3.1526 \(\int \frac {(b+2 c x) (d+e x)^2}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=114 \[ \frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {e \sqrt {b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2}+e x \left (4 d-\frac {b e}{c}\right )+e^2 x^2 \]

[Out]

e*(4*d-b*e/c)*x+e^2*x^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)/c^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)/c^2

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Rubi [A]  time = 0.13, antiderivative size = 114, 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} \[ \frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {e \sqrt {b^2-4 a c} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2}+e x \left (4 d-\frac {b e}{c}\right )+e^2 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.09, size = 111, normalized size = 0.97 \[ \frac {\left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \log (a+x (b+c x))+2 e \sqrt {4 a c-b^2} (b e-2 c d) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )+2 c e x (-b e+4 c d+c e x)}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 282, normalized size = 2.47 \[ \left [\frac {2 \, c^{2} e^{2} x^{2} - {\left (2 \, c d e - b e^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, {\left (4 \, c^{2} d e - b c e^{2}\right )} x + {\left (2 \, c^{2} d^{2} - 2 \, b c d e + {\left (b^{2} - 2 \, a c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}}, \frac {2 \, c^{2} e^{2} x^{2} - 2 \, {\left (2 \, c d e - b e^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (4 \, c^{2} d e - b c e^{2}\right )} x + {\left (2 \, c^{2} d^{2} - 2 \, b c d e + {\left (b^{2} - 2 \, a c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}}\right ] \]

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

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giac [A]  time = 0.18, size = 144, normalized size = 1.26 \[ \frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} + \frac {{\left (2 \, b^{2} c d e - 8 \, a c^{2} d e - b^{3} e^{2} + 4 \, a b c e^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{2}} + \frac {c^{2} x^{2} e^{2} + 4 \, c^{2} d x e - b c x e^{2}}{c^{2}} \]

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]

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

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maple [B]  time = 0.05, size = 264, normalized size = 2.32 \[ \frac {4 a b \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {8 a d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {b^{3} e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}+\frac {2 b^{2} d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+e^{2} x^{2}-\frac {a \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{c}+\frac {b^{2} e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {b d e \ln \left (c \,x^{2}+b x +a \right )}{c}-\frac {b \,e^{2} x}{c}+d^{2} \ln \left (c \,x^{2}+b x +a \right )+4 d e x \]

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]

e^2*x^2-1/c*e^2*b*x+4*d*x*e-1/c*ln(c*x^2+b*x+a)*a*e^2+1/2/c^2*ln(c*x^2+b*x+a)*b^2*e^2-1/c*ln(c*x^2+b*x+a)*b*d*
e+ln(c*x^2+b*x+a)*d^2+4/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b*e^2-8/(4*a*c-b^2)^(1/2)*ar
ctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*d*e-1/c^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/
c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2*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 = 0.26, size = 230, normalized size = 2.02 \[ \ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {\frac {b^2\,e^2}{2}-c\,\left (a\,e^2+b\,d\,e+d\,e\,\sqrt {b^2-4\,a\,c}\right )+\frac {b\,e^2\,\sqrt {b^2-4\,a\,c}}{2}}{c^2}+d^2\right )+x\,\left (\frac {b\,e^2+4\,c\,d\,e}{c}-\frac {2\,b\,e^2}{c}\right )-\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (\frac {c\,\left (a\,e^2+b\,d\,e-d\,e\,\sqrt {b^2-4\,a\,c}\right )-\frac {b^2\,e^2}{2}+\frac {b\,e^2\,\sqrt {b^2-4\,a\,c}}{2}}{c^2}-d^2\right )+e^2\,x^2 \]

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

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sympy [B]  time = 1.80, size = 335, normalized size = 2.94 \[ e^{2} x^{2} + x \left (- \frac {b e^{2}}{c} + 4 d e\right ) + \left (- \frac {e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right ) \log {\left (x + \frac {a e^{2} - c d^{2} + c \left (- \frac {e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right )}{b e^{2} - 2 c d e} \right )} + \left (\frac {e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right ) \log {\left (x + \frac {a e^{2} - c d^{2} + c \left (\frac {e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}}{2 c^{2}}\right )}{b e^{2} - 2 c d e} \right )} \]

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)

[Out]

e**2*x**2 + x*(-b*e**2/c + 4*d*e) + (-e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c**2) - (2*a*c*e**2 - b**2*e**2 +
2*b*c*d*e - 2*c**2*d**2)/(2*c**2))*log(x + (a*e**2 - c*d**2 + c*(-e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c**2
) - (2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**2)))/(b*e**2 - 2*c*d*e)) + (e*sqrt(-4*a*c + b**2)
*(b*e - 2*c*d)/(2*c**2) - (2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**2))*log(x + (a*e**2 - c*d**
2 + c*(e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)/(2*c**2) - (2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c*
*2)))/(b*e**2 - 2*c*d*e))

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