3.1.54 \(\int \frac {x^2}{(a+\frac {c}{x^2}+\frac {b}{x}) (d+e x)} \, dx\)

Optimal. Leaf size=218 \[ -\frac {\left (-2 a b c d+a c^2 e+b^3 d-b^2 c e\right ) \log \left (a x^2+b x+c\right )}{2 a^3 \left (a d^2-e (b d-c e)\right )}-\frac {x (a d+b e)}{a^2 e^2}-\frac {\left (2 a^2 c^2 d-4 a b^2 c d+3 a b c^2 e+b^4 d-b^3 c e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {d^4 \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )}+\frac {x^2}{2 a e} \]

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Rubi [A]  time = 0.40, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1569, 1628, 634, 618, 206, 628} \begin {gather*} -\frac {\left (-2 a b c d+a c^2 e-b^2 c e+b^3 d\right ) \log \left (a x^2+b x+c\right )}{2 a^3 \left (a d^2-e (b d-c e)\right )}-\frac {\left (2 a^2 c^2 d-4 a b^2 c d+3 a b c^2 e-b^3 c e+b^4 d\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{a^3 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {x (a d+b e)}{a^2 e^2}+\frac {d^4 \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )}+\frac {x^2}{2 a e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((a + c/x^2 + b/x)*(d + e*x)),x]

[Out]

-(((a*d + b*e)*x)/(a^2*e^2)) + x^2/(2*a*e) - ((b^4*d - 4*a*b^2*c*d + 2*a^2*c^2*d - b^3*c*e + 3*a*b*c^2*e)*ArcT
anh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(a^3*Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))) + (d^4*Log[d + e*x])/(e^3*
(a*d^2 - e*(b*d - c*e))) - ((b^3*d - 2*a*b*c*d - b^2*c*e + a*c^2*e)*Log[c + b*x + a*x^2])/(2*a^3*(a*d^2 - e*(b
*d - c*e)))

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 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 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]

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 1569

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbo
l] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && E
qQ[mn, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 218, normalized size = 1.00 \begin {gather*} \frac {\left (2 a b c d-a c^2 e+b^3 (-d)+b^2 c e\right ) \log (x (a x+b)+c)}{2 a^3 \left (a d^2+e (c e-b d)\right )}-\frac {x (a d+b e)}{a^2 e^2}+\frac {\left (2 a^2 c^2 d-4 a b^2 c d+3 a b c^2 e+b^4 d-b^3 c e\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a^3 \sqrt {4 a c-b^2} \left (a d^2+e (c e-b d)\right )}+\frac {d^4 \log (d+e x)}{e^3 \left (a d^2+e (c e-b d)\right )}+\frac {x^2}{2 a e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/((a + c/x^2 + b/x)*(d + e*x)),x]

[Out]

-(((a*d + b*e)*x)/(a^2*e^2)) + x^2/(2*a*e) + ((b^4*d - 4*a*b^2*c*d + 2*a^2*c^2*d - b^3*c*e + 3*a*b*c^2*e)*ArcT
an[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]])/(a^3*Sqrt[-b^2 + 4*a*c]*(a*d^2 + e*(-(b*d) + c*e))) + (d^4*Log[d + e*x])/(
e^3*(a*d^2 + e*(-(b*d) + c*e))) + ((-(b^3*d) + 2*a*b*c*d + b^2*c*e - a*c^2*e)*Log[c + x*(b + a*x)])/(2*a^3*(a*
d^2 + e*(-(b*d) + c*e)))

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^2/((a + c/x^2 + b/x)*(d + e*x)),x]

[Out]

IntegrateAlgebraic[x^2/((a + c/x^2 + b/x)*(d + e*x)), x]

