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

Optimal. Leaf size=135 \[ \frac {\left (4 a^2 c e-6 a b c d+b^3 d\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}-\frac {d \log \left (a+b x+c x^2\right )}{2 a^2}+\frac {d \log (x)}{a^2}+\frac {c x (b d-2 a e)-a b e-2 a c d+b^2 d}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {822, 800, 634, 618, 206, 628} \begin {gather*} \frac {\left (4 a^2 c e-6 a b c d+b^3 d\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}-\frac {d \log \left (a+b x+c x^2\right )}{2 a^2}+\frac {d \log (x)}{a^2}+\frac {c x (b d-2 a e)-a b e-2 a c d+b^2 d}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {d+e x}{x \left (a+b x+c x^2\right )^2} \, dx &=\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int \frac {-\left (\left (b^2-4 a c\right ) d\right )-c (b d-2 a e) x}{x \left (a+b x+c x^2\right )} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {\left (-b^2+4 a c\right ) d}{a x}+\frac {b^3 d-5 a b c d+2 a^2 c e+c \left (b^2-4 a c\right ) d x}{a \left (a+b x+c x^2\right )}\right ) \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {d \log (x)}{a^2}-\frac {\int \frac {b^3 d-5 a b c d+2 a^2 c e+c \left (b^2-4 a c\right ) d x}{a+b x+c x^2} \, dx}{a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {d \log (x)}{a^2}-\frac {d \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 a^2}-\frac {\left (b^3 d-6 a b c d+4 a^2 c e\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {d \log (x)}{a^2}-\frac {d \log \left (a+b x+c x^2\right )}{2 a^2}+\frac {\left (b^3 d-6 a b c d+4 a^2 c e\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2 d-2 a c d-a b e+c (b d-2 a e) x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\left (b^3 d-6 a b c d+4 a^2 c e\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}+\frac {d \log (x)}{a^2}-\frac {d \log \left (a+b x+c x^2\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 134, normalized size = 0.99 \begin {gather*} \frac {\frac {2 \left (4 a^2 c e-6 a b c d+b^3 d\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}-\frac {2 a \left (b (a e-c d x)+2 a c (d+e x)+b^2 (-d)\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}-d \log (a+x (b+c x))+2 d \log (x)}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.85, size = 955, normalized size = 7.07 \begin {gather*} \left [-\frac {{\left (4 \, a^{3} c e + {\left (4 \, a^{2} c^{2} e + {\left (b^{3} c - 6 \, a b c^{2}\right )} d\right )} x^{2} + {\left (a b^{3} - 6 \, a^{2} b c\right )} d + {\left (4 \, a^{2} b c e + {\left (b^{4} - 6 \, a b^{2} c\right )} d\right )} x\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 ) - 2 \, {\left (a b^{4} - 6 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} d + 2 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} e - 2 \, {\left ({\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} d - 2 \, {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e\right )} x + {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + {\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{2} + {\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x\right )}}, \frac {2 \, {\left (4 \, a^{3} c e + {\left (4 \, a^{2} c^{2} e + {\left (b^{3} c - 6 \, a b c^{2}\right )} d\right )} x^{2} + {\left (a b^{3} - 6 \, a^{2} b c\right )} d + {\left (4 \, a^{2} b c e + {\left (b^{4} - 6 \, a b^{2} c\right )} d\right )} x\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 ) + 2 \, {\left (a b^{4} - 6 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} d - 2 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} e + 2 \, {\left ({\left (a b^{3} c - 4 \, a^{2} b c^{2}\right )} d - 2 \, {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e\right )} x - {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + {\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{2} + {\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((4*a^3*c*e + (4*a^2*c^2*e + (b^3*c - 6*a*b*c^2)*d)*x^2 + (a*b^3 - 6*a^2*b*c)*d + (4*a^2*b*c*e + (b^4 -
