3.2125 \(\int \frac {(a+b x+c x^2)^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=138 \[ \frac {\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^5}-\frac {\left (a e^2-b d e+c d^2\right )^2}{2 e^5 (d+e x)^2}+\frac {2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)}-\frac {c x (3 c d-2 b e)}{e^4}+\frac {c^2 x^2}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 138, 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} \[ \frac {\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^5}-\frac {\left (a e^2-b d e+c d^2\right )^2}{2 e^5 (d+e x)^2}+\frac {2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)}-\frac {c x (3 c d-2 b e)}{e^4}+\frac {c^2 x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^2/(d + e*x)^3,x]

[Out]

-((c*(3*c*d - 2*b*e)*x)/e^4) + (c^2*x^2)/(2*e^3) - (c*d^2 - b*d*e + a*e^2)^2/(2*e^5*(d + e*x)^2) + (2*(2*c*d -
b*e)*(c*d^2 - b*d*e + a*e^2))/(e^5*(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^
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)^3} \, dx &=\int \left (-\frac {c (3 c d-2 b e)}{e^4}+\frac {c^2 x}{e^3}+\frac {\left (c d^2-b d e+a e^2\right )^2}{e^4 (d+e x)^3}+\frac {2 (-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)^2}+\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 (d+e x)}\right ) \, dx\\ &=-\frac {c (3 c d-2 b e) x}{e^4}+\frac {c^2 x^2}{2 e^3}-\frac {\left (c d^2-b d e+a e^2\right )^2}{2 e^5 (d+e x)^2}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^5 (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^5}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 176, normalized size = 1.28 \[ \frac {2 (d+e x)^2 \log (d+e x) \left (2 c e (a e-3 b d)+b^2 e^2+6 c^2 d^2\right )+2 c e \left (a d e (3 d+4 e x)+b \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+e^2 (b d-a e) (a e+3 b d+4 b e x)+c^2 \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )}{2 e^5 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^2/(d + e*x)^3,x]

[Out]

(e^2*(b*d - a*e)*(3*b*d + a*e + 4*b*e*x) + c^2*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) +
2*c*e*(a*d*e*(3*d + 4*e*x) + b*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + 2*(6*c^2*d^2 + b^2*e^2 + 2*c*
e*(-3*b*d + a*e))*(d + e*x)^2*Log[d + e*x])/(2*e^5*(d + e*x)^2)

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fricas [B]  time = 0.64, size = 286, normalized size = 2.07 \[ \frac {c^{2} e^{4} x^{4} + 7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - 4 \, {\left (c^{2} d e^{3} - b c e^{4}\right )} x^{3} - {\left (11 \, c^{2} d^{2} e^{2} - 8 \, b c d e^{3}\right )} x^{2} + 2 \, {\left (c^{2} d^{3} e - 4 \, b c d^{2} e^{2} - 2 \, a b e^{4} + 2 \, {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x + 2 \, {\left (6 \, c^{2} d^{4} - 6 \, b c d^{3} e + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + {\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 2 \, {\left (6 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(c^2*e^4*x^4 + 7*c^2*d^4 - 10*b*c*d^3*e - 2*a*b*d*e^3 - a^2*e^4 + 3*(b^2 + 2*a*c)*d^2*e^2 - 4*(c^2*d*e^3 -
b*c*e^4)*x^3 - (11*c^2*d^2*e^2 - 8*b*c*d*e^3)*x^2 + 2*(c^2*d^3*e - 4*b*c*d^2*e^2 - 2*a*b*e^4 + 2*(b^2 + 2*a*c
)*d*e^3)*x + 2*(6*c^2*d^4 - 6*b*c*d^3*e + (b^2 + 2*a*c)*d^2*e^2 + (6*c^2*d^2*e^2 - 6*b*c*d*e^3 + (b^2 + 2*a*c)
*e^4)*x^2 + 2*(6*c^2*d^3*e - 6*b*c*d^2*e^2 + (b^2 + 2*a*c)*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*
e^5)

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giac [A]  time = 0.17, size = 176, normalized size = 1.28 \[ {\left (6 \, c^{2} d^{2} - 6 \, b c d e + b^{2} e^{2} + 2 \, a c e^{2}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (c^{2} x^{2} e^{3} - 6 \, c^{2} d x e^{2} + 4 \, b c x e^{3}\right )} e^{\left (-6\right )} + \frac {{\left (7 \, c^{2} d^{4} - 10 \, b c d^{3} e + 3 \, b^{2} d^{2} e^{2} + 6 \, a c d^{2} e^{2} - 2 \, a b d e^{3} - a^{2} e^{4} + 4 \, {\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} + b^{2} d e^{3} + 2 \, a c d e^{3} - a b e^{4}\right )} x\right )} e^{\left (-5\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^3,x, algorithm="giac")

[Out]

(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2 + 2*a*c*e^2)*e^(-5)*log(abs(x*e + d)) + 1/2*(c^2*x^2*e^3 - 6*c^2*d*x*e^2 + 4*
b*c*x*e^3)*e^(-6) + 1/2*(7*c^2*d^4 - 10*b*c*d^3*e + 3*b^2*d^2*e^2 + 6*a*c*d^2*e^2 - 2*a*b*d*e^3 - a^2*e^4 + 4*
(2*c^2*d^3*e - 3*b*c*d^2*e^2 + b^2*d*e^3 + 2*a*c*d*e^3 - a*b*e^4)*x)*e^(-5)/(x*e + d)^2

