Integrand size = 19, antiderivative size = 83 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\frac {2 \sqrt {a x^3} \sqrt {1+x^2}}{3 x}-\frac {\sqrt {a x^3} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{3 x^{3/2} \sqrt {1+x^2}} \] Output:
2/3*(a*x^3)^(1/2)*(x^2+1)^(1/2)/x-1/3*(a*x^3)^(1/2)*(1+x)*((x^2+1)/(1+x)^2 )^(1/2)*InverseJacobiAM(2*arctan(x^(1/2)),1/2*2^(1/2))/x^(3/2)/(x^2+1)^(1/ 2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\frac {2 \sqrt {a x^3} \left (\sqrt {1+x^2}-\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-x^2\right )\right )}{3 x} \] Input:
Integrate[Sqrt[a*x^3]/Sqrt[1 + x^2],x]
Output:
(2*Sqrt[a*x^3]*(Sqrt[1 + x^2] - Hypergeometric2F1[1/4, 1/2, 5/4, -x^2]))/( 3*x)
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {34, 262, 266, 761}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {a x^3}}{\sqrt {x^2+1}} \, dx\) |
\(\Big \downarrow \) 34 |
\(\displaystyle \frac {\sqrt {a x^3} \int \frac {x^{3/2}}{\sqrt {x^2+1}}dx}{x^{3/2}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\sqrt {a x^3} \left (\frac {2}{3} \sqrt {x} \sqrt {x^2+1}-\frac {1}{3} \int \frac {1}{\sqrt {x} \sqrt {x^2+1}}dx\right )}{x^{3/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\sqrt {a x^3} \left (\frac {2}{3} \sqrt {x} \sqrt {x^2+1}-\frac {2}{3} \int \frac {1}{\sqrt {x^2+1}}d\sqrt {x}\right )}{x^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\sqrt {a x^3} \left (\frac {2}{3} \sqrt {x} \sqrt {x^2+1}-\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{3 \sqrt {x^2+1}}\right )}{x^{3/2}}\) |
Input:
Int[Sqrt[a*x^3]/Sqrt[1 + x^2],x]
Output:
(Sqrt[a*x^3]*((2*Sqrt[x]*Sqrt[1 + x^2])/3 - ((1 + x)*Sqrt[(1 + x^2)/(1 + x )^2]*EllipticF[2*ArcTan[Sqrt[x]], 1/2])/(3*Sqrt[1 + x^2])))/x^(3/2)
Int[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F racPart[p]/x^(m*FracPart[p])) Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x ] && !IntegerQ[p]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.27
method | result | size |
meijerg | \(\frac {2 \sqrt {a \,x^{3}}\, x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {9}{4}\right ], -x^{2}\right )}{5}\) | \(22\) |
default | \(-\frac {\sqrt {a \,x^{3}}\, \left (i \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \operatorname {EllipticF}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-2 x^{3}-2 x \right )}{3 x^{2} \sqrt {x^{2}+1}}\) | \(76\) |
risch | \(\frac {2 \sqrt {a \,x^{3}}\, \sqrt {x^{2}+1}}{3 x}-\frac {i \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {i \left (x -i\right )}\, \sqrt {i x}\, \operatorname {EllipticF}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {a \,x^{3}}\, \sqrt {a x \left (x^{2}+1\right )}}{3 \sqrt {a \,x^{3}+x a}\, x^{2} \sqrt {x^{2}+1}}\) | \(104\) |
Input:
int((a*x^3)^(1/2)/(x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/5*(a*x^3)^(1/2)*x*hypergeom([1/2,5/4],[9/4],-x^2)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.37 \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=-\frac {2 \, {\left (\sqrt {a} x {\rm weierstrassPInverse}\left (-4, 0, x\right ) - \sqrt {a x^{3}} \sqrt {x^{2} + 1}\right )}}{3 \, x} \] Input:
integrate((a*x^3)^(1/2)/(x^2+1)^(1/2),x, algorithm="fricas")
Output:
-2/3*(sqrt(a)*x*weierstrassPInverse(-4, 0, x) - sqrt(a*x^3)*sqrt(x^2 + 1)) /x
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\int \frac {\sqrt {a x^{3}}}{\sqrt {x^{2} + 1}}\, dx \] Input:
integrate((a*x**3)**(1/2)/(x**2+1)**(1/2),x)
Output:
Integral(sqrt(a*x**3)/sqrt(x**2 + 1), x)
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\int { \frac {\sqrt {a x^{3}}}{\sqrt {x^{2} + 1}} \,d x } \] Input:
integrate((a*x^3)^(1/2)/(x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a*x^3)/sqrt(x^2 + 1), x)
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\int { \frac {\sqrt {a x^{3}}}{\sqrt {x^{2} + 1}} \,d x } \] Input:
integrate((a*x^3)^(1/2)/(x^2+1)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(a*x^3)/sqrt(x^2 + 1), x)
Timed out. \[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\int \frac {\sqrt {a\,x^3}}{\sqrt {x^2+1}} \,d x \] Input:
int((a*x^3)^(1/2)/(x^2 + 1)^(1/2),x)
Output:
int((a*x^3)^(1/2)/(x^2 + 1)^(1/2), x)
\[ \int \frac {\sqrt {a x^3}}{\sqrt {1+x^2}} \, dx=\frac {\sqrt {a}\, \left (2 \sqrt {x}\, \sqrt {x^{2}+1}-\left (\int \frac {\sqrt {x}\, \sqrt {x^{2}+1}}{x^{3}+x}d x \right )\right )}{3} \] Input:
int((a*x^3)^(1/2)/(x^2+1)^(1/2),x)
Output:
(sqrt(a)*(2*sqrt(x)*sqrt(x**2 + 1) - int((sqrt(x)*sqrt(x**2 + 1))/(x**3 + x),x)))/3