3.2124 \(\int \frac {(a+b x+c x^2)^2}{(d+e x)^2} \, dx\)

Optimal. Leaf size=131 \[ \frac {x \left (-2 c e (2 b d-a e)+b^2 e^2+3 c^2 d^2\right )}{e^4}-\frac {\left (a e^2-b d e+c d^2\right )^2}{e^5 (d+e x)}-\frac {2 (2 c d-b e) \log (d+e x) \left (a e^2-b d e+c d^2\right )}{e^5}-\frac {c x^2 (c d-b e)}{e^3}+\frac {c^2 x^3}{3 e^2} \]

[Out]

(3*c^2*d^2+b^2*e^2-2*c*e*(-a*e+2*b*d))*x/e^4-c*(-b*e+c*d)*x^2/e^3+1/3*c^2*x^3/e^2-(a*e^2-b*d*e+c*d^2)^2/e^5/(e
*x+d)-2*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)*ln(e*x+d)/e^5

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Rubi [A]  time = 0.14, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {698} \[ \frac {x \left (-2 c e (2 b d-a e)+b^2 e^2+3 c^2 d^2\right )}{e^4}-\frac {\left (a e^2-b d e+c d^2\right )^2}{e^5 (d+e x)}-\frac {2 (2 c d-b e) \log (d+e x) \left (a e^2-b d e+c d^2\right )}{e^5}-\frac {c x^2 (c d-b e)}{e^3}+\frac {c^2 x^3}{3 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*c^2*d^2 + b^2*e^2 - 2*c*e*(2*b*d - a*e))*x)/e^4 - (c*(c*d - b*e)*x^2)/e^3 + (c^2*x^3)/(3*e^2) - (c*d^2 - b
*d*e + a*e^2)^2/(e^5*(d + e*x)) - (2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)*Log[d + e*x])/e^5

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 127, normalized size = 0.97 \[ \frac {3 e x \left (2 c e (a e-2 b d)+b^2 e^2+3 c^2 d^2\right )-\frac {3 \left (e (a e-b d)+c d^2\right )^2}{d+e x}-6 (2 c d-b e) \log (d+e x) \left (e (a e-b d)+c d^2\right )+3 c e^2 x^2 (b e-c d)+c^2 e^3 x^3}{3 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*e*(3*c^2*d^2 + b^2*e^2 + 2*c*e*(-2*b*d + a*e))*x + 3*c*e^2*(-(c*d) + b*e)*x^2 + c^2*e^3*x^3 - (3*(c*d^2 + e
*(-(b*d) + a*e))^2)/(d + e*x) - 6*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*Log[d + e*x])/(3*e^5)

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fricas [B]  time = 0.84, size = 259, normalized size = 1.98 \[ \frac {c^{2} e^{4} x^{4} - 3 \, c^{2} d^{4} + 6 \, b c d^{3} e + 6 \, a b d e^{3} - 3 \, a^{2} e^{4} - 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - {\left (2 \, c^{2} d e^{3} - 3 \, b c e^{4}\right )} x^{3} + 3 \, {\left (2 \, c^{2} d^{2} e^{2} - 3 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 3 \, {\left (3 \, c^{2} d^{3} e - 4 \, b c d^{2} e^{2} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x - 6 \, {\left (2 \, c^{2} d^{4} - 3 \, b c d^{3} e - a b d e^{3} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + {\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} - a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{6} x + d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 249, normalized size = 1.90 \[ \frac {1}{3} \, {\left (c^{2} - \frac {3 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} e^{\left (-1\right )}}{x e + d} + \frac {3 \, {\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + b^{2} e^{4} + 2 \, a c e^{4}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}}\right )} {\left (x e + d\right )}^{3} e^{\left (-5\right )} + 2 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2} + 2 \, a c d e^{2} - a b e^{3}\right )} e^{\left (-5\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - {\left (\frac {c^{2} d^{4} e^{3}}{x e + d} - \frac {2 \, b c d^{3} e^{4}}{x e + d} + \frac {b^{2} d^{2} e^{5}}{x e + d} + \frac {2 \, a c d^{2} e^{5}}{x e + d} - \frac {2 \, a b d e^{6}}{x e + d} + \frac {a^{2} e^{7}}{x e + d}\right )} e^{\left (-8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c^2 - 3*(2*c^2*d*e - b*c*e^2)*e^(-1)/(x*e + d) + 3*(6*c^2*d^2*e^2 - 6*b*c*d*e^3 + b^2*e^4 + 2*a*c*e^4)*e^
(-2)/(x*e + d)^2)*(x*e + d)^3*e^(-5) + 2*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2 + 2*a*c*d*e^2 - a*b*e^3)*e^(-5)*
log(abs(x*e + d)*e^(-1)/(x*e + d)^2) - (c^2*d^4*e^3/(x*e + d) - 2*b*c*d^3*e^4/(x*e + d) + b^2*d^2*e^5/(x*e + d
) + 2*a*c*d^2*e^5/(x*e + d) - 2*a*b*d*e^6/(x*e + d) + a^2*e^7/(x*e + d))*e^(-8)

