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

Optimal. Leaf size=50 \[ \frac {(b c-a d) \log \left (a+b x^2\right )}{2 a^2}-\frac {\log (x) (b c-a d)}{a^2}-\frac {c}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 50, 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, 77} \[ \frac {(b c-a d) \log \left (a+b x^2\right )}{2 a^2}-\frac {\log (x) (b c-a d)}{a^2}-\frac {c}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/(x^3*(a + b*x^2)),x]

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.98 \[ \frac {(b c-a d) \log \left (a+b x^2\right )}{2 a^2}+\frac {\log (x) (a d-b c)}{a^2}-\frac {c}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*c/(a*x^2) + ((-(b*c) + a*d)*Log[x])/a^2 + ((b*c - a*d)*Log[a + b*x^2])/(2*a^2)

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fricas [A]  time = 0.47, size = 48, normalized size = 0.96 \[ \frac {{\left (b c - a d\right )} x^{2} \log \left (b x^{2} + a\right ) - 2 \, {\left (b c - a d\right )} x^{2} \log \relax (x) - a c}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 72, normalized size = 1.44 \[ -\frac {{\left (b c - a d\right )} \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2} b} + \frac {b c x^{2} - a d x^{2} - a c}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 56, normalized size = 1.12 \[ \frac {d \ln \relax (x )}{a}-\frac {d \ln \left (b \,x^{2}+a \right )}{2 a}-\frac {b c \ln \relax (x )}{a^{2}}+\frac {b c \ln \left (b \,x^{2}+a \right )}{2 a^{2}}-\frac {c}{2 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.04, size = 48, normalized size = 0.96 \[ \frac {{\left (b c - a d\right )} \log \left (b x^{2} + a\right )}{2 \, a^{2}} - \frac {{\left (b c - a d\right )} \log \left (x^{2}\right )}{2 \, a^{2}} - \frac {c}{2 \, a x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.15, size = 45, normalized size = 0.90 \[ \frac {\ln \relax (x)\,\left (a\,d-b\,c\right )}{a^2}-\frac {\ln \left (b\,x^2+a\right )\,\left (a\,d-b\,c\right )}{2\,a^2}-\frac {c}{2\,a\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/(x^3*(a + b*x^2)),x)

[Out]

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

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sympy [A]  time = 0.69, size = 41, normalized size = 0.82 \[ - \frac {c}{2 a x^{2}} + \frac {\left (a d - b c\right ) \log {\relax (x )}}{a^{2}} - \frac {\left (a d - b c\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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