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

Optimal. Leaf size=66 \[ 3 a^2 A b^2 x^2+4 a^3 A b x+a^4 A \log (x)+\frac{4}{3} a A b^3 x^3+\frac{B (a+b x)^5}{5 b}+\frac{1}{4} A b^4 x^4 \]

[Out]

4*a^3*A*b*x + 3*a^2*A*b^2*x^2 + (4*a*A*b^3*x^3)/3 + (A*b^4*x^4)/4 + (B*(a + b*x)^5)/(5*b) + a^4*A*Log[x]

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Rubi [A]  time = 0.0226127, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {27, 80, 43} \[ 3 a^2 A b^2 x^2+4 a^3 A b x+a^4 A \log (x)+\frac{4}{3} a A b^3 x^3+\frac{B (a+b x)^5}{5 b}+\frac{1}{4} A b^4 x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*a^3*A*b*x + 3*a^2*A*b^2*x^2 + (4*a*A*b^3*x^3)/3 + (A*b^4*x^4)/4 + (B*(a + b*x)^5)/(5*b) + a^4*A*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx &=\int \frac{(a+b x)^4 (A+B x)}{x} \, dx\\ &=\frac{B (a+b x)^5}{5 b}+A \int \frac{(a+b x)^4}{x} \, dx\\ &=\frac{B (a+b x)^5}{5 b}+A \int \left (4 a^3 b+\frac{a^4}{x}+6 a^2 b^2 x+4 a b^3 x^2+b^4 x^3\right ) \, dx\\ &=4 a^3 A b x+3 a^2 A b^2 x^2+\frac{4}{3} a A b^3 x^3+\frac{1}{4} A b^4 x^4+\frac{B (a+b x)^5}{5 b}+a^4 A \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0219544, size = 83, normalized size = 1.26 \[ a^2 b^2 x^2 (3 A+2 B x)+2 a^3 b x (2 A+B x)+a^4 A \log (x)+a^4 B x+\frac{1}{3} a b^3 x^3 (4 A+3 B x)+\frac{1}{20} b^4 x^4 (5 A+4 B x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*B*x + 2*a^3*b*x*(2*A + B*x) + a^2*b^2*x^2*(3*A + 2*B*x) + (a*b^3*x^3*(4*A + 3*B*x))/3 + (b^4*x^4*(5*A + 4*
B*x))/20 + a^4*A*Log[x]

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Maple [A]  time = 0.003, size = 94, normalized size = 1.4 \begin{align*}{\frac{{b}^{4}B{x}^{5}}{5}}+{\frac{A{b}^{4}{x}^{4}}{4}}+B{x}^{4}a{b}^{3}+{\frac{4\,aA{b}^{3}{x}^{3}}{3}}+2\,B{x}^{3}{a}^{2}{b}^{2}+3\,{a}^{2}A{b}^{2}{x}^{2}+2\,B{x}^{2}{a}^{3}b+4\,{a}^{3}Abx+{a}^{4}Bx+{a}^{4}A\ln \left ( x \right ) \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,x)

[Out]

1/5*b^4*B*x^5+1/4*A*b^4*x^4+B*x^4*a*b^3+4/3*a*A*b^3*x^3+2*B*x^3*a^2*b^2+3*a^2*A*b^2*x^2+2*B*x^2*a^3*b+4*a^3*A*
b*x+a^4*B*x+a^4*A*ln(x)

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Maxima [A]  time = 0.992208, size = 126, normalized size = 1.91 \begin{align*} \frac{1}{5} \, B b^{4} x^{5} + A a^{4} \log \left (x\right ) + \frac{1}{4} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + \frac{2}{3} \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} +{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} +{\left (B a^{4} + 4 \, A a^{3} b\right )} 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,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.3206, size = 205, normalized size = 3.11 \begin{align*} \frac{1}{5} \, B b^{4} x^{5} + A a^{4} \log \left (x\right ) + \frac{1}{4} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + \frac{2}{3} \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} +{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} +{\left (B a^{4} + 4 \, A a^{3} b\right )} 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,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.376294, size = 95, normalized size = 1.44 \begin{align*} A a^{4} \log{\left (x \right )} + \frac{B b^{4} x^{5}}{5} + x^{4} \left (\frac{A b^{4}}{4} + B a b^{3}\right ) + x^{3} \left (\frac{4 A a b^{3}}{3} + 2 B a^{2} b^{2}\right ) + x^{2} \left (3 A a^{2} b^{2} + 2 B a^{3} b\right ) + x \left (4 A a^{3} b + B a^{4}\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,x)

[Out]

A*a**4*log(x) + B*b**4*x**5/5 + x**4*(A*b**4/4 + B*a*b**3) + x**3*(4*A*a*b**3/3 + 2*B*a**2*b**2) + x**2*(3*A*a
**2*b**2 + 2*B*a**3*b) + x*(4*A*a**3*b + B*a**4)

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Giac [A]  time = 1.11463, size = 127, normalized size = 1.92 \begin{align*} \frac{1}{5} \, B b^{4} x^{5} + B a b^{3} x^{4} + \frac{1}{4} \, A b^{4} x^{4} + 2 \, B a^{2} b^{2} x^{3} + \frac{4}{3} \, A a b^{3} x^{3} + 2 \, B a^{3} b x^{2} + 3 \, A a^{2} b^{2} x^{2} + B a^{4} x + 4 \, A a^{3} b x + A a^{4} \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,x, algorithm="giac")

[Out]

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