3.5.13 \(\int \frac {x^2}{c+\frac {a}{x^2}+\frac {b}{x}} \, dx\) [413]

Optimal. Leaf size=118 \[ \frac {\left (b^2-a c\right ) x}{c^3}-\frac {b x^2}{2 c^2}+\frac {x^3}{3 c}-\frac {\left (b^4-4 a b^2 c+2 a^2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}-\frac {b \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 c^4} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1368, 715, 648, 632, 212, 642} \begin {gather*} -\frac {\left (2 a^2 c^2-4 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}-\frac {b \left (b^2-2 a c\right ) \log \left (a+b x+c x^2\right )}{2 c^4}+\frac {x \left (b^2-a c\right )}{c^3}-\frac {b x^2}{2 c^2}+\frac {x^3}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(c + a/x^2 + b/x),x]

[Out]

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

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 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 715

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

Rule 1368

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 112, normalized size = 0.95 \begin {gather*} \frac {c x \left (6 b^2-6 a c-3 b c x+2 c^2 x^2\right )+\frac {6 \left (b^4-4 a b^2 c+2 a^2 c^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-3 \left (b^3-2 a b c\right ) \log (a+x (b+c x))}{6 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(c + a/x^2 + b/x),x]

[Out]

(c*x*(6*b^2 - 6*a*c - 3*b*c*x + 2*c^2*x^2) + (6*(b^4 - 4*a*b^2*c + 2*a^2*c^2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4
*a*c]])/Sqrt[-b^2 + 4*a*c] - 3*(b^3 - 2*a*b*c)*Log[a + x*(b + c*x)])/(6*c^4)

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Maple [A]
time = 0.05, size = 128, normalized size = 1.08

