3.2.41 \(\int \frac {x^2 (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [141]

Optimal. Leaf size=59 \[ -\frac {(2 b B-3 A c) \sqrt {b x^2+c x^4}}{3 c^2 x}+\frac {B x \sqrt {b x^2+c x^4}}{3 c} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2064, 1602} \begin {gather*} \frac {B x \sqrt {b x^2+c x^4}}{3 c}-\frac {\sqrt {b x^2+c x^4} (2 b B-3 A c)}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 2064

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Dist[(a*d*(m
+ j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)), Int[(e*x)^m*(a*x^j + b*x^(j + n))^p, x
], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[
m + n + p*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 40, normalized size = 0.68 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-2 b B+3 A c+B c x^2\right )}{3 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.37, size = 42, normalized size = 0.71

method	result	size
trager	\(\frac {\left (B c \,x^{2}+3 A c -2 B b \right ) \sqrt {x^{4} c +b \,x^{2}}}{3 c^{2} x}\)	\(37\)
gosper	\(\frac {\left (c \,x^{2}+b \right ) \left (B c \,x^{2}+3 A c -2 B b \right ) x}{3 c^{2} \sqrt {x^{4} c +b \,x^{2}}}\)	\(42\)
default	\(\frac {\left (c \,x^{2}+b \right ) \left (B c \,x^{2}+3 A c -2 B b \right ) x}{3 c^{2} \sqrt {x^{4} c +b \,x^{2}}}\)	\(42\)
risch	\(\frac {x \left (c \,x^{2}+b \right ) \left (B c \,x^{2}+3 A c -2 B b \right )}{3 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 50, normalized size = 0.85 \begin {gather*} \frac {\sqrt {c x^{2} + b} A}{c} + \frac {{\left (c^{2} x^{4} - b c x^{2} - 2 \, b^{2}\right )} B}{3 \, \sqrt {c x^{2} + b} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.88, size = 36, normalized size = 0.61 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} - 2 \, B b + 3 \, A c\right )}}{3 \, c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.48, size = 67, normalized size = 1.14 \begin {gather*} \frac {{\left (2 \, B b^{\frac {3}{2}} - 3 \, A \sqrt {b} c\right )} \mathrm {sgn}\left (x\right )}{3 \, c^{2}} + \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}} B}{3 \, c^{2} \mathrm {sgn}\left (x\right )} - \frac {\sqrt {c x^{2} + b} {\left (B b - A c\right )}}{c^{2} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.19, size = 41, normalized size = 0.69 \begin {gather*} \frac {\left (\frac {3\,A\,c-2\,B\,b}{3\,c^2}+\frac {B\,x^2}{3\,c}\right )\,\sqrt {c\,x^4+b\,x^2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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