3.10.21 \(\int \frac {\sqrt [4]{a+b x^2}}{\sqrt {c x}} \, dx\) [921]

Optimal. Leaf size=89 \[ \frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}-\frac {\sqrt {a} \sqrt {b} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{c^2 \left (a+b x^2\right )^{3/4}} \]

[Out]

-(1+a/b/x^2)^(3/4)*(c*x)^(3/2)*(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)))
*EllipticF(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*b^(1/2)/c^2/(b*x^2+a)^(3/4)+(b*x^2+a)^(1/4)*(c*
x)^(1/2)/c

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Rubi [A]
time = 0.05, antiderivative size = 89, 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 = {285, 335, 243, 342, 281, 237} \begin {gather*} \frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}-\frac {\sqrt {a} \sqrt {b} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{c^2 \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/Sqrt[c*x],x]

[Out]

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

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

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 285

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

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]

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 [4]{a+b x^2}}{\sqrt {c x}} \, dx &=\frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}+\frac {1}{2} a \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}+\frac {a \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt {c x}\right )}{c}\\ &=\frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}+\frac {\left (a \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {c x}\right )}{c \left (a+b x^2\right )^{3/4}}\\ &=\frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}-\frac {\left (a \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a c^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {c x}}\right )}{c \left (a+b x^2\right )^{3/4}}\\ &=\frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}-\frac {\left (a \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{c x}\right )}{2 c \left (a+b x^2\right )^{3/4}}\\ &=\frac {\sqrt {c x} \sqrt [4]{a+b x^2}}{c}-\frac {\sqrt {a} \sqrt {b} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{c^2 \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 54, normalized size = 0.61 \begin {gather*} \frac {2 x \sqrt [4]{a+b x^2} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )}{\sqrt {c x} \sqrt [4]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(a + b*x^2)^(1/4)*Hypergeometric2F1[-1/4, 1/4, 5/4, -((b*x^2)/a)])/(Sqrt[c*x]*(1 + (b*x^2)/a)^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.70, size = 46, normalized size = 0.52 \begin {gather*} \frac {\sqrt [4]{a} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {c} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/4)*sqrt(x)*gamma(1/4)*hyper((-1/4, 1/4), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(c)*gamma(5/4))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4)/(c*x)^(1/2),x)

[Out]

int((a + b*x^2)^(1/4)/(c*x)^(1/2), x)

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