3.2.57 \(\int \frac {x^5}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=55 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}-\frac {x^2}{c \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.09, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2018, 652, 620, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}-\frac {x^2}{c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x^2/(c*Sqrt[b*x^2 + c*x^4])) + ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]]/c^(3/2)

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 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{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] &&  !IntegerQ[p] && LtQ[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 \frac {x^5}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{2 c}\\ &=-\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{c}\\ &=-\frac {x^2}{c \sqrt {b x^2+c x^4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 66, normalized size = 1.20 \begin {gather*} \frac {\sqrt {b} x \sqrt {\frac {c x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )-\sqrt {c} x^2}{c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-(Sqrt[c]*x^2) + Sqrt[b]*x*Sqrt[1 + (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]])/(c^(3/2)*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.31, size = 74, normalized size = 1.35 \begin {gather*} -\frac {\log \left (-2 c^{3/2} \sqrt {b x^2+c x^4}+b c+2 c^2 x^2\right )}{2 c^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.65, size = 150, normalized size = 2.73 \begin {gather*} \left [\frac {{\left (c x^{2} + b\right )} \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}} c}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}}, -\frac {{\left (c x^{2} + b\right )} \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}} c}{c^{3} x^{2} + b c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{-2,[1]%%%},[2,2]%%%}+%%%{%%{[4,0]:[1,0,%%%{-1,[1]%%%}]%%},[
1,3]%%%}+%%%{-2,[0,4]%%%} / %%%{%%%{1,[2]%%%},[2,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1
]%%%}+%%%{%%%{1,[1]%%%},[0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.01, size = 63, normalized size = 1.15 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (c^{\frac {3}{2}} x -\sqrt {c \,x^{2}+b}\, c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )\right ) x^{3}}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^2)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.47, size = 54, normalized size = 0.98 \begin {gather*} -\frac {x^{2}}{\sqrt {c x^{4} + b x^{2}} c} + \frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.33, size = 55, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,c^{3/2}}-\frac {x^2}{c\,\sqrt {c\,x^4+b\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2 + c*x^4)^(3/2),x)

[Out]

log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2))/(2*c^(3/2)) - x^2/(c*(b*x^2 + c*x^4)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**5/(x**2*(b + c*x**2))**(3/2), x)

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