Integrand size = 23, antiderivative size = 68 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=-\frac {\sqrt {c-a c x}}{2 x^2}-\frac {7 a \sqrt {c-a c x}}{4 x}-\frac {7}{4} a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Output:
-1/2*(-a*c*x+c)^(1/2)/x^2-7/4*a*(-a*c*x+c)^(1/2)/x-7/4*a^2*c^(1/2)*arctanh ((-a*c*x+c)^(1/2)/c^(1/2))
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=-\frac {(2+7 a x) \sqrt {c-a c x}}{4 x^2}-\frac {7}{4} a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Input:
Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^3,x]
Output:
-1/4*((2 + 7*a*x)*Sqrt[c - a*c*x])/x^2 - (7*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a *c*x]/Sqrt[c]])/4
Time = 0.49 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6680, 35, 87, 52, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {(a x+1) \sqrt {c-a c x}}{x^3 (1-a x)}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle c \int \frac {a x+1}{x^3 \sqrt {c-a c x}}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle c \left (\frac {7}{4} a \int \frac {1}{x^2 \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle c \left (\frac {7}{4} a \left (\frac {1}{2} a \int \frac {1}{x \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle c \left (\frac {7}{4} a \left (-\frac {\int \frac {1}{\frac {1}{a}-\frac {c-a c x}{a c}}d\sqrt {c-a c x}}{c}-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle c \left (\frac {7}{4} a \left (-\frac {a \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^3,x]
Output:
c*(-1/2*Sqrt[c - a*c*x]/(c*x^2) + (7*a*(-(Sqrt[c - a*c*x]/(c*x)) - (a*ArcT anh[Sqrt[c - a*c*x]/Sqrt[c]])/Sqrt[c]))/4)
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Int[u*(c + d*x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c , d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(-\frac {7 \left (\operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}}{\sqrt {c}}\right ) a^{2} c \,x^{2}+\sqrt {-c \left (a x -1\right )}\, \sqrt {c}\, \left (a x +\frac {2}{7}\right )\right )}{4 \sqrt {c}\, x^{2}}\) | \(52\) |
risch | \(\frac {\left (7 a^{2} x^{2}-5 a x -2\right ) c}{4 x^{2} \sqrt {-c \left (a x -1\right )}}-\frac {7 a^{2} \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{4}\) | \(54\) |
derivativedivides | \(-2 a^{2} c^{2} \left (\frac {-\frac {7 \left (-a c x +c \right )^{\frac {3}{2}}}{8 c}+\frac {9 \sqrt {-a c x +c}}{8}}{a^{2} c^{2} x^{2}}+\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )\) | \(65\) |
default | \(-2 a^{2} c^{2} \left (\frac {-\frac {7 \left (-a c x +c \right )^{\frac {3}{2}}}{8 c}+\frac {9 \sqrt {-a c x +c}}{8}}{a^{2} c^{2} x^{2}}+\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )\) | \(65\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-7/4*(arctanh((-c*(a*x-1))^(1/2)/c^(1/2))*a^2*c*x^2+(-c*(a*x-1))^(1/2)*c^( 1/2)*(a*x+2/7))/c^(1/2)/x^2
Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.81 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\left [\frac {7 \, a^{2} \sqrt {c} x^{2} \log \left (\frac {a c x + 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, \sqrt {-a c x + c} {\left (7 \, a x + 2\right )}}{8 \, x^{2}}, -\frac {7 \, a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + \sqrt {-a c x + c} {\left (7 \, a x + 2\right )}}{4 \, x^{2}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^3,x, algorithm="fricas ")
Output:
[1/8*(7*a^2*sqrt(c)*x^2*log((a*c*x + 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) - 2*sqrt(-a*c*x + c)*(7*a*x + 2))/x^2, -1/4*(7*a^2*sqrt(-c)*x^2*arctan(sqr t(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + sqrt(-a*c*x + c)*(7*a*x + 2))/x^2]
\[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=- \int \frac {\sqrt {- a c x + c}}{a x^{4} - x^{3}}\, dx - \int \frac {a x \sqrt {- a c x + c}}{a x^{4} - x^{3}}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a*c*x+c)**(1/2)/x**3,x)
Output:
-Integral(sqrt(-a*c*x + c)/(a*x**4 - x**3), x) - Integral(a*x*sqrt(-a*c*x + c)/(a*x**4 - x**3), x)
Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.51 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {1}{8} \, a^{2} c^{2} {\left (\frac {2 \, {\left (7 \, {\left (-a c x + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {-a c x + c} c\right )}}{{\left (a c x - c\right )}^{2} c + 2 \, {\left (a c x - c\right )} c^{2} + c^{3}} + \frac {7 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^3,x, algorithm="maxima ")
Output:
1/8*a^2*c^2*(2*(7*(-a*c*x + c)^(3/2) - 9*sqrt(-a*c*x + c)*c)/((a*c*x - c)^ 2*c + 2*(a*c*x - c)*c^2 + c^3) + 7*log((sqrt(-a*c*x + c) - sqrt(c))/(sqrt( -a*c*x + c) + sqrt(c)))/c^(3/2))
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.12 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {\frac {7 \, a^{3} c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {7 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{3} c - 9 \, \sqrt {-a c x + c} a^{3} c^{2}}{a^{2} c^{2} x^{2}}}{4 \, a} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^3,x, algorithm="giac")
Output:
1/4*(7*a^3*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + (7*(-a*c*x + c)^ (3/2)*a^3*c - 9*sqrt(-a*c*x + c)*a^3*c^2)/(a^2*c^2*x^2))/a
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {7\,{\left (c-a\,c\,x\right )}^{3/2}}{4\,c\,x^2}-\frac {7\,a^2\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c-a\,c\,x}}{\sqrt {c}}\right )}{4}-\frac {9\,\sqrt {c-a\,c\,x}}{4\,x^2} \] Input:
int(-((c - a*c*x)^(1/2)*(a*x + 1)^2)/(x^3*(a^2*x^2 - 1)),x)
Output:
(7*(c - a*c*x)^(3/2))/(4*c*x^2) - (7*a^2*c^(1/2)*atanh((c - a*c*x)^(1/2)/c ^(1/2)))/4 - (9*(c - a*c*x)^(1/2))/(4*x^2)
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {\sqrt {c}\, \left (-14 \sqrt {-a x +1}\, a x -4 \sqrt {-a x +1}+7 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{2} x^{2}-7 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{2} x^{2}\right )}{8 x^{2}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^3,x)
Output:
(sqrt(c)*( - 14*sqrt( - a*x + 1)*a*x - 4*sqrt( - a*x + 1) + 7*log(sqrt( - a*x + 1) - 1)*a**2*x**2 - 7*log(sqrt( - a*x + 1) + 1)*a**2*x**2))/(8*x**2)