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

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

[Out]

arctan(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4)/c^(1/2))/b^(1/4)/c^(1/2)+arctanh(b^(1/4)*(c*x)^(1/2)/(b*x^2+a)^(1/4
)/c^(1/2))/b^(1/4)/c^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {329, 240, 212, 208, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{\sqrt [4]{b} \sqrt {c}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{\sqrt [4]{b} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 208

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

Rule 212

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

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 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]

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

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Mathematica [A]  time = 0.01, size = 65, normalized size = 0.78 \[ \frac {\sqrt {x} \left (\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{\sqrt [4]{b} \sqrt {c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.34, size = 241, normalized size = 2.90 \[ -2 \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} b c \left (\frac {1}{b c^{2}}\right )^{\frac {3}{4}} - {\left (b^{2} c x^{2} + a b c\right )} \sqrt {\frac {\sqrt {b x^{2} + a} c x + {\left (b c^{2} x^{2} + a c^{2}\right )} \sqrt {\frac {1}{b c^{2}}}}{b x^{2} + a}} \left (\frac {1}{b c^{2}}\right )^{\frac {3}{4}}}{b x^{2} + a}\right ) + \frac {1}{2} \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} + {\left (b c x^{2} + a c\right )} \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - \frac {1}{2} \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} - {\left (b c x^{2} + a c\right )} \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) \]

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

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)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 1.51, size = 44, normalized size = 0.53 \[ \frac {\sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \sqrt {c} \Gamma \left (\frac {5}{4}\right )} \]

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(I*pi)/a)/(2*a**(1/4)*sqrt(c)*gamma(5/4))

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