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fricas [A]  time = 52.35, size = 798, normalized size = 3.66 \begin {gather*} \left [\frac {2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{4} \log \left (e x + d\right ) + {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} - {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )} x^{2} + {\left ({\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} d e^{3} - {\left (b^{3} c - 3 \, a b c^{2}\right )} e^{4}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right ) - 2 \, {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{3} e - {\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} d e^{3} + {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} e^{4}\right )} x - {\left ({\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} d e^{3} - {\left (b^{4} c - 5 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} e^{4}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{4} b^{2} - 4 \, a^{5} c\right )} d^{2} e^{3} - {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} d e^{4} + {\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} e^{5}\right )}}, \frac {2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{4} \log \left (e x + d\right ) + {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} - {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )} x^{2} - 2 \, {\left ({\left (b^{4} - 4 \, a b^{2} c + 2 \, a^{2} c^{2}\right )} d e^{3} - {\left (b^{3} c - 3 \, a b c^{2}\right )} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{3} e - {\left (a b^{4} - 5 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} d e^{3} + {\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} e^{4}\right )} x - {\left ({\left (b^{5} - 6 \, a b^{3} c + 8 \, a^{2} b c^{2}\right )} d e^{3} - {\left (b^{4} c - 5 \, a b^{2} c^{2} + 4 \, a^{2} c^{3}\right )} e^{4}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left ({\left (a^{4} b^{2} - 4 \, a^{5} c\right )} d^{2} e^{3} - {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} d e^{4} + {\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} e^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 224, normalized size = 1.03 \begin {gather*} \frac {d^{4} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e^{3} - b d e^{4} + c e^{5}} - \frac {{\left (b^{3} d - 2 \, a b c d - b^{2} c e + a c^{2} e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a^{4} d^{2} - a^{3} b d e + a^{3} c e^{2}\right )}} + \frac {{\left (b^{4} d - 4 \, a b^{2} c d + 2 \, a^{2} c^{2} d - b^{3} c e + 3 \, a b c^{2} e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{4} d^{2} - a^{3} b d e + a^{3} c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {{\left (a x^{2} e - 2 \, a d x - 2 \, b x e\right )} e^{\left (-2\right )}}{2 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^4*log(abs(x*e + d))/(a*d^2*e^3 - b*d*e^4 + c*e^5) - 1/2*(b^3*d - 2*a*b*c*d - b^2*c*e + a*c^2*e)*log(a*x^2 +
b*x + c)/(a^4*d^2 - a^3*b*d*e + a^3*c*e^2) + (b^4*d - 4*a*b^2*c*d + 2*a^2*c^2*d - b^3*c*e + 3*a*b*c^2*e)*arcta
n((2*a*x + b)/sqrt(-b^2 + 4*a*c))/((a^4*d^2 - a^3*b*d*e + a^3*c*e^2)*sqrt(-b^2 + 4*a*c)) + 1/2*(a*x^2*e - 2*a*
d*x - 2*b*x*e)*e^(-2)/a^2

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maple [B]  time = 0.01, size = 512, normalized size = 2.35 \begin {gather*} \frac {2 c^{2} d \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, a}-\frac {4 b^{2} c d \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, a^{2}}+\frac {3 b \,c^{2} e \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, a^{2}}+\frac {b^{4} d \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, a^{3}}-\frac {b^{3} c e \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, a^{3}}+\frac {b c d \ln \left (a \,x^{2}+b x +c \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) a^{2}}-\frac {c^{2} e \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) a^{2}}-\frac {b^{3} d \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) a^{3}}+\frac {b^{2} c e \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) a^{3}}+\frac {d^{4} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) e^{3}}+\frac {x^{2}}{2 a e}-\frac {d x}{a \,e^{2}}-\frac {b x}{a^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+c/x^2+b/x)/(e*x+d),x)

[Out]

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

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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/(a+c/x^2+b/x)/(e*x+d),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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 5.24, size = 2051, normalized size = 9.41

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((d + e*x)*(a + b/x + c/x^2)),x)

[Out]

