Integrand size = 20, antiderivative size = 75 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=-\frac {(A-B x) \sqrt {a+b x^2}}{x}+A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
-B*arctanh((b*x^2+a)^(1/2)/a^(1/2))*a^(1/2)+A*arctanh(x*b^(1/2)/(b*x^2+a)^ (1/2))*b^(1/2)-(-B*x+A)*(b*x^2+a)^(1/2)/x
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=\frac {(-A+B x) \sqrt {a+b x^2}}{x}+2 \sqrt {a} B \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]
((-A + B*x)*Sqrt[a + b*x^2])/x + 2*Sqrt[a]*B*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] - A*Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {536, 538, 224, 219, 243, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {a+b x^2} (A+B x)}{x^2} \, dx\) |
\(\Big \downarrow \) 536 |
\(\displaystyle \int \frac {a B+A b x}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (A-B x)}{x}\) |
\(\Big \downarrow \) 538 |
\(\displaystyle A b \int \frac {1}{\sqrt {b x^2+a}}dx+a B \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (A-B x)}{x}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle A b \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+a B \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (A-B x)}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a B \int \frac {1}{x \sqrt {b x^2+a}}dx+A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (A-B x)}{x}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} a B \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (A-B x)}{x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a B \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (A-B x)}{x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (A-B x)}{x}-\sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\) |
-(((A - B*x)*Sqrt[a + b*x^2])/x) + A*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]] - Sqrt[a]*B*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]
3.1.6.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer Q[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Time = 3.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {A \sqrt {b \,x^{2}+a}}{x}+A \sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )+\sqrt {b \,x^{2}+a}\, B -B \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\) | \(78\) |
default | \(B \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\right )\) | \(103\) |
-A*(b*x^2+a)^(1/2)/x+A*b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+(b*x^2+a)^(1/ 2)*B-B*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
Time = 0.32 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.44 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=\left [\frac {A \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + B \sqrt {a} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt {b x^{2} + a} {\left (B x - A\right )}}{2 \, x}, -\frac {2 \, A \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - B \sqrt {a} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, \sqrt {b x^{2} + a} {\left (B x - A\right )}}{2 \, x}, \frac {2 \, B \sqrt {-a} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + A \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, \sqrt {b x^{2} + a} {\left (B x - A\right )}}{2 \, x}, -\frac {A \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - B \sqrt {-a} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - \sqrt {b x^{2} + a} {\left (B x - A\right )}}{x}\right ] \]
[1/2*(A*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + B*sqrt (a)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*sqrt(b*x^2 + a)*(B*x - A))/x, -1/2*(2*A*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - B*sqrt(a)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*sqrt (b*x^2 + a)*(B*x - A))/x, 1/2*(2*B*sqrt(-a)*x*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + A*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*sqr t(b*x^2 + a)*(B*x - A))/x, -(A*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a )) - B*sqrt(-a)*x*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - sqrt(b*x^2 + a)*(B*x - A))/x]
Time = 1.95 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=- \frac {A \sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + A \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - B \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \]
-A*sqrt(a)/(x*sqrt(1 + b*x**2/a)) + A*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) - A *b*x/(sqrt(a)*sqrt(1 + b*x**2/a)) - B*sqrt(a)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + B*sqrt(b)*x/sqrt(a/(b*x**2) + 1)
Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=A \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \sqrt {b x^{2} + a} B - \frac {\sqrt {b x^{2} + a} A}{x} \]
A*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - B*sqrt(a)*arcsinh(a/(sqrt(a*b)*abs(x))) + sqrt(b*x^2 + a)*B - sqrt(b*x^2 + a)*A/x
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=\frac {2 \, B a \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - A \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \sqrt {b x^{2} + a} B + \frac {2 \, A a \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} \]
2*B*a*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - A*sqrt(b) *log(abs(-sqrt(b)*x + sqrt(b*x^2 + a))) + sqrt(b*x^2 + a)*B + 2*A*a*sqrt(b )/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)
Time = 6.49 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.19 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^2} \, dx=B\,\sqrt {b\,x^2+a}-\frac {A\,\sqrt {b\,x^2+a}}{x}-B\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )-\frac {A\,\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}\,\sqrt {\frac {b\,x^2}{a}+1}} \]