3.2.49 \(\int \frac {x (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [149]

Optimal. Leaf size=37 \[ -\frac {A b-(b B-2 A c) x^2}{b^2 \sqrt {b x^2+c x^4}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2059, 650} \begin {gather*} -\frac {A b-x^2 (b B-2 A c)}{b^2 \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 650

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*((b*d - 2*a*e + (2*c*
d - b*e)*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rule 2059

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 37, normalized size = 1.00 \begin {gather*} \frac {b B x^2-A \left (b+2 c x^2\right )}{b^2 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.39, size = 47, normalized size = 1.27

method	result	size
gosper	\(-\frac {x^{2} \left (c \,x^{2}+b \right ) \left (2 A c \,x^{2}-b B \,x^{2}+A b \right )}{b^{2} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\)	\(47\)
default	\(-\frac {x^{2} \left (c \,x^{2}+b \right ) \left (2 A c \,x^{2}-b B \,x^{2}+A b \right )}{b^{2} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\)	\(47\)
trager	\(-\frac {\left (2 A c \,x^{2}-b B \,x^{2}+A b \right ) \sqrt {x^{4} c +b \,x^{2}}}{\left (c \,x^{2}+b \right ) b^{2} x^{2}}\)	\(49\)
risch	\(-\frac {A \left (c \,x^{2}+b \right )}{b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {x^{2} \left (A c -B b \right )}{b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 65, normalized size = 1.76 \begin {gather*} -A {\left (\frac {2 \, c x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b}\right )} + \frac {B x^{2}}{\sqrt {c x^{4} + b x^{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.86, size = 49, normalized size = 1.32 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2}} {\left ({\left (B b - 2 \, A c\right )} x^{2} - A b\right )}}{b^{2} c x^{4} + b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.46, size = 65, normalized size = 1.76 \begin {gather*} \frac {2 \, A \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )} b \mathrm {sgn}\left (x\right )} + \frac {{\left (B b - A c\right )} x}{\sqrt {c x^{2} + b} b^{2} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.18, size = 53, normalized size = 1.43 \begin {gather*} -\frac {\left (\frac {A}{b}-x^2\,\left (\frac {B}{b}-\frac {2\,A\,c}{b^2}\right )\right )\,\sqrt {c\,x^4+b\,x^2}}{x\,\left (c\,x^3+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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