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

Optimal. Leaf size=89 \[ -\frac{a^3 (a B+4 A b)}{2 x^2}-\frac{2 a^2 b (2 a B+3 A b)}{x}-\frac{a^4 A}{3 x^3}+b^3 x (4 a B+A b)+2 a b^2 \log (x) (3 a B+2 A b)+\frac{1}{2} b^4 B x^2 \]

[Out]

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

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Rubi [A]  time = 0.0543838, antiderivative size = 89, 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} \[ -\frac{a^3 (a B+4 A b)}{2 x^2}-\frac{2 a^2 b (2 a B+3 A b)}{x}-\frac{a^4 A}{3 x^3}+b^3 x (4 a B+A b)+2 a b^2 \log (x) (3 a B+2 A b)+\frac{1}{2} b^4 B x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0360897, size = 86, normalized size = 0.97 \[ -\frac{6 a^2 A b^2}{x}-\frac{2 a^3 b (A+2 B x)}{x^2}-\frac{a^4 (2 A+3 B x)}{6 x^3}+2 a b^2 \log (x) (3 a B+2 A b)+4 a b^3 B x+\frac{1}{2} b^4 x (2 A+B x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 95, normalized size = 1.1 \begin{align*}{\frac{{b}^{4}B{x}^{2}}{2}}+A{b}^{4}x+4\,Ba{b}^{3}x+4\,A\ln \left ( x \right ) a{b}^{3}+6\,B\ln \left ( x \right ){a}^{2}{b}^{2}-{\frac{A{a}^{4}}{3\,{x}^{3}}}-2\,{\frac{A{a}^{3}b}{{x}^{2}}}-{\frac{B{a}^{4}}{2\,{x}^{2}}}-6\,{\frac{A{a}^{2}{b}^{2}}{x}}-4\,{\frac{B{a}^{3}b}{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^4,x)

[Out]

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

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Maxima [A]  time = 0.987805, size = 130, normalized size = 1.46 \begin{align*} \frac{1}{2} \, B b^{4} x^{2} +{\left (4 \, B a b^{3} + A b^{4}\right )} x + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \log \left (x\right ) - \frac{2 \, A a^{4} + 12 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 3 \,{\left (B a^{4} + 4 \, A a^{3} 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)^2/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.33813, size = 221, normalized size = 2.48 \begin{align*} \frac{3 \, B b^{4} x^{5} - 2 \, A a^{4} + 6 \,{\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} \log \left (x\right ) - 12 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 3 \,{\left (B a^{4} + 4 \, A a^{3} 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)^2/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.973026, size = 95, normalized size = 1.07 \begin{align*} \frac{B b^{4} x^{2}}{2} + 2 a b^{2} \left (2 A b + 3 B a\right ) \log{\left (x \right )} + x \left (A b^{4} + 4 B a b^{3}\right ) - \frac{2 A a^{4} + x^{2} \left (36 A a^{2} b^{2} + 24 B a^{3} b\right ) + x \left (12 A a^{3} b + 3 B a^{4}\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)**2/x**4,x)

[Out]

B*b**4*x**2/2 + 2*a*b**2*(2*A*b + 3*B*a)*log(x) + x*(A*b**4 + 4*B*a*b**3) - (2*A*a**4 + x**2*(36*A*a**2*b**2 +
24*B*a**3*b) + x*(12*A*a**3*b + 3*B*a**4))/(6*x**3)

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Giac [A]  time = 1.16351, size = 130, normalized size = 1.46 \begin{align*} \frac{1}{2} \, B b^{4} x^{2} + 4 \, B a b^{3} x + A b^{4} x + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac{2 \, A a^{4} + 12 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 3 \,{\left (B a^{4} + 4 \, A a^{3} 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)^2/x^4,x, algorithm="giac")

[Out]

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