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

Optimal. Leaf size=296 \[ -\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} c^{3/2}}+\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} c^{3/2}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{\sqrt {2} c^{3/2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{\sqrt {2} c^{3/2}}-\frac {2 \sqrt [4]{a-b x^2}}{c \sqrt {c x}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {277, 329, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} c^{3/2}}+\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} c^{3/2}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{\sqrt {2} c^{3/2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{\sqrt {2} c^{3/2}}-\frac {2 \sqrt [4]{a-b x^2}}{c \sqrt {c x}} \]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 297

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

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

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Mathematica [C]  time = 0.01, size = 55, normalized size = 0.19 \[ -\frac {2 x \sqrt [4]{a-b x^2} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};\frac {b x^2}{a}\right )}{(c x)^{3/2} \sqrt [4]{1-\frac {b x^2}{a}}} \]

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

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 = 2.32, size = 51, normalized size = 0.17 \[ \frac {\sqrt [4]{a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{2 c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

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(2*I*pi)/a)/(2*c**(3/2)*sqrt(x)*gamma(3/4))

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