3.1.33 \(\int \frac {A+B x}{x (a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=47 \[ \frac {A+B x}{a \sqrt {a+b x^2}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {823, 12, 266, 63, 208} \begin {gather*} \frac {A+B x}{a \sqrt {a+b x^2}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x*(a + b*x^2)^(3/2)),x]

[Out]

(A + B*x)/(a*Sqrt[a + b*x^2]) - (A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 47, normalized size = 1.00 \begin {gather*} \frac {A+B x}{a \sqrt {a+b x^2}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x*(a + b*x^2)^(3/2)),x]

[Out]

(A + B*x)/(a*Sqrt[a + b*x^2]) - (A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.33, size = 61, normalized size = 1.30 \begin {gather*} \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {A+B x}{a \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x*(a + b*x^2)^(3/2)),x]

[Out]

(A + B*x)/(a*Sqrt[a + b*x^2]) + (2*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a] - Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)

________________________________________________________________________________________

fricas [A]  time = 1.00, size = 146, normalized size = 3.11 \begin {gather*} \left [\frac {{\left (A b x^{2} + A a\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B a x + A a\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a^{2} b x^{2} + a^{3}\right )}}, \frac {{\left (A b x^{2} + A a\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (B a x + A a\right )} \sqrt {b x^{2} + a}}{a^{2} b x^{2} + a^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((A*b*x^2 + A*a)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(B*a*x + A*a)*sqrt(b*x^2
+ a))/(a^2*b*x^2 + a^3), ((A*b*x^2 + A*a)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (B*a*x + A*a)*sqrt(b*x^
2 + a))/(a^2*b*x^2 + a^3)]

________________________________________________________________________________________

giac [A]  time = 0.44, size = 59, normalized size = 1.26 \begin {gather*} \frac {\frac {B x}{a} + \frac {A}{a}}{\sqrt {b x^{2} + a}} + \frac {2 \, A \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x/a + A/a)/sqrt(b*x^2 + a) + 2*A*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 60, normalized size = 1.28 \begin {gather*} \frac {B x}{\sqrt {b \,x^{2}+a}\, a}-\frac {A \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{a^{\frac {3}{2}}}+\frac {A}{\sqrt {b \,x^{2}+a}\, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x/(b*x^2+a)^(3/2),x)

[Out]

B*x/a/(b*x^2+a)^(1/2)+A/a/(b*x^2+a)^(1/2)-A/a^(3/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)

________________________________________________________________________________________

maxima [A]  time = 1.30, size = 48, normalized size = 1.02 \begin {gather*} \frac {B x}{\sqrt {b x^{2} + a} a} - \frac {A \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {3}{2}}} + \frac {A}{\sqrt {b x^{2} + a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x/(sqrt(b*x^2 + a)*a) - A*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) + A/(sqrt(b*x^2 + a)*a)

________________________________________________________________________________________

mupad [B]  time = 1.29, size = 50, normalized size = 1.06 \begin {gather*} \frac {A}{a\,\sqrt {b\,x^2+a}}-\frac {A\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {B\,x}{a\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x*(a + b*x^2)^(3/2)),x)

[Out]

A/(a*(a + b*x^2)^(1/2)) - (A*atanh((a + b*x^2)^(1/2)/a^(1/2)))/a^(3/2) + (B*x)/(a*(a + b*x^2)^(1/2))

________________________________________________________________________________________

sympy [B]  time = 11.29, size = 206, normalized size = 4.38 \begin {gather*} A \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{3} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{2} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{2} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}}\right ) + \frac {B x}{a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(b*x**2+a)**(3/2),x)

[Out]

A*(2*a**3*sqrt(1 + b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**3*log(b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*
x**2) - 2*a**3*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**2*b*x**2*log(b*x**2/a)/(2*a**
(9/2) + 2*a**(7/2)*b*x**2) - 2*a**2*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2)) + B*x
/(a**(3/2)*sqrt(1 + b*x**2/a))

________________________________________________________________________________________