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

Optimal. Leaf size=107 \[ -\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{c^{3/2}}+\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{c^{3/2}}-\frac{2 \sqrt [4]{a+b x^2}}{c \sqrt{c x}} \]

[Out]

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

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Rubi [A]  time = 0.0692396, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {277, 329, 331, 298, 205, 208} \[ -\frac{\sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{c^{3/2}}+\frac{\sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt{c} \sqrt [4]{a+b x^2}}\right )}{c^{3/2}}-\frac{2 \sqrt [4]{a+b x^2}}{c \sqrt{c x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/(c*x)^(3/2),x]

[Out]

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C]  time = 0.0117696, size = 54, normalized size = 0.5 \[ -\frac{2 x \sqrt [4]{a+b x^2} \, _2F_1\left (-\frac{1}{4},-\frac{1}{4};\frac{3}{4};-\frac{b x^2}{a}\right )}{(c x)^{3/2} \sqrt [4]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/4)/(c*x)^(3/2),x]

[Out]

(-2*x*(a + b*x^2)^(1/4)*Hypergeometric2F1[-1/4, -1/4, 3/4, -((b*x^2)/a)])/((c*x)^(3/2)*(1 + (b*x^2)/a)^(1/4))

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [4]{b{x}^{2}+a} \left ( cx \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}{\left (c x\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [C]  time = 3.92682, size = 49, normalized size = 0.46 \begin{align*} \frac{\sqrt [4]{a} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*c**(3/2)*sqrt(x)*gamma(3/4))

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Giac [B]  time = 1.85829, size = 452, normalized size = 4.22 \begin{align*} \frac{2 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} + \frac{2 \,{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \sqrt{{\left | c \right |}}}{\sqrt{c x}}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}}}\right ) + 2 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} - \frac{2 \,{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \sqrt{{\left | c \right |}}}{\sqrt{c x}}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}}}\right ) + \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \log \left (\frac{\sqrt{2}{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}{\left | c \right |}}{\sqrt{c x}} + \sqrt{-b}{\left | c \right |} + \frac{\sqrt{b c^{2} x^{2} + a c^{2}}{\left | c \right |}}{c x}\right ) - \sqrt{2} \left (-b\right )^{\frac{1}{4}} \sqrt{{\left | c \right |}} \log \left (-\frac{\sqrt{2}{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}{\left | c \right |}}{\sqrt{c x}} + \sqrt{-b}{\left | c \right |} + \frac{\sqrt{b c^{2} x^{2} + a c^{2}}{\left | c \right |}}{c x}\right ) - \frac{8 \,{\left (b c^{2} x^{2} + a c^{2}\right )}^{\frac{1}{4}} \sqrt{{\left | c \right |}}}{\sqrt{c x}}}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(2)*(-b)^(1/4)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4)*sqrt(abs(c)) + 2*(b*c^2*x^2 + a*
c^2)^(1/4)*sqrt(abs(c))/sqrt(c*x))/((-b)^(1/4)*sqrt(abs(c)))) + 2*sqrt(2)*(-b)^(1/4)*sqrt(abs(c))*arctan(-1/2*
sqrt(2)*(sqrt(2)*(-b)^(1/4)*sqrt(abs(c)) - 2*(b*c^2*x^2 + a*c^2)^(1/4)*sqrt(abs(c))/sqrt(c*x))/((-b)^(1/4)*sqr
t(abs(c)))) + sqrt(2)*(-b)^(1/4)*sqrt(abs(c))*log(sqrt(2)*(b*c^2*x^2 + a*c^2)^(1/4)*(-b)^(1/4)*abs(c)/sqrt(c*x
) + sqrt(-b)*abs(c) + sqrt(b*c^2*x^2 + a*c^2)*abs(c)/(c*x)) - sqrt(2)*(-b)^(1/4)*sqrt(abs(c))*log(-sqrt(2)*(b*
c^2*x^2 + a*c^2)^(1/4)*(-b)^(1/4)*abs(c)/sqrt(c*x) + sqrt(-b)*abs(c) + sqrt(b*c^2*x^2 + a*c^2)*abs(c)/(c*x)) -
8*(b*c^2*x^2 + a*c^2)^(1/4)*sqrt(abs(c))/sqrt(c*x))/c^2