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]

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

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Rubi [A]  time = 0.07, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.50 \[ -\frac {2 x \sqrt [4]{a+b x^2} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};-\frac {b x^2}{a}\right )}{(c x)^{3/2} \sqrt [4]{\frac {b x^2}{a}+1}} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {3}{2}}}\,{d x} \]

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]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (c x \right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\left (c x\right )^{\frac {3}{2}}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4)/(c*x)^(3/2),x)

[Out]

int((a + b*x^2)^(1/4)/(c*x)^(3/2), x)

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sympy [C]  time = 2.31, size = 49, normalized size = 0.46 \[ \frac {\sqrt [4]{a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

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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