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

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

[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)

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Rubi [A]  time = 0.198658, antiderivative size = 135, normalized size of antiderivative = 1., 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} \[ \frac{\left (4 a^2 c e-6 a b c d+b^3 d\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}-\frac{d \log \left (a+b x+c x^2\right )}{2 a^2}+\frac{d \log (x)}{a^2}+\frac{c x (b d-2 a e)-a b e-2 a c d+b^2 d}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

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 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])

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 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 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 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 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]

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 (b^2-4 a c\right ) d-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*}

Mathematica [A]  time = 0.223841, size = 134, normalized size = 0.99 \[ \frac{\frac{2 \left (4 a^2 c e-6 a b c d+b^3 d\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}-\frac{2 a \left (b (a e-c d x)+2 a c (d+e x)+b^2 (-d)\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}-d \log (a+x (b+c x))+2 d \log (x)}{2 a^2} \]

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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Maple [B]  time = 0.011, size = 337, normalized size = 2.5 \begin{align*}{\frac{d\ln \left ( x \right ) }{{a}^{2}}}+2\,{\frac{cxe}{ \left ( c{x}^{2}+bx+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{cxbd}{a \left ( c{x}^{2}+bx+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{be}{ \left ( c{x}^{2}+bx+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+2\,{\frac{cd}{ \left ( c{x}^{2}+bx+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{b}^{2}d}{a \left ( c{x}^{2}+bx+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-2\,{\frac{c\ln \left ( c{x}^{2}+bx+a \right ) d}{a \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}d}{2\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }}+4\,{\frac{ce}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-6\,{\frac{bcd}{a \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{3}d}{{a}^{2}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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]

d*ln(x)/a^2+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

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

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

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Fricas [B]  time = 3.08388, size = 2043, normalized size = 15.13 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.14716, size = 216, normalized size = 1.6 \begin{align*} -\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{align*}

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)