\(\int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx\) [368]

Optimal result

Integrand size = 19, antiderivative size = 24 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\frac {2 \sqrt {a x^3} \text {arcsinh}\left (x^{5/2}\right )}{5 x^{3/2}} \]

[Out]

2/5*arcsinh(x^(5/2))*(a*x^3)^(1/2)/x^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {15, 335, 281, 221} \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\frac {2 \sqrt {a x^3} \text {arcsinh}\left (x^{5/2}\right )}{5 x^{3/2}} \]

[In]

Int[Sqrt[a*x^3]/Sqrt[1 + x^5],x]

[Out]

(2*Sqrt[a*x^3]*ArcSinh[x^(5/2)])/(5*x^(3/2))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\frac {2 \sqrt {a x^3} \log \left (x^{5/2}+\sqrt {1+x^5}\right )}{5 x^{3/2}} \]

[In]

Integrate[Sqrt[a*x^3]/Sqrt[1 + x^5],x]

[Out]

(2*Sqrt[a*x^3]*Log[x^(5/2) + Sqrt[1 + x^5]])/(5*x^(3/2))

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71

method	result	size
meijerg	\(\frac {2 \,\operatorname {arcsinh}\left (x^{\frac {5}{2}}\right ) \sqrt {a \,x^{3}}}{5 x^{\frac {3}{2}}}\)	\(17\)

[In]

int((a*x^3)^(1/2)/(x^5+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/5*arcsinh(x^(5/2))*(a*x^3)^(1/2)/x^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (16) = 32\).

Time = 0.36 (sec) , antiderivative size = 98, normalized size of antiderivative = 4.08 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\left [\frac {1}{10} \, \sqrt {a} \log \left (-8 \, a x^{10} - 8 \, a x^{5} - 4 \, {\left (2 \, x^{6} + x\right )} \sqrt {x^{5} + 1} \sqrt {a x^{3}} \sqrt {a} - a\right ), -\frac {1}{5} \, \sqrt {-a} \arctan \left (\frac {{\left (2 \, x^{5} + 1\right )} \sqrt {x^{5} + 1} \sqrt {a x^{3}} \sqrt {-a}}{2 \, {\left (a x^{9} + a x^{4}\right )}}\right )\right ] \]

[In]

integrate((a*x^3)^(1/2)/(x^5+1)^(1/2),x, algorithm="fricas")

[Out]

[1/10*sqrt(a)*log(-8*a*x^10 - 8*a*x^5 - 4*(2*x^6 + x)*sqrt(x^5 + 1)*sqrt(a*x^3)*sqrt(a) - a), -1/5*sqrt(-a)*ar
ctan(1/2*(2*x^5 + 1)*sqrt(x^5 + 1)*sqrt(a*x^3)*sqrt(-a)/(a*x^9 + a*x^4))]

Sympy [F]

\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\int \frac {\sqrt {a x^{3}}}{\sqrt {\left (x + 1\right ) \left (x^{4} - x^{3} + x^{2} - x + 1\right )}}\, dx \]

[In]

integrate((a*x**3)**(1/2)/(x**5+1)**(1/2),x)

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\int { \frac {\sqrt {a x^{3}}}{\sqrt {x^{5} + 1}} \,d x } \]

[In]

integrate((a*x^3)^(1/2)/(x^5+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*x^3)/sqrt(x^5 + 1), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (16) = 32\).

Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=-\frac {2 \, a^{\frac {3}{2}} \log \left (-\sqrt {a x} a^{\frac {5}{2}} x^{2} + \sqrt {a^{6} x^{5} + a^{6}}\right ) \mathrm {sgn}\left (x\right )}{5 \, {\left | a \right |}} + \frac {2 \, a^{\frac {3}{2}} \log \left (a^{2} {\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{5 \, {\left | a \right |}} \]

[In]

integrate((a*x^3)^(1/2)/(x^5+1)^(1/2),x, algorithm="giac")

[Out]

-2/5*a^(3/2)*log(-sqrt(a*x)*a^(5/2)*x^2 + sqrt(a^6*x^5 + a^6))*sgn(x)/abs(a) + 2/5*a^(3/2)*log(a^2*abs(a))*sgn
(x)/abs(a)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^5}} \, dx=\int \frac {\sqrt {a\,x^3}}{\sqrt {x^5+1}} \,d x \]

[In]

int((a*x^3)^(1/2)/(x^5 + 1)^(1/2),x)

[Out]

int((a*x^3)^(1/2)/(x^5 + 1)^(1/2), x)