3.1530 \(\int \frac {b+2 c x}{(d+e x)^2 (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=210 \[ \frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac {\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac {e \sqrt {b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac {2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {800, 634, 618, 206, 628} \[ \frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac {\log (d+e x) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac {e \sqrt {b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (a e^2-b d e+c d^2\right )^2}+\frac {2 c d-b e}{(d+e x) \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 177, normalized size = 0.84 \[ \frac {\log (d+e x) \left (4 c e (a e+b d)-2 b^2 e^2-4 c^2 d^2\right )+\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log (a+x (b+c x))-2 e \sqrt {4 a c-b^2} (b e-2 c d) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )+\frac {2 (2 c d-b e) \left (e (a e-b d)+c d^2\right )}{d+e x}}{2 \left (e (a e-b d)+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 4.88, size = 745, normalized size = 3.55 \[ \left [\frac {4 \, c^{2} d^{3} - 6 \, b c d^{2} e - 2 \, a b e^{3} + 2 \, {\left (b^{2} + 2 \, a c\right )} d e^{2} - {\left (2 \, c d^{2} e - b d e^{2} + {\left (2 \, c d e^{2} - b e^{3}\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 ) + {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}, \frac {4 \, c^{2} d^{3} - 6 \, b c d^{2} e - 2 \, a b e^{3} + 2 \, {\left (b^{2} + 2 \, a c\right )} d e^{2} + 2 \, {\left (2 \, c d^{2} e - b d e^{2} + {\left (2 \, c d e^{2} - b e^{3}\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 ) + {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c^{2} d^{3} - 2 \, b c d^{2} e + {\left (b^{2} - 2 \, a c\right )} d e^{2} + {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + {\left (b^{2} - 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 362, normalized size = 1.72 \[ -\frac {{\left (2 \, b^{2} c d e^{3} - 8 \, a c^{2} d e^{3} - b^{3} e^{4} + 4 \, a b c e^{4}\right )} \arctan \left (\frac {{\left (2 \, c d - \frac {2 \, c d^{2}}{x e + d} - b e + \frac {2 \, b d e}{x e + d} - \frac {2 \, a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}} + \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}} + \frac {\frac {2 \, c d e^{2}}{x e + d} - \frac {b e^{3}}{x e + d}}{c d^{2} e^{2} - b d e^{3} + a e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 560, normalized size = 2.67 \[ -\frac {4 a b c \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}+\frac {8 a \,c^{2} d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}+\frac {b^{3} e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}-\frac {2 b^{2} c d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \sqrt {4 a c -b^{2}}}+\frac {2 a c \,e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {a c \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {b^{2} e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {b^{2} e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {2 b c d e \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {b c d e \ln \left (c \,x^{2}+b x +a \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {2 c^{2} d^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}+\frac {c^{2} d^{2} \ln \left (c \,x^{2}+b x +a \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2}}-\frac {b e}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (e x +d \right )}+\frac {2 c d}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (e x +d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*a*c*e^2-1/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*b^2*e^2+2/(a*e^2-b*d*e+c*d^2)^2*ln
(e*x+d)*b*c*d*e-2/(a*e^2-b*d*e+c*d^2)^2*ln(e*x+d)*c^2*d^2-1/(a*e^2-b*d*e+c*d^2)/(e*x+d)*b*e+2/(a*e^2-b*d*e+c*d
