3.367 \(\int \frac{\sqrt{a x^{13}}}{\sqrt{1+x^5}} \, dx\)

Optimal. Leaf size=50 \[ \frac{\sqrt{x^5+1} \sqrt{a x^{13}}}{5 x^4}-\frac{\sqrt{a x^{13}} \sinh ^{-1}\left (x^{5/2}\right )}{5 x^{13/2}} \]

[Out]

(Sqrt[a*x^13]*Sqrt[1 + x^5])/(5*x^4) - (Sqrt[a*x^13]*ArcSinh[x^(5/2)])/(5*x^(13/2))

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Rubi [A]  time = 0.012267, antiderivative size = 50, normalized size of antiderivative = 1., 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 = {15, 321, 329, 275, 215} \[ \frac{\sqrt{x^5+1} \sqrt{a x^{13}}}{5 x^4}-\frac{\sqrt{a x^{13}} \sinh ^{-1}\left (x^{5/2}\right )}{5 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^13]/Sqrt[1 + x^5],x]

[Out]

(Sqrt[a*x^13]*Sqrt[1 + x^5])/(5*x^4) - (Sqrt[a*x^13]*ArcSinh[x^(5/2)])/(5*x^(13/2))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 321

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 329

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]

Rule 275

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 215

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]

Rubi steps

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

Mathematica [A]  time = 0.0104912, size = 42, normalized size = 0.84 \[ \frac{\sqrt{a x^{13}} \left (x^{5/2} \sqrt{x^5+1}-\sinh ^{-1}\left (x^{5/2}\right )\right )}{5 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^13]/Sqrt[1 + x^5],x]

[Out]

(Sqrt[a*x^13]*(x^(5/2)*Sqrt[1 + x^5] - ArcSinh[x^(5/2)]))/(5*x^(13/2))

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Maple [A]  time = 0.039, size = 57, normalized size = 1.1 \begin{align*}{\frac{1}{5\,{x}^{4}}\sqrt{a{x}^{13}}\sqrt{{x}^{5}+1}}-{\frac{1}{5\,{x}^{7}}{\it Arcsinh} \left ({x}^{{\frac{5}{2}}} \right ) \sqrt{a{x}^{13}}\sqrt{ax \left ({x}^{5}+1 \right ) }{\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{x}^{5}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^13)^(1/2)/(x^5+1)^(1/2),x)

[Out]

1/5*(a*x^13)^(1/2)*(x^5+1)^(1/2)/x^4-1/5/a^(1/2)*arcsinh(x^(5/2))*(a*x^13)^(1/2)/x^7*(a*x*(x^5+1))^(1/2)/(x^5+
1)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a x^{13}}}{\sqrt{x^{5} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x^13)/sqrt(x^5 + 1), x)

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Fricas [B]  time = 1.86198, size = 381, normalized size = 7.62 \begin{align*} \left [\frac{\sqrt{a} x^{4} \log \left (-\frac{8 \, a x^{14} + 8 \, a x^{9} + a x^{4} - 4 \, \sqrt{a x^{13}}{\left (2 \, x^{5} + 1\right )} \sqrt{x^{5} + 1} \sqrt{a}}{x^{4}}\right ) + 4 \, \sqrt{a x^{13}} \sqrt{x^{5} + 1}}{20 \, x^{4}}, \frac{\sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{a x^{13}}{\left (2 \, x^{5} + 1\right )} \sqrt{x^{5} + 1} \sqrt{-a}}{2 \,{\left (a x^{14} + a x^{9}\right )}}\right ) + 2 \, \sqrt{a x^{13}} \sqrt{x^{5} + 1}}{10 \, x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/20*(sqrt(a)*x^4*log(-(8*a*x^14 + 8*a*x^9 + a*x^4 - 4*sqrt(a*x^13)*(2*x^5 + 1)*sqrt(x^5 + 1)*sqrt(a))/x^4) +
4*sqrt(a*x^13)*sqrt(x^5 + 1))/x^4, 1/10*(sqrt(-a)*x^4*arctan(1/2*sqrt(a*x^13)*(2*x^5 + 1)*sqrt(x^5 + 1)*sqrt(
-a)/(a*x^14 + a*x^9)) + 2*sqrt(a*x^13)*sqrt(x^5 + 1))/x^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a x^{13}}}{\sqrt{\left (x + 1\right ) \left (x^{4} - x^{3} + x^{2} - x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x**13)/sqrt((x + 1)*(x**4 - x**3 + x**2 - x + 1)), x)

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Giac [A]  time = 1.21185, size = 92, normalized size = 1.84 \begin{align*} \frac{a^{\frac{11}{2}} \log \left (-\sqrt{a x} a^{\frac{5}{2}} x^{2} + \sqrt{a^{6} x^{5} + a^{6}}\right )}{5 \,{\left | a \right |}^{5}} + \frac{\sqrt{a^{6} x^{5} + a^{6}} \sqrt{a x} x^{2}}{5 \, a^{2}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^(11/2)*log(-sqrt(a*x)*a^(5/2)*x^2 + sqrt(a^6*x^5 + a^6))/abs(a)^5 + 1/5*sqrt(a^6*x^5 + a^6)*sqrt(a*x)*x^
2/(a^2*abs(a))