3.1.4 \(\int \frac {\csc (x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) [4]

Optimal. Leaf size=129 \[ -\frac {\left (b^2-2 a c-2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c) (a+b+c) \sqrt {b^2-4 a c}}+\frac {\log (1-\cos (x))}{2 (a+b+c)}-\frac {\log (1+\cos (x))}{2 (a-b+c)}+\frac {b \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 (a-b+c) (a+b+c)} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3340, 995, 648, 632, 212, 642, 647, 31} \begin {gather*} -\frac {\left (-2 a c+b^2-2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c \cos (x)}{\sqrt {b^2-4 a c}}\right )}{(a-b+c) (a+b+c) \sqrt {b^2-4 a c}}+\frac {b \log \left (a+b \cos (x)+c \cos ^2(x)\right )}{2 (a-b+c) (a+b+c)}+\frac {\log (1-\cos (x))}{2 (a+b+c)}-\frac {\log (\cos (x)+1)}{2 (a-b+c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b^2 - 2*a*c - 2*c^2)*ArcTanh[(b + 2*c*Cos[x])/Sqrt[b^2 - 4*a*c]])/((a - b + c)*(a + b + c)*Sqrt[b^2 - 4*a*
c])) + Log[1 - Cos[x]]/(2*(a + b + c)) - Log[1 + Cos[x]]/(2*(a - b + c)) + (b*Log[a + b*Cos[x] + c*Cos[x]^2])/
(2*(a - b + c)*(a + b + c))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 212

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

Rule 632

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 642

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]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 648

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 995

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

Rule 3340

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]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 126, normalized size = 0.98 \begin {gather*} -\frac {\left (-2 b^2+4 c (a+c)\right ) \text {ArcTan}\left (\frac {b+2 c \cos (x)}{\sqrt {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} \left (-((a-b+c) \log (1-\cos (x)))+(a+b+c) \log (1+\cos (x))-b \log \left (a+b \cos (x)+c \cos ^2(x)\right )\right )}{2 (a-b+c) (a+b+c) \sqrt {-b^2+4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-2*b^2 + 4*c*(a + c))*ArcTan[(b + 2*c*Cos[x])/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*(-((a - b + c)*L
og[1 - Cos[x]]) + (a + b + c)*Log[1 + Cos[x]] - b*Log[a + b*Cos[x] + c*Cos[x]^2]))/((a - b + c)*(a + b + c)*Sq
rt[-b^2 + 4*a*c])

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Maple [A]
time = 0.46, size = 121, normalized size = 0.94

