3.383 \(\int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx\)

Optimal. Leaf size=159 \[ \frac {2 \sqrt {x^2+1} x^2 \sqrt {\frac {a}{x^3}}}{x+1}-2 \sqrt {x^2+1} x \sqrt {\frac {a}{x^3}}+\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} x^{3/2} \sqrt {\frac {a}{x^3}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}}-\frac {2 (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} x^{3/2} \sqrt {\frac {a}{x^3}} E\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}} \]

[Out]

-2*x*(a/x^3)^(1/2)*(x^2+1)^(1/2)+2*x^2*(a/x^3)^(1/2)*(x^2+1)^(1/2)/(1+x)-2*x^(3/2)*(1+x)*(cos(2*arctan(x^(1/2)
))^2)^(1/2)/cos(2*arctan(x^(1/2)))*EllipticE(sin(2*arctan(x^(1/2))),1/2*2^(1/2))*(a/x^3)^(1/2)*((x^2+1)/(1+x)^
2)^(1/2)/(x^2+1)^(1/2)+x^(3/2)*(1+x)*(cos(2*arctan(x^(1/2)))^2)^(1/2)/cos(2*arctan(x^(1/2)))*EllipticF(sin(2*a
rctan(x^(1/2))),1/2*2^(1/2))*(a/x^3)^(1/2)*((x^2+1)/(1+x)^2)^(1/2)/(x^2+1)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {15, 325, 329, 305, 220, 1196} \[ \frac {2 \sqrt {x^2+1} x^2 \sqrt {\frac {a}{x^3}}}{x+1}-2 \sqrt {x^2+1} x \sqrt {\frac {a}{x^3}}+\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} x^{3/2} \sqrt {\frac {a}{x^3}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}}-\frac {2 (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} x^{3/2} \sqrt {\frac {a}{x^3}} E\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a/x^3]/Sqrt[1 + x^2],x]

[Out]

-2*Sqrt[a/x^3]*x*Sqrt[1 + x^2] + (2*Sqrt[a/x^3]*x^2*Sqrt[1 + x^2])/(1 + x) - (2*Sqrt[a/x^3]*x^(3/2)*(1 + x)*Sq
rt[(1 + x^2)/(1 + x)^2]*EllipticE[2*ArcTan[Sqrt[x]], 1/2])/Sqrt[1 + x^2] + (Sqrt[a/x^3]*x^(3/2)*(1 + x)*Sqrt[(
1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTan[Sqrt[x]], 1/2])/Sqrt[1 + x^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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {1+x^2}} \, dx &=\left (\sqrt {\frac {a}{x^3}} x^{3/2}\right ) \int \frac {1}{x^{3/2} \sqrt {1+x^2}} \, dx\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^2}+\left (\sqrt {\frac {a}{x^3}} x^{3/2}\right ) \int \frac {\sqrt {x}}{\sqrt {1+x^2}} \, dx\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^2}+\left (2 \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^2}+\left (2 \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt {x}\right )-\left (2 \sqrt {\frac {a}{x^3}} x^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt {x}\right )\\ &=-2 \sqrt {\frac {a}{x^3}} x \sqrt {1+x^2}+\frac {2 \sqrt {\frac {a}{x^3}} x^2 \sqrt {1+x^2}}{1+x}-\frac {2 \sqrt {\frac {a}{x^3}} x^{3/2} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} E\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {1+x^2}}+\frac {\sqrt {\frac {a}{x^3}} x^{3/2} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} F\left (2 \tan ^{-1}\left (\sqrt {x}\right )|\frac {1}{2}\right )}{\sqrt {1+x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.17 \[ -2 x \sqrt {\frac {a}{x^3}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a/x^3]/Sqrt[1 + x^2],x]

[Out]

-2*Sqrt[a/x^3]*x*Hypergeometric2F1[-1/4, 1/2, 3/4, -x^2]

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a/x^3)/sqrt(x^2 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a/x^3)/sqrt(x^2 + 1), x)

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maple [C]  time = 0.04, size = 116, normalized size = 0.73 \[ \frac {\sqrt {\frac {a}{x^{3}}}\, \left (-2 x^{2}+2 \sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \EllipticE \left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {-i \left (x +i\right )}\, \sqrt {2}\, \sqrt {-i \left (-x +i\right )}\, \sqrt {i x}\, \EllipticF \left (\sqrt {-i \left (x +i\right )}, \frac {\sqrt {2}}{2}\right )-2\right ) x}{\sqrt {x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/x^3)^(1/2)/(x^2+1)^(1/2),x)

[Out]

(a/x^3)^(1/2)*x*(2*(-I*(x+I))^(1/2)*2^(1/2)*(-I*(-x+I))^(1/2)*(I*x)^(1/2)*EllipticE((-I*(x+I))^(1/2),1/2*2^(1/
2))-(-I*(x+I))^(1/2)*2^(1/2)*(-I*(-x+I))^(1/2)*(I*x)^(1/2)*EllipticF((-I*(x+I))^(1/2),1/2*2^(1/2))-2*x^2-2)/(x
^2+1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a/x^3)/sqrt(x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {a}{x^3}}}{\sqrt {x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/x^3)^(1/2)/(x^2 + 1)^(1/2),x)

[Out]

int((a/x^3)^(1/2)/(x^2 + 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a}{x^{3}}}}{\sqrt {x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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