(d^4*log(d + e*x))/(c*e^5 + a*d^2*e^3 - b*d*e^4) - (log(4*a^4*c*d^6 - 2*a*c^4*e^6 - a^3*b^2*d^6 + b^2*c^3*e^6
- b^5*d^3*e^3 - 3*b^3*c^2*d*e^5 + 3*b^4*c*d^2*e^4 + b^4*d^3*e^3*(b^2 - 4*a*c)^(1/2) + 6*a^2*c^3*d^2*e^4 - 4*a^
3*c^2*d^4*e^2 + a^3*b*d^6*(b^2 - 4*a*c)^(1/2) - b*c^3*e^6*(b^2 - 4*a*c)^(1/2) + 2*a^4*d^6*x*(b^2 - 4*a*c)^(1/2
) + 9*a*b*c^3*d*e^5 + a^2*c^2*d^3*e^3*(b^2 - 4*a*c)^(1/2) + a*b*c^3*e^6*x + 8*a^4*c*d^5*e*x - 3*a*c^3*d*e^5*(b
^2 - 4*a*c)^(1/2) - 4*a^3*c*d^5*e*(b^2 - 4*a*c)^(1/2) - a*c^3*e^6*x*(b^2 - 4*a*c)^(1/2) + 5*a*b^3*c*d^3*e^3 -
a*b^4*d^3*e^3*x - 2*a^3*b^2*d^5*e*x + 6*a^2*c^3*d*e^5*x + 3*b^2*c^2*d*e^5*(b^2 - 4*a*c)^(1/2) - 3*b^3*c*d^2*e^
4*(b^2 - 4*a*c)^(1/2) - 12*a*b^2*c^2*d^2*e^4 - 5*a^2*b*c^2*d^3*e^3 + a^2*b^2*c*d^4*e^2 + a^2*b^3*d^4*e^2*x - 2
*a^3*c^2*d^3*e^3*x + 6*a*b*c^2*d^2*e^4*(b^2 - 4*a*c)^(1/2) - 3*a*b^2*c*d^3*e^3*(b^2 - 4*a*c)^(1/2) + a^2*b*c*d
^4*e^2*(b^2 - 4*a*c)^(1/2) + a*b^3*d^3*e^3*x*(b^2 - 4*a*c)^(1/2) - 2*a^3*c*d^4*e^2*x*(b^2 - 4*a*c)^(1/2) - 9*a
^2*b*c^2*d^2*e^4*x + 4*a^2*b^2*c*d^3*e^3*x + a^2*b^2*d^4*e^2*x*(b^2 - 4*a*c)^(1/2) + 3*a^2*c^2*d^2*e^4*x*(b^2
- 4*a*c)^(1/2) - 2*a^3*b*d^5*e*x*(b^2 - 4*a*c)^(1/2) - 3*a*b^2*c^2*d*e^5*x + 3*a*b^3*c*d^2*e^4*x - 4*a^3*b*c*d
^4*e^2*x + 3*a*b*c^2*d*e^5*x*(b^2 - 4*a*c)^(1/2) - 3*a*b^2*c*d^2*e^4*x*(b^2 - 4*a*c)^(1/2) - 2*a^2*b*c*d^3*e^3
*x*(b^2 - 4*a*c)^(1/2))*(b^4*d*(b^2 - 4*a*c)^(1/2) - b^5*d + 4*a^2*c^3*e + b^4*c*e + 6*a*b^3*c*d - b^3*c*e*(b^
2 - 4*a*c)^(1/2) - 8*a^2*b*c^2*d - 5*a*b^2*c^2*e + 2*a^2*c^2*d*(b^2 - 4*a*c)^(1/2) - 4*a*b^2*c*d*(b^2 - 4*a*c)
^(1/2) + 3*a*b*c^2*e*(b^2 - 4*a*c)^(1/2)))/(2*(4*a^5*c*d^2 - a^4*b^2*d^2 + 4*a^4*c^2*e^2 - a^3*b^2*c*e^2 + a^3
*b^3*d*e - 4*a^4*b*c*d*e)) + (log(2*a*c^4*e^6 - 4*a^4*c*d^6 + a^3*b^2*d^6 - b^2*c^3*e^6 + b^5*d^3*e^3 + 3*b^3*
