Integrand size = 20, antiderivative size = 80 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}+\sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {A b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]
-1/2*A*b*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(1/2)+B*arctanh(x*b^(1/2)/(b*x ^2+a)^(1/2))*b^(1/2)-1/2*(2*B*x+A)*(b*x^2+a)^(1/2)/x^2
Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=-\frac {(A+2 B x) \sqrt {a+b x^2}}{2 x^2}+\frac {A b \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\sqrt {b} B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]
-1/2*((A + 2*B*x)*Sqrt[a + b*x^2])/x^2 + (A*b*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]])/Sqrt[a] - Sqrt[b]*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2] ]
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {537, 25, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^2} (A+B x)}{x^3} \, dx\) |
\(\Big \downarrow \) 537 |
\(\displaystyle -\frac {1}{2} b \int -\frac {A+2 B x}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} b \int \frac {A+2 B x}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} b \left (A \int \frac {1}{x \sqrt {b x^2+a}}dx+2 B \int \frac {1}{\sqrt {b x^2+a}}dx\right )-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} b \left (A \int \frac {1}{x \sqrt {b x^2+a}}dx+2 B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} b \left (A \int \frac {1}{x \sqrt {b x^2+a}}dx+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{2} A \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} b \left (\frac {A \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} b \left (\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )-\frac {\sqrt {a+b x^2} (A+2 B x)}{2 x^2}\) |
-1/2*((A + 2*B*x)*Sqrt[a + b*x^2])/x^2 + (b*((2*B*ArcTanh[(Sqrt[b]*x)/Sqrt [a + b*x^2]])/Sqrt[b] - (A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/2
3.1.7.3.1 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[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 0] && IntegerQ[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 = 3.43 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\left (2 B x +A \right ) \sqrt {b \,x^{2}+a}}{2 x^{2}}+\sqrt {b}\, B \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )-\frac {b A \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 \sqrt {a}}\) | \(73\) |
default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )\) | \(127\) |
-1/2*(2*B*x+A)*(b*x^2+a)^(1/2)/x^2+b^(1/2)*B*ln(x*b^(1/2)+(b*x^2+a)^(1/2)) -1/2*b*A/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
Time = 0.28 (sec) , antiderivative size = 377, normalized size of antiderivative = 4.71 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=\left [\frac {2 \, B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{4 \, a x^{2}}, -\frac {4 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{4 \, a x^{2}}, \frac {A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{2 \, a x^{2}}, -\frac {2 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, B a x + A a\right )} \sqrt {b x^{2} + a}}{2 \, a x^{2}}\right ] \]
[1/4*(2*B*a*sqrt(b)*x^2*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + A*sqrt(a)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(2 *B*a*x + A*a)*sqrt(b*x^2 + a))/(a*x^2), -1/4*(4*B*a*sqrt(-b)*x^2*arctan(sq rt(-b)*x/sqrt(b*x^2 + a)) - A*sqrt(a)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a )*sqrt(a) + 2*a)/x^2) + 2*(2*B*a*x + A*a)*sqrt(b*x^2 + a))/(a*x^2), 1/2*(A *sqrt(-a)*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + B*a*sqrt(b)*x^2*log(-2* b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - (2*B*a*x + A*a)*sqrt(b*x^2 + a) )/(a*x^2), -1/2*(2*B*a*sqrt(-b)*x^2*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - A *sqrt(-a)*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (2*B*a*x + A*a)*sqrt(b* x^2 + a))/(a*x^2)]
Time = 1.99 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {a}} - \frac {B \sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + B \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {B b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]
-A*sqrt(b)*sqrt(a/(b*x**2) + 1)/(2*x) - A*b*asinh(sqrt(a)/(sqrt(b)*x))/(2* sqrt(a)) - B*sqrt(a)/(x*sqrt(1 + b*x**2/a)) + B*sqrt(b)*asinh(sqrt(b)*x/sq rt(a)) - B*b*x/(sqrt(a)*sqrt(1 + b*x**2/a))
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=B \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \frac {\sqrt {b x^{2} + a} A b}{2 \, a} - \frac {\sqrt {b x^{2} + a} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{2 \, a x^{2}} \]
B*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - 1/2*A*b*arcsinh(a/(sqrt(a*b)*abs(x)))/s qrt(a) + 1/2*sqrt(b*x^2 + a)*A*b/a - sqrt(b*x^2 + a)*B/x - 1/2*(b*x^2 + a) ^(3/2)*A/(a*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (62) = 124\).
Time = 0.31 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.04 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=\frac {A b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - B \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2}} \]
A*b*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - B*sqrt(b)*l og(abs(-sqrt(b)*x + sqrt(b*x^2 + a))) + ((sqrt(b)*x - sqrt(b*x^2 + a))^3*A *b + 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a*sqrt(b) + (sqrt(b)*x - sqrt(b*x ^2 + a))*A*a*b - 2*B*a^2*sqrt(b))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^2
Time = 6.45 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^3} \, dx=-\frac {A\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {B\,\sqrt {b\,x^2+a}}{x}-\frac {A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,\sqrt {a}}-\frac {B\,\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}\,\sqrt {\frac {b\,x^2}{a}+1}} \]