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

Optimal. Leaf size=100 \[ \frac{\left (-4 a B c-4 A b c+3 b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac{\sqrt{a+b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{8 c^2} \]

[Out]

-((3*b*B - 4*A*c - 2*B*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c^2) + ((3*b^2*B - 4*A*b*c - 4*a*B*c)*ArcTanh[(b + 2
*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(5/2))

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Rubi [A]  time = 0.0927491, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1251, 779, 621, 206} \[ \frac{\left (-4 a B c-4 A b c+3 b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac{\sqrt{a+b x^2+c x^4} \left (-4 A c+3 b B-2 B c x^2\right )}{8 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*B - 4*A*c - 2*B*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*c^2) + ((3*b^2*B - 4*A*b*c - 4*a*B*c)*ArcTanh[(b + 2
*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(5/2))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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

Mathematica [A]  time = 0.0476248, size = 101, normalized size = 1.01 \[ \frac{\left (-4 a B c-4 A b c+3 b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )+2 \sqrt{c} \sqrt{a+b x^2+c x^4} \left (4 A c-3 b B+2 B c x^2\right )}{16 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*(-3*b*B + 4*A*c + 2*B*c*x^2)*Sqrt[a + b*x^2 + c*x^4] + (3*b^2*B - 4*A*b*c - 4*a*B*c)*ArcTanh[(b + 2
*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*c^(5/2))

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Maple [B]  time = 0.015, size = 176, normalized size = 1.8 \begin{align*}{\frac{B{x}^{2}}{4\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,bB}{8\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{2}B}{16}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{aB}{4}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{A}{2\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{Ab}{4}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{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+a)^(1/2),x)

[Out]

1/4*B*x^2/c*(c*x^4+b*x^2+a)^(1/2)-3/8*B*b/c^2*(c*x^4+b*x^2+a)^(1/2)+3/16*B*b^2/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2
)+(c*x^4+b*x^2+a)^(1/2))-1/4*B*a/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+1/2*A/c*(c*x^4+b*x^2+
a)^(1/2)-1/4*A*b/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))

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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+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48283, size = 545, normalized size = 5.45 \begin{align*} \left [-\frac{{\left (3 \, B b^{2} - 4 \,{\left (B a + A b\right )} c\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (2 \, B c^{2} x^{2} - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{32 \, c^{3}}, -\frac{{\left (3 \, B b^{2} - 4 \,{\left (B a + A b\right )} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\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} + a}}{16 \, 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+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/32*((3*B*b^2 - 4*(B*a + A*b)*c)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 + 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*
x^2 + b)*sqrt(c) - 4*a*c) - 4*(2*B*c^2*x^2 - 3*B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^3, -1/16*((3*B*b^2
- 4*(B*a + A*b)*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c
)) - 2*(2*B*c^2*x^2 - 3*B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2 + a))/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{a + b x^{2} + c x^{4}}}\, 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+a)**(1/2),x)

[Out]

Integral(x**3*(A + B*x**2)/sqrt(a + b*x**2 + c*x**4), x)

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Giac [A]  time = 1.15289, size = 132, normalized size = 1.32 \begin{align*} \frac{1}{8} \, \sqrt{c x^{4} + b x^{2} + a}{\left (\frac{2 \, B x^{2}}{c} - \frac{3 \, B b - 4 \, A c}{c^{2}}\right )} - \frac{{\left (3 \, B b^{2} - 4 \, B a c - 4 \, A b c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\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+a)^(1/2),x, algorithm="giac")

[Out]

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