3.98 \(\int \frac{(1+(a+b x)^2)^2}{x} \, dx\)

Optimal. Leaf size=59 \[ \frac{1}{2} \left (a^2+2\right ) (a+b x)^2+a \left (a^2+2\right ) b x+\left (a^2+1\right )^2 \log (x)+\frac{1}{4} (a+b x)^4+\frac{1}{3} a (a+b x)^3 \]

[Out]

a*(2 + a^2)*b*x + ((2 + a^2)*(a + b*x)^2)/2 + (a*(a + b*x)^3)/3 + (a + b*x)^4/4 + (1 + a^2)^2*Log[x]

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Rubi [A]  time = 0.0553684, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {371, 697} \[ \frac{1}{2} \left (a^2+2\right ) (a+b x)^2+a \left (a^2+2\right ) b x+\left (a^2+1\right )^2 \log (x)+\frac{1}{4} (a+b x)^4+\frac{1}{3} a (a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[(1 + (a + b*x)^2)^2/x,x]

[Out]

a*(2 + a^2)*b*x + ((2 + a^2)*(a + b*x)^2)/2 + (a*(a + b*x)^3)/3 + (a + b*x)^4/4 + (1 + a^2)^2*Log[x]

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0214168, size = 64, normalized size = 1.08 \[ \frac{1}{2} \left (a^2+2\right ) (a+b x)^2+a \left (a^2+2\right ) (a+b x)+\left (a^2+1\right )^2 \log (b x)+\frac{1}{4} (a+b x)^4+\frac{1}{3} a (a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + (a + b*x)^2)^2/x,x]

[Out]

a*(2 + a^2)*(a + b*x) + ((2 + a^2)*(a + b*x)^2)/2 + (a*(a + b*x)^3)/3 + (a + b*x)^4/4 + (1 + a^2)^2*Log[b*x]

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Maple [A]  time = 0.003, size = 64, normalized size = 1.1 \begin{align*}{\frac{{b}^{4}{x}^{4}}{4}}+{\frac{4\,a{b}^{3}{x}^{3}}{3}}+3\,{x}^{2}{a}^{2}{b}^{2}+4\,{a}^{3}bx+{b}^{2}{x}^{2}+4\,abx+\ln \left ( x \right ){a}^{4}+2\,\ln \left ( x \right ){a}^{2}+\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+(b*x+a)^2)^2/x,x)

[Out]

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

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Maxima [A]  time = 1.12995, size = 73, normalized size = 1.24 \begin{align*} \frac{1}{4} \, b^{4} x^{4} + \frac{4}{3} \, a b^{3} x^{3} +{\left (3 \, a^{2} + 1\right )} b^{2} x^{2} + 4 \,{\left (a^{3} + a\right )} b x +{\left (a^{4} + 2 \, a^{2} + 1\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+(b*x+a)^2)^2/x,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77001, size = 130, normalized size = 2.2 \begin{align*} \frac{1}{4} \, b^{4} x^{4} + \frac{4}{3} \, a b^{3} x^{3} +{\left (3 \, a^{2} + 1\right )} b^{2} x^{2} + 4 \,{\left (a^{3} + a\right )} b x +{\left (a^{4} + 2 \, a^{2} + 1\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+(b*x+a)^2)^2/x,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.312202, size = 58, normalized size = 0.98 \begin{align*} \frac{4 a b^{3} x^{3}}{3} + \frac{b^{4} x^{4}}{4} + x^{2} \left (3 a^{2} b^{2} + b^{2}\right ) + x \left (4 a^{3} b + 4 a b\right ) + \left (a^{2} + 1\right )^{2} \log{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+(b*x+a)**2)**2/x,x)

[Out]

4*a*b**3*x**3/3 + b**4*x**4/4 + x**2*(3*a**2*b**2 + b**2) + x*(4*a**3*b + 4*a*b) + (a**2 + 1)**2*log(x)

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Giac [A]  time = 1.12301, size = 84, normalized size = 1.42 \begin{align*} \frac{1}{4} \, b^{4} x^{4} + \frac{4}{3} \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + b^{2} x^{2} + 4 \, a b x +{\left (a^{4} + 2 \, a^{2} + 1\right )} \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+(b*x+a)^2)^2/x,x, algorithm="giac")

[Out]

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