3.10.51 \(\int \frac {\sqrt {c x}}{\sqrt [4]{a+b x^2}} \, dx\) [951]

Optimal. Leaf size=83 \[ \frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}+\frac {\sqrt {a} \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}} \]

[Out]

x*(c*x)^(1/2)/(b*x^2+a)^(1/4)+(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*
b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/b^
(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {320, 290, 342, 202} \begin {gather*} \frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}+\frac {\sqrt {a} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/(a + b*x^2)^(1/4),x]

[Out]

(x*Sqrt[c*x])/(a + b*x^2)^(1/4) + (Sqrt[a]*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a
]]/2, 2])/(Sqrt[b]*(a + b*x^2)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 290

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[Sqrt[c*x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b
*x^2)^(1/4))), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 320

Int[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[x*(Sqrt[c*x]/(a + b*x^2)^(1/4)), x] - Dist[
a/2, Int[Sqrt[c*x]/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sqrt {c x}}{\sqrt [4]{a+b x^2}} \, dx &=\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}-\frac {1}{2} a \int \frac {\sqrt {c x}}{\left (a+b x^2\right )^{5/4}} \, dx\\ &=\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}-\frac {\left (a \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{2 b \sqrt [4]{a+b x^2}}\\ &=\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}+\frac {\left (a \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{2 b \sqrt [4]{a+b x^2}}\\ &=\frac {x \sqrt {c x}}{\sqrt [4]{a+b x^2}}+\frac {\sqrt {a} \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 56, normalized size = 0.67 \begin {gather*} \frac {2 x \sqrt {c x} \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )}{3 \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x]/(a + b*x^2)^(1/4),x]

[Out]

(2*x*Sqrt[c*x]*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 3/4, 7/4, -((b*x^2)/a)])/(3*(a + b*x^2)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {c x}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(1/2)/(b*x^2+a)^(1/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x)/(b*x^2 + a)^(1/4), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral(sqrt(c*x)/(b*x^2 + a)^(1/4), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.52, size = 44, normalized size = 0.53 \begin {gather*} \frac {\sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(1/2)/(b*x**2+a)**(1/4),x)

[Out]

sqrt(c)*x**(3/2)*gamma(3/4)*hyper((1/4, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(1/4)*gamma(7/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(sqrt(c*x)/(b*x^2 + a)^(1/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x}}{{\left (b\,x^2+a\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(a + b*x^2)^(1/4),x)

[Out]

int((c*x)^(1/2)/(a + b*x^2)^(1/4), x)

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