Integrand size = 24, antiderivative size = 80 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}} \]
-arctanh((d*x^2+c)^(1/2)/c^(1/2))/a/c^(1/2)+arctanh(b^(1/2)*(d*x^2+c)^(1/2 )/(-a*d+b*c)^(1/2))*b^(1/2)/a/(-a*d+b*c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}}{a} \]
-(((Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/Sqrt[-(b *c) + a*d] + ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]/Sqrt[c])/a)
Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {354, 97, 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.
\(\displaystyle \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2\) |
\(\Big \downarrow \) 97 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2}{a}-\frac {b \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{a}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{a d}-\frac {2 b \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{a d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}\right )\) |
((-2*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(a*Sqrt[c]) + (2*Sqrt[b]*ArcTanh[(S qrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/2
3.8.6.3.1 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[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[b/(b*c - a*d) Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && !IntegerQ[p]
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 3.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98
method | result | size |
pseudoelliptic | \(-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \sqrt {c}+\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) \sqrt {\left (a d -b c \right ) b}}{a \sqrt {\left (a d -b c \right ) b}\, \sqrt {c}}\) | \(78\) |
default | \(-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{a \sqrt {c}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 a \sqrt {-\frac {a d -b c}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 a \sqrt {-\frac {a d -b c}{b}}}\) | \(331\) |
-(b*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^(1/2)+arctanh((d*x^2+c )^(1/2)/c^(1/2))*((a*d-b*c)*b)^(1/2))/a/((a*d-b*c)*b)^(1/2)/c^(1/2)
Time = 0.32 (sec) , antiderivative size = 603, normalized size of antiderivative = 7.54 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\left [\frac {c \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right )}{4 \, a c}, \frac {c \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right )}{4 \, a c}, -\frac {c \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) - \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right )}{2 \, a c}, -\frac {c \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) - 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right )}{2 \, a c}\right ] \]
[1/4*(c*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2 *d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2 *c)/x^2))/(a*c), 1/4*(c*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(2*b^2*c^2 - 3*a* b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a *d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)))/(a*c), -1/2*(c*sqrt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) - sqrt(c)*log(- (d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2))/(a*c), -1/2*(c*sqrt(-b/(b* c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) - 2*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)))/(a *c)]
Time = 4.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\begin {cases} \frac {2 \left (- \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 a \sqrt {\frac {a d - b c}{b}}} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{2 a \sqrt {- c}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\operatorname {atan}{\left (\frac {2 \left (\frac {a}{2 b} + x^{2}\right )}{\sqrt {- \frac {a^{2}}{b^{2}}}} \right )}}{b \sqrt {c} \sqrt {- \frac {a^{2}}{b^{2}}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-d*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*a*sqrt((a*d - b*c)/b)) + d*atan(sqrt(c + d*x**2)/sqrt(-c))/(2*a*sqrt(-c)))/d, Ne(d, 0 )), (atan(2*(a/(2*b) + x**2)/sqrt(-a**2/b**2))/(b*sqrt(c)*sqrt(-a**2/b**2) ), True))
\[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} \sqrt {d x^{2} + c} x} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {b \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a} + \frac {\arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} \]
-b*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a) + arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a*sqrt(-c))
Time = 5.52 (sec) , antiderivative size = 651, normalized size of antiderivative = 8.14 \[ \int \frac {1}{x \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{a\,\sqrt {c}}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{a^2\,d-a\,b\,c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{a^2\,d-a\,b\,c}}{\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )}{a^2\,d-a\,b\,c}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,b^3\,d^2\,\sqrt {d\,x^2+c}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (2\,a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^2+c}\,\sqrt {b^2\,c-a\,b\,d}}{4\,\left (a^2\,d-a\,b\,c\right )}\right )}{2\,\left (a^2\,d-a\,b\,c\right )}\right )}{a^2\,d-a\,b\,c}}\right )\,\sqrt {b^2\,c-a\,b\,d}\,1{}\mathrm {i}}{a^2\,d-a\,b\,c} \]
- atanh((c + d*x^2)^(1/2)/c^(1/2))/(a*c^(1/2)) - (atan((((b^2*c - a*b*d)^( 1/2)*(2*b^3*d^2*(c + d*x^2)^(1/2) - ((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 - ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x^2)^(1/2)*(b^2*c - a*b*d)^(1 /2))/(4*(a^2*d - a*b*c))))/(2*(a^2*d - a*b*c)))*1i)/(a^2*d - a*b*c) + ((b^ 2*c - a*b*d)^(1/2)*(2*b^3*d^2*(c + d*x^2)^(1/2) + ((b^2*c - a*b*d)^(1/2)*( 2*a^2*b^2*d^3 + ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x^2)^(1/2)*(b^2 *c - a*b*d)^(1/2))/(4*(a^2*d - a*b*c))))/(2*(a^2*d - a*b*c)))*1i)/(a^2*d - a*b*c))/(((b^2*c - a*b*d)^(1/2)*(2*b^3*d^2*(c + d*x^2)^(1/2) - ((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 - ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x ^2)^(1/2)*(b^2*c - a*b*d)^(1/2))/(4*(a^2*d - a*b*c))))/(2*(a^2*d - a*b*c)) ))/(a^2*d - a*b*c) - ((b^2*c - a*b*d)^(1/2)*(2*b^3*d^2*(c + d*x^2)^(1/2) + ((b^2*c - a*b*d)^(1/2)*(2*a^2*b^2*d^3 + ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^ 2)*(c + d*x^2)^(1/2)*(b^2*c - a*b*d)^(1/2))/(4*(a^2*d - a*b*c))))/(2*(a^2* d - a*b*c))))/(a^2*d - a*b*c)))*(b^2*c - a*b*d)^(1/2)*1i)/(a^2*d - a*b*c)