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

Optimal. Leaf size=272 \[ \frac {\sqrt {c} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} b^{3/4}}-\frac {\sqrt {c} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} b^{3/4}}-\frac {\sqrt {c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{\sqrt {2} b^{3/4}}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{\sqrt {2} b^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {329, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {\sqrt {c} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} b^{3/4}}-\frac {\sqrt {c} \log \left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a-b x^2}}+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a-b x^2}}+\sqrt {c}\right )}{2 \sqrt {2} b^{3/4}}-\frac {\sqrt {c} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}\right )}{\sqrt {2} b^{3/4}}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a-b x^2}}+1\right )}{\sqrt {2} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/(a - b*x^2)^(3/4),x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 75, normalized size = 0.28 \[ \frac {\sqrt {c x} \left (\tanh ^{-1}\left (\frac {\sqrt [4]{-b} \sqrt {x}}{\sqrt [4]{a-b x^2}}\right )-\tan ^{-1}\left (\frac {\sqrt [4]{-b} \sqrt {x}}{\sqrt [4]{a-b x^2}}\right )\right )}{(-b)^{3/4} \sqrt {x}} \]

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

[Out]

Timed out

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

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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maple [F]  time = 0.28, 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 \[ \int \frac {\sqrt {c x}}{{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

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

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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sympy [C]  time = 1.56, size = 46, normalized size = 0.17 \[ \frac {\sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} \]

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(2*I*pi)/a)/(2*a**(3/4)*gamma(7/4))

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