method	result	size
default	\(\frac {\ln \left (-1+\cos \left (x \right )\right )}{2 a +2 b +2 c}-\frac {\ln \left (\cos \left (x \right )+1\right )}{2 a -2 b +2 c}+\frac {\frac {b \ln \left (a +b \cos \left (x \right )+c \left (\cos ^{2}\left (x \right )\right )\right )}{2}+\frac {2 \left (-a c +\frac {1}{2} b^{2}-c^{2}\right ) \arctan \left (\frac {b +2 c \cos \left (x \right )}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a -b +c \right ) \left (a +b +c \right )}\)	\(121\)
risch	\(-\frac {i x}{a +b +c}+\frac {i x}{a -b +c}-\frac {2 i x a b c}{a^{3} c -\frac {1}{4} a^{2} b^{2}+2 a^{2} c^{2}-\frac {3}{2} a \,b^{2} c +a \,c^{3}+\frac {1}{4} b^{4}-\frac {1}{4} b^{2} c^{2}}+\frac {i x \,b^{3}}{2 a^{3} c -\frac {1}{2} a^{2} b^{2}+4 a^{2} c^{2}-3 a \,b^{2} c +2 a \,c^{3}+\frac {1}{2} b^{4}-\frac {1}{2} b^{2} c^{2}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{a +b +c}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{a -b +c}+\frac {2 \ln \left ({\mathrm e}^{2 i x}+\frac {\left (2 c a b -b^{3}+2 c^{2} b -i \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}\right ) {\mathrm e}^{i x}}{c \left (2 a c -b^{2}+2 c^{2}\right )}+1\right ) a b c}{4 a^{3} c -a^{2} b^{2}+8 a^{2} c^{2}-6 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {\left (2 c a b -b^{3}+2 c^{2} b -i \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}\right ) {\mathrm e}^{i x}}{c \left (2 a c -b^{2}+2 c^{2}\right )}+1\right ) b^{3}}{2 \left (4 a^{3} c -a^{2} b^{2}+8 a^{2} c^{2}-6 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}\right )}+\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {\left (2 c a b -b^{3}+2 c^{2} b -i \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}\right ) {\mathrm e}^{i x}}{c \left (2 a c -b^{2}+2 c^{2}\right )}+1\right ) \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}}{8 a^{3} c -2 a^{2} b^{2}+16 a^{2} c^{2}-12 a \,b^{2} c +8 a \,c^{3}+2 b^{4}-2 b^{2} c^{2}}+\frac {2 \ln \left ({\mathrm e}^{2 i x}+\frac {\left (2 c a b -b^{3}+2 c^{2} b +i \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}\right ) {\mathrm e}^{i x}}{c \left (2 a c -b^{2}+2 c^{2}\right )}+1\right ) a b c}{4 a^{3} c -a^{2} b^{2}+8 a^{2} c^{2}-6 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {\left (2 c a b -b^{3}+2 c^{2} b +i \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}\right ) {\mathrm e}^{i x}}{c \left (2 a c -b^{2}+2 c^{2}\right )}+1\right ) b^{3}}{2 \left (4 a^{3} c -a^{2} b^{2}+8 a^{2} c^{2}-6 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}\right )}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {\left (2 c a b -b^{3}+2 c^{2} b +i \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}\right ) {\mathrm e}^{i x}}{c \left (2 a c -b^{2}+2 c^{2}\right )}+1\right ) \sqrt {16 a^{3} c^{3}-20 b^{2} a^{2} c^{2}+32 a^{2} c^{4}+8 a \,b^{4} c -24 a \,b^{2} c^{3}+16 a \,c^{5}-b^{6}+4 b^{4} c^{2}-4 b^{2} c^{4}}}{2 \left (4 a^{3} c -a^{2} b^{2}+8 a^{2} c^{2}-6 a \,b^{2} c +4 a \,c^{3}+b^{4}-b^{2} c^{2}\right )}\)	\(1397\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(a+b*cos(x)+c*cos(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

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Fricas [A]
time = 0.82, size = 470, normalized size = 3.64 \begin {gather*} \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 \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 ) - {\left (b^{3} - 4 \, a b c\right )} \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right ) + {\left (a b^{2} + b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} + 4 \, a b - b^{2}\right )} c\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (a b^{2} - b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} - 4 \, a b - b^{2}\right )} c\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b^{2} - b^{4} - 4 \, a c^{3} - {\left (8 \, a^{2} - b^{2}\right )} c^{2} - 2 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} c\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 \cos \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 \cos \left (x\right ) + a\right ) + {\left (a b^{2} + b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} + 4 \, a b - b^{2}\right )} c\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (a b^{2} - b^{3} - 4 \, a c^{2} - {\left (4 \, a^{2} - 4 \, a b - b^{2}\right )} c\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{2} b^{2} - b^{4} - 4 \, a c^{3} - {\left (8 \, a^{2} - b^{2}\right )} c^{2} - 2 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} c\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cos(x)+c*cos(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*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)) - (b^3 - 4*a*b*c)*log(c*cos(x)^2 + b*cos(x) + a) + (a*b^
2 + b^3 - 4*a*c^2 - (4*a^2 + 4*a*b - b^2)*c)*log(1/2*cos(x) + 1/2) - (a*b^2 - b^3 - 4*a*c^2 - (4*a^2 - 4*a*b -
b^2)*c)*log(-1/2*cos(x) + 1/2))/(a^2*b^2 - b^4 - 4*a*c^3 - (8*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c), -1/2*(
2*(b^2 - 2*a*c - 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)) - (b^3 -
4*a*b*c)*log(c*cos(x)^2 + b*cos(x) + a) + (a*b^2 + b^3 - 4*a*c^2 - (4*a^2 + 4*a*b - b^2)*c)*log(1/2*cos(x) +
1/2) - (a*b^2 - b^3 - 4*a*c^2 - (4*a^2 - 4*a*b - b^2)*c)*log(-1/2*cos(x) + 1/2))/(a^2*b^2 - b^4 - 4*a*c^3 - (8
*a^2 - b^2)*c^2 - 2*(2*a^3 - 3*a*b^2)*c)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc {\left (x \right )}}{a + b \cos {\left (x \right )} + c \cos ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cos(x)+c*cos(x)**2),x)

