3.6.73 \(\int \frac {\sqrt {2+3 x}}{a-b x^2} \, dx\)

Optimal. Leaf size=132 \[ \frac {\sqrt {3 \sqrt {a}+2 \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {3 x+2}}{\sqrt {3 \sqrt {a}+2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}}-\frac {\sqrt {3 \sqrt {a}-2 \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {3 x+2}}{\sqrt {3 \sqrt {a}-2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}} \]

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Rubi [A]  time = 0.13, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {700, 1130, 208, 205} \begin {gather*} \frac {\sqrt {3 \sqrt {a}+2 \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {3 x+2}}{\sqrt {3 \sqrt {a}+2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}}-\frac {\sqrt {3 \sqrt {a}-2 \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {3 x+2}}{\sqrt {3 \sqrt {a}-2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 3*x]/(a - b*x^2),x]

[Out]

-((Sqrt[3*Sqrt[a] - 2*Sqrt[b]]*ArcTan[(b^(1/4)*Sqrt[2 + 3*x])/Sqrt[3*Sqrt[a] - 2*Sqrt[b]]])/(Sqrt[a]*b^(3/4)))
+ (Sqrt[3*Sqrt[a] + 2*Sqrt[b]]*ArcTanh[(b^(1/4)*Sqrt[2 + 3*x])/Sqrt[3*Sqrt[a] + 2*Sqrt[b]]])/(Sqrt[a]*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 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1130

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

Rubi steps

\begin {align*} \int \frac {\sqrt {2+3 x}}{a-b x^2} \, dx &=6 \operatorname {Subst}\left (\int \frac {x^2}{9 a-4 b+4 b x^2-b x^4} \, dx,x,\sqrt {2+3 x}\right )\\ &=\left (3-\frac {2 \sqrt {b}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {1}{-3 \sqrt {a} \sqrt {b}+2 b-b x^2} \, dx,x,\sqrt {2+3 x}\right )+\left (3+\frac {2 \sqrt {b}}{\sqrt {a}}\right ) \operatorname {Subst}\left (\int \frac {1}{3 \sqrt {a} \sqrt {b}+2 b-b x^2} \, dx,x,\sqrt {2+3 x}\right )\\ &=-\frac {\sqrt {3 \sqrt {a}-2 \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {2+3 x}}{\sqrt {3 \sqrt {a}-2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}}+\frac {\sqrt {3 \sqrt {a}+2 \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {2+3 x}}{\sqrt {3 \sqrt {a}+2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 123, normalized size = 0.93 \begin {gather*} \frac {\sqrt {3 \sqrt {a}+2 \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {3 x+2}}{\sqrt {3 \sqrt {a}+2 \sqrt {b}}}\right )-\sqrt {3 \sqrt {a}-2 \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {3 x+2}}{\sqrt {3 \sqrt {a}-2 \sqrt {b}}}\right )}{\sqrt {a} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 3*x]/(a - b*x^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 200, normalized size = 1.52 \begin {gather*} \frac {\left (3 \sqrt {a}+2 \sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {3 x+2} \sqrt {-3 \sqrt {a} \sqrt {b}-2 b}}{3 \sqrt {a}+2 \sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \sqrt {-\left (\sqrt {b} \left (3 \sqrt {a}+2 \sqrt {b}\right )\right )}}+\frac {\left (2 \sqrt {b}-3 \sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {3 x+2} \sqrt {3 \sqrt {a} \sqrt {b}-2 b}}{3 \sqrt {a}-2 \sqrt {b}}\right )}{\sqrt {a} \sqrt {b} \sqrt {\sqrt {b} \left (3 \sqrt {a}-2 \sqrt {b}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[2 + 3*x]/(a - b*x^2),x]

[Out]

((3*Sqrt[a] + 2*Sqrt[b])*ArcTan[(Sqrt[-3*Sqrt[a]*Sqrt[b] - 2*b]*Sqrt[2 + 3*x])/(3*Sqrt[a] + 2*Sqrt[b])])/(Sqrt
[a]*Sqrt[-((3*Sqrt[a] + 2*Sqrt[b])*Sqrt[b])]*Sqrt[b]) + ((-3*Sqrt[a] + 2*Sqrt[b])*ArcTan[(Sqrt[3*Sqrt[a]*Sqrt[
b] - 2*b]*Sqrt[2 + 3*x])/(3*Sqrt[a] - 2*Sqrt[b])])/(Sqrt[a]*Sqrt[(3*Sqrt[a] - 2*Sqrt[b])*Sqrt[b]]*Sqrt[b])