method	result	size
default	\(-\frac {-\frac {1}{3} c^{2} x^{3}+\frac {1}{2} b c \,x^{2}+a c x -b^{2} x}{c^{3}}+\frac {\frac {\left (2 a b c -b^{3}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (a^{2} c -a \,b^{2}-\frac {\left (2 a b c -b^{3}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{3}}\)	\(128\)
risch	\(\frac {x^{3}}{3 c}-\frac {b \,x^{2}}{2 c^{2}}-\frac {a x}{c^{2}}+\frac {b^{2} x}{c^{3}}+\frac {4 \ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) a^{2} b}{c^{2} \left (4 a c -b^{2}\right )}-\frac {3 \ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) a \,b^{3}}{c^{3} \left (4 a c -b^{2}\right )}+\frac {\ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) b^{5}}{2 c^{4} \left (4 a c -b^{2}\right )}+\frac {\ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}-2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x -\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}}{2 c^{4} \left (4 a c -b^{2}\right )}+\frac {4 \ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) a^{2} b}{c^{2} \left (4 a c -b^{2}\right )}-\frac {3 \ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) a \,b^{3}}{c^{3} \left (4 a c -b^{2}\right )}+\frac {\ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) b^{5}}{2 c^{4} \left (4 a c -b^{2}\right )}-\frac {\ln \left (8 a^{3} c^{3}-18 a^{2} b^{2} c^{2}+8 a \,b^{4} c -b^{6}+2 \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, c x +\sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}\, b \right ) \sqrt {-\left (4 a c -b^{2}\right ) \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )^{2}}}{2 c^{4} \left (4 a c -b^{2}\right )}\)	\(1138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c^3*(-1/3*c^2*x^3+1/2*b*c*x^2+a*c*x-b^2*x)+1/c^3*(1/2*(2*a*b*c-b^3)/c*ln(c*x^2+b*x+a)+2*(a^2*c-a*b^2-1/2*(2
*a*b*c-b^3)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(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(x^2/(c+a/x^2+b/x),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 = 0.37, size = 383, normalized size = 3.25 \begin {gather*} \left [\frac {2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} - 3 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + 3 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 6 \, {\left (b^{4} c - 5 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} x - 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} \log \left (c x^{2} + b x + 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 )} x^{3} - 3 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} - 6 \, {\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 6 \, {\left (b^{4} c - 5 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} x - 3 \, {\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{6 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (110) = 220\).
time = 0.65, size = 498, normalized size = 4.22 \begin {gather*} - \frac {b x^{2}}{2 c^{2}} + x \left (- \frac {a}{c^{2}} + \frac {b^{2}}{c^{3}}\right ) + \left (\frac {b \left (2 a c - b^{2}\right )}{2 c^{4}} - \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a^{2} c^{2} - 4 a b^{2} c + b^{4}\right )}{2 c^{4} \cdot \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 3 a^{2} b c + a b^{3} + 4 a c^{4} \left (\frac {b \left (2 a c - b^{2}\right )}{2 c^{4}} - \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a^{2} c^{2} - 4 a b^{2} c + b^{4}\right )}{2 c^{4} \cdot \left (4 a c - b^{2}\right )}\right ) - b^{2} c^{3} \left (\frac {b \left (2 a c - b^{2}\right )}{2 c^{4}} - \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a^{2} c^{2} - 4 a b^{2} c + b^{4}\right )}{2 c^{4} \cdot \left (4 a c - b^{2}\right )}\right )}{2 a^{2} c^{2} - 4 a b^{2} c + b^{4}} \right )} + \left (\frac {b \left (2 a c - b^{2}\right )}{2 c^{4}} + \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a^{2} c^{2} - 4 a b^{2} c + b^{4}\right )}{2 c^{4} \cdot \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 3 a^{2} b c + a b^{3} + 4 a c^{4} \left (\frac {b \left (2 a c - b^{2}\right )}{2 c^{4}} + \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a^{2} c^{2} - 4 a b^{2} c + b^{4}\right )}{2 c^{4} \cdot \left (4 a c - b^{2}\right )}\right ) - b^{2} c^{3} \left (\frac {b \left (2 a c - b^{2}\right )}{2 c^{4}} + \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a^{2} c^{2} - 4 a b^{2} c + b^{4}\right )}{2 c^{4} \cdot \left (4 a c - b^{2}\right )}\right )}{2 a^{2} c^{2} - 4 a b^{2} c + b^{4}} \right )} + \frac {x^{3}}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*x**2/(2*c**2) + x*(-a/c**2 + b**2/c**3) + (b*(2*a*c - b**2)/(2*c**4) - sqrt(-4*a*c + b**2)*(2*a**2*c**2 - 4
*a*b**2*c + b**4)/(2*c**4*(4*a*c - b**2)))*log(x + (-3*a**2*b*c + a*b**3 + 4*a*c**4*(b*(2*a*c - b**2)/(2*c**4)
- sqrt(-4*a*c + b**2)*(2*a**2*c**2 - 4*a*b**2*c + b**4)/(2*c**4*(4*a*c - b**2))) - b**2*c**3*(b*(2*a*c - b**2
)/(2*c**4) - sqrt(-4*a*c + b**2)*(2*a**2*c**2 - 4*a*b**2*c + b**4)/(2*c**4*(4*a*c - b**2))))/(2*a**2*c**2 - 4*
a*b**2*c + b**4)) + (b*(2*a*c - b**2)/(2*c**4) + sqrt(-4*a*c + b**2)*(2*a**2*c**2 - 4*a*b**2*c + b**4)/(2*c**4
*(4*a*c - b**2)))*log(x + (-3*a**2*b*c + a*b**3 + 4*a*c**4*(b*(2*a*c - b**2)/(2*c**4) + sqrt(-4*a*c + b**2)*(2
*a**2*c**2 - 4*a*b**2*c + b**4)/(2*c**4*(4*a*c - b**2))) - b**2*c**3*(b*(2*a*c - b**2)/(2*c**4) + sqrt(-4*a*c
+ b**2)*(2*a**2*c**2 - 4*a*b**2*c + b**4)/(2*c**4*(4*a*c - b**2))))/(2*a**2*c**2 - 4*a*b**2*c + b**4)) + x**3/
(3*c)

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Giac [A]
time = 3.35, size = 113, normalized size = 0.96 \begin {gather*} \frac {2 \, c^{2} x^{3} - 3 \, b c x^{2} + 6 \, b^{2} x - 6 \, a c x}{6 \, c^{3}} - \frac {{\left (b^{3} - 2 \, a b c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{4}} + \frac {{\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*c^2*x^3 - 3*b*c*x^2 + 6*b^2*x - 6*a*c*x)/c^3 - 1/2*(b^3 - 2*a*b*c)*log(c*x^2 + b*x + a)/c^4 + (b^4 - 4*
a*b^2*c + 2*a^2*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^4)

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Mupad [B]
time = 1.40, size = 151, normalized size = 1.28 \begin {gather*} \frac {x^3}{3\,c}-x\,\left (\frac {a}{c^2}-\frac {b^2}{c^3}\right )-\frac {b\,x^2}{2\,c^2}+\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (8\,a^2\,b\,c^2-6\,a\,b^3\,c+b^5\right )}{2\,\left (4\,a\,c^5-b^2\,c^4\right )}+\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (2\,a^2\,c^2-4\,a\,b^2\,c+b^4\right )}{c^4\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(c + a/x^2 + b/x),x)

[Out]

x^3/(3*c) - x*(a/c^2 - b^2/c^3) - (b*x^2)/(2*c^2) + (log(a + b*x + c*x^2)*(b^5 + 8*a^2*b*c^2 - 6*a*b^3*c))/(2*
(4*a*c^5 - b^2*c^4)) + (atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2))*(b^4 + 2*a^2*c^2 - 4*a*b^2*c
))/(c^4*(4*a*c - b^2)^(1/2))

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