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

Optimal. Leaf size=86 \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c} \]

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Rubi [A]  time = 0.10, antiderivative size = 86, 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, 670, 640, 620, 206} \begin {gather*} \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[b*x^2 + c*x^4],x]

[Out]

(-3*b*Sqrt[b*x^2 + c*x^4])/(8*c^2) + (x^2*Sqrt[b*x^2 + c*x^4])/(4*c) + (3*b^2*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 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 \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac {x^2 \sqrt {b x^2+c x^4}}{4 c}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{8 c}\\ &=-\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=-\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c}+\frac {\left (3 b^2\right ) \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 {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c}+\frac {3 b^2 \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 = 0.04, size = 89, normalized size = 1.03 \begin {gather*} \frac {x \left (\sqrt {c} x \left (-3 b^2-b c x^2+2 c^2 x^4\right )+3 b^2 \sqrt {b+c x^2} \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^5/Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*(Sqrt[c]*x*(-3*b^2 - b*c*x^2 + 2*c^2*x^4) + 3*b^2*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 = 0.22, size = 82, normalized size = 0.95 \begin {gather*} \frac {\left (2 c x^2-3 b\right ) \sqrt {b x^2+c x^4}}{8 c^2}-\frac {3 b^2 \log \left (-2 c^{5/2} \sqrt {b x^2+c x^4}+b c^2+2 c^3 x^2\right )}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/Sqrt[b*x^2 + c*x^4],x]

[Out]

((-3*b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(8*c^2) - (3*b^2*Log[b*c^2 + 2*c^3*x^2 - 2*c^(5/2)*Sqrt[b*x^2 + c*x^4]]
)/(16*c^(5/2))

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fricas [A]  time = 0.78, size = 145, normalized size = 1.69 \begin {gather*} \left [\frac {3 \, b^{2} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{16 \, c^{3}}, -\frac {3 \, b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} - 3 \, b c\right )}}{8 \, c^{3}}\right ] \end {gather*}

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]

[1/16*(3*b^2*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*sqrt(c*x^4 + b*x^2)*(2*c^2*x^2 - 3*
b*c))/c^3, -1/8*(3*b^2*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - sqrt(c*x^4 + b*x^2)*(2*c^2*
x^2 - 3*b*c))/c^3]

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giac [A]  time = 0.22, size = 73, normalized size = 0.85 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{4} + b x^{2}} {\left (\frac {2 \, x^{2}}{c} - \frac {3 \, b}{c^{2}}\right )} - \frac {3 \, b^{2} \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^5/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/8*sqrt(c*x^4 + b*x^2)*(2*x^2/c - 3*b/c^2) - 3/16*b^2*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 = 0.01, size = 85, normalized size = 0.99 \begin {gather*} \frac {\sqrt {c \,x^{2}+b}\, \left (2 \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x^{3}+3 b^{2} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-3 \sqrt {c \,x^{2}+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^5/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/8*x*(c*x^2+b)^(1/2)*(2*x^3*(c*x^2+b)^(1/2)*c^(5/2)-3*c^(3/2)*(c*x^2+b)^(1/2)*x*b+3*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 = 1.45, size = 76, normalized size = 0.88 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2}} x^{2}}{4 \, c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{8 \, c^{2}} \end {gather*}

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{\sqrt {c\,x^4+b\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^5/(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^{5}}{\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**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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