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

Optimal. Leaf size=61 \[ -\frac {(b c-a d)^2}{2 b^3 \left (a+b x^2\right )}+\frac {d (b c-a d) \log \left (a+b x^2\right )}{b^3}+\frac {d^2 x^2}{2 b^2} \]

[Out]

1/2*d^2*x^2/b^2-1/2*(-a*d+b*c)^2/b^3/(b*x^2+a)+d*(-a*d+b*c)*ln(b*x^2+a)/b^3

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Rubi [A]  time = 0.06, antiderivative size = 61, 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 = {444, 43} \[ -\frac {(b c-a d)^2}{2 b^3 \left (a+b x^2\right )}+\frac {d (b c-a d) \log \left (a+b x^2\right )}{b^3}+\frac {d^2 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*x^2)/(2*b^2) - (b*c - a*d)^2/(2*b^3*(a + b*x^2)) + (d*(b*c - a*d)*Log[a + b*x^2])/b^3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.92 \[ \frac {-\frac {(b c-a d)^2}{a+b x^2}+2 d (b c-a d) \log \left (a+b x^2\right )+b d^2 x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^2*x^2 - (b*c - a*d)^2/(a + b*x^2) + 2*d*(b*c - a*d)*Log[a + b*x^2])/(2*b^3)

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fricas [A]  time = 0.43, size = 101, normalized size = 1.66 \[ \frac {b^{2} d^{2} x^{4} + a b d^{2} x^{2} - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} + 2 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*d^2*x^4 + a*b*d^2*x^2 - b^2*c^2 + 2*a*b*c*d - a^2*d^2 + 2*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*d^2)*x^
2)*log(b*x^2 + a))/(b^4*x^2 + a*b^3)

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giac [A]  time = 0.34, size = 111, normalized size = 1.82 \[ \frac {{\left (b x^{2} + a\right )} d^{2}}{2 \, b^{3}} - \frac {{\left (b c d - a d^{2}\right )} \log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b^{3}} - \frac {\frac {b^{3} c^{2}}{b x^{2} + a} - \frac {2 \, a b^{2} c d}{b x^{2} + a} + \frac {a^{2} b d^{2}}{b x^{2} + a}}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + a)*d^2/b^3 - (b*c*d - a*d^2)*log(abs(b*x^2 + a)/((b*x^2 + a)^2*abs(b)))/b^3 - 1/2*(b^3*c^2/(b*x^2
+ a) - 2*a*b^2*c*d/(b*x^2 + a) + a^2*b*d^2/(b*x^2 + a))/b^4

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maple [A]  time = 0.01, size = 97, normalized size = 1.59 \[ \frac {d^{2} x^{2}}{2 b^{2}}-\frac {a^{2} d^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {a c d}{\left (b \,x^{2}+a \right ) b^{2}}-\frac {a \,d^{2} \ln \left (b \,x^{2}+a \right )}{b^{3}}-\frac {c^{2}}{2 \left (b \,x^{2}+a \right ) b}+\frac {c d \ln \left (b \,x^{2}+a \right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*d^2*x^2/b^2-1/2/b^3/(b*x^2+a)*a^2*d^2+1/b^2/(b*x^2+a)*a*d*c-1/2/b/(b*x^2+a)*c^2-1/b^3*ln(b*x^2+a)*d^2*a+1/
b^2*ln(b*x^2+a)*d*c

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maxima [A]  time = 1.04, size = 73, normalized size = 1.20 \[ \frac {d^{2} x^{2}}{2 \, b^{2}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {{\left (b c d - a d^{2}\right )} \log \left (b x^{2} + a\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*d^2*x^2/b^2 - 1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(b^4*x^2 + a*b^3) + (b*c*d - a*d^2)*log(b*x^2 + a)/b^3

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mupad [B]  time = 0.26, size = 77, normalized size = 1.26 \[ \frac {d^2\,x^2}{2\,b^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{2\,b\,\left (b^3\,x^2+a\,b^2\right )}-\frac {\ln \left (b\,x^2+a\right )\,\left (a\,d^2-b\,c\,d\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*x^2)/(2*b^2) - (a^2*d^2 + b^2*c^2 - 2*a*b*c*d)/(2*b*(a*b^2 + b^3*x^2)) - (log(a + b*x^2)*(a*d^2 - b*c*d))
/b^3

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sympy [A]  time = 0.76, size = 68, normalized size = 1.11 \[ \frac {- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {d^{2} x^{2}}{2 b^{2}} - \frac {d \left (a d - b c\right ) \log {\left (a + b x^{2} \right )}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a**2*d**2 + 2*a*b*c*d - b**2*c**2)/(2*a*b**3 + 2*b**4*x**2) + d**2*x**2/(2*b**2) - d*(a*d - b*c)*log(a + b*x
**2)/b**3

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