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maple [A]  time = 0.05, size = 266, normalized size = 1.93 \[ -\frac {a^{2}}{2 \left (e x +d \right )^{2} e}+\frac {a b d}{\left (e x +d \right )^{2} e^{2}}-\frac {a c \,d^{2}}{\left (e x +d \right )^{2} e^{3}}-\frac {b^{2} d^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {b c \,d^{3}}{\left (e x +d \right )^{2} e^{4}}-\frac {c^{2} d^{4}}{2 \left (e x +d \right )^{2} e^{5}}+\frac {c^{2} x^{2}}{2 e^{3}}-\frac {2 a b}{\left (e x +d \right ) e^{2}}+\frac {4 a c d}{\left (e x +d \right ) e^{3}}+\frac {2 a c \ln \left (e x +d \right )}{e^{3}}+\frac {2 b^{2} d}{\left (e x +d \right ) e^{3}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {6 b c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {6 b c d \ln \left (e x +d \right )}{e^{4}}+\frac {2 b c x}{e^{3}}+\frac {4 c^{2} d^{3}}{\left (e x +d \right ) e^{5}}+\frac {6 c^{2} d^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {3 c^{2} d x}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^2/(e*x+d)^3,x)

[Out]

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

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maxima [A]  time = 1.06, size = 185, normalized size = 1.34 \[ \frac {7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 4 \, {\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} - a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac {c^{2} e x^{2} - 2 \, {\left (3 \, c^{2} d - 2 \, b c e\right )} x}{2 \, e^{4}} + \frac {{\left (6 \, c^{2} d^{2} - 6 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*(7*c^2*d^4 - 10*b*c*d^3*e - 2*a*b*d*e^3 - a^2*e^4 + 3*(b^2 + 2*a*c)*d^2*e^2 + 4*(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)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5) + 1/2*(c^2*e*x^2 - 2*(3*c^2*d - 2*b*c*e
)*x)/e^4 + (6*c^2*d^2 - 6*b*c*d*e + (b^2 + 2*a*c)*e^2)*log(e*x + d)/e^5

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mupad [B]  time = 0.09, size = 200, normalized size = 1.45 \[ \frac {\ln \left (d+e\,x\right )\,\left (b^2\,e^2-6\,b\,c\,d\,e+6\,c^2\,d^2+2\,a\,c\,e^2\right )}{e^5}-\frac {\frac {a^2\,e^4+2\,a\,b\,d\,e^3-6\,a\,c\,d^2\,e^2-3\,b^2\,d^2\,e^2+10\,b\,c\,d^3\,e-7\,c^2\,d^4}{2\,e}-x\,\left (2\,b^2\,d\,e^2-6\,b\,c\,d^2\,e-2\,a\,b\,e^3+4\,c^2\,d^3+4\,a\,c\,d\,e^2\right )}{d^2\,e^4+2\,d\,e^5\,x+e^6\,x^2}-x\,\left (\frac {3\,c^2\,d}{e^4}-\frac {2\,b\,c}{e^3}\right )+\frac {c^2\,x^2}{2\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^2/(d + e*x)^3,x)

[Out]

(log(d + e*x)*(b^2*e^2 + 6*c^2*d^2 + 2*a*c*e^2 - 6*b*c*d*e))/e^5 - ((a^2*e^4 - 7*c^2*d^4 - 3*b^2*d^2*e^2 + 2*a
*b*d*e^3 + 10*b*c*d^3*e - 6*a*c*d^2*e^2)/(2*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))/(d^2*e^4 + e^6*x^2 + 2*d*e^5*x) - x*((3*c^2*d)/e^4 - (2*b*c)/e^3) + (c^2*x^2)/(2*e^3)

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sympy [A]  time = 2.93, size = 211, normalized size = 1.53 \[ \frac {c^{2} x^{2}}{2 e^{3}} + x \left (\frac {2 b c}{e^{3}} - \frac {3 c^{2} d}{e^{4}}\right ) + \frac {- a^{2} e^{4} - 2 a b d e^{3} + 6 a c d^{2} e^{2} + 3 b^{2} d^{2} e^{2} - 10 b c d^{3} e + 7 c^{2} d^{4} + x \left (- 4 a b e^{4} + 8 a c d e^{3} + 4 b^{2} d e^{3} - 12 b c d^{2} e^{2} + 8 c^{2} d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} + \frac {\left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right ) \log {\left (d + e x \right )}}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**2/(e*x+d)**3,x)

[Out]

c**2*x**2/(2*e**3) + x*(2*b*c/e**3 - 3*c**2*d/e**4) + (-a**2*e**4 - 2*a*b*d*e**3 + 6*a*c*d**2*e**2 + 3*b**2*d*
*2*e**2 - 10*b*c*d**3*e + 7*c**2*d**4 + x*(-4*a*b*e**4 + 8*a*c*d*e**3 + 4*b**2*d*e**3 - 12*b*c*d**2*e**2 + 8*c
**2*d**3*e))/(2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + (2*a*c*e**2 + b**2*e**2 - 6*b*c*d*e + 6*c**2*d**2)*log
(d + e*x)/e**5

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