^2)/(e*x+d)*c*d-1/(a*e^2-b*d*e+c*d^2)^2*c*ln(c*x^2+b*x+a)*a*e^2+1/2/(a*e^2-b*d*e+c*d^2)^2*ln(c*x^2+b*x+a)*b^2*
e^2-1/(a*e^2-b*d*e+c*d^2)^2*c*ln(c*x^2+b*x+a)*b*d*e+1/(a*e^2-b*d*e+c*d^2)^2*c^2*ln(c*x^2+b*x+a)*d^2-4/(a*e^2-b
*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b*c*e^2+8/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b
^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*c^2*d*e+1/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*
x+b)/(4*a*c-b^2)^(1/2))*b^3*e^2-2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*
b^2*c*d*e

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)/(e*x+d)^2/(c*x^2+b*x+a),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 = 4.08, size = 1637, normalized size = 7.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(d + e*x)*(e^2*(2*a*c - b^2) - 2*c^2*d^2 + 2*b*c*d*e))/(a^2*e^4 + c^2*d^4 + b^2*d^2*e^2 - 2*a*b*d*e^3 - 2*
b*c*d^3*e + 2*a*c*d^2*e^2) - (log(3*b^2*c^3*d^4 - 12*a*c^4*d^4 - 2*b^5*e^4*x - 12*a^3*c^2*e^4 - 2*a*b^4*e^4 +
2*b^4*e^4*x*(b^2 - 4*a*c)^(1/2) + 6*c^4*d^4*x*(b^2 - 4*a*c)^(1/2) + 11*a^2*b^2*c*e^4 - 2*b^3*c^2*d^3*e + b^4*c
*d^2*e^2 + 40*a^2*c^3*d^2*e^2 + 2*a*b^3*e^4*(b^2 - 4*a*c)^(1/2) + 3*b*c^3*d^4*(b^2 - 4*a*c)^(1/2) + 8*a*b*c^3*
d^3*e + 6*a*b^3*c*d*e^3 + 12*a*b^3*c*e^4*x - 32*a*c^4*d^3*e*x + 8*b^4*c*d*e^3*x - 5*a^2*b*c*e^4*(b^2 - 4*a*c)^
(1/2) - 16*a*c^3*d^3*e*(b^2 - 4*a*c)^(1/2) - 24*a^2*b*c^2*d*e^3 - 16*a^2*b*c^2*e^4*x + 32*a^2*c^3*d*e^3*x + 8*
b^2*c^3*d^3*e*x + 16*a^2*c^2*d*e^3*(b^2 - 4*a*c)^(1/2) - 2*b^2*c^2*d^3*e*(b^2 - 4*a*c)^(1/2) + b^3*c*d^2*e^2*(
b^2 - 4*a*c)^(1/2) + 6*a^2*c^2*e^4*x*(b^2 - 4*a*c)^(1/2) - 14*a*b^2*c^2*d^2*e^2 - 12*b^3*c^2*d^2*e^2*x + 14*a*
b*c^2*d^2*e^2*(b^2 - 4*a*c)^(1/2) - 20*a*c^3*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) + 14*b^2*c^2*d^2*e^2*x*(b^2 - 4*a*c
)^(1/2) - 10*a*b^2*c*d*e^3*(b^2 - 4*a*c)^(1/2) - 8*a*b^2*c*e^4*x*(b^2 - 4*a*c)^(1/2) - 12*b*c^3*d^3*e*x*(b^2 -
4*a*c)^(1/2) - 8*b^3*c*d*e^3*x*(b^2 - 4*a*c)^(1/2) + 48*a*b*c^3*d^2*e^2*x - 40*a*b^2*c^2*d*e^3*x + 20*a*b*c^2
*d*e^3*x*(b^2 - 4*a*c)^(1/2))*(e^2*(a*c + (b*(b^2 - 4*a*c)^(1/2))/2 - b^2/2) + e*(b*c*d - c*d*(b^2 - 4*a*c)^(1
/2)) - c^2*d^2))/(a^2*e^4 + c^2*d^4 + b^2*d^2*e^2 - 2*a*b*d*e^3 - 2*b*c*d^3*e + 2*a*c*d^2*e^2) + (log(2*a*b^4*
e^4 + 12*a*c^4*d^4 + 2*b^5*e^4*x + 12*a^3*c^2*e^4 - 3*b^2*c^3*d^4 + 2*b^4*e^4*x*(b^2 - 4*a*c)^(1/2) + 6*c^4*d^
4*x*(b^2 - 4*a*c)^(1/2) - 11*a^2*b^2*c*e^4 + 2*b^3*c^2*d^3*e - b^4*c*d^2*e^2 - 40*a^2*c^3*d^2*e^2 + 2*a*b^3*e^
4*(b^2 - 4*a*c)^(1/2) + 3*b*c^3*d^4*(b^2 - 4*a*c)^(1/2) - 8*a*b*c^3*d^3*e - 6*a*b^3*c*d*e^3 - 12*a*b^3*c*e^4*x
+ 32*a*c^4*d^3*e*x - 8*b^4*c*d*e^3*x - 5*a^2*b*c*e^4*(b^2 - 4*a*c)^(1/2) - 16*a*c^3*d^3*e*(b^2 - 4*a*c)^(1/2)
+ 24*a^2*b*c^2*d*e^3 + 16*a^2*b*c^2*e^4*x - 32*a^2*c^3*d*e^3*x - 8*b^2*c^3*d^3*e*x + 16*a^2*c^2*d*e^3*(b^2 -
4*a*c)^(1/2) - 2*b^2*c^2*d^3*e*(b^2 - 4*a*c)^(1/2) + b^3*c*d^2*e^2*(b^2 - 4*a*c)^(1/2) + 6*a^2*c^2*e^4*x*(b^2
- 4*a*c)^(1/2) + 14*a*b^2*c^2*d^2*e^2 + 12*b^3*c^2*d^2*e^2*x + 14*a*b*c^2*d^2*e^2*(b^2 - 4*a*c)^(1/2) - 20*a*c
^3*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) + 14*b^2*c^2*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) - 10*a*b^2*c*d*e^3*(b^2 - 4*a*c)^(
1/2) - 8*a*b^2*c*e^4*x*(b^2 - 4*a*c)^(1/2) - 12*b*c^3*d^3*e*x*(b^2 - 4*a*c)^(1/2) - 8*b^3*c*d*e^3*x*(b^2 - 4*a
*c)^(1/2) - 48*a*b*c^3*d^2*e^2*x + 40*a*b^2*c^2*d*e^3*x + 20*a*b*c^2*d*e^3*x*(b^2 - 4*a*c)^(1/2))*(e^2*((b*(b^
2 - 4*a*c)^(1/2))/2 - a*c + b^2/2) - e*(b*c*d + c*d*(b^2 - 4*a*c)^(1/2)) + c^2*d^2))/(a^2*e^4 + c^2*d^4 + b^2*
d^2*e^2 - 2*a*b*d*e^3 - 2*b*c*d^3*e + 2*a*c*d^2*e^2) - (b*e - 2*c*d)/((d + e*x)*(a*e^2 + c*d^2 - b*d*e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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