3.518 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0240986, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {27, 76} \[ -\frac{a^2 A}{3 x^3}-\frac{a (a B+2 A b)}{2 x^2}-\frac{b (2 a B+A b)}{x}+b^2 B \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^4,x]

[Out]

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

Mathematica [A]  time = 0.0254893, size = 48, normalized size = 0.98 \[ b^2 B \log (x)-\frac{a^2 (2 A+3 B x)+6 a b x (A+2 B x)+6 A b^2 x^2}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/x^4,x]

[Out]

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

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Maple [A]  time = 0.006, size = 52, normalized size = 1.1 \begin{align*}{b}^{2}B\ln \left ( x \right ) -{\frac{A{a}^{2}}{3\,{x}^{3}}}-{\frac{Aab}{{x}^{2}}}-{\frac{B{a}^{2}}{2\,{x}^{2}}}-{\frac{A{b}^{2}}{x}}-2\,{\frac{abB}{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)/x^4,x)

[Out]

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

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Maxima [A]  time = 0.991728, size = 68, normalized size = 1.39 \begin{align*} B b^{2} \log \left (x\right ) - \frac{2 \, A a^{2} + 6 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, x^{3}} \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)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.23697, size = 122, normalized size = 2.49 \begin{align*} \frac{6 \, B b^{2} x^{3} \log \left (x\right ) - 2 \, A a^{2} - 6 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} - 3 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, x^{3}} \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)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.739027, size = 51, normalized size = 1.04 \begin{align*} B b^{2} \log{\left (x \right )} - \frac{2 A a^{2} + x^{2} \left (6 A b^{2} + 12 B a b\right ) + x \left (6 A a b + 3 B a^{2}\right )}{6 x^{3}} \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)/x**4,x)

[Out]

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

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Giac [A]  time = 1.13386, size = 69, normalized size = 1.41 \begin{align*} B b^{2} \log \left ({\left | x \right |}\right ) - \frac{2 \, A a^{2} + 6 \,{\left (2 \, B a b + A b^{2}\right )} x^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} x}{6 \, x^{3}} \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)/x^4,x, algorithm="giac")

[Out]

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