c^2*d*e^5 - 3*b^4*c*d^2*e^4 + b^4*d^3*e^3*(b^2 - 4*a*c)^(1/2) - 6*a^2*c^3*d^2*e^4 + 4*a^3*c^2*d^4*e^2 + a^3*b*
d^6*(b^2 - 4*a*c)^(1/2) - b*c^3*e^6*(b^2 - 4*a*c)^(1/2) + 2*a^4*d^6*x*(b^2 - 4*a*c)^(1/2) - 9*a*b*c^3*d*e^5 +
a^2*c^2*d^3*e^3*(b^2 - 4*a*c)^(1/2) - a*b*c^3*e^6*x - 8*a^4*c*d^5*e*x - 3*a*c^3*d*e^5*(b^2 - 4*a*c)^(1/2) - 4*
a^3*c*d^5*e*(b^2 - 4*a*c)^(1/2) - a*c^3*e^6*x*(b^2 - 4*a*c)^(1/2) - 5*a*b^3*c*d^3*e^3 + a*b^4*d^3*e^3*x + 2*a^
3*b^2*d^5*e*x - 6*a^2*c^3*d*e^5*x + 3*b^2*c^2*d*e^5*(b^2 - 4*a*c)^(1/2) - 3*b^3*c*d^2*e^4*(b^2 - 4*a*c)^(1/2)
+ 12*a*b^2*c^2*d^2*e^4 + 5*a^2*b*c^2*d^3*e^3 - a^2*b^2*c*d^4*e^2 - a^2*b^3*d^4*e^2*x + 2*a^3*c^2*d^3*e^3*x + 6
*a*b*c^2*d^2*e^4*(b^2 - 4*a*c)^(1/2) - 3*a*b^2*c*d^3*e^3*(b^2 - 4*a*c)^(1/2) + a^2*b*c*d^4*e^2*(b^2 - 4*a*c)^(
1/2) + a*b^3*d^3*e^3*x*(b^2 - 4*a*c)^(1/2) - 2*a^3*c*d^4*e^2*x*(b^2 - 4*a*c)^(1/2) + 9*a^2*b*c^2*d^2*e^4*x - 4
*a^2*b^2*c*d^3*e^3*x + a^2*b^2*d^4*e^2*x*(b^2 - 4*a*c)^(1/2) + 3*a^2*c^2*d^2*e^4*x*(b^2 - 4*a*c)^(1/2) - 2*a^3
*b*d^5*e*x*(b^2 - 4*a*c)^(1/2) + 3*a*b^2*c^2*d*e^5*x - 3*a*b^3*c*d^2*e^4*x + 4*a^3*b*c*d^4*e^2*x + 3*a*b*c^2*d
*e^5*x*(b^2 - 4*a*c)^(1/2) - 3*a*b^2*c*d^2*e^4*x*(b^2 - 4*a*c)^(1/2) - 2*a^2*b*c*d^3*e^3*x*(b^2 - 4*a*c)^(1/2)
)*(b^5*d + b^4*d*(b^2 - 4*a*c)^(1/2) - 4*a^2*c^3*e - b^4*c*e - 6*a*b^3*c*d - b^3*c*e*(b^2 - 4*a*c)^(1/2) + 8*a
^2*b*c^2*d + 5*a*b^2*c^2*e + 2*a^2*c^2*d*(b^2 - 4*a*c)^(1/2) - 4*a*b^2*c*d*(b^2 - 4*a*c)^(1/2) + 3*a*b*c^2*e*(
b^2 - 4*a*c)^(1/2)))/(2*(4*a^5*c*d^2 - a^4*b^2*d^2 + 4*a^4*c^2*e^2 - a^3*b^2*c*e^2 + a^3*b^3*d*e - 4*a^4*b*c*d
*e)) + x^2/(2*a*e) - (x*(a*d + b*e))/(a^2*e^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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