3.1665 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{d+e x} \, dx\)

Optimal. Leaf size=92 \[ -\frac{(a+b x)^2 (B d-A e)}{2 e^2}+\frac{b x (b d-a e) (B d-A e)}{e^3}-\frac{(b d-a e)^2 (B d-A e) \log (d+e x)}{e^4}+\frac{B (a+b x)^3}{3 b e} \]

[Out]

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

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Rubi [A]  time = 0.0628983, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {27, 77} \[ -\frac{(a+b x)^2 (B d-A e)}{2 e^2}+\frac{b x (b d-a e) (B d-A e)}{e^3}-\frac{(b d-a e)^2 (B d-A e) \log (d+e x)}{e^4}+\frac{B (a+b x)^3}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0570592, size = 102, normalized size = 1.11 \[ \frac{e x \left (6 a^2 B e^2+6 a b e (2 A e-2 B d+B e x)+b^2 \left (3 A e (e x-2 d)+B \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )\right )-6 (b d-a e)^2 (B d-A e) \log (d+e x)}{6 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.003, size = 197, normalized size = 2.1 \begin{align*}{\frac{{b}^{2}B{x}^{3}}{3\,e}}+{\frac{A{b}^{2}{x}^{2}}{2\,e}}+{\frac{B{x}^{2}ab}{e}}-{\frac{{b}^{2}B{x}^{2}d}{2\,{e}^{2}}}+2\,{\frac{aAbx}{e}}-{\frac{Ad{b}^{2}x}{{e}^{2}}}+{\frac{{a}^{2}Bx}{e}}-2\,{\frac{abBdx}{{e}^{2}}}+{\frac{B{b}^{2}{d}^{2}x}{{e}^{3}}}+{\frac{\ln \left ( ex+d \right ) A{a}^{2}}{e}}-2\,{\frac{\ln \left ( ex+d \right ) Aabd}{{e}^{2}}}+{\frac{\ln \left ( ex+d \right ) A{b}^{2}{d}^{2}}{{e}^{3}}}-{\frac{\ln \left ( ex+d \right ) B{a}^{2}d}{{e}^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) Bab{d}^{2}}{{e}^{3}}}-{\frac{\ln \left ( ex+d \right ) B{b}^{2}{d}^{3}}{{e}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.22968, size = 205, normalized size = 2.23 \begin{align*} \frac{2 \, B b^{2} e^{2} x^{3} - 3 \,{\left (B b^{2} d e -{\left (2 \, B a b + A b^{2}\right )} e^{2}\right )} x^{2} + 6 \,{\left (B b^{2} d^{2} -{\left (2 \, B a b + A b^{2}\right )} d e +{\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} x}{6 \, e^{3}} - \frac{{\left (B b^{2} d^{3} - A a^{2} e^{3} -{\left (2 \, B a b + A b^{2}\right )} d^{2} e +{\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^2*e^2*x^3 - 3*(B*b^2*d*e - (2*B*a*b + A*b^2)*e^2)*x^2 + 6*(B*b^2*d^2 - (2*B*a*b + A*b^2)*d*e + (B*a
^2 + 2*A*a*b)*e^2)*x)/e^3 - (B*b^2*d^3 - A*a^2*e^3 - (2*B*a*b + A*b^2)*d^2*e + (B*a^2 + 2*A*a*b)*d*e^2)*log(e*
x + d)/e^4

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Fricas [A]  time = 1.43392, size = 319, normalized size = 3.47 \begin{align*} \frac{2 \, B b^{2} e^{3} x^{3} - 3 \,{\left (B b^{2} d e^{2} -{\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 6 \,{\left (B b^{2} d^{2} e -{\left (2 \, B a b + A b^{2}\right )} d e^{2} +{\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x - 6 \,{\left (B b^{2} d^{3} - A a^{2} e^{3} -{\left (2 \, B a b + A b^{2}\right )} d^{2} e +{\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^2*e^3*x^3 - 3*(B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 6*(B*b^2*d^2*e - (2*B*a*b + A*b^2)*d*e^2
+ (B*a^2 + 2*A*a*b)*e^3)*x - 6*(B*b^2*d^3 - A*a^2*e^3 - (2*B*a*b + A*b^2)*d^2*e + (B*a^2 + 2*A*a*b)*d*e^2)*log
(e*x + d))/e^4

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Sympy [A]  time = 0.730056, size = 117, normalized size = 1.27 \begin{align*} \frac{B b^{2} x^{3}}{3 e} + \frac{x^{2} \left (A b^{2} e + 2 B a b e - B b^{2} d\right )}{2 e^{2}} + \frac{x \left (2 A a b e^{2} - A b^{2} d e + B a^{2} e^{2} - 2 B a b d e + B b^{2} d^{2}\right )}{e^{3}} - \frac{\left (- A e + B d\right ) \left (a e - b d\right )^{2} \log{\left (d + e x \right )}}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14308, size = 219, normalized size = 2.38 \begin{align*} -{\left (B b^{2} d^{3} - 2 \, B a b d^{2} e - A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} - A a^{2} e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{6} \,{\left (2 \, B b^{2} x^{3} e^{2} - 3 \, B b^{2} d x^{2} e + 6 \, B b^{2} d^{2} x + 6 \, B a b x^{2} e^{2} + 3 \, A b^{2} x^{2} e^{2} - 12 \, B a b d x e - 6 \, A b^{2} d x e + 6 \, B a^{2} x e^{2} + 12 \, A a b x e^{2}\right )} e^{\left (-3\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*b^2*d^3 - 2*B*a*b*d^2*e - A*b^2*d^2*e + B*a^2*d*e^2 + 2*A*a*b*d*e^2 - A*a^2*e^3)*e^(-4)*log(abs(x*e + d))
+ 1/6*(2*B*b^2*x^3*e^2 - 3*B*b^2*d*x^2*e + 6*B*b^2*d^2*x + 6*B*a*b*x^2*e^2 + 3*A*b^2*x^2*e^2 - 12*B*a*b*d*x*e
- 6*A*b^2*d*x*e + 6*B*a^2*x*e^2 + 12*A*a*b*x*e^2)*e^(-3)