3.14.14 \(\int \frac {x^{13/2}}{\sqrt {1+x^5}} \, dx\) [1314]

Optimal. Leaf size=29 \[ \frac {1}{5} x^{5/2} \sqrt {1+x^5}-\frac {1}{5} \sinh ^{-1}\left (x^{5/2}\right ) \]

[Out]

-1/5*arcsinh(x^(5/2))+1/5*x^(5/2)*(x^5+1)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 29, 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 = {327, 335, 281, 221} \begin {gather*} \frac {1}{5} x^{5/2} \sqrt {x^5+1}-\frac {1}{5} \sinh ^{-1}\left (x^{5/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(13/2)/Sqrt[1 + x^5],x]

[Out]

(x^(5/2)*Sqrt[1 + x^5])/5 - ArcSinh[x^(5/2)]/5

Rule 221

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

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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

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

Rubi steps

\begin {align*} \int \frac {x^{13/2}}{\sqrt {1+x^5}} \, dx &=\frac {1}{5} x^{5/2} \sqrt {1+x^5}-\frac {1}{2} \int \frac {x^{3/2}}{\sqrt {1+x^5}} \, dx\\ &=\frac {1}{5} x^{5/2} \sqrt {1+x^5}-\text {Subst}\left (\int \frac {x^4}{\sqrt {1+x^{10}}} \, dx,x,\sqrt {x}\right )\\ &=\frac {1}{5} x^{5/2} \sqrt {1+x^5}-\frac {1}{5} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^{5/2}\right )\\ &=\frac {1}{5} x^{5/2} \sqrt {1+x^5}-\frac {1}{5} \sinh ^{-1}\left (x^{5/2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 39, normalized size = 1.34 \begin {gather*} \frac {1}{5} x^{5/2} \sqrt {1+x^5}-\frac {1}{5} \tanh ^{-1}\left (\frac {x^{5/2}}{\sqrt {1+x^5}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(13/2)/Sqrt[1 + x^5],x]

[Out]

(x^(5/2)*Sqrt[1 + x^5])/5 - ArcTanh[x^(5/2)/Sqrt[1 + x^5]]/5

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Maple [A]
time = 0.17, size = 30, normalized size = 1.03

method	result	size
meijerg	\(\frac {\sqrt {\pi }\, x^{\frac {5}{2}} \sqrt {x^{5}+1}-\sqrt {\pi }\, \arcsinh \left (x^{\frac {5}{2}}\right )}{5 \sqrt {\pi }}\)	\(30\)
risch	\(\frac {x^{\frac {5}{2}} \sqrt {x^{5}+1}}{5}-\frac {\arcsinh \left (x^{\frac {5}{2}}\right ) \sqrt {x \left (x^{5}+1\right )}}{5 \sqrt {x}\, \sqrt {x^{5}+1}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(x^5+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/5/Pi^(1/2)*(Pi^(1/2)*x^(5/2)*(x^5+1)^(1/2)-Pi^(1/2)*arcsinh(x^(5/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (19) = 38\).
time = 0.30, size = 58, normalized size = 2.00 \begin {gather*} \frac {\sqrt {x^{5} + 1}}{5 \, x^{\frac {5}{2}} {\left (\frac {x^{5} + 1}{x^{5}} - 1\right )}} - \frac {1}{10} \, \log \left (\frac {\sqrt {x^{5} + 1}}{x^{\frac {5}{2}}} + 1\right ) + \frac {1}{10} \, \log \left (\frac {\sqrt {x^{5} + 1}}{x^{\frac {5}{2}}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(x^5+1)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(x^5 + 1)/(x^(5/2)*((x^5 + 1)/x^5 - 1)) - 1/10*log(sqrt(x^5 + 1)/x^(5/2) + 1) + 1/10*log(sqrt(x^5 + 1)
/x^(5/2) - 1)

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Fricas [A]
time = 0.39, size = 35, normalized size = 1.21 \begin {gather*} \frac {1}{5} \, \sqrt {x^{5} + 1} x^{\frac {5}{2}} + \frac {1}{10} \, \log \left (-2 \, x^{5} + 2 \, \sqrt {x^{5} + 1} x^{\frac {5}{2}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(x^5+1)^(1/2),x, algorithm="fricas")

[Out]

1/5*sqrt(x^5 + 1)*x^(5/2) + 1/10*log(-2*x^5 + 2*sqrt(x^5 + 1)*x^(5/2) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(13/2)/(x**5+1)**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3655 deep

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Giac [A]
time = 0.96, size = 29, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, \sqrt {x^{5} + 1} x^{\frac {5}{2}} + \frac {1}{5} \, \log \left (-x^{\frac {5}{2}} + \sqrt {x^{5} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(x^5+1)^(1/2),x, algorithm="giac")

[Out]

1/5*sqrt(x^5 + 1)*x^(5/2) + 1/5*log(-x^(5/2) + sqrt(x^5 + 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^{13/2}}{\sqrt {x^5+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(x^5 + 1)^(1/2),x)

[Out]

int(x^(13/2)/(x^5 + 1)^(1/2), x)

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