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

Optimal. Leaf size=94 \[ -\frac{x \sqrt{b x^2+c x^4} (4 b B-5 A c)}{15 c^2}+\frac{2 b \sqrt{b x^2+c x^4} (4 b B-5 A c)}{15 c^3 x}+\frac{B x^3 \sqrt{b x^2+c x^4}}{5 c} \]

[Out]

(2*b*(4*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(15*c^3*x) - ((4*b*B - 5*A*c)*x*Sqrt[b*x^2 + c*x^4])/(15*c^2) + (B*x
^3*Sqrt[b*x^2 + c*x^4])/(5*c)

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Rubi [A]  time = 0.193394, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2039, 2016, 1588} \[ -\frac{x \sqrt{b x^2+c x^4} (4 b B-5 A c)}{15 c^2}+\frac{2 b \sqrt{b x^2+c x^4} (4 b B-5 A c)}{15 c^3 x}+\frac{B x^3 \sqrt{b x^2+c x^4}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(4*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(15*c^3*x) - ((4*b*B - 5*A*c)*x*Sqrt[b*x^2 + c*x^4])/(15*c^2) + (B*x
^3*Sqrt[b*x^2 + c*x^4])/(5*c)

Rule 2039

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])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 1588

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]

Rubi steps

\begin{align*} \int \frac{x^4 \left (A+B x^2\right )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{B x^3 \sqrt{b x^2+c x^4}}{5 c}-\frac{(4 b B-5 A c) \int \frac{x^4}{\sqrt{b x^2+c x^4}} \, dx}{5 c}\\ &=-\frac{(4 b B-5 A c) x \sqrt{b x^2+c x^4}}{15 c^2}+\frac{B x^3 \sqrt{b x^2+c x^4}}{5 c}+\frac{(2 b (4 b B-5 A c)) \int \frac{x^2}{\sqrt{b x^2+c x^4}} \, dx}{15 c^2}\\ &=\frac{2 b (4 b B-5 A c) \sqrt{b x^2+c x^4}}{15 c^3 x}-\frac{(4 b B-5 A c) x \sqrt{b x^2+c x^4}}{15 c^2}+\frac{B x^3 \sqrt{b x^2+c x^4}}{5 c}\\ \end{align*}

Mathematica [A]  time = 0.042267, size = 63, normalized size = 0.67 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (-2 b c \left (5 A+2 B x^2\right )+c^2 x^2 \left (5 A+3 B x^2\right )+8 b^2 B\right )}{15 c^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(8*b^2*B - 2*b*c*(5*A + 2*B*x^2) + c^2*x^2*(5*A + 3*B*x^2)))/(15*c^3*x)

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Maple [A]  time = 0.005, size = 65, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -3\,B{c}^{2}{x}^{4}-5\,A{x}^{2}{c}^{2}+4\,B{x}^{2}bc+10\,Abc-8\,B{b}^{2} \right ) x}{15\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(c*x^2+b)*(-3*B*c^2*x^4-5*A*c^2*x^2+4*B*b*c*x^2+10*A*b*c-8*B*b^2)*x/c^3/(c*x^4+b*x^2)^(1/2)

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Maxima [A]  time = 1.31643, size = 112, normalized size = 1.19 \begin{align*} \frac{{\left (c^{2} x^{4} - b c x^{2} - 2 \, b^{2}\right )} A}{3 \, \sqrt{c x^{2} + b} c^{2}} + \frac{{\left (3 \, c^{3} x^{6} - b c^{2} x^{4} + 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} B}{15 \, \sqrt{c x^{2} + b} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c^2*x^4 - b*c*x^2 - 2*b^2)*A/(sqrt(c*x^2 + b)*c^2) + 1/15*(3*c^3*x^6 - b*c^2*x^4 + 4*b^2*c*x^2 + 8*b^3)*B
/(sqrt(c*x^2 + b)*c^3)

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Fricas [A]  time = 1.01806, size = 128, normalized size = 1.36 \begin{align*} \frac{{\left (3 \, B c^{2} x^{4} + 8 \, B b^{2} - 10 \, A b c -{\left (4 \, B b c - 5 \, A c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{15 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*B*c^2*x^4 + 8*B*b^2 - 10*A*b*c - (4*B*b*c - 5*A*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^3*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (A + B x^{2}\right )}{\sqrt{x^{2} \left (b + c x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*(A + B*x**2)/sqrt(x**2*(b + c*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{4}}{\sqrt{c x^{4} + b x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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