3.961 \(\int \frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}} \, dx\)

Optimal. Leaf size=90 \[ \frac {\sqrt {a} \sqrt {c x} \sqrt [4]{1-\frac {a}{b x^2}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a-b x^2}}-\frac {c \left (a-b x^2\right )^{3/4}}{b \sqrt {c x}} \]

[Out]

-c*(-b*x^2+a)^(3/4)/b/(c*x)^(1/2)+(1-a/b/x^2)^(1/4)*(cos(1/2*arccsc(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccs
c(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccsc(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 = 0.04, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {315, 317, 335, 228} \[ \frac {\sqrt {a} \sqrt {c x} \sqrt [4]{1-\frac {a}{b x^2}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b} \sqrt [4]{a-b x^2}}-\frac {c \left (a-b x^2\right )^{3/4}}{b \sqrt {c x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/(a - b*x^2)^(1/4),x]

[Out]

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

Rule 228

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

Rule 315

Int[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[(c*(a + b*x^2)^(3/4))/(b*Sqrt[c*x]), x] + D
ist[(a*c^2)/(2*b), Int[1/((c*x)^(3/2)*(a + b*x^2)^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && NegQ[b/a]

Rule 317

Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[(Sqrt[c*x]*(1 + a/(b*x^2))^(1/4))/(c
^2*(a + b*x^2)^(1/4)), Int[1/(x^2*(1 + a/(b*x^2))^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && NegQ[b/a]

Rule 335

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 {c \left (a-b x^2\right )^{3/4}}{b \sqrt {c x}}-\frac {\left (a c^2\right ) \int \frac {1}{(c x)^{3/2} \sqrt [4]{a-b x^2}} \, dx}{2 b}\\ &=-\frac {c \left (a-b x^2\right )^{3/4}}{b \sqrt {c x}}-\frac {\left (a \sqrt [4]{1-\frac {a}{b x^2}} \sqrt {c x}\right ) \int \frac {1}{\sqrt [4]{1-\frac {a}{b x^2}} x^2} \, dx}{2 b \sqrt [4]{a-b x^2}}\\ &=-\frac {c \left (a-b x^2\right )^{3/4}}{b \sqrt {c x}}+\frac {\left (a \sqrt [4]{1-\frac {a}{b x^2}} \sqrt {c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^2}{b}}} \, dx,x,\frac {1}{x}\right )}{2 b \sqrt [4]{a-b x^2}}\\ &=-\frac {c \left (a-b x^2\right )^{3/4}}{b \sqrt {c x}}+\frac {\sqrt {a} \sqrt [4]{1-\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \csc ^{-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]  time = 0.01, size = 57, normalized size = 0.63 \[ \frac {2 x \sqrt {c x} \sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {b x^2}{a}\right )}{3 \sqrt [4]{a-b x^2}} \]

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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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}}{b x^{2} - a}, x\right ) \]

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(-(-b*x^2 + a)^(3/4)*sqrt(c*x)/(b*x^2 - a), x)

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

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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maple [F]  time = 0.28, 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 \[ \int \frac {\sqrt {c x}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}\,{d x} \]

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

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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sympy [C]  time = 1.04, size = 46, normalized size = 0.51 \[ \frac {\sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac {7}{4}\right )} \]

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(2*I*pi)/a)/(2*a**(1/4)*gamma(7/4))

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