3.10.71 \(\int \frac {\sqrt {c x}}{(a+b x^2)^{3/4}} \, dx\) [971]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 84, 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 = {335, 338, 304, 211, 214} \begin {gather*} \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} \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{b^{3/4}} \end {gather*}

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 211

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

Rule 214

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

Rule 304

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 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 338

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {c x}}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac {2 \text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.21, size = 67, normalized size = 0.80 \begin {gather*} \frac {\sqrt {c 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 )}{b^{3/4} \sqrt {x}} \end {gather*}

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 = 0.00 \[\int \frac {\sqrt {c x}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx\]

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 = 0.00, 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((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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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Sympy [C] Result contains complex when optimal does not.
time = 0.75, size = 44, normalized size = 0.52 \begin {gather*} \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 {gather*}

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 = 0.00, 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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x}}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(a + b*x^2)^(3/4),x)

[Out]

int((c*x)^(1/2)/(a + b*x^2)^(3/4), x)

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