3.221 \(\int x^5 \sqrt {b x^2+c x^4} \, dx\)

Optimal. Leaf size=119 \[ -\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}+\frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c} \]

[Out]

-5/48*b*(c*x^4+b*x^2)^(3/2)/c^2+1/8*x^2*(c*x^4+b*x^2)^(3/2)/c-5/128*b^4*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2
))/c^(7/2)+5/128*b^2*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^3

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Rubi [A]  time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2018, 670, 640, 612, 620, 206} \[ \frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[b*x^2 + c*x^4],x]

[Out]

(5*b^2*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(128*c^3) - (5*b*(b*x^2 + c*x^4)^(3/2))/(48*c^2) + (x^2*(b*x^2 + c*x
^4)^(3/2))/(8*c) - (5*b^4*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(128*c^(7/2))

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 670

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

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^5 \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {(5 b) \operatorname {Subst}\left (\int x \sqrt {b x+c x^2} \, dx,x,x^2\right )}{16 c}\\ &=-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{32 c^2}\\ &=\frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {\left (5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{256 c^3}\\ &=\frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {\left (5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^3}\\ &=\frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 114, normalized size = 0.96 \[ \frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (15 b^3-10 b^2 c x^2+8 b c^2 x^4+48 c^3 x^6\right )-15 b^4 \log \left (\sqrt {c} \sqrt {b+c x^2}+c x\right )\right )}{384 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(15*b^3 - 10*b^2*c*x^2 + 8*b*c^2*x^4 + 48*c^3*x^6) - 15*b^4*Log[
c*x + Sqrt[c]*Sqrt[b + c*x^2]]))/(384*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 0.60, size = 188, normalized size = 1.58 \[ \left [\frac {15 \, b^{4} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} - 10 \, b^{2} c^{2} x^{2} + 15 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, c^{4}}, \frac {15 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} - 10 \, b^{2} c^{2} x^{2} + 15 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(15*b^4*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(48*c^4*x^6 + 8*b*c^3*x^4 - 10*b^
2*c^2*x^2 + 15*b^3*c)*sqrt(c*x^4 + b*x^2))/c^4, 1/384*(15*b^4*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*
x^2 + b)) + (48*c^4*x^6 + 8*b*c^3*x^4 - 10*b^2*c^2*x^2 + 15*b^3*c)*sqrt(c*x^4 + b*x^2))/c^4]

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giac [A]  time = 0.19, size = 101, normalized size = 0.85 \[ \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, x^{2} \mathrm {sgn}\relax (x) + \frac {b \mathrm {sgn}\relax (x)}{c}\right )} x^{2} - \frac {5 \, b^{2} \mathrm {sgn}\relax (x)}{c^{2}}\right )} x^{2} + \frac {15 \, b^{3} \mathrm {sgn}\relax (x)}{c^{3}}\right )} \sqrt {c x^{2} + b} x + \frac {5 \, b^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\relax (x)}{128 \, c^{\frac {7}{2}}} - \frac {5 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{256 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/384*(2*(4*(6*x^2*sgn(x) + b*sgn(x)/c)*x^2 - 5*b^2*sgn(x)/c^2)*x^2 + 15*b^3*sgn(x)/c^3)*sqrt(c*x^2 + b)*x + 5
/128*b^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/c^(7/2) - 5/256*b^4*log(abs(b))*sgn(x)/c^(7/2)

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maple [A]  time = 0.02, size = 124, normalized size = 1.04 \[ \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (48 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{5}-40 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,c^{\frac {3}{2}} x^{3}-15 b^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 \sqrt {c \,x^{2}+b}\, b^{3} \sqrt {c}\, x +30 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}\, x \right )}{384 \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*x^4+b*x^2)^(1/2),x)

[Out]

1/384*(c*x^4+b*x^2)^(1/2)*(48*x^5*(c*x^2+b)^(3/2)*c^(5/2)-40*(c*x^2+b)^(3/2)*c^(3/2)*x^3*b+30*(c*x^2+b)^(3/2)*
c^(1/2)*x*b^2-15*(c*x^2+b)^(1/2)*c^(1/2)*x*b^3-15*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^4)/x/(c*x^2+b)^(1/2)/c^(7/2)

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maxima [A]  time = 1.46, size = 121, normalized size = 1.02 \[ \frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{64 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{8 \, c} - \frac {5 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{128 \, c^{3}} - \frac {5 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{48 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

5/64*sqrt(c*x^4 + b*x^2)*b^2*x^2/c^2 + 1/8*(c*x^4 + b*x^2)^(3/2)*x^2/c - 5/256*b^4*log(2*c*x^2 + b + 2*sqrt(c*
x^4 + b*x^2)*sqrt(c))/c^(7/2) + 5/128*sqrt(c*x^4 + b*x^2)*b^3/c^3 - 5/48*(c*x^4 + b*x^2)^(3/2)*b/c^2

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mupad [B]  time = 4.68, size = 105, normalized size = 0.88 \[ \frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {2\,c\,x^2+b}{\sqrt {c}}+2\,\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{16\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2 + c*x^4)^(1/2),x)

[Out]

(x^2*(b*x^2 + c*x^4)^(3/2))/(8*c) - (5*b*((b^3*log((b + 2*c*x^2)/c^(1/2) + 2*(b*x^2 + c*x^4)^(1/2)))/(16*c^(5/
2)) + ((b*x^2 + c*x^4)^(1/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(24*c^2)))/(16*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{5} \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**5*sqrt(x**2*(b + c*x**2)), x)

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