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

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

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Rubi [A]  time = 0.19, antiderivative size = 94, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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])

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*}

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Mathematica [A]  time = 0.07, size = 63, normalized size = 0.67 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.09, size = 63, normalized size = 0.67 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-10 A b c+5 A c^2 x^2+8 b^2 B-4 b B c x^2+3 B c^2 x^4\right )}{15 c^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 59, normalized size = 0.63 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{2} + A\right )} x^{4}}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \end {gather*}

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)

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maple [A]  time = 0.05, size = 65, normalized size = 0.69 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-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}\right ) x}{15 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{3}} \end {gather*}

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 = 1.54, size = 83, normalized size = 0.88 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.23, size = 64, normalized size = 0.68 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {8\,B\,b^2-10\,A\,b\,c}{15\,c^3}+\frac {x^2\,\left (5\,A\,c^2-4\,B\,b\,c\right )}{15\,c^3}+\frac {B\,x^4}{5\,c}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \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**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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