6*a*b^2*c)*d)*x)*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)) - 2*(a*b^4 - 6*a^2*b^2*c + 8*a^3*c^2)*d + 2*(a^2*b^3 - 4*a^3*b*c)*e - 2*((a*b^3*c - 4*a^2*b*c^
2)*d - 2*(a^2*b^2*c - 4*a^3*c^2)*e)*x + ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*
b*c^2)*d*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*d)*log(c*x^2 + b*x + a) - 2*((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3
)*d*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*d)*log(x))/(a^3*b^4 - 8*a^
4*b^2*c + 16*a^5*c^2 + (a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*x^2 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x
), 1/2*(2*(4*a^3*c*e + (4*a^2*c^2*e + (b^3*c - 6*a*b*c^2)*d)*x^2 + (a*b^3 - 6*a^2*b*c)*d + (4*a^2*b*c*e + (b^4
- 6*a*b^2*c)*d)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(a*b^4 - 6*a^
2*b^2*c + 8*a^3*c^2)*d - 2*(a^2*b^3 - 4*a^3*b*c)*e + 2*((a*b^3*c - 4*a^2*b*c^2)*d - 2*(a^2*b^2*c - 4*a^3*c^2)*
e)*x - ((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d*x + (a*b^4 - 8*a^2*b^2*c
+ 16*a^3*c^2)*d)*log(c*x^2 + b*x + a) + 2*((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*x^2 + (b^5 - 8*a*b^3*c + 16*a
^2*b*c^2)*d*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*d)*log(x))/(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 + (a^2*b^4*c
- 8*a^3*b^2*c^2 + 16*a^4*c^3)*x^2 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x)]

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giac [A]  time = 0.16, size = 160, normalized size = 1.19 \begin {gather*} -\frac {{\left (b^{3} d - 6 \, a b c d + 4 \, a^{2} c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {d \log \left (c x^{2} + b x + a\right )}{2 \, a^{2}} + \frac {d \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {a b^{2} d - 2 \, a^{2} c d - a^{2} b e + {\left (a b c d - 2 \, a^{2} c e\right )} x}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 337, normalized size = 2.50 \begin {gather*} -\frac {b c d x}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right ) a}-\frac {6 b c d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} a}+\frac {b^{3} d \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} a^{2}}+\frac {2 c e x}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right )}+\frac {4 c e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}-\frac {b^{2} d}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right ) a}-\frac {2 c d \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right ) a}+\frac {b^{2} d \ln \left (c \,x^{2}+b x +a \right )}{2 \left (4 a c -b^{2}\right ) a^{2}}+\frac {b e}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right )}+\frac {2 c d}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right )}+\frac {d \ln \relax (x )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(c*x^2+b*x+a)*c/(4*a*c-b^2)*x*e-1/a/(c*x^2+b*x+a)*c/(4*a*c-b^2)*x*b*d+1/(c*x^2+b*x+a)/(4*a*c-b^2)*b*e+2/(c*x
^2+b*x+a)/(4*a*c-b^2)*c*d-1/a/(c*x^2+b*x+a)/(4*a*c-b^2)*b^2*d-2/a/(4*a*c-b^2)*c*ln(c*x^2+b*x+a)*d+1/2/a^2/(4*a
*c-b^2)*ln(c*x^2+b*x+a)*b^2*d+4/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c*e-6/a/(4*a*c-b^2)^(3/2
)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*c*d+1/a^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3*d+
1/a^2*d*ln(x)

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

