Integrand size = 35, antiderivative size = 161 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=-\frac {(a+2 b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 x^2 (a+b x)}+\frac {b \sqrt {d} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{a+b x}-\frac {a d \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c} (a+b x)} \]
-1/2*a*d*arctanh((d*x^2+c)^(1/2)/c^(1/2))*((b*x+a)^2)^(1/2)/(b*x+a)/c^(1/2 )+b*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))*d^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)-1 /2*(2*b*x+a)*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x^2/(b*x+a)
Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=\frac {\sqrt {(a+b x)^2} \left (2 a d x^2 \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+d x^2}}{\sqrt {c}}\right )-\sqrt {c} \left ((a+2 b x) \sqrt {c+d x^2}+2 b \sqrt {d} x^2 \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )\right )}{2 \sqrt {c} x^2 (a+b x)} \]
(Sqrt[(a + b*x)^2]*(2*a*d*x^2*ArcTanh[(Sqrt[d]*x - Sqrt[c + d*x^2])/Sqrt[c ]] - Sqrt[c]*((a + 2*b*x)*Sqrt[c + d*x^2] + 2*b*Sqrt[d]*x^2*Log[-(Sqrt[d]* x) + Sqrt[c + d*x^2]])))/(2*Sqrt[c]*x^2*(a + b*x))
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1334, 27, 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.
| \(\displaystyle \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx\) |
| \(\Big \downarrow \) 1334 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {2 b (a+b x) \sqrt {d x^2+c}}{x^3}dx}{2 b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) \sqrt {d x^2+c}}{x^3}dx}{a+b x}\) |
| \(\Big \downarrow \) 537 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {1}{2} d \int -\frac {a+2 b x}{x \sqrt {d x^2+c}}dx-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \int \frac {a+2 b x}{x \sqrt {d x^2+c}}dx-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \left (a \int \frac {1}{x \sqrt {d x^2+c}}dx+2 b \int \frac {1}{\sqrt {d x^2+c}}dx\right )-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \left (a \int \frac {1}{x \sqrt {d x^2+c}}dx+2 b \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}\right )-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \left (a \int \frac {1}{x \sqrt {d x^2+c}}dx+\frac {2 b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right )-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \left (\frac {1}{2} a \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2+\frac {2 b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right )-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \left (\frac {a \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{d}+\frac {2 b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}\right )-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{2} d \left (\frac {2 b \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}}-\frac {a \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right )-\frac {(a+2 b x) \sqrt {c+d x^2}}{2 x^2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/2*((a + 2*b*x)*Sqrt[c + d*x^2])/x^2 + ( d*((2*b*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/Sqrt[d] - (a*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/Sqrt[c]))/2))/(a + b*x)
3.1.45.3.1 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[(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]
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 = 0.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.66
| method | result | size |
| risch | \(-\frac {\left (2 b x +a \right ) \sqrt {\left (b x +a \right )^{2}}\, \sqrt {d \,x^{2}+c}}{2 x^{2} \left (b x +a \right )}+\frac {\left (\sqrt {d}\, b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )-\frac {d a \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 \sqrt {c}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\) | \(107\) |
| default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (-2 b \,d^{\frac {3}{2}} x^{3} \sqrt {d \,x^{2}+c}+\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) d^{\frac {3}{2}} a \,x^{2}+2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b x -a \,d^{\frac {3}{2}} x^{2} \sqrt {d \,x^{2}+c}-2 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b c d \,x^{2}+a \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\right )}{2 c \,x^{2} \sqrt {d}}\) | \(141\) |
-1/2*(2*b*x+a)*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/x^2/(b*x+a)+(d^(1/2)*b*ln (d^(1/2)*x+(d*x^2+c)^(1/2))-1/2*d*a/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 = 0.28 (sec) , antiderivative size = 377, normalized size of antiderivative = 2.34 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=\left [\frac {2 \, b c \sqrt {d} x^{2} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + a \sqrt {c} d x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{4 \, c x^{2}}, -\frac {4 \, b c \sqrt {-d} x^{2} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {c} d x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{4 \, c x^{2}}, \frac {a \sqrt {-c} d x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + b c \sqrt {d} x^{2} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{2 \, c x^{2}}, -\frac {2 \, b c \sqrt {-d} x^{2} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - a \sqrt {-c} d x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b c x + a c\right )} \sqrt {d x^{2} + c}}{2 \, c x^{2}}\right ] \]
[1/4*(2*b*c*sqrt(d)*x^2*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + a*sqrt(c)*d*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(2 *b*c*x + a*c)*sqrt(d*x^2 + c))/(c*x^2), -1/4*(4*b*c*sqrt(-d)*x^2*arctan(sq rt(-d)*x/sqrt(d*x^2 + c)) - a*sqrt(c)*d*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c )*sqrt(c) + 2*c)/x^2) + 2*(2*b*c*x + a*c)*sqrt(d*x^2 + c))/(c*x^2), 1/2*(a *sqrt(-c)*d*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + b*c*sqrt(d)*x^2*log(-2* d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - (2*b*c*x + a*c)*sqrt(d*x^2 + c) )/(c*x^2), -1/2*(2*b*c*sqrt(-d)*x^2*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - a *sqrt(-c)*d*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (2*b*c*x + a*c)*sqrt(d* x^2 + c))/(c*x^2)]
\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=\int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}}{x^{3}}\, dx \]
\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=\int { \frac {\sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}}}{x^{3}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=\frac {a d \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-c}} - b \sqrt {d} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{3} a d \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c \sqrt {d} \mathrm {sgn}\left (b x + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )} a c d \mathrm {sgn}\left (b x + a\right ) - 2 \, b c^{2} \sqrt {d} \mathrm {sgn}\left (b x + a\right )}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{2}} \]
a*d*arctan(-(sqrt(d)*x - sqrt(d*x^2 + c))/sqrt(-c))*sgn(b*x + a)/sqrt(-c) - b*sqrt(d)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a) + ((sqrt(d )*x - sqrt(d*x^2 + c))^3*a*d*sgn(b*x + a) + 2*(sqrt(d)*x - sqrt(d*x^2 + c) )^2*b*c*sqrt(d)*sgn(b*x + a) + (sqrt(d)*x - sqrt(d*x^2 + c))*a*c*d*sgn(b*x + a) - 2*b*c^2*sqrt(d)*sgn(b*x + a))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c )^2
Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{x^3} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c}}{x^3} \,d x \]