3.21.57 \(\int \frac {(d+e x)^2 (f+g x)}{(c f^2-b f g-b g^2 x-c g^2 x^2)^2} \, dx\)

Optimal. Leaf size=143 \[ \frac {(-b e g+c d g+c e f)^2}{c^2 g^3 (2 c f-b g) (-b g+c f-c g x)}+\frac {(-b e g-c d g+3 c e f) (-b e g+c d g+c e f) \log (-b g+c f-c g x)}{c^2 g^3 (2 c f-b g)^2}+\frac {(e f-d g)^2 \log (f+g x)}{g^3 (2 c f-b g)^2} \]

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Rubi [A]  time = 0.21, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {798, 88} \begin {gather*} \frac {(-b e g+c d g+c e f)^2}{c^2 g^3 (2 c f-b g) (-b g+c f-c g x)}+\frac {(-b e g-c d g+3 c e f) (-b e g+c d g+c e f) \log (-b g+c f-c g x)}{c^2 g^3 (2 c f-b g)^2}+\frac {(e f-d g)^2 \log (f+g x)}{g^3 (2 c f-b g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*e*f + c*d*g - b*e*g)^2/(c^2*g^3*(2*c*f - b*g)*(c*f - b*g - c*g*x)) + ((e*f - d*g)^2*Log[f + g*x])/(g^3*(2*c
*f - b*g)^2) + ((3*c*e*f - c*d*g - b*e*g)*(c*e*f + c*d*g - b*e*g)*Log[c*f - b*g - c*g*x])/(c^2*g^3*(2*c*f - b*
g)^2)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 798

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[
(d + e*x)^m*(f + g*x)^(p + 1)*(a/f + (c*x)/g)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c,
0] && EqQ[c*f^2 - b*f*g + a*g^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 153, normalized size = 1.07 \begin {gather*} \frac {\frac {\left (b^2 e^2 g^2-4 b c e^2 f g+c^2 \left (-d^2 g^2+2 d e f g+3 e^2 f^2\right )\right ) \log (-b g+c f-c g x)}{c^2 (b g-2 c f)^2}+\frac {(-b e g+c d g+c e f)^2}{c^2 (2 c f-b g) (c (f-g x)-b g)}+\frac {(e f-d g)^2 \log (f+g x)}{(b g-2 c f)^2}}{g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 448, normalized size = 3.13 \begin {gather*} \frac {2 \, c^{3} e^{2} f^{3} + {\left (4 \, c^{3} d e - 5 \, b c^{2} e^{2}\right )} f^{2} g + 2 \, {\left (c^{3} d^{2} - 3 \, b c^{2} d e + 2 \, b^{2} c e^{2}\right )} f g^{2} - {\left (b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} g^{3} + {\left (3 \, c^{3} e^{2} f^{3} + {\left (2 \, c^{3} d e - 7 \, b c^{2} e^{2}\right )} f^{2} g - {\left (c^{3} d^{2} + 2 \, b c^{2} d e - 5 \, b^{2} c e^{2}\right )} f g^{2} + {\left (b c^{2} d^{2} - b^{3} e^{2}\right )} g^{3} - {\left (3 \, c^{3} e^{2} f^{2} g + 2 \, {\left (c^{3} d e - 2 \, b c^{2} e^{2}\right )} f g^{2} - {\left (c^{3} d^{2} - b^{2} c e^{2}\right )} g^{3}\right )} x\right )} \log \left (c g x - c f + b g\right ) + {\left (c^{3} e^{2} f^{3} - b c^{2} d^{2} g^{3} - {\left (2 \, c^{3} d e + b c^{2} e^{2}\right )} f^{2} g + {\left (c^{3} d^{2} + 2 \, b c^{2} d e\right )} f g^{2} - {\left (c^{3} e^{2} f^{2} g - 2 \, c^{3} d e f g^{2} + c^{3} d^{2} g^{3}\right )} x\right )} \log \left (g x + f\right )}{4 \, c^{5} f^{3} g^{3} - 8 \, b c^{4} f^{2} g^{4} + 5 \, b^{2} c^{3} f g^{5} - b^{3} c^{2} g^{6} - {\left (4 \, c^{5} f^{2} g^{4} - 4 \, b c^{4} f g^{5} + b^{2} c^{3} g^{6}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 297, normalized size = 2.08 \begin {gather*} -\frac {{\left (c^{2} d^{2} g^{2} - 2 \, c^{2} d f g e - 3 \, c^{2} f^{2} e^{2} + 4 \, b c f g e^{2} - b^{2} g^{2} e^{2}\right )} \log \left ({\left | c g x - c f + b g \right |}\right )}{4 \, c^{4} f^{2} g^{3} - 4 \, b c^{3} f g^{4} + b^{2} c^{2} g^{5}} + \frac {{\left (d^{2} g^{2} - 2 \, d f g e + f^{2} e^{2}\right )} \log \left ({\left | g x + f \right |}\right )}{4 \, c^{2} f^{2} g^{3} - 4 \, b c f g^{4} + b^{2} g^{5}} - \frac {2 \, c^{3} d^{2} f g^{2} - b c^{2} d^{2} g^{3} + 4 \, c^{3} d f^{2} g e - 6 \, b c^{2} d f g^{2} e + 2 \, b^{2} c d g^{3} e + 2 \, c^{3} f^{3} e^{2} - 5 \, b c^{2} f^{2} g e^{2} + 4 \, b^{2} c f g^{2} e^{2} - b^{3} g^{3} e^{2}}{{\left (c g x - c f + b g\right )} {\left (2 \, c f - b g\right )}^{2} c^{2} g^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 449, normalized size = 3.14 \begin {gather*} \frac {b^{2} e^{2} \ln \left (c g x +b g -c f \right )}{\left (b g -2 c f \right )^{2} c^{2} g}-\frac {4 b \,e^{2} f \ln \left (c g x +b g -c f \right )}{\left (b g -2 c f \right )^{2} c \,g^{2}}+\frac {d^{2} \ln \left (g x +f \right )}{\left (b g -2 c f \right )^{2} g}-\frac {d^{2} \ln \left (c g x +b g -c f \right )}{\left (b g -2 c f \right )^{2} g}-\frac {2 d e f \ln \left (g x +f \right )}{\left (b g -2 c f \right )^{2} g^{2}}+\frac {2 d e f \ln \left (c g x +b g -c f \right )}{\left (b g -2 c f \right )^{2} g^{2}}+\frac {e^{2} f^{2} \ln \left (g x +f \right )}{\left (b g -2 c f \right )^{2} g^{3}}+\frac {3 e^{2} f^{2} \ln \left (c g x +b g -c f \right )}{\left (b g -2 c f \right )^{2} g^{3}}+\frac {b^{2} e^{2}}{\left (b g -2 c f \right ) \left (c g x +b g -c f \right ) c^{2} g}-\frac {2 b d e}{\left (b g -2 c f \right ) \left (c g x +b g -c f \right ) c g}-\frac {2 b \,e^{2} f}{\left (b g -2 c f \right ) \left (c g x +b g -c f \right ) c \,g^{2}}+\frac {d^{2}}{\left (b g -2 c f \right ) \left (c g x +b g -c f \right ) g}+\frac {2 d e f}{\left (b g -2 c f \right ) \left (c g x +b g -c f \right ) g^{2}}+\frac {e^{2} f^{2}}{\left (b g -2 c f \right ) \left (c g x +b g -c f \right ) g^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(g*x+f)/(-c*g^2*x^2-b*g^2*x-b*f*g+c*f^2)^2,x)

