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

Optimal. Leaf size=86 \[ 2 a^2 b x (2 a B+3 A b)+a^3 \log (x) (a B+4 A b)-\frac{a^4 A}{x}+a b^2 x^2 (3 a B+2 A b)+\frac{1}{3} b^3 x^3 (4 a B+A b)+\frac{1}{4} b^4 B x^4 \]

[Out]

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

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Rubi [A]  time = 0.0506458, antiderivative size = 86, 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} \[ 2 a^2 b x (2 a B+3 A b)+a^3 \log (x) (a B+4 A b)-\frac{a^4 A}{x}+a b^2 x^2 (3 a B+2 A b)+\frac{1}{3} b^3 x^3 (4 a B+A b)+\frac{1}{4} b^4 B x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0356784, size = 85, normalized size = 0.99 \[ 3 a^2 b^2 x (2 A+B x)+a^3 \log (x) (a B+4 A b)-\frac{a^4 A}{x}+4 a^3 b B x+\frac{2}{3} a b^3 x^2 (3 A+2 B x)+\frac{1}{12} b^4 x^3 (4 A+3 B x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^4*A)/x) + 4*a^3*b*B*x + 3*a^2*b^2*x*(2*A + B*x) + (2*a*b^3*x^2*(3*A + 2*B*x))/3 + (b^4*x^3*(4*A + 3*B*x))
/12 + a^3*(4*A*b + a*B)*Log[x]

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Maple [A]  time = 0.005, size = 95, normalized size = 1.1 \begin{align*}{\frac{{b}^{4}B{x}^{4}}{4}}+{\frac{A{x}^{3}{b}^{4}}{3}}+{\frac{4\,B{x}^{3}a{b}^{3}}{3}}+2\,A{x}^{2}a{b}^{3}+3\,B{x}^{2}{a}^{2}{b}^{2}+6\,A{a}^{2}{b}^{2}x+4\,B{a}^{3}bx+4\,A\ln \left ( x \right ){a}^{3}b+{a}^{4}B\ln \left ( x \right ) -{\frac{A{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^2,x)

[Out]

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

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Maxima [A]  time = 1.00816, size = 127, normalized size = 1.48 \begin{align*} \frac{1}{4} \, B b^{4} x^{4} - \frac{A a^{4}}{x} + \frac{1}{3} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{3} +{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{2} + 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x +{\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left (x\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)^2/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.27581, size = 223, normalized size = 2.59 \begin{align*} \frac{3 \, B b^{4} x^{5} - 12 \, A a^{4} + 4 \,{\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} + 24 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 12 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x \log \left (x\right )}{12 \, x} \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^2,x, algorithm="fricas")

[Out]

1/12*(3*B*b^4*x^5 - 12*A*a^4 + 4*(4*B*a*b^3 + A*b^4)*x^4 + 12*(3*B*a^2*b^2 + 2*A*a*b^3)*x^3 + 24*(2*B*a^3*b +
3*A*a^2*b^2)*x^2 + 12*(B*a^4 + 4*A*a^3*b)*x*log(x))/x

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Sympy [A]  time = 0.425191, size = 94, normalized size = 1.09 \begin{align*} - \frac{A a^{4}}{x} + \frac{B b^{4} x^{4}}{4} + a^{3} \left (4 A b + B a\right ) \log{\left (x \right )} + x^{3} \left (\frac{A b^{4}}{3} + \frac{4 B a b^{3}}{3}\right ) + x^{2} \left (2 A a b^{3} + 3 B a^{2} b^{2}\right ) + x \left (6 A a^{2} b^{2} + 4 B a^{3} b\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)**2/x**2,x)

[Out]

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

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Giac [A]  time = 1.16837, size = 128, normalized size = 1.49 \begin{align*} \frac{1}{4} \, B b^{4} x^{4} + \frac{4}{3} \, B a b^{3} x^{3} + \frac{1}{3} \, A b^{4} x^{3} + 3 \, B a^{2} b^{2} x^{2} + 2 \, A a b^{3} x^{2} + 4 \, B a^{3} b x + 6 \, A a^{2} b^{2} x - \frac{A a^{4}}{x} +{\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | x \right |}\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)^2/x^2,x, algorithm="giac")

[Out]

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