3.1.34 \(\int \frac {A+B x}{x^2 (a+b x^2)^{3/2}} \, dx\) [34]

Optimal. Leaf size=70 \[ \frac {A+B x}{a x \sqrt {a+b x^2}}-\frac {2 A \sqrt {a+b x^2}}{a^2 x}-\frac {B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 70, 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 = {837, 821, 272, 65, 214} \begin {gather*} -\frac {B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2 A \sqrt {a+b x^2}}{a^2 x}+\frac {A+B x}{a x \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 214

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

Rule 272

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 821

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

Rule 837

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

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Mathematica [A]
time = 0.23, size = 71, normalized size = 1.01 \begin {gather*} \frac {-a A+a B x-2 A b x^2}{a^2 x \sqrt {a+b x^2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 82, normalized size = 1.17

method	result	size
risch	\(-\frac {A \sqrt {b \,x^{2}+a}}{a^{2} x}-\frac {A b x}{a^{2} \sqrt {b \,x^{2}+a}}+\frac {B}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) B}{a^{\frac {3}{2}}}\)	\(80\)
default	\(B \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )+A \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 68, normalized size = 0.97 \begin {gather*} -\frac {2 \, A b x}{\sqrt {b x^{2} + a} a^{2}} - \frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {3}{2}}} + \frac {B}{\sqrt {b x^{2} + a} a} - \frac {A}{\sqrt {b x^{2} + a} a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.52, size = 169, normalized size = 2.41 \begin {gather*} \left [\frac {{\left (B b x^{3} + B a x\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, A b x^{2} - B a x + A a\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a^{2} b x^{3} + a^{3} x\right )}}, \frac {{\left (B b x^{3} + B a x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, A b x^{2} - B a x + A a\right )} \sqrt {b x^{2} + a}}{a^{2} b x^{3} + a^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (60) = 120\).
time = 5.22, size = 235, normalized size = 3.36 \begin {gather*} A \left (- \frac {1}{a \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{2}} + 1}}\right ) + B \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1))) + B*(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))

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Giac [A]
time = 0.71, size = 96, normalized size = 1.37 \begin {gather*} -\frac {\frac {A b x}{a^{2}} - \frac {B}{a}}{\sqrt {b x^{2} + a}} + \frac {2 \, B \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {2 \, A \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.45, size = 70, normalized size = 1.00 \begin {gather*} \frac {B}{a\,\sqrt {b\,x^2+a}}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {A}{a\,x\,\sqrt {b\,x^2+a}}-\frac {2\,A\,b\,x}{a^2\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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