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

Optimal. Leaf size=176 \[ \frac {\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (-3 a b c d+2 a c^2 e+b^3 d-b^2 c e\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {x}{a e} \]

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Rubi [A]  time = 0.29, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1569, 1628, 634, 618, 206, 628} \begin {gather*} \frac {\left (-3 a b c d+2 a c^2 e-b^2 c e+b^3 d\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {x}{a e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(a*e) + ((b^3*d - 3*a*b*c*d - b^2*c*e + 2*a*c^2*e)*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4
*a*c]*(a*d^2 - e*(b*d - c*e))) - (d^3*Log[d + e*x])/(e^2*(a*d^2 - e*(b*d - c*e))) + ((b^2*d - a*c*d - b*c*e)*L
og[c + b*x + a*x^2])/(2*a^2*(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}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) (d+e x)} \, dx &=\int \frac {x^3}{(d+e x) \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac {1}{a e}+\frac {d^3}{e \left (-a d^2+e (b d-c e)\right ) (d+e x)}+\frac {c (b d-c e)+\left (b^2 d-a c d-b c e\right ) x}{a \left (a d^2-e (b d-c e)\right ) \left (c+b x+a x^2\right )}\right ) \, dx\\ &=\frac {x}{a e}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {\int \frac {c (b d-c e)+\left (b^2 d-a c d-b c e\right ) x}{c+b x+a x^2} \, dx}{a \left (a d^2-b d e+c e^2\right )}\\ &=\frac {x}{a e}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \int \frac {b+2 a x}{c+b x+a x^2} \, dx}{2 a^2 \left (a d^2-e (b d-c e)\right )}-\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a c^2 e\right ) \int \frac {1}{c+b x+a x^2} \, dx}{2 a^2 \left (a d^2-e (b d-c e)\right )}\\ &=\frac {x}{a e}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a 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^2 \left (a d^2-e (b d-c e)\right )}\\ &=\frac {x}{a e}+\frac {\left (b^3 d-3 a b c d-b^2 c e+2 a c^2 e\right ) \tanh ^{-1}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac {\left (b^2 d-a c d-b c e\right ) \log \left (c+b x+a x^2\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 178, normalized size = 1.01 \begin {gather*} \frac {\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-b d e+c e^2\right )}+\frac {\left (-3 a b c d+2 a c^2 e+b^3 d-b^2 c e\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{a^2 \sqrt {4 a c-b^2} \left (-a d^2+b d e-c e^2\right )}-\frac {d^3 \log (d+e x)}{e^2 \left (a d^2-b d e+c e^2\right )}+\frac {x}{a e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [A]  time = 16.04, size = 596, normalized size = 3.39 \begin {gather*} \left [-\frac {2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - {\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} - {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\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^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )} x - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\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 )}}, -\frac {2 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - 2 \, {\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} - {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\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^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e - {\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )} x - {\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\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 )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.40, size = 185, normalized size = 1.05 \begin {gather*} -\frac {d^{3} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e^{2} - b d e^{3} + c e^{4}} + \frac {x e^{\left (-1\right )}}{a} + \frac {{\left (b^{2} d - a c d - b c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )}} - \frac {{\left (b^{3} d - 3 \, a b c d - b^{2} c e + 2 \, a c^{2} e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 388, normalized size = 2.20 \begin {gather*} \frac {3 b 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}-\frac {2 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}-\frac {b^{3} 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 {b^{2} 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^{2}}-\frac {c d \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) a}+\frac {b^{2} d \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 c 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 {d^{3} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) e^{2}}+\frac {x}{a e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a/e-1/2/(a*d^2-b*d*e+c*e^2)/a*ln(a*x^2+b*x+c)*c*d+1/2/(a*d^2-b*d*e+c*e^2)/a^2*ln(a*x^2+b*x+c)*b^2*d-1/2/(a*d
^2-b*d*e+c*e^2)/a^2*ln(a*x^2+b*x+c)*b*c*e+3/(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))*b*c*d-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))*e*c^2-1/(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^3*d+1/(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*e-1/e^2*d^3/(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/(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 = 4.34, size = 1367, normalized size = 7.77 \begin {gather*} \frac {x}{a\,e}-\frac {\ln \left (c^3\,e^5\,\sqrt {b^2-4\,a\,c}-b\,c^3\,e^5-4\,a^3\,c\,d^5+a^2\,b^2\,d^5+b^4\,d^3\,e^2+3\,b^2\,c^2\,d\,e^4-3\,b^3\,c\,d^2\,e^3-b^3\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^3\,e^2-6\,a\,c^3\,d\,e^4-2\,a\,c^3\,e^5\,x-a^2\,b\,d^5\,\sqrt {b^2-4\,a\,c}-2\,a^3\,d^5\,x\,\sqrt {b^2-4\,a\,c}-8\,a^3\,c\,d^4\,e\,x+4\,a^2\,c\,d^4\,e\,\sqrt {b^2-4\,a\,c}-3\,b\,c^2\,d\,e^4\,\sqrt {b^2-4\,a\,c}+9\,a\,b\,c^2\,d^2\,e^3-5\,a\,b^2\,c\,d^3\,e^2+2\,a^2\,b^2\,d^4\,e\,x-3\,a\,c^2\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}+3\,b^2\,c\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^2\,e^3\,x-2\,a\,b^2\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}+3\,a^2\,c\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c^2\,d\,e^4\,x+a\,b\,c\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}+2\,a^2\,b\,d^4\,e\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,c^2\,d\,e^4\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,b^2\,c\,d^2\,e^3\,x+a^2\,b\,c\,d^3\,e^2\,x+3\,a\,b\,c\,d^2\,e^3\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^4\,d-b^3\,d\,\sqrt {b^2-4\,a\,c}+4\,a^2\,c^2\,d-b^3\,c\,e-5\,a\,b^2\,c\,d+4\,a\,b\,c^2\,e-2\,a\,c^2\,e\,\sqrt {b^2-4\,a\,c}+b^2\,c\,e\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c\,d\,\sqrt {b^2-4\,a\,c}\right )}{2\,\left (4\,a^4\,c\,d^2-a^3\,b^2\,d^2-4\,a^3\,b\,c\,d\,e+4\,a^3\,c^2\,e^2+a^2\,b^3\,d\,e-a^2\,b^2\,c\,e^2\right )}-\frac {\ln \left (a^2\,b^2\,d^5-b\,c^3\,e^5-c^3\,e^5\,\sqrt {b^2-4\,a\,c}-4\,a^3\,c\,d^5+b^4\,d^3\,e^2+3\,b^2\,c^2\,d\,e^4-3\,b^3\,c\,d^2\,e^3+b^3\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^3\,e^2-6\,a\,c^3\,d\,e^4-2\,a\,c^3\,e^5\,x+a^2\,b\,d^5\,\sqrt {b^2-4\,a\,c}+2\,a^3\,d^5\,x\,\sqrt {b^2-4\,a\,c}-8\,a^3\,c\,d^4\,e\,x-4\,a^2\,c\,d^4\,e\,\sqrt {b^2-4\,a\,c}+3\,b\,c^2\,d\,e^4\,\sqrt {b^2-4\,a\,c}+9\,a\,b\,c^2\,d^2\,e^3-5\,a\,b^2\,c\,d^3\,e^2+2\,a^2\,b^2\,d^4\,e\,x+3\,a\,c^2\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}-3\,b^2\,c\,d^2\,e^3\,\sqrt {b^2-4\,a\,c}+6\,a^2\,c^2\,d^2\,e^3\,x+2\,a\,b^2\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}-3\,a^2\,c\,d^3\,e^2\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,b\,c^2\,d\,e^4\,x-a\,b\,c\,d^3\,e^2\,\sqrt {b^2-4\,a\,c}-2\,a^2\,b\,d^4\,e\,x\,\sqrt {b^2-4\,a\,c}+3\,a\,c^2\,d\,e^4\,x\,\sqrt {b^2-4\,a\,c}-3\,a\,b^2\,c\,d^2\,e^3\,x+a^2\,b\,c\,d^3\,e^2\,x-3\,a\,b\,c\,d^2\,e^3\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^4\,d+b^3\,d\,\sqrt {b^2-4\,a\,c}+4\,a^2\,c^2\,d-b^3\,c\,e-5\,a\,b^2\,c\,d+4\,a\,b\,c^2\,e+2\,a\,c^2\,e\,\sqrt {b^2-4\,a\,c}-b^2\,c\,e\,\sqrt {b^2-4\,a\,c}-3\,a\,b\,c\,d\,\sqrt {b^2-4\,a\,c}\right )}{2\,\left (4\,a^4\,c\,d^2-a^3\,b^2\,d^2-4\,a^3\,b\,c\,d\,e+4\,a^3\,c^2\,e^2+a^2\,b^3\,d\,e-a^2\,b^2\,c\,e^2\right )}-\frac {d^3\,\ln \left (d+e\,x\right )}{a\,d^2\,e^2-b\,d\,e^3+c\,e^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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