3.132 \(\int \frac{x^3 (A+B x^2)}{\sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=83 \[ \frac{b (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac{\sqrt{b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{8 c^2} \]

[Out]

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

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Rubi [A]  time = 0.171246, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2034, 779, 620, 206} \[ \frac{b (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac{\sqrt{b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{8 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2034

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]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^3 \left (A+B x^2\right )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (A+B x)}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (3 b B-4 A c-2 B c x^2\right ) \sqrt{b x^2+c x^4}}{8 c^2}+\frac{(b (3 b B-4 A c)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac{\left (3 b B-4 A c-2 B c x^2\right ) \sqrt{b x^2+c x^4}}{8 c^2}+\frac{(b (3 b B-4 A c)) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^2}\\ &=-\frac{\left (3 b B-4 A c-2 B c x^2\right ) \sqrt{b x^2+c x^4}}{8 c^2}+\frac{b (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0901898, size = 97, normalized size = 1.17 \[ \frac{x \left (\sqrt{c} x \left (b+c x^2\right ) \left (4 A c-3 b B+2 B c x^2\right )+b \sqrt{b+c x^2} (3 b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b+c x^2}}\right )\right )}{8 c^{5/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 127, normalized size = 1.5 \begin{align*}{\frac{x}{8}\sqrt{c{x}^{2}+b} \left ( 2\,B{c}^{5/2}\sqrt{c{x}^{2}+b}{x}^{3}+4\,A{c}^{5/2}\sqrt{c{x}^{2}+b}x-3\,B{c}^{3/2}\sqrt{c{x}^{2}+b}xb-4\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) b{c}^{2}+3\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{2}c \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(c*x^2+b)^(1/2)*(2*B*c^(5/2)*(c*x^2+b)^(1/2)*x^3+4*A*c^(5/2)*(c*x^2+b)^(1/2)*x-3*B*c^(3/2)*(c*x^2+b)^(1/
2)*x*b-4*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b*c^2+3*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^2*c)/(c*x^4+b*x^2)^(1/2)/c^
(7/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.0668, size = 404, normalized size = 4.87 \begin{align*} \left [-\frac{{\left (3 \, B b^{2} - 4 \, A b c\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \,{\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{16 \, c^{3}}, -\frac{{\left (3 \, B b^{2} - 4 \, A b c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) -{\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{8 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*((3*B*b^2 - 4*A*b*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(2*B*c^2*x^2 - 3*B*b
*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^3, -1/8*((3*B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)
/(c*x^2 + b)) - (2*B*c^2*x^2 - 3*B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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**3*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.23462, size = 123, normalized size = 1.48 \begin{align*} \frac{1}{8} \, \sqrt{c x^{4} + b x^{2}}{\left (\frac{2 \, B x^{2}}{c} - \frac{3 \, B b - 4 \, A c}{c^{2}}\right )} - \frac{{\left (3 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2}}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

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