3.1.62 \(\int \frac {(a+b x^2) (A+B x+C x^2+D x^3)}{x} \, dx\)

Optimal. Leaf size=56 \[ \frac {1}{2} x^2 (a C+A b)+a A \log (x)+\frac {1}{3} x^3 (a D+b B)+a B x+\frac {1}{4} b C x^4+\frac {1}{5} b D x^5 \]

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1802} \begin {gather*} \frac {1}{2} x^2 (a C+A b)+a A \log (x)+\frac {1}{3} x^3 (a D+b B)+a B x+\frac {1}{4} b C x^4+\frac {1}{5} b D x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x,x]

[Out]

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

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 56, normalized size = 1.00 \begin {gather*} \frac {1}{2} x^2 (a C+A b)+a A \log (x)+\frac {1}{3} x^3 (a D+b B)+a B x+\frac {1}{4} b C x^4+\frac {1}{5} b D x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x,x]

[Out]

IntegrateAlgebraic[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x, x]

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fricas [A]  time = 0.82, size = 48, normalized size = 0.86 \begin {gather*} \frac {1}{5} \, D b x^{5} + \frac {1}{4} \, C b x^{4} + \frac {1}{3} \, {\left (D a + B b\right )} x^{3} + B a x + \frac {1}{2} \, {\left (C a + A b\right )} x^{2} + A a \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.41, size = 53, normalized size = 0.95 \begin {gather*} \frac {1}{5} \, D b x^{5} + \frac {1}{4} \, C b x^{4} + \frac {1}{3} \, D a x^{3} + \frac {1}{3} \, B b x^{3} + \frac {1}{2} \, C a x^{2} + \frac {1}{2} \, A b x^{2} + B a x + A a \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 53, normalized size = 0.95 \begin {gather*} \frac {D b \,x^{5}}{5}+\frac {C b \,x^{4}}{4}+\frac {B b \,x^{3}}{3}+\frac {D a \,x^{3}}{3}+\frac {A b \,x^{2}}{2}+\frac {C a \,x^{2}}{2}+A a \ln \relax (x )+B a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x,x)

[Out]

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

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maxima [A]  time = 1.35, size = 48, normalized size = 0.86 \begin {gather*} \frac {1}{5} \, D b x^{5} + \frac {1}{4} \, C b x^{4} + \frac {1}{3} \, {\left (D a + B b\right )} x^{3} + B a x + \frac {1}{2} \, {\left (C a + A b\right )} x^{2} + A a \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.17, size = 52, normalized size = 0.93 \begin {gather*} \frac {a\,x^3\,D}{3}+\frac {b\,x^5\,D}{5}+B\,a\,x+\frac {A\,b\,x^2}{2}+\frac {C\,a\,x^2}{2}+\frac {B\,b\,x^3}{3}+\frac {C\,b\,x^4}{4}+A\,a\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(A + B*x + C*x^2 + x^3*D))/x,x)

[Out]

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

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sympy [A]  time = 0.32, size = 54, normalized size = 0.96 \begin {gather*} A a \log {\relax (x )} + B a x + \frac {C b x^{4}}{4} + \frac {D b x^{5}}{5} + x^{3} \left (\frac {B b}{3} + \frac {D a}{3}\right ) + x^{2} \left (\frac {A b}{2} + \frac {C a}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(D*x**3+C*x**2+B*x+A)/x,x)

[Out]

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

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