3.9.57 \(\int \frac {x^5}{(a+b x^4)^{3/2}} \, dx\) [857]

Optimal. Leaf size=52 \[ -\frac {x^2}{2 b \sqrt {a+b x^4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {281, 294, 223, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {x^2}{2 b \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^4)^(3/2),x]

[Out]

-1/2*x^2/(b*Sqrt[a + b*x^4]) + ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]/(2*b^(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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 52, normalized size = 1.00 \begin {gather*} -\frac {x^2}{2 b \sqrt {a+b x^4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {b} x^2}\right )}{2 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^4)^(3/2),x]

[Out]

-1/2*x^2/(b*Sqrt[a + b*x^4]) + ArcTanh[Sqrt[a + b*x^4]/(Sqrt[b]*x^2)]/(2*b^(3/2))

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Maple [A]
time = 0.15, size = 42, normalized size = 0.81

method	result	size
default	\(-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\)	\(42\)
elliptic	\(-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 63, normalized size = 1.21 \begin {gather*} -\frac {x^{2}}{2 \, \sqrt {b x^{4} + a} b} - \frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{4} + a}}{x^{2}}}{\sqrt {b} + \frac {\sqrt {b x^{4} + a}}{x^{2}}}\right )}{4 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 138, normalized size = 2.65 \begin {gather*} \left [-\frac {2 \, \sqrt {b x^{4} + a} b x^{2} - {\left (b x^{4} + a\right )} \sqrt {b} \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right )}{4 \, {\left (b^{3} x^{4} + a b^{2}\right )}}, -\frac {\sqrt {b x^{4} + a} b x^{2} + {\left (b x^{4} + a\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{2}}{\sqrt {b x^{4} + a}}\right )}{2 \, {\left (b^{3} x^{4} + a b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.79, size = 44, normalized size = 0.85 \begin {gather*} \frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh(sqrt(b)*x**2/sqrt(a))/(2*b**(3/2)) - x**2/(2*sqrt(a)*b*sqrt(1 + b*x**4/a))

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Giac [A]
time = 1.68, size = 43, normalized size = 0.83 \begin {gather*} -\frac {x^{2}}{2 \, \sqrt {b x^{4} + a} b} - \frac {\log \left ({\left | -\sqrt {b} x^{2} + \sqrt {b x^{4} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^5}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^4)^(3/2),x)

[Out]

int(x^5/(a + b*x^4)^(3/2), x)

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