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

Optimal. Leaf size=51 \[ -\frac {a^2 A}{4 x^4}-\frac {a (a B+2 A b)}{2 x^2}+b \log (x) (2 a B+A b)+\frac {1}{2} b^2 B x^2 \]

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 76} \begin {gather*} -\frac {a^2 A}{4 x^4}-\frac {a (a B+2 A b)}{2 x^2}+b \log (x) (2 a B+A b)+\frac {1}{2} b^2 B x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.98 \begin {gather*} b \log (x) (2 a B+A b)-\frac {a^2 \left (A+2 B x^2\right )+4 a A b x^2-2 b^2 B x^6}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(4*a*A*b*x^2 - 2*b^2*B*x^6 + a^2*(A + 2*B*x^2))/x^4 + b*(A*b + 2*a*B)*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 )^2 \left (A+B x^2\right )}{x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 55, normalized size = 1.08 \begin {gather*} \frac {2 \, B b^{2} x^{6} + 4 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} \log \relax (x) - A a^{2} - 2 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^5,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.34, size = 72, normalized size = 1.41 \begin {gather*} \frac {1}{2} \, B b^{2} x^{2} + \frac {1}{2} \, {\left (2 \, B a b + A b^{2}\right )} \log \left (x^{2}\right ) - \frac {6 \, B a b x^{4} + 3 \, A b^{2} x^{4} + 2 \, B a^{2} x^{2} + 4 \, A a b x^{2} + A a^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 51, normalized size = 1.00 \begin {gather*} \frac {B \,b^{2} x^{2}}{2}+A \,b^{2} \ln \relax (x )+2 B a b \ln \relax (x )-\frac {A a b}{x^{2}}-\frac {B \,a^{2}}{2 x^{2}}-\frac {A \,a^{2}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(B*x^2+A)/x^5,x)

[Out]

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

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maxima [A]  time = 1.39, size = 54, normalized size = 1.06 \begin {gather*} \frac {1}{2} \, B b^{2} x^{2} + \frac {1}{2} \, {\left (2 \, B a b + A b^{2}\right )} \log \left (x^{2}\right ) - \frac {A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.08, size = 51, normalized size = 1.00 \begin {gather*} \ln \relax (x)\,\left (A\,b^2+2\,B\,a\,b\right )-\frac {x^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )+\frac {A\,a^2}{4}}{x^4}+\frac {B\,b^2\,x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^2)/x^5,x)

[Out]

log(x)*(A*b^2 + 2*B*a*b) - (x^2*((B*a^2)/2 + A*a*b) + (A*a^2)/4)/x^4 + (B*b^2*x^2)/2

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sympy [A]  time = 0.52, size = 51, normalized size = 1.00 \begin {gather*} \frac {B b^{2} x^{2}}{2} + b \left (A b + 2 B a\right ) \log {\relax (x )} + \frac {- A a^{2} + x^{2} \left (- 4 A a b - 2 B a^{2}\right )}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(B*x**2+A)/x**5,x)

[Out]

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

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