3.2.75 \(\int \frac {(a+b x^2)^2}{x^3 (c+d x^2)} \, dx\)

Optimal. Leaf size=58 \[ -\frac {a^2}{2 c x^2}+\frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d}+\frac {a \log (x) (2 b c-a d)}{c^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {446, 88} \begin {gather*} -\frac {a^2}{2 c x^2}+\frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d}+\frac {a \log (x) (2 b c-a d)}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^3*(c + d*x^2)),x]

[Out]

-a^2/(2*c*x^2) + (a*(2*b*c - a*d)*Log[x])/c^2 + ((b*c - a*d)^2*Log[c + d*x^2])/(2*c^2*d)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -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}{x^3 \left (c+d x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^2 (c+d x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {a^2}{c x^2}-\frac {a (-2 b c+a d)}{c^2 x}+\frac {(b c-a d)^2}{c^2 (c+d x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {a^2}{2 c x^2}+\frac {a (2 b c-a d) \log (x)}{c^2}+\frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 60, normalized size = 1.03 \begin {gather*} \frac {a^2 (-c) d-2 a d x^2 \log (x) (a d-2 b c)+x^2 (b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^3*(c + d*x^2)),x]

[Out]

(-(a^2*c*d) - 2*a*d*(-2*b*c + a*d)*x^2*Log[x] + (b*c - a*d)^2*x^2*Log[c + d*x^2])/(2*c^2*d*x^2)

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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}{x^3 \left (c+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x^2)^2/(x^3*(c + d*x^2)),x]

[Out]

IntegrateAlgebraic[(a + b*x^2)^2/(x^3*(c + d*x^2)), x]

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fricas [A]  time = 0.87, size = 74, normalized size = 1.28 \begin {gather*} -\frac {a^{2} c d - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \log \left (d x^{2} + c\right ) - 2 \, {\left (2 \, a b c d - a^{2} d^{2}\right )} x^{2} \log \relax (x)}{2 \, c^{2} d x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2*c*d - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*log(d*x^2 + c) - 2*(2*a*b*c*d - a^2*d^2)*x^2*log(x))/(c^2*
d*x^2)

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giac [A]  time = 0.40, size = 91, normalized size = 1.57 \begin {gather*} \frac {{\left (2 \, a b c - a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{2} d} - \frac {2 \, a b c x^{2} - a^{2} d x^{2} + a^{2} c}{2 \, c^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*b*c - a^2*d)*log(x^2)/c^2 + 1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(abs(d*x^2 + c))/(c^2*d) - 1/2*(2*
a*b*c*x^2 - a^2*d*x^2 + a^2*c)/(c^2*x^2)

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maple [A]  time = 0.01, size = 81, normalized size = 1.40 \begin {gather*} -\frac {a^{2} d \ln \relax (x )}{c^{2}}+\frac {a^{2} d \ln \left (d \,x^{2}+c \right )}{2 c^{2}}+\frac {2 a b \ln \relax (x )}{c}-\frac {a b \ln \left (d \,x^{2}+c \right )}{c}+\frac {b^{2} \ln \left (d \,x^{2}+c \right )}{2 d}-\frac {a^{2}}{2 c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^3/(d*x^2+c),x)

[Out]

1/2/c^2*d*ln(d*x^2+c)*a^2-1/c*ln(d*x^2+c)*a*b+1/2/d*ln(d*x^2+c)*b^2-1/2*a^2/c/x^2-a^2/c^2*ln(x)*d+2*a/c*ln(x)*
b

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maxima [A]  time = 1.04, size = 70, normalized size = 1.21 \begin {gather*} \frac {{\left (2 \, a b c - a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{2}} - \frac {a^{2}}{2 \, c x^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*b*c - a^2*d)*log(x^2)/c^2 - 1/2*a^2/(c*x^2) + 1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(d*x^2 + c)/(c^2
*d)

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mupad [B]  time = 0.17, size = 67, normalized size = 1.16 \begin {gather*} \frac {\ln \left (d\,x^2+c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,c^2\,d}-\frac {a^2}{2\,c\,x^2}-\frac {\ln \relax (x)\,\left (a^2\,d-2\,a\,b\,c\right )}{c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^3*(c + d*x^2)),x)

[Out]

(log(c + d*x^2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(2*c^2*d) - a^2/(2*c*x^2) - (log(x)*(a^2*d - 2*a*b*c))/c^2

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sympy [A]  time = 1.40, size = 49, normalized size = 0.84 \begin {gather*} - \frac {a^{2}}{2 c x^{2}} - \frac {a \left (a d - 2 b c\right ) \log {\relax (x )}}{c^{2}} + \frac {\left (a d - b c\right )^{2} \log {\left (\frac {c}{d} + x^{2} \right )}}{2 c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**3/(d*x**2+c),x)

[Out]

-a**2/(2*c*x**2) - a*(a*d - 2*b*c)*log(x)/c**2 + (a*d - b*c)**2*log(c/d + x**2)/(2*c**2*d)

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