3.1529 \(\int \frac {b+2 c x}{(d+e x) (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=130 \[ \frac {e \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac {(2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {(2 c d-b e) \log (d+e x)}{a e^2-b d e+c d^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 130, 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} \[ \frac {e \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac {(2 c d-b e) \log \left (a+b x+c x^2\right )}{2 \left (a e^2-b d e+c d^2\right )}-\frac {(2 c d-b e) \log (d+e x)}{a e^2-b d e+c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.12, size = 116, normalized size = 0.89 \[ \frac {\sqrt {4 a c-b^2} (2 c d-b e) (2 \log (d+e x)-\log (a+x (b+c x)))+2 e \left (b^2-4 a c\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{2 \sqrt {4 a c-b^2} \left (e (b d-a e)-c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 229, normalized size = 1.76 \[ \left [\frac {\sqrt {b^{2} - 4 \, a c} e \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (2 \, c d - b e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c d - b e\right )} \log \left (e x + d\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} e \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (2 \, c d - b e\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (2 \, c d - b e\right )} \log \left (e x + d\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 149, normalized size = 1.15 \[ \frac {{\left (2 \, c d - b e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} - b d e + a e^{2}\right )}} - \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} - \frac {{\left (b^{2} e - 4 \, a c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 233, normalized size = 1.79 \[ \frac {4 a c e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}-\frac {b^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {b e \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}-\frac {b e \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}-\frac {2 c d \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}+\frac {c d \ln \left (c \,x^{2}+b x +a \right )}{a \,e^{2}-b d e +c \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(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 = 4.38, size = 515, normalized size = 3.96 \[ \frac {\ln \left (2\,b\,c^2\,e-\frac {\left (c\,d-\frac {b\,e}{2}+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (2\,a\,c^2\,e^2-b^2\,c\,e^2+b\,c^2\,d\,e-c^2\,e\,x\,\left (b\,e-2\,c\,d\right )+\frac {c\,e\,\left (c\,d-\frac {b\,e}{2}+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{c\,d^2-b\,d\,e+a\,e^2}\right )}{c\,d^2-b\,d\,e+a\,e^2}+4\,c^3\,e\,x\right )\,\left (c\,d-e\,\left (\frac {b}{2}-\frac {\sqrt {b^2-4\,a\,c}}{2}\right )\right )}{c\,d^2-b\,d\,e+a\,e^2}+\frac {\ln \left (2\,b\,c^2\,e-\frac {\left (\frac {b\,e}{2}-c\,d+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (b^2\,c\,e^2-2\,a\,c^2\,e^2-b\,c^2\,d\,e+c^2\,e\,x\,\left (b\,e-2\,c\,d\right )+\frac {c\,e\,\left (\frac {b\,e}{2}-c\,d+\frac {e\,\sqrt {b^2-4\,a\,c}}{2}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{c\,d^2-b\,d\,e+a\,e^2}\right )}{c\,d^2-b\,d\,e+a\,e^2}+4\,c^3\,e\,x\right )\,\left (c\,d-e\,\left (\frac {b}{2}+\frac {\sqrt {b^2-4\,a\,c}}{2}\right )\right )}{c\,d^2-b\,d\,e+a\,e^2}+\frac {\ln \left (d+e\,x\right )\,\left (b\,e-2\,c\,d\right )}{c\,d^2-b\,d\,e+a\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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