Integrand size = 20, antiderivative size = 79 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=\frac {1}{2} (2 A+B x) \sqrt {a+c x^2}+\frac {a B \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}}-\sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \]
-A*arctanh((c*x^2+a)^(1/2)/a^(1/2))*a^(1/2)+1/2*a*B*arctanh(x*c^(1/2)/(c*x ^2+a)^(1/2))/c^(1/2)+1/2*(B*x+2*A)*(c*x^2+a)^(1/2)
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=\frac {1}{2} \left ((2 A+B x) \sqrt {a+c x^2}+4 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {a B \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{\sqrt {c}}\right ) \]
((2*A + B*x)*Sqrt[a + c*x^2] + 4*Sqrt[a]*A*ArcTanh[(Sqrt[c]*x - Sqrt[a + c *x^2])/Sqrt[a]] - (a*B*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/Sqrt[c])/2
Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {535, 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.
\(\displaystyle \int \frac {\sqrt {a+c x^2} (A+B x)}{x} \, dx\) |
\(\Big \downarrow \) 535 |
\(\displaystyle \frac {1}{2} a \int \frac {2 A+B x}{x \sqrt {c x^2+a}}dx+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} a \left (2 A \int \frac {1}{x \sqrt {c x^2+a}}dx+B \int \frac {1}{\sqrt {c x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} a \left (2 A \int \frac {1}{x \sqrt {c x^2+a}}dx+B \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}\right )+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} a \left (2 A \int \frac {1}{x \sqrt {c x^2+a}}dx+\frac {B \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}}\right )+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} a \left (A \int \frac {1}{x^2 \sqrt {c x^2+a}}dx^2+\frac {B \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}}\right )+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} a \left (\frac {2 A \int \frac {1}{\frac {x^4}{c}-\frac {a}{c}}d\sqrt {c x^2+a}}{c}+\frac {B \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}}\right )+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} a \left (\frac {B \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {1}{2} \sqrt {a+c x^2} (2 A+B x)\) |
((2*A + B*x)*Sqrt[a + c*x^2])/2 + (a*((B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^ 2]])/Sqrt[c] - (2*A*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/Sqrt[a]))/2
3.4.19.3.1 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_), x_Symbol] :> Sim p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p + 1) Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free Q[{a, b, c, d}, x] && GtQ[p, 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 = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00
method | result | size |
default | \(B \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )+A \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )\) | \(79\) |
B*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2)))+A*(( c*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x))
Time = 0.27 (sec) , antiderivative size = 341, normalized size of antiderivative = 4.32 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=\left [\frac {B a \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, A \sqrt {a} c \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{4 \, c}, -\frac {B a \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - A \sqrt {a} c \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{2 \, c}, \frac {4 \, A \sqrt {-a} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + B a \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{4 \, c}, -\frac {B a \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 2 \, A \sqrt {-a} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a}}{2 \, c}\right ] \]
[1/4*(B*a*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*A*sq rt(a)*c*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(B*c*x + 2 *A*c)*sqrt(c*x^2 + a))/c, -1/2*(B*a*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - A*sqrt(a)*c*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - (B*c*x + 2*A*c)*sqrt(c*x^2 + a))/c, 1/4*(4*A*sqrt(-a)*c*arctan(sqrt(-a)/sq rt(c*x^2 + a)) + B*a*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(B*c*x + 2*A*c)*sqrt(c*x^2 + a))/c, -1/2*(B*a*sqrt(-c)*arctan(sqrt( -c)*x/sqrt(c*x^2 + a)) - 2*A*sqrt(-a)*c*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (B*c*x + 2*A*c)*sqrt(c*x^2 + a))/c]
Time = 2.03 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=- A \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )} + \frac {A a}{\sqrt {c} x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {A \sqrt {c} x}{\sqrt {\frac {a}{c x^{2}} + 1}} + B \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + c x^{2}}}{2} & \text {for}\: c \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \]
-A*sqrt(a)*asinh(sqrt(a)/(sqrt(c)*x)) + A*a/(sqrt(c)*x*sqrt(a/(c*x**2) + 1 )) + A*sqrt(c)*x/sqrt(a/(c*x**2) + 1) + B*Piecewise((a*Piecewise((log(2*sq rt(c)*sqrt(a + c*x**2) + 2*c*x)/sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2) , True))/2 + x*sqrt(a + c*x**2)/2, Ne(c, 0)), (sqrt(a)*x, True))
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + a} B x + \frac {B a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c}} - A \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \sqrt {c x^{2} + a} A \]
1/2*sqrt(c*x^2 + a)*B*x + 1/2*B*a*arcsinh(c*x/sqrt(a*c))/sqrt(c) - A*sqrt( a)*arcsinh(a/(sqrt(a*c)*abs(x))) + sqrt(c*x^2 + a)*A
Exception generated. \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Time = 10.85 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \sqrt {a+c x^2}}{x} \, dx=A\,\sqrt {c\,x^2+a}-A\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )+\frac {B\,x\,\sqrt {c\,x^2+a}}{2}+\frac {B\,a\,\ln \left (\sqrt {c}\,x+\sqrt {c\,x^2+a}\right )}{2\,\sqrt {c}} \]