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

Optimal. Leaf size=90 \[ -\frac{a^3 (a B+4 A b)}{x}+2 a^2 b \log (x) (2 a B+3 A b)-\frac{a^4 A}{2 x^2}+\frac{1}{2} b^3 x^2 (4 a B+A b)+2 a b^2 x (3 a B+2 A b)+\frac{1}{3} b^4 B x^3 \]

[Out]

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

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Rubi [A]  time = 0.0504487, 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, 76} \[ -\frac{a^3 (a B+4 A b)}{x}+2 a^2 b \log (x) (2 a B+3 A b)-\frac{a^4 A}{2 x^2}+\frac{1}{2} b^3 x^2 (4 a B+A b)+2 a b^2 x (3 a B+2 A b)+\frac{1}{3} b^4 B x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^3} \, dx &=\int \frac{(a+b x)^4 (A+B x)}{x^3} \, dx\\ &=\int \left (2 a b^2 (2 A b+3 a B)+\frac{a^4 A}{x^3}+\frac{a^3 (4 A b+a B)}{x^2}+\frac{2 a^2 b (3 A b+2 a B)}{x}+b^3 (A b+4 a B) x+b^4 B x^2\right ) \, dx\\ &=-\frac{a^4 A}{2 x^2}-\frac{a^3 (4 A b+a B)}{x}+2 a b^2 (2 A b+3 a B) x+\frac{1}{2} b^3 (A b+4 a B) x^2+\frac{1}{3} b^4 B x^3+2 a^2 b (3 A b+2 a B) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0328443, size = 86, normalized size = 0.96 \[ 2 a^2 b \log (x) (2 a B+3 A b)-\frac{4 a^3 A b}{x}-\frac{a^4 (A+2 B x)}{2 x^2}+6 a^2 b^2 B x+2 a b^3 x (2 A+B x)+\frac{1}{6} b^4 x^2 (3 A+2 B x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 96, normalized size = 1.1 \begin{align*}{\frac{{b}^{4}B{x}^{3}}{3}}+{\frac{A{x}^{2}{b}^{4}}{2}}+2\,B{x}^{2}a{b}^{3}+4\,Aa{b}^{3}x+6\,B{a}^{2}{b}^{2}x+6\,A\ln \left ( x \right ){a}^{2}{b}^{2}+4\,B\ln \left ( x \right ){a}^{3}b-{\frac{A{a}^{4}}{2\,{x}^{2}}}-4\,{\frac{A{a}^{3}b}{x}}-{\frac{B{a}^{4}}{x}} \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)^2/x^3,x)

[Out]

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

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Maxima [A]  time = 0.987165, size = 130, normalized size = 1.44 \begin{align*} \frac{1}{3} \, B b^{4} x^{3} + \frac{1}{2} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{2} + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x + 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \log \left (x\right ) - \frac{A a^{4} + 2 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x}{2 \, x^{2}} \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)^2/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.32512, size = 221, normalized size = 2.46 \begin{align*} \frac{2 \, B b^{4} x^{5} - 3 \, A a^{4} + 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 12 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 12 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} \log \left (x\right ) - 6 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x}{6 \, x^{2}} \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)^2/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.598278, size = 95, normalized size = 1.06 \begin{align*} \frac{B b^{4} x^{3}}{3} + 2 a^{2} b \left (3 A b + 2 B a\right ) \log{\left (x \right )} + x^{2} \left (\frac{A b^{4}}{2} + 2 B a b^{3}\right ) + x \left (4 A a b^{3} + 6 B a^{2} b^{2}\right ) - \frac{A a^{4} + x \left (8 A a^{3} b + 2 B a^{4}\right )}{2 x^{2}} \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)**2/x**3,x)

[Out]

B*b**4*x**3/3 + 2*a**2*b*(3*A*b + 2*B*a)*log(x) + x**2*(A*b**4/2 + 2*B*a*b**3) + x*(4*A*a*b**3 + 6*B*a**2*b**2
) - (A*a**4 + x*(8*A*a**3*b + 2*B*a**4))/(2*x**2)

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Giac [A]  time = 1.1056, size = 130, normalized size = 1.44 \begin{align*} \frac{1}{3} \, B b^{4} x^{3} + 2 \, B a b^{3} x^{2} + \frac{1}{2} \, A b^{4} x^{2} + 6 \, B a^{2} b^{2} x + 4 \, A a b^{3} x + 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac{A a^{4} + 2 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x}{2 \, x^{2}} \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)^2/x^3,x, algorithm="giac")

[Out]

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