Integrand size = 17, antiderivative size = 131 \[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\frac {2 \sqrt {a x} \sqrt {1+x^2}}{1+x}-\frac {2 \sqrt {a} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} E\left (2 \arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right )|\frac {1}{2}\right )}{\sqrt {1+x^2}}+\frac {\sqrt {a} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right ),\frac {1}{2}\right )}{\sqrt {1+x^2}} \] Output:
2*(a*x)^(1/2)*(x^2+1)^(1/2)/(1+x)-2*a^(1/2)*(1+x)*((x^2+1)/(1+x)^2)^(1/2)* EllipticE(sin(2*arctan((a*x)^(1/2)/a^(1/2))),1/2*2^(1/2))/(x^2+1)^(1/2)+a^ (1/2)*(1+x)*((x^2+1)/(1+x)^2)^(1/2)*InverseJacobiAM(2*arctan((a*x)^(1/2)/a ^(1/2)),1/2*2^(1/2))/(x^2+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.21 \[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\frac {2}{3} x \sqrt {a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-x^2\right ) \] Input:
Integrate[Sqrt[a*x]/Sqrt[1 + x^2],x]
Output:
(2*x*Sqrt[a*x]*Hypergeometric2F1[1/2, 3/4, 7/4, -x^2])/3
Time = 0.42 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {266, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {\sqrt {a x}}{\sqrt {x^2+1}} \, dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 \int \frac {a x}{\sqrt {x^2+1}}d\sqrt {a x}}{a}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {2 \left (a \int \frac {1}{\sqrt {x^2+1}}d\sqrt {a x}-a \int \frac {a-a x}{a \sqrt {x^2+1}}d\sqrt {a x}\right )}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \left (a \int \frac {1}{\sqrt {x^2+1}}d\sqrt {a x}-\int \frac {a-a x}{\sqrt {x^2+1}}d\sqrt {a x}\right )}{a}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 \left (\frac {\sqrt {a} (a x+a) \sqrt {\frac {a^2 x^2+a^2}{(a x+a)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right ),\frac {1}{2}\right )}{2 \sqrt {x^2+1}}-\int \frac {a-a x}{\sqrt {x^2+1}}d\sqrt {a x}\right )}{a}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {2 \left (\frac {\sqrt {a} (a x+a) \sqrt {\frac {a^2 x^2+a^2}{(a x+a)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right ),\frac {1}{2}\right )}{2 \sqrt {x^2+1}}-\frac {\sqrt {a} (a x+a) \sqrt {\frac {a^2 x^2+a^2}{(a x+a)^2}} E\left (2 \arctan \left (\frac {\sqrt {a x}}{\sqrt {a}}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}}+\frac {a^2 \sqrt {x^2+1} \sqrt {a x}}{a x+a}\right )}{a}\) |
Input:
Int[Sqrt[a*x]/Sqrt[1 + x^2],x]
Output:
(2*((a^2*Sqrt[a*x]*Sqrt[1 + x^2])/(a + a*x) - (Sqrt[a]*(a + a*x)*Sqrt[(a^2 + a^2*x^2)/(a + a*x)^2]*EllipticE[2*ArcTan[Sqrt[a*x]/Sqrt[a]], 1/2])/Sqrt [1 + x^2] + (Sqrt[a]*(a + a*x)*Sqrt[(a^2 + a^2*x^2)/(a + a*x)^2]*EllipticF [2*ArcTan[Sqrt[a*x]/Sqrt[a]], 1/2])/(2*Sqrt[1 + x^2])))/a
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.15
method | result | size |
meijerg | \(\frac {2 \sqrt {x a}\, x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{2}\right )}{3}\) | \(20\) |
default | \(\frac {\sqrt {x a}\, \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \left (2 \operatorname {EllipticE}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {x^{2}+1}\, x}\) | \(81\) |
elliptic | \(\frac {i \sqrt {x a}\, \sqrt {a x \left (x^{2}+1\right )}\, \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {i \left (x -i\right )}\, \sqrt {i x}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {x^{2}+1}\, x \sqrt {a \,x^{3}+x a}}\) | \(104\) |
Input:
int((x*a)^(1/2)/(x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*(x*a)^(1/2)*x*hypergeom([1/2,3/4],[7/4],-x^2)
Time = 0.11 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=-2 \, \sqrt {a} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, x\right )\right ) \] Input:
integrate((a*x)^(1/2)/(x^2+1)^(1/2),x, algorithm="fricas")
Output:
-2*sqrt(a)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, x))
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\frac {\sqrt {a} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {x^{2} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((a*x)**(1/2)/(x**2+1)**(1/2),x)
Output:
sqrt(a)*x**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), x**2*exp_polar(I*pi) )/(2*gamma(7/4))
\[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\int { \frac {\sqrt {a x}}{\sqrt {x^{2} + 1}} \,d x } \] Input:
integrate((a*x)^(1/2)/(x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a*x)/sqrt(x^2 + 1), x)
\[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\int { \frac {\sqrt {a x}}{\sqrt {x^{2} + 1}} \,d x } \] Input:
integrate((a*x)^(1/2)/(x^2+1)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(a*x)/sqrt(x^2 + 1), x)
Timed out. \[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\int \frac {\sqrt {a\,x}}{\sqrt {x^2+1}} \,d x \] Input:
int((a*x)^(1/2)/(x^2 + 1)^(1/2),x)
Output:
int((a*x)^(1/2)/(x^2 + 1)^(1/2), x)
\[ \int \frac {\sqrt {a x}}{\sqrt {1+x^2}} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {x}\, \sqrt {x^{2}+1}}{x^{2}+1}d x \right ) \] Input:
int((a*x)^(1/2)/(x^2+1)^(1/2),x)
Output:
sqrt(a)*int((sqrt(x)*sqrt(x**2 + 1))/(x**2 + 1),x)