out]

-1/6*(2*c^2*cos(x)^3 - 3*b*c*cos(x)^2 + 6*b^2*cos(x) - 6*a*c*cos(x) - 12*c^2*cos(x))/c^3 + 1/2*(b^3 - 2*a*b*c
- 2*b*c^2)*log(c*cos(x)^2 + b*cos(x) + a)/c^4 - (b^4 - 4*a*b^2*c + 2*a^2*c^2 - 2*b^2*c^2 + 4*a*c^3 + 2*c^4)*ar
ctan((2*c*cos(x) + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^4)