3.48 \(\int \frac {x^3 (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=35 \[ \frac {B x^2}{2 c}-\frac {(b B-A c) \log \left (b+c x^2\right )}{2 c^2} \]

[Out]

1/2*B*x^2/c-1/2*(-A*c+B*b)*ln(c*x^2+b)/c^2

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1584, 444, 43} \[ \frac {B x^2}{2 c}-\frac {(b B-A c) \log \left (b+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(B*x^2)/(2*c) - ((b*B - A*c)*Log[b + c*x^2])/(2*c^2)

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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^3 \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac {x \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{b+c x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {B}{c}+\frac {-b B+A c}{c (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {B x^2}{2 c}-\frac {(b B-A c) \log \left (b+c x^2\right )}{2 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.89 \[ \frac {(A c-b B) \log \left (b+c x^2\right )+B c x^2}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(B*c*x^2 + (-(b*B) + A*c)*Log[b + c*x^2])/(2*c^2)

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fricas [A]  time = 0.76, size = 30, normalized size = 0.86 \[ \frac {B c x^{2} - {\left (B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/2*(B*c*x^2 - (B*b - A*c)*log(c*x^2 + b))/c^2

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giac [A]  time = 0.23, size = 32, normalized size = 0.91 \[ \frac {B x^{2}}{2 \, c} - \frac {{\left (B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/2*B*x^2/c - 1/2*(B*b - A*c)*log(abs(c*x^2 + b))/c^2

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maple [A]  time = 0.04, size = 40, normalized size = 1.14 \[ \frac {B \,x^{2}}{2 c}+\frac {A \ln \left (c \,x^{2}+b \right )}{2 c}-\frac {B b \ln \left (c \,x^{2}+b \right )}{2 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

1/2*B*x^2/c+1/2/c*ln(c*x^2+b)*A-1/2/c^2*ln(c*x^2+b)*b*B

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maxima [A]  time = 1.30, size = 31, normalized size = 0.89 \[ \frac {B x^{2}}{2 \, c} - \frac {{\left (B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/2*B*x^2/c - 1/2*(B*b - A*c)*log(c*x^2 + b)/c^2

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mupad [B]  time = 0.06, size = 31, normalized size = 0.89 \[ \frac {B\,x^2}{2\,c}+\frac {\ln \left (c\,x^2+b\right )\,\left (A\,c-B\,b\right )}{2\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

(B*x^2)/(2*c) + (log(b + c*x^2)*(A*c - B*b))/(2*c^2)

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sympy [A]  time = 0.25, size = 27, normalized size = 0.77 \[ \frac {B x^{2}}{2 c} - \frac {\left (- A c + B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

B*x**2/(2*c) - (-A*c + B*b)*log(b + c*x**2)/(2*c**2)

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