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

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.84, size = 54, normalized size = 0.92 \[ \frac {1}{4} \, b^{4} x^{4} + \frac {4}{3} \, a b^{3} x^{3} + {\left (3 \, a^{2} + 1\right )} b^{2} x^{2} + 4 \, {\left (a^{3} + a\right )} b x + {\left (a^{4} + 2 \, a^{2} + 1\right )} \log \relax (x) \]

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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giac [A]  time = 0.31, size = 62, normalized size = 1.05 \[ \frac {1}{4} \, b^{4} x^{4} + \frac {4}{3} \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + b^{2} x^{2} + 4 \, a b x + {\left (a^{4} + 2 \, a^{2} + 1\right )} \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.00, size = 64, normalized size = 1.08 \[ \frac {b^{4} x^{4}}{4}+\frac {4 a \,b^{3} x^{3}}{3}+3 a^{2} b^{2} x^{2}+a^{4} \ln \relax (x )+4 a^{3} b x +b^{2} x^{2}+2 a^{2} \ln \relax (x )+4 a b x +\ln \relax (x ) \]

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 = 0.54, size = 54, normalized size = 0.92 \[ \frac {1}{4} \, b^{4} x^{4} + \frac {4}{3} \, a b^{3} x^{3} + {\left (3 \, a^{2} + 1\right )} b^{2} x^{2} + 4 \, {\left (a^{3} + a\right )} b x + {\left (a^{4} + 2 \, a^{2} + 1\right )} \log \relax (x) \]

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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mupad [B]  time = 0.05, size = 55, normalized size = 0.93 \[ \ln \relax (x)\,\left (a^4+2\,a^2+1\right )+\frac {b^4\,x^4}{4}+\frac {4\,a\,b^3\,x^3}{3}+b^2\,x^2\,\left (3\,a^2+1\right )+4\,a\,b\,x\,\left (a^2+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2 + 1)^2/x,x)

[Out]

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

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sympy [A]  time = 0.17, size = 58, normalized size = 0.98 \[ \frac {4 a b^{3} x^{3}}{3} + \frac {b^{4} x^{4}}{4} + x^{2} \left (3 a^{2} b^{2} + b^{2}\right ) + x \left (4 a^{3} b + 4 a b\right ) + \left (a^{2} + 1\right )^{2} \log {\relax (x )} \]

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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