3.222 \(\int x^3 \sqrt {b x^2+c x^4} \, dx\)

Optimal. Leaf size=91 \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c} \]

[Out]

1/6*(c*x^4+b*x^2)^(3/2)/c+1/16*b^3*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(5/2)-1/16*b*(2*c*x^2+b)*(c*x^4+
b*x^2)^(1/2)/c^2

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Rubi [A]  time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2018, 640, 612, 620, 206} \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[b*x^2 + c*x^4],x]

[Out]

-(b*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(16*c^2) + (b*x^2 + c*x^4)^(3/2)/(6*c) + (b^3*ArcTanh[(Sqrt[c]*x^2)/Sqr
t[b*x^2 + c*x^4]])/(16*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 612

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

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 640

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

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int x^3 \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c}-\frac {b \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{32 c^2}\\ &=-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^2}\\ &=-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {\left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 103, normalized size = 1.13 \[ \frac {x \sqrt {b+c x^2} \left (3 b^3 \log \left (\sqrt {c} \sqrt {b+c x^2}+c x\right )+\sqrt {c} x \sqrt {b+c x^2} \left (-3 b^2+2 b c x^2+8 c^2 x^4\right )\right )}{48 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(-3*b^2 + 2*b*c*x^2 + 8*c^2*x^4) + 3*b^3*Log[c*x + Sqrt[c]*Sqrt[
b + c*x^2]]))/(48*c^(5/2)*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 0.49, size = 167, normalized size = 1.84 \[ \left [\frac {3 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{3}}, -\frac {3 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (8 \, c^{3} x^{4} + 2 \, b c^{2} x^{2} - 3 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*b^3*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(8*c^3*x^4 + 2*b*c^2*x^2 - 3*b^2*c)
*sqrt(c*x^4 + b*x^2))/c^3, -1/48*(3*b^3*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (8*c^3*x^4
+ 2*b*c^2*x^2 - 3*b^2*c)*sqrt(c*x^4 + b*x^2))/c^3]

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giac [A]  time = 0.19, size = 85, normalized size = 0.93 \[ \frac {1}{48} \, {\left (2 \, {\left (4 \, x^{2} \mathrm {sgn}\relax (x) + \frac {b \mathrm {sgn}\relax (x)}{c}\right )} x^{2} - \frac {3 \, b^{2} \mathrm {sgn}\relax (x)}{c^{2}}\right )} \sqrt {c x^{2} + b} x - \frac {b^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\relax (x)}{16 \, c^{\frac {5}{2}}} + \frac {b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{32 \, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*x^2*sgn(x) + b*sgn(x)/c)*x^2 - 3*b^2*sgn(x)/c^2)*sqrt(c*x^2 + b)*x - 1/16*b^3*log(abs(-sqrt(c)*x +
sqrt(c*x^2 + b)))*sgn(x)/c^(5/2) + 1/32*b^3*log(abs(b))*sgn(x)/c^(5/2)

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maple [A]  time = 0.01, size = 104, normalized size = 1.14 \[ \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (8 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} x^{3}+3 b^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 \sqrt {c \,x^{2}+b}\, b^{2} \sqrt {c}\, x -6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \sqrt {c}\, x \right )}{48 \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(c*x^4+b*x^2)^(1/2)*(8*x^3*(c*x^2+b)^(3/2)*c^(3/2)-6*c^(1/2)*(c*x^2+b)^(3/2)*x*b+3*c^(1/2)*(c*x^2+b)^(1/2
)*x*b^2+3*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3)/x/(c*x^2+b)^(1/2)/c^(5/2)

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maxima [A]  time = 1.45, size = 97, normalized size = 1.07 \[ -\frac {\sqrt {c x^{4} + b x^{2}} b x^{2}}{8 \, c} + \frac {b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {5}{2}}} - \frac {\sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{6 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(c*x^4 + b*x^2)*b*x^2/c + 1/32*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2) - 1/16*sq
rt(c*x^4 + b*x^2)*b^2/c^2 + 1/6*(c*x^4 + b*x^2)^(3/2)/c

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mupad [B]  time = 4.36, size = 77, normalized size = 0.85 \[ \frac {b^3\,\ln \left (\frac {2\,c\,x^2+b}{\sqrt {c}}+2\,\sqrt {c\,x^4+b\,x^2}\right )}{32\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{48\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2 + c*x^4)^(1/2),x)

[Out]

(b^3*log((b + 2*c*x^2)/c^(1/2) + 2*(b*x^2 + c*x^4)^(1/2)))/(32*c^(5/2)) + ((b*x^2 + c*x^4)^(1/2)*(8*c^2*x^4 -
3*b^2 + 2*b*c*x^2))/(48*c^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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