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

Optimal. Leaf size=61 \[ \frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 d^3}-\frac {b x^2 (b c-a d)}{2 d^2}+\frac {\left (a+b x^2\right )^2}{4 d} \]

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Rubi [A]  time = 0.05, 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} \begin {gather*} -\frac {b x^2 (b c-a d)}{2 d^2}+\frac {(b c-a d)^2 \log \left (c+d x^2\right )}{2 d^3}+\frac {\left (a+b x^2\right )^2}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*x^2)^2)/(c + d*x^2),x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.80 \begin {gather*} \frac {b d x^2 \left (4 a d-2 b c+b d x^2\right )+2 (b c-a d)^2 \log \left (c+d x^2\right )}{4 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x^2*(-2*b*c + 4*a*d + b*d*x^2) + 2*(b*c - a*d)^2*Log[c + d*x^2])/(4*d^3)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 66, normalized size = 1.08 \begin {gather*} \frac {b^{2} d^{2} x^{4} - 2 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{2} + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{4 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 67, normalized size = 1.10 \begin {gather*} \frac {b^{2} d x^{4} - 2 \, b^{2} c x^{2} + 4 \, a b d x^{2}}{4 \, d^{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 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 85, normalized size = 1.39 \begin {gather*} \frac {b^{2} x^{4}}{4 d}+\frac {a b \,x^{2}}{d}-\frac {b^{2} c \,x^{2}}{2 d^{2}}+\frac {a^{2} \ln \left (d \,x^{2}+c \right )}{2 d}-\frac {a b c \ln \left (d \,x^{2}+c \right )}{d^{2}}+\frac {b^{2} c^{2} \ln \left (d \,x^{2}+c \right )}{2 d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.03, size = 65, normalized size = 1.07 \begin {gather*} \frac {b^{2} d x^{4} - 2 \, {\left (b^{2} c - 2 \, a b d\right )} x^{2}}{4 \, d^{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 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 68, normalized size = 1.11 \begin {gather*} \frac {b^2\,x^4}{4\,d}-x^2\,\left (\frac {b^2\,c}{2\,d^2}-\frac {a\,b}{d}\right )+\frac {\ln \left (d\,x^2+c\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{2\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*x^2)^2)/(c + d*x^2),x)

[Out]

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

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sympy [A]  time = 0.36, size = 49, normalized size = 0.80 \begin {gather*} \frac {b^{2} x^{4}}{4 d} + x^{2} \left (\frac {a b}{d} - \frac {b^{2} c}{2 d^{2}}\right ) + \frac {\left (a d - b c\right )^{2} \log {\left (c + d x^{2} \right )}}{2 d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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