3.532 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^2}{x^5} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0472656, 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} \[ -\frac{a^3 (a B+4 A b)}{3 x^3}-\frac{a^2 b (2 a B+3 A b)}{x^2}-\frac{a^4 A}{4 x^4}-\frac{2 a b^2 (3 a B+2 A b)}{x}+b^3 \log (x) (4 a B+A b)+b^4 B x \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^5,x]

[Out]

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

Mathematica [A]  time = 0.042099, size = 85, normalized size = 0.99 \[ -\frac{3 a^2 b^2 (A+2 B x)}{x^2}-\frac{2 a^3 b (2 A+3 B x)}{3 x^3}-\frac{a^4 (3 A+4 B x)}{12 x^4}+b^3 \log (x) (4 a B+A b)-\frac{4 a A b^3}{x}+b^4 B x \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^5,x]

[Out]

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

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Maple [A]  time = 0.006, size = 96, normalized size = 1.1 \begin{align*}{b}^{4}Bx+A\ln \left ( x \right ){b}^{4}+4\,B\ln \left ( x \right ) a{b}^{3}-{\frac{4\,A{a}^{3}b}{3\,{x}^{3}}}-{\frac{B{a}^{4}}{3\,{x}^{3}}}-{\frac{A{a}^{4}}{4\,{x}^{4}}}-3\,{\frac{A{a}^{2}{b}^{2}}{{x}^{2}}}-2\,{\frac{B{a}^{3}b}{{x}^{2}}}-4\,{\frac{Aa{b}^{3}}{x}}-6\,{\frac{B{a}^{2}{b}^{2}}{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^5,x)

[Out]

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

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Maxima [A]  time = 0.95723, size = 128, normalized size = 1.49 \begin{align*} B b^{4} x +{\left (4 \, B a b^{3} + A b^{4}\right )} \log \left (x\right ) - \frac{3 \, A a^{4} + 24 \,{\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} + 4 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.36633, size = 225, normalized size = 2.62 \begin{align*} \frac{12 \, B b^{4} x^{5} + 12 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} \log \left (x\right ) - 3 \, A a^{4} - 24 \,{\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} - 4 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.54207, size = 94, normalized size = 1.09 \begin{align*} B b^{4} x + b^{3} \left (A b + 4 B a\right ) \log{\left (x \right )} - \frac{3 A a^{4} + x^{3} \left (48 A a b^{3} + 72 B a^{2} b^{2}\right ) + x^{2} \left (36 A a^{2} b^{2} + 24 B a^{3} b\right ) + x \left (16 A a^{3} b + 4 B a^{4}\right )}{12 x^{4}} \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**5,x)

[Out]

B*b**4*x + b**3*(A*b + 4*B*a)*log(x) - (3*A*a**4 + x**3*(48*A*a*b**3 + 72*B*a**2*b**2) + x**2*(36*A*a**2*b**2
+ 24*B*a**3*b) + x*(16*A*a**3*b + 4*B*a**4))/(12*x**4)

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Giac [A]  time = 1.14627, size = 130, normalized size = 1.51 \begin{align*} B b^{4} x +{\left (4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | x \right |}\right ) - \frac{3 \, A a^{4} + 24 \,{\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} + 4 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

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