3.3.75 \(\int \frac {x^5}{(b x^2+c x^4)^{3/2}} \, dx\) [275]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {2043, 666, 634, 212} \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 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 666

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 2043

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} \text {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 {\text {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 {\text {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 = 0.05, size = 64, normalized size = 1.16 \begin {gather*} -\frac {x \left (\sqrt {c} x+\sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{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]

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

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Maple [A]
time = 0.09, size = 63, normalized size = 1.15

method	result	size
default	\(-\frac {x^{3} \left (c \,x^{2}+b \right ) \left (x \,c^{\frac {3}{2}}-\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) c \sqrt {c \,x^{2}+b}\right )}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {5}{2}}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3*(c*x^2+b)*(x*c^(3/2)-ln(x*c^(1/2)+(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 = 0.34, 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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Fricas [A]
time = 0.38, 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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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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Giac [A]
time = 4.20, size = 57, normalized size = 1.04 \begin {gather*} \frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{2 \, c^{\frac {3}{2}}} - \frac {x}{\sqrt {c x^{2} + b} c \mathrm {sgn}\left (x\right )} - \frac {\log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (x\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="giac")

[Out]

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

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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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