[Out]

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

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maxima [A]  time = 0.65, size = 268, normalized size = 1.87 \begin {gather*} \frac {{\left (3 \, c^{2} e^{2} f^{2} + 2 \, {\left (c^{2} d e - 2 \, b c e^{2}\right )} f g - {\left (c^{2} d^{2} - b^{2} e^{2}\right )} g^{2}\right )} \log \left (c g x - c f + b g\right )}{4 \, c^{4} f^{2} g^{3} - 4 \, b c^{3} f g^{4} + b^{2} c^{2} g^{5}} + \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (g x + f\right )}{4 \, c^{2} f^{2} g^{3} - 4 \, b c f g^{4} + b^{2} g^{5}} + \frac {c^{2} e^{2} f^{2} + 2 \, {\left (c^{2} d e - b c e^{2}\right )} f g + {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} g^{2}}{2 \, c^{4} f^{2} g^{3} - 3 \, b c^{3} f g^{4} + b^{2} c^{2} g^{5} - {\left (2 \, c^{4} f g^{4} - b c^{3} g^{5}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.73, size = 224, normalized size = 1.57 \begin {gather*} \frac {\ln \left (f+g\,x\right )\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )}{b^2\,g^5-4\,b\,c\,f\,g^4+4\,c^2\,f^2\,g^3}+\frac {b^2\,e^2\,g^2-2\,b\,c\,d\,e\,g^2-2\,b\,c\,e^2\,f\,g+c^2\,d^2\,g^2+2\,c^2\,d\,e\,f\,g+c^2\,e^2\,f^2}{c^2\,g^3\,\left (b\,g-2\,c\,f\right )\,\left (b\,g-c\,f+c\,g\,x\right )}+\frac {\ln \left (b\,g-c\,f+c\,g\,x\right )\,\left (c^2\,\left (-d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )+b^2\,e^2\,g^2-4\,b\,c\,e^2\,f\,g\right )}{c^2\,g^3\,{\left (b\,g-2\,c\,f\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^2)/(c*g^2*x^2 - c*f^2 + b*f*g + b*g^2*x)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(g*x+f)/(-c*g**2*x**2-b*g**2*x-b*f*g+c*f**2)**2,x)

[Out]

Timed out

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