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fricas [B]  time = 0.42, size = 299, normalized size = 2.27 \begin {gather*} \frac {1}{2} \, \sqrt {\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} + 2}{a b}} \log \left (a b^{2} \sqrt {\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} + 2}{a b}} \sqrt {\frac {1}{a b^{3}}} + \sqrt {3 \, x + 2}\right ) - \frac {1}{2} \, \sqrt {\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} + 2}{a b}} \log \left (-a b^{2} \sqrt {\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} + 2}{a b}} \sqrt {\frac {1}{a b^{3}}} + \sqrt {3 \, x + 2}\right ) - \frac {1}{2} \, \sqrt {-\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} - 2}{a b}} \log \left (a b^{2} \sqrt {-\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} - 2}{a b}} \sqrt {\frac {1}{a b^{3}}} + \sqrt {3 \, x + 2}\right ) + \frac {1}{2} \, \sqrt {-\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} - 2}{a b}} \log \left (-a b^{2} \sqrt {-\frac {3 \, a b \sqrt {\frac {1}{a b^{3}}} - 2}{a b}} \sqrt {\frac {1}{a b^{3}}} + \sqrt {3 \, x + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt((3*a*b*sqrt(1/(a*b^3)) + 2)/(a*b))*log(a*b^2*sqrt((3*a*b*sqrt(1/(a*b^3)) + 2)/(a*b))*sqrt(1/(a*b^3))
+ sqrt(3*x + 2)) - 1/2*sqrt((3*a*b*sqrt(1/(a*b^3)) + 2)/(a*b))*log(-a*b^2*sqrt((3*a*b*sqrt(1/(a*b^3)) + 2)/(a*
b))*sqrt(1/(a*b^3)) + sqrt(3*x + 2)) - 1/2*sqrt(-(3*a*b*sqrt(1/(a*b^3)) - 2)/(a*b))*log(a*b^2*sqrt(-(3*a*b*sqr
t(1/(a*b^3)) - 2)/(a*b))*sqrt(1/(a*b^3)) + sqrt(3*x + 2)) + 1/2*sqrt(-(3*a*b*sqrt(1/(a*b^3)) - 2)/(a*b))*log(-
a*b^2*sqrt(-(3*a*b*sqrt(1/(a*b^3)) - 2)/(a*b))*sqrt(1/(a*b^3)) + sqrt(3*x + 2))

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giac [B]  time = 0.41, size = 216, normalized size = 1.64 \begin {gather*} \frac {{\left (4 \, \sqrt {a b} \sqrt {-2 \, b^{2} - 3 \, \sqrt {a b} b} a + 17 \, \sqrt {a b} \sqrt {-2 \, b^{2} - 3 \, \sqrt {a b} b} b\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {3 \, x + 2}}{\sqrt {-\frac {2 \, b + \sqrt {{\left (9 \, a - 4 \, b\right )} b + 4 \, b^{2}}}{b}}}\right )}{4 \, a^{2} b^{3} + 17 \, a b^{4}} - \frac {{\left (4 \, \sqrt {a b} \sqrt {-2 \, b^{2} + 3 \, \sqrt {a b} b} a + 17 \, \sqrt {a b} \sqrt {-2 \, b^{2} + 3 \, \sqrt {a b} b} b\right )} {\left | b \right |} \arctan \left (\frac {\sqrt {3 \, x + 2}}{\sqrt {-\frac {2 \, b - \sqrt {{\left (9 \, a - 4 \, b\right )} b + 4 \, b^{2}}}{b}}}\right )}{4 \, a^{2} b^{3} + 17 \, a b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*sqrt(a*b)*sqrt(-2*b^2 - 3*sqrt(a*b)*b)*a + 17*sqrt(a*b)*sqrt(-2*b^2 - 3*sqrt(a*b)*b)*b)*abs(b)*arctan(sqrt(
3*x + 2)/sqrt(-(2*b + sqrt((9*a - 4*b)*b + 4*b^2))/b))/(4*a^2*b^3 + 17*a*b^4) - (4*sqrt(a*b)*sqrt(-2*b^2 + 3*s
qrt(a*b)*b)*a + 17*sqrt(a*b)*sqrt(-2*b^2 + 3*sqrt(a*b)*b)*b)*abs(b)*arctan(sqrt(3*x + 2)/sqrt(-(2*b - sqrt((9*
a - 4*b)*b + 4*b^2))/b))/(4*a^2*b^3 + 17*a*b^4)

