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

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

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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 {(b c-a d)^2 \log \left (a+b x^2\right )}{2 a^2 b}-\frac {c \log (x) (b c-2 a d)}{a^2}-\frac {c^2}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 73, normalized size = 1.26 \begin {gather*} -\frac {a b c^{2} - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \log \left (b x^{2} + a\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d\right )} x^{2} \log \relax (x)}{2 \, a^{2} b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 90, normalized size = 1.55 \begin {gather*} -\frac {{\left (b c^{2} - 2 \, a c d\right )} \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2} b} + \frac {b c^{2} x^{2} - 2 \, a c d x^{2} - a c^{2}}{2 \, a^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 69, normalized size = 1.19 \begin {gather*} -\frac {{\left (b c^{2} - 2 \, a c d\right )} \log \left (x^{2}\right )}{2 \, a^{2}} - \frac {c^{2}}{2 \, a x^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 67, normalized size = 1.16 \begin {gather*} \frac {\ln \left (b\,x^2+a\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,a^2\,b}-\frac {c^2}{2\,a\,x^2}-\frac {\ln \relax (x)\,\left (b\,c^2-2\,a\,c\,d\right )}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.39, size = 49, normalized size = 0.84 \begin {gather*} - \frac {c^{2}}{2 a x^{2}} + \frac {c \left (2 a d - b c\right ) \log {\relax (x )}}{a^{2}} + \frac {\left (a d - b c\right )^{2} \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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