3.1458 \(\int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=59 \[ \frac {2 b (b d-a e)}{e^3 (d+e x)}-\frac {(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac {b^2 \log (d+e x)}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 43} \[ \frac {2 b (b d-a e)}{e^3 (d+e x)}-\frac {(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac {b^2 \log (d+e x)}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.81 \[ \frac {\frac {(b d-a e) (a e+3 b d+4 b e x)}{(d+e x)^2}+2 b^2 \log (d+e x)}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 100, normalized size = 1.69 \[ \frac {3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 69, normalized size = 1.17 \[ b^{2} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {{\left (4 \, {\left (b^{2} d - a b e\right )} x + {\left (3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2}\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 92, normalized size = 1.56 \[ -\frac {a^{2}}{2 \left (e x +d \right )^{2} e}+\frac {a b d}{\left (e x +d \right )^{2} e^{2}}-\frac {b^{2} d^{2}}{2 \left (e x +d \right )^{2} e^{3}}-\frac {2 a b}{\left (e x +d \right ) e^{2}}+\frac {2 b^{2} d}{\left (e x +d \right ) e^{3}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 80, normalized size = 1.36 \[ \frac {3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {b^{2} \log \left (e x + d\right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 77, normalized size = 1.31 \[ \frac {b^2\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {a^2\,e^2+2\,a\,b\,d\,e-3\,b^2\,d^2}{2\,e^3}+\frac {2\,b\,x\,\left (a\,e-b\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.46, size = 80, normalized size = 1.36 \[ \frac {b^{2} \log {\left (d + e x \right )}}{e^{3}} + \frac {- a^{2} e^{2} - 2 a b d e + 3 b^{2} d^{2} + x \left (- 4 a b e^{2} + 4 b^{2} d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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