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maple [A]  time = 0.10, size = 182, normalized size = 1.38 \begin {gather*} \frac {2 b \arctanh \left (\frac {\sqrt {3 x +2}\, b}{\sqrt {\left (2 b +3 \sqrt {a b}\right ) b}}\right )}{\sqrt {a b}\, \sqrt {\left (2 b +3 \sqrt {a b}\right ) b}}+\frac {2 b \arctan \left (\frac {\sqrt {3 x +2}\, b}{\sqrt {\left (-2 b +3 \sqrt {a b}\right ) b}}\right )}{\sqrt {a b}\, \sqrt {\left (-2 b +3 \sqrt {a b}\right ) b}}+\frac {3 \arctanh \left (\frac {\sqrt {3 x +2}\, b}{\sqrt {\left (2 b +3 \sqrt {a b}\right ) b}}\right )}{\sqrt {\left (2 b +3 \sqrt {a b}\right ) b}}-\frac {3 \arctan \left (\frac {\sqrt {3 x +2}\, b}{\sqrt {\left (-2 b +3 \sqrt {a b}\right ) b}}\right )}{\sqrt {\left (-2 b +3 \sqrt {a b}\right ) b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {\sqrt {3 \, x + 2}}{b x^{2} - a}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(sqrt(3*x + 2)/(b*x^2 - a), x)

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mupad [B]  time = 0.76, size = 255, normalized size = 1.93 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {2\,\left (\left (576\,b^3+1296\,a\,b^2\right )\,\sqrt {3\,x+2}+\frac {288\,b\,\sqrt {3\,x+2}\,\left (3\,\sqrt {a^3\,b^3}-2\,a\,b^2\right )}{a}\right )\,\sqrt {-\frac {3\,\sqrt {a^3\,b^3}-2\,a\,b^2}{4\,a^2\,b^3}}}{3888\,a\,b-1728\,b^2}\right )\,\sqrt {-\frac {3\,\sqrt {a^3\,b^3}-2\,a\,b^2}{4\,a^2\,b^3}}+2\,\mathrm {atanh}\left (\frac {2\,\left (\left (576\,b^3+1296\,a\,b^2\right )\,\sqrt {3\,x+2}-\frac {288\,b\,\sqrt {3\,x+2}\,\left (3\,\sqrt {a^3\,b^3}+2\,a\,b^2\right )}{a}\right )\,\sqrt {\frac {3\,\sqrt {a^3\,b^3}+2\,a\,b^2}{4\,a^2\,b^3}}}{3888\,a\,b-1728\,b^2}\right )\,\sqrt {\frac {3\,\sqrt {a^3\,b^3}+2\,a\,b^2}{4\,a^2\,b^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 2)^(1/2)/(a - b*x^2),x)

[Out]

2*atanh((2*((1296*a*b^2 + 576*b^3)*(3*x + 2)^(1/2) + (288*b*(3*x + 2)^(1/2)*(3*(a^3*b^3)^(1/2) - 2*a*b^2))/a)*
(-(3*(a^3*b^3)^(1/2) - 2*a*b^2)/(4*a^2*b^3))^(1/2))/(3888*a*b - 1728*b^2))*(-(3*(a^3*b^3)^(1/2) - 2*a*b^2)/(4*
a^2*b^3))^(1/2) + 2*atanh((2*((1296*a*b^2 + 576*b^3)*(3*x + 2)^(1/2) - (288*b*(3*x + 2)^(1/2)*(3*(a^3*b^3)^(1/
2) + 2*a*b^2))/a)*((3*(a^3*b^3)^(1/2) + 2*a*b^2)/(4*a^2*b^3))^(1/2))/(3888*a*b - 1728*b^2))*((3*(a^3*b^3)^(1/2
) + 2*a*b^2)/(4*a^2*b^3))^(1/2)

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sympy [A]  time = 6.80, size = 58, normalized size = 0.44 \begin {gather*} - 6 \operatorname {RootSum} {\left (20736 t^{4} a^{2} b^{3} - 576 t^{2} a b^{2} - 9 a + 4 b, \left (t \mapsto t \log {\left (- 576 t^{3} a b^{2} + 8 t b + \sqrt {3 x + 2} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*RootSum(20736*_t**4*a**2*b**3 - 576*_t**2*a*b**2 - 9*a + 4*b, Lambda(_t, _t*log(-576*_t**3*a*b**2 + 8*_t*b
+ sqrt(3*x + 2))))

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