Integrand size = 35, antiderivative size = 156 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=-\frac {(a-b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x (a+b x)}+\frac {a \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {b \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a+b x} \]
-b*arctanh((d*x^2+c)^(1/2)/c^(1/2))*c^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)+a*ar ctanh(x*d^(1/2)/(d*x^2+c)^(1/2))*d^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)-(-b*x+a )*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x/(b*x+a)
Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left ((-a+b x) \sqrt {c+d x^2}+2 b \sqrt {c} x \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+d x^2}}{\sqrt {c}}\right )-a \sqrt {d} x \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{x (a+b x)} \]
(Sqrt[(a + b*x)^2]*((-a + b*x)*Sqrt[c + d*x^2] + 2*b*Sqrt[c]*x*ArcTanh[(Sq rt[d]*x - Sqrt[c + d*x^2])/Sqrt[c]] - a*Sqrt[d]*x*Log[-(Sqrt[d]*x) + Sqrt[ c + d*x^2]]))/(x*(a + b*x))
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {1334, 27, 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.
| \(\displaystyle \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 1334 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {2 b (a+b x) \sqrt {d x^2+c}}{x^2}dx}{2 b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) \sqrt {d x^2+c}}{x^2}dx}{a+b x}\) |
| \(\Big \downarrow \) 536 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\int \frac {b c+a d x}{x \sqrt {d x^2+c}}dx-\frac {(a-b x) \sqrt {c+d x^2}}{x}\right )}{a+b x}\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a d \int \frac {1}{\sqrt {d x^2+c}}dx+b c \int \frac {1}{x \sqrt {d x^2+c}}dx-\frac {(a-b x) \sqrt {c+d x^2}}{x}\right )}{a+b x}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a d \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+b c \int \frac {1}{x \sqrt {d x^2+c}}dx-\frac {(a-b x) \sqrt {c+d x^2}}{x}\right )}{a+b x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (b c \int \frac {1}{x \sqrt {d x^2+c}}dx+a \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {(a-b x) \sqrt {c+d x^2}}{x}\right )}{a+b x}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} b c \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2+a \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {(a-b x) \sqrt {c+d x^2}}{x}\right )}{a+b x}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b c \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{d}+a \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {(a-b x) \sqrt {c+d x^2}}{x}\right )}{a+b x}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )-\frac {(a-b x) \sqrt {c+d x^2}}{x}-b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-(((a - b*x)*Sqrt[c + d*x^2])/x) + a*Sqrt[ d]*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]] - b*Sqrt[c]*ArcTanh[Sqrt[c + d*x^2 ]/Sqrt[c]]))/(a + b*x)
3.1.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ ) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/((4 *c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])) Int[(g + h*x)^m*(b + 2*c*x)^( 2*p)*(d + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, f, g, h, m, p, q}, x] && E qQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.72
| method | result | size |
| risch | \(-\frac {a \sqrt {d \,x^{2}+c}\, \sqrt {\left (b x +a \right )^{2}}}{x \left (b x +a \right )}+\frac {\left (a \sqrt {d}\, \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )+\sqrt {d \,x^{2}+c}\, b -b \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\) | \(112\) |
| default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (-a \,d^{\frac {3}{2}} x^{2} \sqrt {d \,x^{2}+c}+c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) \sqrt {d}\, b x +a \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}-\sqrt {d \,x^{2}+c}\, \sqrt {d}\, b c x -\ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) a c d x \right )}{c x \sqrt {d}}\) | \(120\) |
-a*(d*x^2+c)^(1/2)/x*((b*x+a)^2)^(1/2)/(b*x+a)+(a*d^(1/2)*ln(d^(1/2)*x+(d* x^2+c)^(1/2))+(d*x^2+c)^(1/2)*b-b*c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2 ))/x))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.29 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.13 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=\left [\frac {a \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + b \sqrt {c} x \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} {\left (b x - a\right )}}{2 \, x}, -\frac {2 \, a \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - b \sqrt {c} x \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {d x^{2} + c} {\left (b x - a\right )}}{2 \, x}, \frac {2 \, b \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + a \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, \sqrt {d x^{2} + c} {\left (b x - a\right )}}{2 \, x}, -\frac {a \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - b \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - \sqrt {d x^{2} + c} {\left (b x - a\right )}}{x}\right ] \]
[1/2*(a*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + b*sqrt (c)*x*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*sqrt(d*x^2 + c)*(b*x - a))/x, -1/2*(2*a*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - b*sqrt(c)*x*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*sqrt (d*x^2 + c)*(b*x - a))/x, 1/2*(2*b*sqrt(-c)*x*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + a*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*sqr t(d*x^2 + c)*(b*x - a))/x, -(a*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c )) - b*sqrt(-c)*x*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - sqrt(d*x^2 + c)*(b*x - a))/x]
\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=\int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x^{2}}\, dx \]
\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=\int { \frac {\sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}}}{x^{2}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=\frac {2 \, b c \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-c}} - a \sqrt {d} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \sqrt {d x^{2} + c} b \mathrm {sgn}\left (b x + a\right ) + \frac {2 \, a c \sqrt {d} \mathrm {sgn}\left (b x + a\right )}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} \]
2*b*c*arctan(-(sqrt(d)*x - sqrt(d*x^2 + c))/sqrt(-c))*sgn(b*x + a)/sqrt(-c ) - a*sqrt(d)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a) + sqrt(d *x^2 + c)*b*sgn(b*x + a) + 2*a*c*sqrt(d)*sgn(b*x + a)/((sqrt(d)*x - sqrt(d *x^2 + c))^2 - c)
Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^2} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x^2} \,d x \]