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

Optimal. Leaf size=102 \[ \frac {\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^4}+\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^4 (d+e x)}-\frac {c x (4 c d-3 b e)}{e^3}+\frac {c^2 x^2}{e^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {771} \begin {gather*} \frac {\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^4}+\frac {(2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^4 (d+e x)}-\frac {c x (4 c d-3 b e)}{e^3}+\frac {c^2 x^2}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(4*c*d - 3*b*e)*x)/e^3) + (c^2*x^2)/e^2 + ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^4*(d + e*x)) + ((6*c
^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))*Log[d + e*x])/e^4

Rule 771

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

Rubi steps

\begin {align*} \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^2} \, dx &=\int \left (-\frac {c (4 c d-3 b e)}{e^3}+\frac {2 c^2 x}{e^2}+\frac {(-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^3 (d+e x)^2}+\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^3 (d+e x)}\right ) \, dx\\ &=-\frac {c (4 c d-3 b e) x}{e^3}+\frac {c^2 x^2}{e^2}+\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)}+\frac {\left (6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)\right ) \log (d+e x)}{e^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 97, normalized size = 0.95 \begin {gather*} \frac {\log (d+e x) \left (2 c e (a e-3 b d)+b^2 e^2+6 c^2 d^2\right )+\frac {(2 c d-b e) \left (e (a e-b d)+c d^2\right )}{d+e x}-c e x (4 c d-3 b e)+c^2 e^2 x^2}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(c*e*(4*c*d - 3*b*e)*x) + c^2*e^2*x^2 + ((2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e)))/(d + e*x) + (6*c^2*d^2 +
b^2*e^2 + 2*c*e*(-3*b*d + a*e))*Log[d + e*x])/e^4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 172, normalized size = 1.69 \begin {gather*} \frac {c^{2} e^{3} x^{3} + 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} - 3 \, {\left (c^{2} d e^{2} - b c e^{3}\right )} x^{2} - {\left (4 \, c^{2} d^{2} e - 3 \, b c d e^{2}\right )} x + {\left (6 \, c^{2} d^{3} - 6 \, b c d^{2} e + {\left (b^{2} + 2 \, a c\right )} d e^{2} + {\left (6 \, c^{2} d^{2} e - 6 \, b c d e^{2} + {\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{e^{5} x + d e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 177, normalized size = 1.74 \begin {gather*} {\left (c^{2} - \frac {3 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )} {\left (x e + d\right )}^{2} e^{\left (-4\right )} - {\left (6 \, c^{2} d^{2} - 6 \, b c d e + b^{2} e^{2} + 2 \, a c e^{2}\right )} e^{\left (-4\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (\frac {2 \, c^{2} d^{3} e^{2}}{x e + d} - \frac {3 \, b c d^{2} e^{3}}{x e + d} + \frac {b^{2} d e^{4}}{x e + d} + \frac {2 \, a c d e^{4}}{x e + d} - \frac {a b e^{5}}{x e + d}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 166, normalized size = 1.63 \begin {gather*} \frac {c^{2} x^{2}}{e^{2}}-\frac {a b}{\left (e x +d \right ) e}+\frac {2 a c d}{\left (e x +d \right ) e^{2}}+\frac {2 a c \ln \left (e x +d \right )}{e^{2}}+\frac {b^{2} d}{\left (e x +d \right ) e^{2}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{2}}-\frac {3 b c \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {6 b c d \ln \left (e x +d \right )}{e^{3}}+\frac {3 b c x}{e^{2}}+\frac {2 c^{2} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {6 c^{2} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {4 c^{2} d x}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 117, normalized size = 1.15 \begin {gather*} \frac {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}}{e^{5} x + d e^{4}} + \frac {c^{2} e x^{2} - {\left (4 \, c^{2} d - 3 \, b c e\right )} x}{e^{3}} + \frac {{\left (6 \, c^{2} d^{2} - 6 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 127, normalized size = 1.25 \begin {gather*} \frac {b^2\,d\,e^2-3\,b\,c\,d^2\,e-a\,b\,e^3+2\,c^2\,d^3+2\,a\,c\,d\,e^2}{e\,\left (x\,e^4+d\,e^3\right )}-x\,\left (\frac {4\,c^2\,d}{e^3}-\frac {3\,b\,c}{e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (b^2\,e^2-6\,b\,c\,d\,e+6\,c^2\,d^2+2\,a\,c\,e^2\right )}{e^4}+\frac {c^2\,x^2}{e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.67, size = 126, normalized size = 1.24 \begin {gather*} \frac {c^{2} x^{2}}{e^{2}} + x \left (\frac {3 b c}{e^{2}} - \frac {4 c^{2} d}{e^{3}}\right ) + \frac {- a b e^{3} + 2 a c d e^{2} + b^{2} d e^{2} - 3 b c d^{2} e + 2 c^{2} d^{3}}{d e^{4} + e^{5} x} + \frac {\left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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