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

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

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Rubi [A]  time = 0.05, 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}{2 x^2}+\frac {1}{2} b x^2 (2 a B+A b)+a \log (x) (a B+2 A b)+\frac {1}{4} b^2 B x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.96 \begin {gather*} \frac {1}{4} \left (-\frac {2 a^2 A}{x^2}+2 b x^2 (2 a B+A b)+4 a \log (x) (a B+2 A b)+b^2 B x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 54, normalized size = 1.06 \begin {gather*} \frac {B b^{2} x^{6} + 2 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 4 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2} \log \relax (x) - 2 \, A a^{2}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 70, normalized size = 1.37 \begin {gather*} \frac {1}{4} \, B b^{2} x^{4} + B a b x^{2} + \frac {1}{2} \, A b^{2} x^{2} + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} \log \left (x^{2}\right ) - \frac {B a^{2} x^{2} + 2 \, A a b x^{2} + A a^{2}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 50, normalized size = 0.98 \begin {gather*} \frac {B \,b^{2} x^{4}}{4}+\frac {A \,b^{2} x^{2}}{2}+B a b \,x^{2}+2 A a b \ln \relax (x )+B \,a^{2} \ln \relax (x )-\frac {A \,a^{2}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 52, normalized size = 1.02 \begin {gather*} \frac {1}{4} \, B b^{2} x^{4} + \frac {1}{2} \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} \log \left (x^{2}\right ) - \frac {A a^{2}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 48, normalized size = 0.94 \begin {gather*} x^2\,\left (\frac {A\,b^2}{2}+B\,a\,b\right )+\ln \relax (x)\,\left (B\,a^2+2\,A\,b\,a\right )-\frac {A\,a^2}{2\,x^2}+\frac {B\,b^2\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.24, size = 48, normalized size = 0.94 \begin {gather*} - \frac {A a^{2}}{2 x^{2}} + \frac {B b^{2} x^{4}}{4} + a \left (2 A b + B a\right ) \log {\relax (x )} + x^{2} \left (\frac {A b^{2}}{2} + B a b\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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