3.624 \(\int \frac{x^3 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=90 \[ \frac{a^3 (A b-a B)}{b^5 (a+b x)}+\frac{a^2 (3 A b-4 a B) \log (a+b x)}{b^5}+\frac{x^2 (A b-2 a B)}{2 b^3}-\frac{a x (2 A b-3 a B)}{b^4}+\frac{B x^3}{3 b^2} \]

[Out]

-((a*(2*A*b - 3*a*B)*x)/b^4) + ((A*b - 2*a*B)*x^2)/(2*b^3) + (B*x^3)/(3*b^2) + (a^3*(A*b - a*B))/(b^5*(a + b*x
)) + (a^2*(3*A*b - 4*a*B)*Log[a + b*x])/b^5

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Rubi [A]  time = 0.0821095, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {27, 77} \[ \frac{a^3 (A b-a B)}{b^5 (a+b x)}+\frac{a^2 (3 A b-4 a B) \log (a+b x)}{b^5}+\frac{x^2 (A b-2 a B)}{2 b^3}-\frac{a x (2 A b-3 a B)}{b^4}+\frac{B x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

-((a*(2*A*b - 3*a*B)*x)/b^4) + ((A*b - 2*a*B)*x^2)/(2*b^3) + (B*x^3)/(3*b^2) + (a^3*(A*b - a*B))/(b^5*(a + b*x
)) + (a^2*(3*A*b - 4*a*B)*Log[a + b*x])/b^5

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{x^3 (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{x^3 (A+B x)}{(a+b x)^2} \, dx\\ &=\int \left (\frac{a (-2 A b+3 a B)}{b^4}+\frac{(A b-2 a B) x}{b^3}+\frac{B x^2}{b^2}+\frac{a^3 (-A b+a B)}{b^4 (a+b x)^2}-\frac{a^2 (-3 A b+4 a B)}{b^4 (a+b x)}\right ) \, dx\\ &=-\frac{a (2 A b-3 a B) x}{b^4}+\frac{(A b-2 a B) x^2}{2 b^3}+\frac{B x^3}{3 b^2}+\frac{a^3 (A b-a B)}{b^5 (a+b x)}+\frac{a^2 (3 A b-4 a B) \log (a+b x)}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0550972, size = 87, normalized size = 0.97 \[ \frac{\frac{6 a^3 (A b-a B)}{a+b x}+6 a^2 (3 A b-4 a B) \log (a+b x)+3 b^2 x^2 (A b-2 a B)+6 a b x (3 a B-2 A b)+2 b^3 B x^3}{6 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(6*a*b*(-2*A*b + 3*a*B)*x + 3*b^2*(A*b - 2*a*B)*x^2 + 2*b^3*B*x^3 + (6*a^3*(A*b - a*B))/(a + b*x) + 6*a^2*(3*A
*b - 4*a*B)*Log[a + b*x])/(6*b^5)

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Maple [A]  time = 0.009, size = 109, normalized size = 1.2 \begin{align*}{\frac{B{x}^{3}}{3\,{b}^{2}}}+{\frac{A{x}^{2}}{2\,{b}^{2}}}-{\frac{B{x}^{2}a}{{b}^{3}}}-2\,{\frac{aAx}{{b}^{3}}}+3\,{\frac{{a}^{2}Bx}{{b}^{4}}}+{\frac{A{a}^{3}}{{b}^{4} \left ( bx+a \right ) }}-{\frac{B{a}^{4}}{{b}^{5} \left ( bx+a \right ) }}+3\,{\frac{{a}^{2}\ln \left ( bx+a \right ) A}{{b}^{4}}}-4\,{\frac{{a}^{3}\ln \left ( bx+a \right ) B}{{b}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x)

[Out]

1/3*B*x^3/b^2+1/2/b^2*A*x^2-1/b^3*B*x^2*a-2/b^3*A*a*x+3/b^4*B*a^2*x+a^3/b^4/(b*x+a)*A-a^4/b^5/(b*x+a)*B+3*a^2/
b^4*ln(b*x+a)*A-4*a^3/b^5*ln(b*x+a)*B

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Maxima [A]  time = 1.03563, size = 136, normalized size = 1.51 \begin{align*} -\frac{B a^{4} - A a^{3} b}{b^{6} x + a b^{5}} + \frac{2 \, B b^{2} x^{3} - 3 \,{\left (2 \, B a b - A b^{2}\right )} x^{2} + 6 \,{\left (3 \, B a^{2} - 2 \, A a b\right )} x}{6 \, b^{4}} - \frac{{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x + a\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*a^4 - A*a^3*b)/(b^6*x + a*b^5) + 1/6*(2*B*b^2*x^3 - 3*(2*B*a*b - A*b^2)*x^2 + 6*(3*B*a^2 - 2*A*a*b)*x)/b^4
- (4*B*a^3 - 3*A*a^2*b)*log(b*x + a)/b^5

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Fricas [A]  time = 1.3095, size = 297, normalized size = 3.3 \begin{align*} \frac{2 \, B b^{4} x^{4} - 6 \, B a^{4} + 6 \, A a^{3} b -{\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{3} + 3 \,{\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{2} + 6 \,{\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x - 6 \,{\left (4 \, B a^{4} - 3 \, A a^{3} b +{\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \,{\left (b^{6} x + a b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^4*x^4 - 6*B*a^4 + 6*A*a^3*b - (4*B*a*b^3 - 3*A*b^4)*x^3 + 3*(4*B*a^2*b^2 - 3*A*a*b^3)*x^2 + 6*(3*B*
a^3*b - 2*A*a^2*b^2)*x - 6*(4*B*a^4 - 3*A*a^3*b + (4*B*a^3*b - 3*A*a^2*b^2)*x)*log(b*x + a))/(b^6*x + a*b^5)

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Sympy [A]  time = 0.609767, size = 90, normalized size = 1. \begin{align*} \frac{B x^{3}}{3 b^{2}} - \frac{a^{2} \left (- 3 A b + 4 B a\right ) \log{\left (a + b x \right )}}{b^{5}} - \frac{- A a^{3} b + B a^{4}}{a b^{5} + b^{6} x} - \frac{x^{2} \left (- A b + 2 B a\right )}{2 b^{3}} + \frac{x \left (- 2 A a b + 3 B a^{2}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x+A)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

B*x**3/(3*b**2) - a**2*(-3*A*b + 4*B*a)*log(a + b*x)/b**5 - (-A*a**3*b + B*a**4)/(a*b**5 + b**6*x) - x**2*(-A*
b + 2*B*a)/(2*b**3) + x*(-2*A*a*b + 3*B*a**2)/b**4

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Giac [A]  time = 1.14334, size = 140, normalized size = 1.56 \begin{align*} -\frac{{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac{2 \, B b^{4} x^{3} - 6 \, B a b^{3} x^{2} + 3 \, A b^{4} x^{2} + 18 \, B a^{2} b^{2} x - 12 \, A a b^{3} x}{6 \, b^{6}} - \frac{B a^{4} - A a^{3} b}{{\left (b x + a\right )} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*B*a^3 - 3*A*a^2*b)*log(abs(b*x + a))/b^5 + 1/6*(2*B*b^4*x^3 - 6*B*a*b^3*x^2 + 3*A*b^4*x^2 + 18*B*a^2*b^2*x
- 12*A*a*b^3*x)/b^6 - (B*a^4 - A*a^3*b)/((b*x + a)*b^5)