[Out]

Integral(csc(x)/(a + b*cos(x) + c*cos(x)**2), x)

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Giac [A]
time = 0.50, size = 130, normalized size = 1.01 \begin {gather*} \frac {b \log \left (c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a\right )}{2 \, {\left (a^{2} - b^{2} + 2 \, a c + c^{2}\right )}} + \frac {{\left (b^{2} - 2 \, a c - 2 \, c^{2}\right )} \arctan \left (\frac {2 \, c \cos \left (x\right ) + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} - b^{2} + 2 \, a c + c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {\log \left (\cos \left (x\right ) + 1\right )}{2 \, {\left (a - b + c\right )}} + \frac {\log \left (-\cos \left (x\right ) + 1\right )}{2 \, {\left (a + b + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.97, size = 1003, normalized size = 7.78 \begin {gather*} \frac {\ln \left (\cos \left (x\right )-1\right )}{2\,\left (a+b+c\right )}-\frac {\ln \left (\cos \left (x\right )+1\right )}{2\,\left (a-b+c\right )}-\frac {\ln \left (b\,c^2+4\,c^3\,\cos \left (x\right )+\frac {\left (a\,\left (4\,b\,c-2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3+b^2\,\sqrt {b^2-4\,a\,c}-2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (8\,a\,c^3+\cos \left (x\right )\,\left (-3\,b^3\,c+12\,b\,c^3+12\,a\,b\,c^2\right )+4\,c^4+4\,a^2\,c^2+3\,b^2\,c^2-\frac {\left (a\,\left (4\,b\,c-2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3+b^2\,\sqrt {b^2-4\,a\,c}-2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (4\,b\,c^4+4\,b^3\,c^2+\cos \left (x\right )\,\left (-8\,a^3\,c^2+2\,a^2\,b^2\,c-8\,a^2\,c^3-20\,a\,b^2\,c^2+8\,a\,c^4+6\,b^4\,c-6\,b^2\,c^3+8\,c^5\right )-28\,a^2\,b\,c^2-24\,a\,b\,c^3+8\,a\,b^3\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}-a\,b^2\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}\right )\,\left (a\,\left (4\,b\,c-2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3+b^2\,\sqrt {b^2-4\,a\,c}-2\,c^2\,\sqrt {b^2-4\,a\,c}\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}-\frac {\ln \left (b\,c^2+4\,c^3\,\cos \left (x\right )+\frac {\left (a\,\left (4\,b\,c+2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3-b^2\,\sqrt {b^2-4\,a\,c}+2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (8\,a\,c^3+\cos \left (x\right )\,\left (-3\,b^3\,c+12\,b\,c^3+12\,a\,b\,c^2\right )+4\,c^4+4\,a^2\,c^2+3\,b^2\,c^2-\frac {\left (a\,\left (4\,b\,c+2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3-b^2\,\sqrt {b^2-4\,a\,c}+2\,c^2\,\sqrt {b^2-4\,a\,c}\right )\,\left (4\,b\,c^4+4\,b^3\,c^2+\cos \left (x\right )\,\left (-8\,a^3\,c^2+2\,a^2\,b^2\,c-8\,a^2\,c^3-20\,a\,b^2\,c^2+8\,a\,c^4+6\,b^4\,c-6\,b^2\,c^3+8\,c^5\right )-28\,a^2\,b\,c^2-24\,a\,b\,c^3+8\,a\,b^3\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}-a\,b^2\,c\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )}\right )\,\left (a\,\left (4\,b\,c+2\,c\,\sqrt {b^2-4\,a\,c}\right )-b^3-b^2\,\sqrt {b^2-4\,a\,c}+2\,c^2\,\sqrt {b^2-4\,a\,c}\right )}{b^2\,\left (2\,a^2+12\,a\,c-2\,b^2+2\,c^2\right )-4\,a\,c\,\left (2\,a^2+4\,a\,c+2\,c^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a + b*cos(x) + c*cos(x)^2)),x)