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maple [A]  time = 0.06, size = 246, normalized size = 1.88 \[ \frac {c^{2} x^{3}}{3 e^{2}}+\frac {b c \,x^{2}}{e^{2}}-\frac {c^{2} d \,x^{2}}{e^{3}}-\frac {a^{2}}{\left (e x +d \right ) e}+\frac {2 a b d}{\left (e x +d \right ) e^{2}}+\frac {2 a b \ln \left (e x +d \right )}{e^{2}}-\frac {2 a c \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {4 a c d \ln \left (e x +d \right )}{e^{3}}+\frac {2 a c x}{e^{2}}-\frac {b^{2} d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 b^{2} d \ln \left (e x +d \right )}{e^{3}}+\frac {b^{2} x}{e^{2}}+\frac {2 b c \,d^{3}}{\left (e x +d \right ) e^{4}}+\frac {6 b c \,d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {4 b c d x}{e^{3}}-\frac {c^{2} d^{4}}{\left (e x +d \right ) e^{5}}-\frac {4 c^{2} d^{3} \ln \left (e x +d \right )}{e^{5}}+\frac {3 c^{2} d^{2} x}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*c^2/e^2*x^3+1/e^2*x^2*b*c-c^2*d/e^3*x^2+2/e^2*a*c*x+b^2/e^2*x-4/e^3*b*c*d*x+3/e^4*c^2*d^2*x-1/e/(e*x+d)*a^
2+2/e^2/(e*x+d)*a*b*d-2/e^3/(e*x+d)*a*c*d^2-1/e^3/(e*x+d)*b^2*d^2+2/e^4/(e*x+d)*b*c*d^3-1/e^5/(e*x+d)*c^2*d^4+
2/e^2*ln(e*x+d)*a*b-4/e^3*ln(e*x+d)*a*c*d-2/e^3*ln(e*x+d)*b^2*d+6/e^4*ln(e*x+d)*b*c*d^2-4/e^5*ln(e*x+d)*c^2*d^
3

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maxima [A]  time = 1.08, size = 175, normalized size = 1.34 \[ -\frac {c^{2} d^{4} - 2 \, b c d^{3} e - 2 \, a b d e^{3} + a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2}}{e^{6} x + d e^{5}} + \frac {c^{2} e^{2} x^{3} - 3 \, {\left (c^{2} d e - b c e^{2}\right )} x^{2} + 3 \, {\left (3 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} x}{3 \, e^{4}} - \frac {2 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 203, normalized size = 1.55 \[ x\,\left (\frac {b^2+2\,a\,c}{e^2}+\frac {2\,d\,\left (\frac {2\,c^2\,d}{e^3}-\frac {2\,b\,c}{e^2}\right )}{e}-\frac {c^2\,d^2}{e^4}\right )-x^2\,\left (\frac {c^2\,d}{e^3}-\frac {b\,c}{e^2}\right )-\frac {a^2\,e^4-2\,a\,b\,d\,e^3+2\,a\,c\,d^2\,e^2+b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4}{e\,\left (x\,e^5+d\,e^4\right )}-\frac {\ln \left (d+e\,x\right )\,\left (2\,b^2\,d\,e^2-6\,b\,c\,d^2\,e-2\,a\,b\,e^3+4\,c^2\,d^3+4\,a\,c\,d\,e^2\right )}{e^5}+\frac {c^2\,x^3}{3\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((2*a*c + b^2)/e^2 + (2*d*((2*c^2*d)/e^3 - (2*b*c)/e^2))/e - (c^2*d^2)/e^4) - x^2*((c^2*d)/e^3 - (b*c)/e^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)/(e*(d*e^4 + e^5*x)) - (log(d +
e*x)*(4*c^2*d^3 + 2*b^2*d*e^2 - 2*a*b*e^3 + 4*a*c*d*e^2 - 6*b*c*d^2*e))/e^5 + (c^2*x^3)/(3*e^2)

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sympy [A]  time = 0.99, size = 170, normalized size = 1.30 \[ \frac {c^{2} x^{3}}{3 e^{2}} + x^{2} \left (\frac {b c}{e^{2}} - \frac {c^{2} d}{e^{3}}\right ) + x \left (\frac {2 a c}{e^{2}} + \frac {b^{2}}{e^{2}} - \frac {4 b c d}{e^{3}} + \frac {3 c^{2} d^{2}}{e^{4}}\right ) + \frac {- a^{2} e^{4} + 2 a b d e^{3} - 2 a c d^{2} e^{2} - b^{2} d^{2} e^{2} + 2 b c d^{3} e - c^{2} d^{4}}{d e^{5} + e^{6} x} + \frac {2 \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (d + e x \right )}}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*x**3/(3*e**2) + x**2*(b*c/e**2 - c**2*d/e**3) + x*(2*a*c/e**2 + b**2/e**2 - 4*b*c*d/e**3 + 3*c**2*d**2/e*
*4) + (-a**2*e**4 + 2*a*b*d*e**3 - 2*a*c*d**2*e**2 - b**2*d**2*e**2 + 2*b*c*d**3*e - c**2*d**4)/(d*e**5 + e**6
*x) + 2*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2)*log(d + e*x)/e**5

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