3.971 \(\int \frac{\sqrt{c x}}{(a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=84 \[ \frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}}-\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}} \]

[Out]

-((Sqrt[c]*ArcTan[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/b^(3/4)) + (Sqrt[c]*ArcTanh[(b^(1/4)*Sqrt[
c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/b^(3/4)

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Rubi [A]  time = 0.0557165, antiderivative size = 84, normalized size of antiderivative = 1., 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, 331, 298, 205, 208} \[ \frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}}-\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/(a + b*x^2)^(3/4),x]

[Out]

-((Sqrt[c]*ArcTan[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/b^(3/4)) + (Sqrt[c]*ArcTanh[(b^(1/4)*Sqrt[
c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/b^(3/4)

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 298

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

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{c x}}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+\frac{b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt{c x}\right )}{c}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{b x^4}{c^2}} \, dx,x,\frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}}\right )}{c}\\ &=\frac{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 )}{\sqrt{b}}-\frac{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 )}{\sqrt{b}}\\ &=-\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0131137, size = 67, normalized size = 0.8 \[ \frac{\sqrt{c x} \left (\tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )-\tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )\right )}{b^{3/4} \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x]/(a + b*x^2)^(3/4),x]

[Out]

(Sqrt[c*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^(
3/4)*Sqrt[x])

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{cx} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(b*x^2+a)^(3/4),x)

[Out]

int((c*x)^(1/2)/(b*x^2+a)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [C]  time = 1.84899, size = 44, normalized size = 0.52 \begin{align*} \frac{\sqrt{c} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(1/2)/(b*x**2+a)**(3/4),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

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