3.312 \(\int \frac {\sqrt {x}}{a-b x^2} \, dx\)

Optimal. Leaf size=58 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a} b^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a} b^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {329, 298, 205, 208} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a} b^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a - b*x^2),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.83 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )-\tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{a} b^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a - b*x^2),x]

[Out]

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

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fricas [B]  time = 0.94, size = 117, normalized size = 2.02 \[ 2 \, \left (\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {a b \sqrt {\frac {1}{a b^{3}}} + x} b \left (\frac {1}{a b^{3}}\right )^{\frac {1}{4}} - b \sqrt {x} \left (\frac {1}{a b^{3}}\right )^{\frac {1}{4}}\right ) + \frac {1}{2} \, \left (\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \log \left (a b^{2} \left (\frac {1}{a b^{3}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - \frac {1}{2} \, \left (\frac {1}{a b^{3}}\right )^{\frac {1}{4}} \log \left (-a b^{2} \left (\frac {1}{a b^{3}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.63, size = 194, normalized size = 3.34 \[ \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a b^{3}} + \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a b^{3}} - \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {-\frac {a}{b}}\right )}{4 \, a b^{3}} + \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {-\frac {a}{b}}\right )}{4 \, a b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(-a*b^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) + 2*sqrt(x))/(-a/b)^(1/4))/(a*b^3) + 1/2*s
qrt(2)*(-a*b^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) - 2*sqrt(x))/(-a/b)^(1/4))/(a*b^3) - 1/4*sqrt(
2)*(-a*b^3)^(3/4)*log(sqrt(2)*sqrt(x)*(-a/b)^(1/4) + x + sqrt(-a/b))/(a*b^3) + 1/4*sqrt(2)*(-a*b^3)^(3/4)*log(
-sqrt(2)*sqrt(x)*(-a/b)^(1/4) + x + sqrt(-a/b))/(a*b^3)

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maple [A]  time = 0.01, size = 66, normalized size = 1.14 \[ -\frac {\arctan \left (\frac {\sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}} b}+\frac {\ln \left (\frac {\sqrt {x}+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sqrt {x}-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(-b*x^2+a),x)

[Out]

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

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maxima [B]  time = 2.97, size = 86, normalized size = 1.48 \[ -\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\log \left (\frac {\sqrt {b} \sqrt {x} - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sqrt {x} + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{2 \, \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(sqrt(b)*sqrt(x)/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - 1/2*log((sqrt(b)*sqrt(x) - sq
rt(sqrt(a)*sqrt(b)))/(sqrt(b)*sqrt(x) + sqrt(sqrt(a)*sqrt(b))))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b))

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mupad [B]  time = 0.08, size = 33, normalized size = 0.57 \[ -\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{a^{1/4}}\right )-\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{a^{1/4}}\right )}{a^{1/4}\,b^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a - b*x^2),x)

[Out]

-(atan((b^(1/4)*x^(1/2))/a^(1/4)) - atanh((b^(1/4)*x^(1/2))/a^(1/4)))/(a^(1/4)*b^(3/4))

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sympy [A]  time = 3.10, size = 122, normalized size = 2.10 \[ \begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2}{b \sqrt {x}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a} & \text {for}\: b = 0 \\- \frac {\log {\left (- \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 \sqrt [4]{a} b \sqrt [4]{\frac {1}{b}}} + \frac {\log {\left (\sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 \sqrt [4]{a} b \sqrt [4]{\frac {1}{b}}} - \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{\sqrt [4]{a} b \sqrt [4]{\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(-b*x**2+a),x)

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (2/(b*sqrt(x)), Eq(a, 0)), (2*x**(3/2)/(3*a), Eq(b, 0)), (-log(-
a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(1/4)*b*(1/b)**(1/4)) + log(a**(1/4)*(1/b)**(1/4) + sqrt(x))/(2*a**(1/4
)*b*(1/b)**(1/4)) - atan(sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/(a**(1/4)*b*(1/b)**(1/4)), True))

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