3.1929 \(\int \frac{(a+b x) (d+e x)^2}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=49 \[ \frac{e x (b d-a e)}{b^2}+\frac{(b d-a e)^2 \log (a+b x)}{b^3}+\frac{(d+e x)^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0201425, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {27, 43} \[ \frac{e x (b d-a e)}{b^2}+\frac{(b d-a e)^2 \log (a+b x)}{b^3}+\frac{(d+e x)^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0160117, size = 43, normalized size = 0.88 \[ \frac{b e x (-2 a e+4 b d+b e x)+2 (b d-a e)^2 \log (a+b x)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.001, size = 74, normalized size = 1.5 \begin{align*}{\frac{{e}^{2}{x}^{2}}{2\,b}}-{\frac{a{e}^{2}x}{{b}^{2}}}+2\,{\frac{edx}{b}}+{\frac{\ln \left ( bx+a \right ){a}^{2}{e}^{2}}{{b}^{3}}}-2\,{\frac{\ln \left ( bx+a \right ) ade}{{b}^{2}}}+{\frac{\ln \left ( bx+a \right ){d}^{2}}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.954582, size = 82, normalized size = 1.67 \begin{align*} \frac{b e^{2} x^{2} + 2 \,{\left (2 \, b d e - a e^{2}\right )} x}{2 \, b^{2}} + \frac{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (b x + a\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5328, size = 135, normalized size = 2.76 \begin{align*} \frac{b^{2} e^{2} x^{2} + 2 \,{\left (2 \, b^{2} d e - a b e^{2}\right )} x + 2 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.386651, size = 44, normalized size = 0.9 \begin{align*} \frac{e^{2} x^{2}}{2 b} - \frac{x \left (a e^{2} - 2 b d e\right )}{b^{2}} + \frac{\left (a e - b d\right )^{2} \log{\left (a + b x \right )}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10599, size = 80, normalized size = 1.63 \begin{align*} \frac{b x^{2} e^{2} + 4 \, b d x e - 2 \, a x e^{2}}{2 \, b^{2}} + \frac{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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