[Out]

log(cos(x) - 1)/(2*(a + b + c)) - log(cos(x) + 1)/(2*(a - b + c)) - (log(b*c^2 + 4*c^3*cos(x) + ((a*(4*b*c - 2
*c*(b^2 - 4*a*c)^(1/2)) - b^3 + b^2*(b^2 - 4*a*c)^(1/2) - 2*c^2*(b^2 - 4*a*c)^(1/2))*(8*a*c^3 + cos(x)*(12*b*c
^3 - 3*b^3*c + 12*a*b*c^2) + 4*c^4 + 4*a^2*c^2 + 3*b^2*c^2 - ((a*(4*b*c - 2*c*(b^2 - 4*a*c)^(1/2)) - b^3 + b^2
*(b^2 - 4*a*c)^(1/2) - 2*c^2*(b^2 - 4*a*c)^(1/2))*(4*b*c^4 + 4*b^3*c^2 + cos(x)*(8*a*c^4 + 6*b^4*c + 8*c^5 - 8
*a^2*c^3 - 8*a^3*c^2 - 6*b^2*c^3 - 20*a*b^2*c^2 + 2*a^2*b^2*c) - 28*a^2*b*c^2 - 24*a*b*c^3 + 8*a*b^3*c))/(b^2*
(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)) - a*b^2*c))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2
*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)))*(a*(4*b*c - 2*c*(b^2 - 4*a*c)^(1/2)) - b^3 + b^2*(b^2 - 4*a*c)^(1/2) -
2*c^2*(b^2 - 4*a*c)^(1/2)))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)) - (log(b*c
^2 + 4*c^3*cos(x) + ((a*(4*b*c + 2*c*(b^2 - 4*a*c)^(1/2)) - b^3 - b^2*(b^2 - 4*a*c)^(1/2) + 2*c^2*(b^2 - 4*a*c
)^(1/2))*(8*a*c^3 + cos(x)*(12*b*c^3 - 3*b^3*c + 12*a*b*c^2) + 4*c^4 + 4*a^2*c^2 + 3*b^2*c^2 - ((a*(4*b*c + 2*
c*(b^2 - 4*a*c)^(1/2)) - b^3 - b^2*(b^2 - 4*a*c)^(1/2) + 2*c^2*(b^2 - 4*a*c)^(1/2))*(4*b*c^4 + 4*b^3*c^2 + cos
(x)*(8*a*c^4 + 6*b^4*c + 8*c^5 - 8*a^2*c^3 - 8*a^3*c^2 - 6*b^2*c^3 - 20*a*b^2*c^2 + 2*a^2*b^2*c) - 28*a^2*b*c^
2 - 24*a*b*c^3 + 8*a*b^3*c))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)) - a*b^2*c)
)/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(4*a*c + 2*a^2 + 2*c^2)))*(a*(4*b*c + 2*c*(b^2 - 4*a*c)^(1/2))
- b^3 - b^2*(b^2 - 4*a*c)^(1/2) + 2*c^2*(b^2 - 4*a*c)^(1/2)))/(b^2*(12*a*c + 2*a^2 - 2*b^2 + 2*c^2) - 4*a*c*(
4*a*c + 2*a^2 + 2*c^2))

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