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

Optimal. Leaf size=76 \[ \frac{\left (b^2-2 c (a+c)\right ) \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 c^2}-\frac{\sin (x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.142085, antiderivative size = 76, 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 = {3258, 1657, 634, 618, 206, 628} \[ \frac{\left (b^2-2 c (a+c)\right ) \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 c^2}-\frac{\sin (x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3258

Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*sin[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*sin[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_.), x_Symbol] :> Module[{g = FreeFactors[Sin[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, Sin[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{\cos ^3(x)}{a+b \sin (x)+c \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1-x^2}{a+b x+c x^2} \, dx,x,\sin (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{1}{c}+\frac{a+c+b x}{c \left (a+b x+c x^2\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\frac{\sin (x)}{c}+\frac{\operatorname{Subst}\left (\int \frac{a+c+b x}{a+b x+c x^2} \, dx,x,\sin (x)\right )}{c}\\ &=-\frac{\sin (x)}{c}+\frac{b \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,\sin (x)\right )}{2 c^2}-\frac{\left (b^2-2 c (a+c)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,\sin (x)\right )}{2 c^2}\\ &=\frac{b \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 c^2}-\frac{\sin (x)}{c}+\frac{\left (b^2-2 c (a+c)\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c \sin (x)\right )}{c^2}\\ &=\frac{\left (b^2-2 c (a+c)\right ) \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b \sin (x)+c \sin ^2(x)\right )}{2 c^2}-\frac{\sin (x)}{c}\\ \end{align*}

Mathematica [A]  time = 0.114902, size = 73, normalized size = 0.96 \[ \frac{\frac{2 \left (b^2-2 c (a+c)\right ) \tanh ^{-1}\left (\frac{b+2 c \sin (x)}{\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c}}+b \log \left (a+b \sin (x)+c \sin ^2(x)\right )-2 c \sin (x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.179, size = 143, normalized size = 1.9 \begin{align*} -{\frac{\sin \left ( x \right ) }{c}}+{\frac{b\ln \left ( a+b\sin \left ( x \right ) +c \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }{2\,{c}^{2}}}+2\,{\frac{a}{c\sqrt{4\,ca-{b}^{2}}}\arctan \left ({\frac{b+2\,c\sin \left ( x \right ) }{\sqrt{4\,ca-{b}^{2}}}} \right ) }+2\,{\frac{1}{\sqrt{4\,ca-{b}^{2}}}\arctan \left ({\frac{b+2\,c\sin \left ( x \right ) }{\sqrt{4\,ca-{b}^{2}}}} \right ) }-{\frac{{b}^{2}}{{c}^{2}}\arctan \left ({(b+2\,c\sin \left ( x \right ) ){\frac{1}{\sqrt{4\,ca-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ca-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a+b*sin(x)+c*sin(x)^2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.91011, size = 670, normalized size = 8.82 \begin{align*} \left [-\frac{{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \sqrt{b^{2} - 4 \, a c} \log \left (-\frac{2 \, c^{2} \cos \left (x\right )^{2} - 2 \, b c \sin \left (x\right ) - b^{2} + 2 \, a c - 2 \, c^{2} + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c \sin \left (x\right ) + b\right )}}{c \cos \left (x\right )^{2} - b \sin \left (x\right ) - a - c}\right ) -{\left (b^{3} - 4 \, a b c\right )} \log \left (-c \cos \left (x\right )^{2} + b \sin \left (x\right ) + a + c\right ) + 2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} \sin \left (x\right )}{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac{2 \,{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c \sin \left (x\right ) + b\right )}}{b^{2} - 4 \, a c}\right ) +{\left (b^{3} - 4 \, a b c\right )} \log \left (-c \cos \left (x\right )^{2} + b \sin \left (x\right ) + a + c\right ) - 2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} \sin \left (x\right )}{2 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="fricas")

[Out]

[-1/2*((b^2 - 2*a*c - 2*c^2)*sqrt(b^2 - 4*a*c)*log(-(2*c^2*cos(x)^2 - 2*b*c*sin(x) - b^2 + 2*a*c - 2*c^2 + sqr
t(b^2 - 4*a*c)*(2*c*sin(x) + b))/(c*cos(x)^2 - b*sin(x) - a - c)) - (b^3 - 4*a*b*c)*log(-c*cos(x)^2 + b*sin(x)
+ a + c) + 2*(b^2*c - 4*a*c^2)*sin(x))/(b^2*c^2 - 4*a*c^3), 1/2*(2*(b^2 - 2*a*c - 2*c^2)*sqrt(-b^2 + 4*a*c)*a
rctan(-sqrt(-b^2 + 4*a*c)*(2*c*sin(x) + b)/(b^2 - 4*a*c)) + (b^3 - 4*a*b*c)*log(-c*cos(x)^2 + b*sin(x) + a + c
) - 2*(b^2*c - 4*a*c^2)*sin(x))/(b^2*c^2 - 4*a*c^3)]

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

[Out]

Timed out

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Giac [A]  time = 1.26818, size = 105, normalized size = 1.38 \begin{align*} \frac{b \log \left (c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a\right )}{2 \, c^{2}} - \frac{\sin \left (x\right )}{c} - \frac{{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \arctan \left (\frac{2 \, c \sin \left (x\right ) + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*sin(x)+c*sin(x)^2),x, algorithm="giac")

[Out]

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