3.956 \(\int \frac{1}{\sqrt{c x} \sqrt [4]{a-b x^2}} \, dx\)

Optimal. Leaf size=272 \[ -\frac{\log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}+1\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}} \]

[Out]

-(ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))]/(Sqrt[2]*b^(1/4)*Sqrt[c])) + ArcTan[1 +
(Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))]/(Sqrt[2]*b^(1/4)*Sqrt[c]) - Log[Sqrt[c] + (Sqrt[b]*Sq
rt[c]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)]/(2*Sqrt[2]*b^(1/4)*Sqrt[c]) + Log[Sq
rt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a - b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)]/(2*Sqrt[2]*b^(1/4
)*Sqrt[c])

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Rubi [A]  time = 0.227996, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {329, 240, 211, 1165, 628, 1162, 617, 204} \[ -\frac{\log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\log \left (\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}+\sqrt{c}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}+1\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))]/(Sqrt[2]*b^(1/4)*Sqrt[c])) + ArcTan[1 +
(Sqrt[2]*b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a - b*x^2)^(1/4))]/(Sqrt[2]*b^(1/4)*Sqrt[c]) - Log[Sqrt[c] + (Sqrt[b]*Sq
rt[c]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)]/(2*Sqrt[2]*b^(1/4)*Sqrt[c]) + Log[Sq
rt[c] + (Sqrt[b]*Sqrt[c]*x)/Sqrt[a - b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[c*x])/(a - b*x^2)^(1/4)]/(2*Sqrt[2]*b^(1/4
)*Sqrt[c])

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 240

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{c x} \sqrt [4]{a-b x^2}} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a-\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{c}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+\frac{b x^4}{c^2}} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{c}\\ &=\frac{\operatorname{Subst}\left (\int \frac{c-\sqrt{b} x^2}{1+\frac{b x^4}{c^2}} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{c^2}+\frac{\operatorname{Subst}\left (\int \frac{c+\sqrt{b} x^2}{1+\frac{b x^4}{c^2}} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{c^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{c}{\sqrt{b}}-\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{b}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{c}{\sqrt{b}}+\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{b}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{c}}{\sqrt [4]{b}}+2 x}{-\frac{c}{\sqrt{b}}-\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{c}}{\sqrt [4]{b}}-2 x}{-\frac{c}{\sqrt{b}}+\frac{\sqrt{2} \sqrt{c} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}\\ &=-\frac{\log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}}\\ &=-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a-b x^2}}\right )}{\sqrt{2} \sqrt [4]{b} \sqrt{c}}-\frac{\log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}+\frac{\log \left (\sqrt{c}+\frac{\sqrt{b} \sqrt{c} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a-b x^2}}\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.0417946, size = 197, normalized size = 0.72 \[ \frac{\sqrt{x} \left (-\log \left (\frac{\sqrt{b} x}{\sqrt{a-b x^2}}-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a-b x^2}}+1\right )+\log \left (\frac{\sqrt{b} x}{\sqrt{a-b x^2}}+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a-b x^2}}+1\right )-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a-b x^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a-b x^2}}+1\right )\right )}{2 \sqrt{2} \sqrt [4]{b} \sqrt{c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-2*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/(a - b*x^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/
(a - b*x^2)^(1/4)] - Log[1 + (Sqrt[b]*x)/Sqrt[a - b*x^2] - (Sqrt[2]*b^(1/4)*Sqrt[x])/(a - b*x^2)^(1/4)] + Log[
1 + (Sqrt[b]*x)/Sqrt[a - b*x^2] + (Sqrt[2]*b^(1/4)*Sqrt[x])/(a - b*x^2)^(1/4)]))/(2*Sqrt[2]*b^(1/4)*Sqrt[c*x])

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{cx}}}{\frac{1}{\sqrt [4]{-b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68594, size = 594, normalized size = 2.18 \begin{align*} -2 \, \left (-\frac{1}{b c^{2}}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (-b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} b c \left (-\frac{1}{b c^{2}}\right )^{\frac{3}{4}} -{\left (b^{2} c x^{2} - a b c\right )} \sqrt{-\frac{\sqrt{-b x^{2} + a} c x -{\left (b c^{2} x^{2} - a c^{2}\right )} \sqrt{-\frac{1}{b c^{2}}}}{b x^{2} - a}} \left (-\frac{1}{b c^{2}}\right )^{\frac{3}{4}}}{b x^{2} - a}\right ) - \frac{1}{2} \, \left (-\frac{1}{b c^{2}}\right )^{\frac{1}{4}} \log \left (\frac{{\left (-b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} +{\left (b c x^{2} - a c\right )} \left (-\frac{1}{b c^{2}}\right )^{\frac{1}{4}}}{b x^{2} - a}\right ) + \frac{1}{2} \, \left (-\frac{1}{b c^{2}}\right )^{\frac{1}{4}} \log \left (\frac{{\left (-b x^{2} + a\right )}^{\frac{3}{4}} \sqrt{c x} -{\left (b c x^{2} - a c\right )} \left (-\frac{1}{b c^{2}}\right )^{\frac{1}{4}}}{b x^{2} - a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-1/(b*c^2))^(1/4)*arctan(-((-b*x^2 + a)^(3/4)*sqrt(c*x)*b*c*(-1/(b*c^2))^(3/4) - (b^2*c*x^2 - a*b*c)*sqrt(
-(sqrt(-b*x^2 + a)*c*x - (b*c^2*x^2 - a*c^2)*sqrt(-1/(b*c^2)))/(b*x^2 - a))*(-1/(b*c^2))^(3/4))/(b*x^2 - a)) -
1/2*(-1/(b*c^2))^(1/4)*log(((-b*x^2 + a)^(3/4)*sqrt(c*x) + (b*c*x^2 - a*c)*(-1/(b*c^2))^(1/4))/(b*x^2 - a)) +
1/2*(-1/(b*c^2))^(1/4)*log(((-b*x^2 + a)^(3/4)*sqrt(c*x) - (b*c*x^2 - a*c)*(-1/(b*c^2))^(1/4))/(b*x^2 - a))

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Sympy [C]  time = 1.81993, size = 46, normalized size = 0.17 \begin{align*} \frac{\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^{2 i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \sqrt{c} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{2} + a\right )}^{\frac{1}{4}} \sqrt{c x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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