3.46 \(\int \frac {x^5 (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=54 \[ \frac {b (b B-A c) \log \left (b+c x^2\right )}{2 c^3}-\frac {x^2 (b B-A c)}{2 c^2}+\frac {B x^4}{4 c} \]

[Out]

-1/2*(-A*c+B*b)*x^2/c^2+1/4*B*x^4/c+1/2*b*(-A*c+B*b)*ln(c*x^2+b)/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 54, 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, 446, 77} \[ -\frac {x^2 (b B-A c)}{2 c^2}+\frac {b (b B-A c) \log \left (b+c x^2\right )}{2 c^3}+\frac {B x^4}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

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

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]]

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^5 \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac {x^3 \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x (A+B x)}{b+c x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {-b B+A c}{c^2}+\frac {B x}{c}+\frac {b (b B-A c)}{c^2 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) x^2}{2 c^2}+\frac {B x^4}{4 c}+\frac {b (b B-A c) \log \left (b+c x^2\right )}{2 c^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 47, normalized size = 0.87 \[ \frac {c x^2 \left (2 A c-2 b B+B c x^2\right )+2 b (b B-A c) \log \left (b+c x^2\right )}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 51, normalized size = 0.94 \[ \frac {B c^{2} x^{4} - 2 \, {\left (B b c - A c^{2}\right )} x^{2} + 2 \, {\left (B b^{2} - A b c\right )} \log \left (c x^{2} + b\right )}{4 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/4*(B*c^2*x^4 - 2*(B*b*c - A*c^2)*x^2 + 2*(B*b^2 - A*b*c)*log(c*x^2 + b))/c^3

________________________________________________________________________________________

giac [A]  time = 0.15, size = 52, normalized size = 0.96 \[ \frac {B c x^{4} - 2 \, B b x^{2} + 2 \, A c x^{2}}{4 \, c^{2}} + \frac {{\left (B b^{2} - A b c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 62, normalized size = 1.15 \[ \frac {B \,x^{4}}{4 c}+\frac {A \,x^{2}}{2 c}-\frac {B b \,x^{2}}{2 c^{2}}-\frac {A b \ln \left (c \,x^{2}+b \right )}{2 c^{2}}+\frac {B \,b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

1/4*B*x^4/c+1/2/c*A*x^2-1/2/c^2*B*x^2*b-1/2*b/c^2*ln(c*x^2+b)*A+1/2*b^2/c^3*ln(c*x^2+b)*B

________________________________________________________________________________________

maxima [A]  time = 1.33, size = 50, normalized size = 0.93 \[ \frac {B c x^{4} - 2 \, {\left (B b - A c\right )} x^{2}}{4 \, c^{2}} + \frac {{\left (B b^{2} - A b c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/4*(B*c*x^4 - 2*(B*b - A*c)*x^2)/c^2 + 1/2*(B*b^2 - A*b*c)*log(c*x^2 + b)/c^3

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 52, normalized size = 0.96 \[ x^2\,\left (\frac {A}{2\,c}-\frac {B\,b}{2\,c^2}\right )+\frac {\ln \left (c\,x^2+b\right )\,\left (B\,b^2-A\,b\,c\right )}{2\,c^3}+\frac {B\,x^4}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

x^2*(A/(2*c) - (B*b)/(2*c^2)) + (log(b + c*x^2)*(B*b^2 - A*b*c))/(2*c^3) + (B*x^4)/(4*c)

________________________________________________________________________________________

sympy [A]  time = 0.29, size = 46, normalized size = 0.85 \[ \frac {B x^{4}}{4 c} + \frac {b \left (- A c + B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{3}} + x^{2} \left (\frac {A}{2 c} - \frac {B b}{2 c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

B*x**4/(4*c) + b*(-A*c + B*b)*log(b + c*x**2)/(2*c**3) + x**2*(A/(2*c) - B*b/(2*c**2))

________________________________________________________________________________________