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

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

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Rubi [A]  time = 0.17, antiderivative size = 83, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.13, size = 97, normalized size = 1.17 \begin {gather*} \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 )}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.47, size = 91, normalized size = 1.10 \begin {gather*} \frac {\left (4 A b c-3 b^2 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{16 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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.43, size = 177, normalized size = 2.13 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.22, size = 91, normalized size = 1.10 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.05, size = 127, normalized size = 1.53 \begin {gather*} \frac {\sqrt {c \,x^{2}+b}\, \left (2 \sqrt {c \,x^{2}+b}\, B \,c^{\frac {5}{2}} x^{3}-4 A b \,c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 B \,b^{2} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+4 \sqrt {c \,x^{2}+b}\, A \,c^{\frac {5}{2}} x -3 \sqrt {c \,x^{2}+b}\, B b \,c^{\frac {3}{2}} x \right ) x}{8 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {7}{2}}} \end {gather*}

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(c^(1/2)*x+(c*x^2+b)^(1/2))*b*c^2+3*B*ln(c^(1/2)*x+(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 [A]  time = 1.46, size = 134, normalized size = 1.61 \begin {gather*} \frac {1}{16} \, {\left (\frac {4 \, \sqrt {c x^{4} + b x^{2}} x^{2}}{c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{c^{2}}\right )} B - \frac {1}{4} \, A {\left (\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} - \frac {2 \, \sqrt {c x^{4} + b x^{2}}}{c}\right )} \end {gather*}

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (B\,x^2+A\right )}{\sqrt {c\,x^4+b\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \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**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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