3.2.32 \(\int \frac {x^3 (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [132]

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

[Out]

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

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Rubi [A]
time = 0.11, 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 = {2059, 793, 634, 212} \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]

-1/8*((3*b*B - 4*A*c - 2*B*c*x^2)*Sqrt[b*x^2 + c*x^4])/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 212

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

Rule 634

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 793

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 2059

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} \text {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)) \text {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)) \text {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 = 0.10, size = 99, normalized size = 1.19 \begin {gather*} \frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (-3 b B+4 A c+2 B c x^2\right )+b (-3 b B+4 A c) \sqrt {b+c x^2} \log \left (-\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]*Log[-(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 = 0.39, size = 127, normalized size = 1.53

method	result	size
risch	\(\frac {x^{2} \left (2 B c \,x^{2}+4 A c -3 B b \right ) \left (c \,x^{2}+b \right )}{8 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (-\frac {b \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) A}{2 c^{\frac {3}{2}}}+\frac {3 b^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) B}{8 c^{\frac {5}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(119\)
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \left (2 B \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x^{3}+4 A \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x -3 B \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b x -4 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b \,c^{2}+3 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c \right )}{8 \sqrt {x^{4} c +b \,x^{2}}\, c^{\frac {7}{2}}}\)	\(127\)

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,method=_RETURNVERBOSE)

[Out]

1/8*x*(c*x^2+b)^(1/2)*(2*B*(c*x^2+b)^(1/2)*c^(5/2)*x^3+4*A*(c*x^2+b)^(1/2)*c^(5/2)*x-3*B*(c*x^2+b)^(1/2)*c^(3/
2)*b*x-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 = 0.29, 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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Fricas [A]
time = 1.67, 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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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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Giac [A]
time = 0.62, size = 112, normalized size = 1.35 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{2} + b} x {\left (\frac {2 \, B x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {3 \, B b c \mathrm {sgn}\left (x\right ) - 4 \, A c^{2} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} + \frac {{\left (3 \, B b^{2} \log \left ({\left | b \right |}\right ) - 4 \, A b c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {5}{2}}} - \frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\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="giac")

[Out]

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

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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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