3.139 \(\int \frac{x^6 (A+B x^2)}{\sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{8 b^2 \sqrt{b x^2+c x^4} (6 b B-7 A c)}{105 c^4 x}-\frac{x^3 \sqrt{b x^2+c x^4} (6 b B-7 A c)}{35 c^2}+\frac{4 b x \sqrt{b x^2+c x^4} (6 b B-7 A c)}{105 c^3}+\frac{B x^5 \sqrt{b x^2+c x^4}}{7 c} \]

[Out]

(-8*b^2*(6*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(105*c^4*x) + (4*b*(6*b*B - 7*A*c)*x*Sqrt[b*x^2 + c*x^4])/(105*c^
3) - ((6*b*B - 7*A*c)*x^3*Sqrt[b*x^2 + c*x^4])/(35*c^2) + (B*x^5*Sqrt[b*x^2 + c*x^4])/(7*c)

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Rubi [A]  time = 0.239866, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2039, 2016, 1588} \[ -\frac{8 b^2 \sqrt{b x^2+c x^4} (6 b B-7 A c)}{105 c^4 x}-\frac{x^3 \sqrt{b x^2+c x^4} (6 b B-7 A c)}{35 c^2}+\frac{4 b x \sqrt{b x^2+c x^4} (6 b B-7 A c)}{105 c^3}+\frac{B x^5 \sqrt{b x^2+c x^4}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(-8*b^2*(6*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(105*c^4*x) + (4*b*(6*b*B - 7*A*c)*x*Sqrt[b*x^2 + c*x^4])/(105*c^
3) - ((6*b*B - 7*A*c)*x^3*Sqrt[b*x^2 + c*x^4])/(35*c^2) + (B*x^5*Sqrt[b*x^2 + c*x^4])/(7*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^6 \left (A+B x^2\right )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{B x^5 \sqrt{b x^2+c x^4}}{7 c}-\frac{(6 b B-7 A c) \int \frac{x^6}{\sqrt{b x^2+c x^4}} \, dx}{7 c}\\ &=-\frac{(6 b B-7 A c) x^3 \sqrt{b x^2+c x^4}}{35 c^2}+\frac{B x^5 \sqrt{b x^2+c x^4}}{7 c}+\frac{(4 b (6 b B-7 A c)) \int \frac{x^4}{\sqrt{b x^2+c x^4}} \, dx}{35 c^2}\\ &=\frac{4 b (6 b B-7 A c) x \sqrt{b x^2+c x^4}}{105 c^3}-\frac{(6 b B-7 A c) x^3 \sqrt{b x^2+c x^4}}{35 c^2}+\frac{B x^5 \sqrt{b x^2+c x^4}}{7 c}-\frac{\left (8 b^2 (6 b B-7 A c)\right ) \int \frac{x^2}{\sqrt{b x^2+c x^4}} \, dx}{105 c^3}\\ &=-\frac{8 b^2 (6 b B-7 A c) \sqrt{b x^2+c x^4}}{105 c^4 x}+\frac{4 b (6 b B-7 A c) x \sqrt{b x^2+c x^4}}{105 c^3}-\frac{(6 b B-7 A c) x^3 \sqrt{b x^2+c x^4}}{35 c^2}+\frac{B x^5 \sqrt{b x^2+c x^4}}{7 c}\\ \end{align*}

Mathematica [A]  time = 0.0599342, size = 85, normalized size = 0.65 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (8 b^2 c \left (7 A+3 B x^2\right )-2 b c^2 x^2 \left (14 A+9 B x^2\right )+3 c^3 x^4 \left (7 A+5 B x^2\right )-48 b^3 B\right )}{105 c^4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-48*b^3*B + 8*b^2*c*(7*A + 3*B*x^2) + 3*c^3*x^4*(7*A + 5*B*x^2) - 2*b*c^2*x^2*(14*A +
9*B*x^2)))/(105*c^4*x)

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Maple [A]  time = 0.005, size = 89, normalized size = 0.7 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ) \left ( 15\,B{c}^{3}{x}^{6}+21\,A{x}^{4}{c}^{3}-18\,B{x}^{4}b{c}^{2}-28\,A{x}^{2}b{c}^{2}+24\,B{x}^{2}{b}^{2}c+56\,A{b}^{2}c-48\,B{b}^{3} \right ) x}{105\,{c}^{4}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/105*(c*x^2+b)*(15*B*c^3*x^6+21*A*c^3*x^4-18*B*b*c^2*x^4-28*A*b*c^2*x^2+24*B*b^2*c*x^2+56*A*b^2*c-48*B*b^3)*x
/c^4/(c*x^4+b*x^2)^(1/2)

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Maxima [A]  time = 1.18615, size = 143, normalized size = 1.09 \begin{align*} \frac{{\left (3 \, c^{3} x^{6} - b c^{2} x^{4} + 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} A}{15 \, \sqrt{c x^{2} + b} c^{3}} + \frac{{\left (5 \, c^{4} x^{8} - b c^{3} x^{6} + 2 \, b^{2} c^{2} x^{4} - 8 \, b^{3} c x^{2} - 16 \, b^{4}\right )} B}{35 \, \sqrt{c x^{2} + b} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*c^3*x^6 - b*c^2*x^4 + 4*b^2*c*x^2 + 8*b^3)*A/(sqrt(c*x^2 + b)*c^3) + 1/35*(5*c^4*x^8 - b*c^3*x^6 + 2*b
^2*c^2*x^4 - 8*b^3*c*x^2 - 16*b^4)*B/(sqrt(c*x^2 + b)*c^4)

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Fricas [A]  time = 1.10809, size = 184, normalized size = 1.4 \begin{align*} \frac{{\left (15 \, B c^{3} x^{6} - 3 \,{\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} - 48 \, B b^{3} + 56 \, A b^{2} c + 4 \,{\left (6 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{105 \, c^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*B*c^3*x^6 - 3*(6*B*b*c^2 - 7*A*c^3)*x^4 - 48*B*b^3 + 56*A*b^2*c + 4*(6*B*b^2*c - 7*A*b*c^2)*x^2)*sqr
t(c*x^4 + b*x^2)/(c^4*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6} \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**6*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**6*(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^{6}}{\sqrt{c x^{4} + b x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^6/sqrt(c*x^4 + b*x^2), x)