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mupad [B]  time = 2.42, size = 920, normalized size = 6.81 \begin {gather*} \frac {\frac {-d\,b^2+a\,e\,b+2\,a\,c\,d}{a\,\left (4\,a\,c-b^2\right )}+\frac {c\,x\,\left (2\,a\,e-b\,d\right )}{a\,\left (4\,a\,c-b^2\right )}}{c\,x^2+b\,x+a}-\ln \left (96\,a^4\,c^3\,d-2\,a\,b^6\,d-2\,b^7\,d\,x-84\,a^3\,b^2\,c^2\,d+2\,a\,b^3\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+23\,a^2\,b^4\,c\,d-2\,a^3\,b^3\,c\,e+8\,a^4\,b\,c^2\,e+2\,a^3\,c\,e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+2\,b^4\,d\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-16\,a^4\,c^3\,e\,x-9\,a^2\,b\,c\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+120\,a^3\,b\,c^3\,d\,x-2\,a^2\,b^4\,c\,e\,x-94\,a^2\,b^3\,c^2\,d\,x+12\,a^2\,c^2\,d\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+12\,a^3\,b^2\,c^2\,e\,x+24\,a\,b^5\,c\,d\,x-12\,a\,b^2\,c\,d\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+2\,a^2\,b\,c\,e\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (\frac {d}{2\,a^2}-\frac {\frac {b^3\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{2}+2\,a^2\,c\,e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-3\,a\,b\,c\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{-64\,a^5\,c^3+48\,a^4\,b^2\,c^2-12\,a^3\,b^4\,c+a^2\,b^6}\right )-\ln \left (2\,a\,b^6\,d-96\,a^4\,c^3\,d+2\,b^7\,d\,x+84\,a^3\,b^2\,c^2\,d+2\,a\,b^3\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-23\,a^2\,b^4\,c\,d+2\,a^3\,b^3\,c\,e-8\,a^4\,b\,c^2\,e+2\,a^3\,c\,e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+2\,b^4\,d\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+16\,a^4\,c^3\,e\,x-9\,a^2\,b\,c\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-120\,a^3\,b\,c^3\,d\,x+2\,a^2\,b^4\,c\,e\,x+94\,a^2\,b^3\,c^2\,d\,x+12\,a^2\,c^2\,d\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-12\,a^3\,b^2\,c^2\,e\,x-24\,a\,b^5\,c\,d\,x-12\,a\,b^2\,c\,d\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+2\,a^2\,b\,c\,e\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (\frac {d}{2\,a^2}+\frac {\frac {b^3\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{2}+2\,a^2\,c\,e\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-3\,a\,b\,c\,d\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{-64\,a^5\,c^3+48\,a^4\,b^2\,c^2-12\,a^3\,b^4\,c+a^2\,b^6}\right )+\frac {d\,\ln \relax (x)}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*b*e - b^2*d + 2*a*c*d)/(a*(4*a*c - b^2)) + (c*x*(2*a*e - b*d))/(a*(4*a*c - b^2)))/(a + b*x + c*x^2) - log(
96*a^4*c^3*d - 2*a*b^6*d - 2*b^7*d*x - 84*a^3*b^2*c^2*d + 2*a*b^3*d*(-(4*a*c - b^2)^3)^(1/2) + 23*a^2*b^4*c*d
- 2*a^3*b^3*c*e + 8*a^4*b*c^2*e + 2*a^3*c*e*(-(4*a*c - b^2)^3)^(1/2) + 2*b^4*d*x*(-(4*a*c - b^2)^3)^(1/2) - 16
*a^4*c^3*e*x - 9*a^2*b*c*d*(-(4*a*c - b^2)^3)^(1/2) + 120*a^3*b*c^3*d*x - 2*a^2*b^4*c*e*x - 94*a^2*b^3*c^2*d*x
+ 12*a^2*c^2*d*x*(-(4*a*c - b^2)^3)^(1/2) + 12*a^3*b^2*c^2*e*x + 24*a*b^5*c*d*x - 12*a*b^2*c*d*x*(-(4*a*c - b
^2)^3)^(1/2) + 2*a^2*b*c*e*x*(-(4*a*c - b^2)^3)^(1/2))*(d/(2*a^2) - ((b^3*d*(-(4*a*c - b^2)^3)^(1/2))/2 + 2*a^
2*c*e*(-(4*a*c - b^2)^3)^(1/2) - 3*a*b*c*d*(-(4*a*c - b^2)^3)^(1/2))/(a^2*b^6 - 64*a^5*c^3 - 12*a^3*b^4*c + 48
*a^4*b^2*c^2)) - log(2*a*b^6*d - 96*a^4*c^3*d + 2*b^7*d*x + 84*a^3*b^2*c^2*d + 2*a*b^3*d*(-(4*a*c - b^2)^3)^(1
/2) - 23*a^2*b^4*c*d + 2*a^3*b^3*c*e - 8*a^4*b*c^2*e + 2*a^3*c*e*(-(4*a*c - b^2)^3)^(1/2) + 2*b^4*d*x*(-(4*a*c
- b^2)^3)^(1/2) + 16*a^4*c^3*e*x - 9*a^2*b*c*d*(-(4*a*c - b^2)^3)^(1/2) - 120*a^3*b*c^3*d*x + 2*a^2*b^4*c*e*x
+ 94*a^2*b^3*c^2*d*x + 12*a^2*c^2*d*x*(-(4*a*c - b^2)^3)^(1/2) - 12*a^3*b^2*c^2*e*x - 24*a*b^5*c*d*x - 12*a*b
^2*c*d*x*(-(4*a*c - b^2)^3)^(1/2) + 2*a^2*b*c*e*x*(-(4*a*c - b^2)^3)^(1/2))*(d/(2*a^2) + ((b^3*d*(-(4*a*c - b^
2)^3)^(1/2))/2 + 2*a^2*c*e*(-(4*a*c - b^2)^3)^(1/2) - 3*a*b*c*d*(-(4*a*c - b^2)^3)^(1/2))/(a^2*b^6 - 64*a^5*c^
3 - 12*a^3*b^4*c + 48*a^4*b^2*c^2)) + (d*log(x))/a^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((e*x+d)/x/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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