Integrand size = 25, antiderivative size = 80 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=-\frac {7 \sqrt {c-\frac {c}{a x}} x}{4 a}-\frac {1}{2} \sqrt {c-\frac {c}{a x}} x^2-\frac {7 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{4 a^2} \] Output:
-7/4*(c-c/a/x)^(1/2)*x/a-1/2*(c-c/a/x)^(1/2)*x^2-7/4*c^(1/2)*arctanh((c-c/ a/x)^(1/2)/c^(1/2))/a^2
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=-\frac {\sqrt {c-\frac {c}{a x}} \left (a \sqrt {1-\frac {1}{a x}} x (7+2 a x)+7 \text {arctanh}\left (\sqrt {1-\frac {1}{a x}}\right )\right )}{4 a^2 \sqrt {1-\frac {1}{a x}}} \] Input:
Integrate[E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)]*x,x]
Output:
-1/4*(Sqrt[c - c/(a*x)]*(a*Sqrt[1 - 1/(a*x)]*x*(7 + 2*a*x) + 7*ArcTanh[Sqr t[1 - 1/(a*x)]]))/(a^2*Sqrt[1 - 1/(a*x)])
Time = 0.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6683, 1070, 281, 948, 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.
\(\displaystyle \int x e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} \, dx\) |
\(\Big \downarrow \) 6683 |
\(\displaystyle \int \frac {x (a x+1) \sqrt {c-\frac {c}{a x}}}{1-a x}dx\) |
\(\Big \downarrow \) 1070 |
\(\displaystyle \int \frac {x \left (a+\frac {1}{x}\right ) \sqrt {c-\frac {c}{a x}}}{\frac {1}{x}-a}dx\) |
\(\Big \downarrow \) 281 |
\(\displaystyle -\frac {c \int \frac {\left (a+\frac {1}{x}\right ) x}{\sqrt {c-\frac {c}{a x}}}dx}{a}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {c \int \frac {\left (a+\frac {1}{x}\right ) x^3}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {c \left (\frac {7}{4} \int \frac {x^2}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}-\frac {a x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{a}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {c \left (\frac {7}{4} \left (\frac {\int \frac {x}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{2 a}-\frac {x \sqrt {c-\frac {c}{a x}}}{c}\right )-\frac {a x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{a}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {c \left (\frac {7}{4} \left (-\frac {\int \frac {1}{a-\frac {a}{c x^2}}d\sqrt {c-\frac {c}{a x}}}{c}-\frac {x \sqrt {c-\frac {c}{a x}}}{c}\right )-\frac {a x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {c \left (\frac {7}{4} \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}}-\frac {x \sqrt {c-\frac {c}{a x}}}{c}\right )-\frac {a x^2 \sqrt {c-\frac {c}{a x}}}{2 c}\right )}{a}\) |
Input:
Int[E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)]*x,x]
Output:
(c*(-1/2*(a*Sqrt[c - c/(a*x)]*x^2)/c + (7*(-((Sqrt[c - c/(a*x)]*x)/c) - Ar cTanh[Sqrt[c - c/(a*x)]/Sqrt[c]]/(a*Sqrt[c])))/4))/a
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[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_.))^ (p_.)*((e_) + (f_.)*(x_)^(n_.))^(r_.), x_Symbol] :> Int[x^(m + n*(p + r))*( b + a/x^n)^p*(c + d/x^n)^q*(f + e/x^n)^r, x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && EqQ[mn, -n] && IntegerQ[p] && IntegerQ[r]
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, p}, x] && EqQ[c^2 - a^2*d^2, 0] && !IntegerQ[p] && IntegerQ[n/2] && !G tQ[c, 0]
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.39
method | result | size |
risch | \(-\frac {\left (2 a x +7\right ) x \sqrt {\frac {c \left (a x -1\right )}{a x}}}{4 a}-\frac {7 \ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-a c x}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {c \left (a x -1\right ) a x}}{8 a \sqrt {a^{2} c}\, \left (a x -1\right )}\) | \(111\) |
default | \(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (4 \sqrt {a \,x^{2}-x}\, a^{\frac {5}{2}} x -2 \sqrt {a \,x^{2}-x}\, a^{\frac {3}{2}}+16 a^{\frac {3}{2}} \sqrt {x \left (a x -1\right )}+8 a \ln \left (\frac {2 \sqrt {x \left (a x -1\right )}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )-\ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \right )}{8 \sqrt {x \left (a x -1\right )}\, a^{\frac {5}{2}}}\) | \(139\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x,x,method=_RETURNVERBOSE)
Output:
-1/4*(2*a*x+7)/a*x*(c*(a*x-1)/a/x)^(1/2)-7/8/a*ln((-1/2*a*c+a^2*c*x)/(a^2* c)^(1/2)+(a^2*c*x^2-a*c*x)^(1/2))/(a^2*c)^(1/2)/(a*x-1)*(c*(a*x-1)/a/x)^(1 /2)*(c*(a*x-1)*a*x)^(1/2)
Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.95 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\left [-\frac {2 \, {\left (2 \, a^{2} x^{2} + 7 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 7 \, \sqrt {c} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right )}{8 \, a^{2}}, -\frac {{\left (2 \, a^{2} x^{2} + 7 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 7 \, \sqrt {-c} \arctan \left (\frac {a \sqrt {-c} x \sqrt {\frac {a c x - c}{a x}}}{a c x - c}\right )}{4 \, a^{2}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x,x, algorithm="fricas")
Output:
[-1/8*(2*(2*a^2*x^2 + 7*a*x)*sqrt((a*c*x - c)/(a*x)) - 7*sqrt(c)*log(-2*a* c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c))/a^2, -1/4*((2*a^2*x^2 + 7*a*x)*sqrt((a*c*x - c)/(a*x)) - 7*sqrt(-c)*arctan(a*sqrt(-c)*x*sqrt((a*c* x - c)/(a*x))/(a*c*x - c)))/a^2]
\[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=- \int \frac {x \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx - \int \frac {a x^{2} \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(c-c/a/x)**(1/2)*x,x)
Output:
-Integral(x*sqrt(c - c/(a*x))/(a*x - 1), x) - Integral(a*x**2*sqrt(c - c/( a*x))/(a*x - 1), x)
\[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\int { -\frac {{\left (a x + 1\right )}^{2} \sqrt {c - \frac {c}{a x}} x}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x,x, algorithm="maxima")
Output:
-integrate((a*x + 1)^2*sqrt(c - c/(a*x))*x/(a^2*x^2 - 1), x)
Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.40 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=-\frac {1}{4} \, \sqrt {a^{2} c x^{2} - a c x} {\left (\frac {2 \, x {\left | a \right |}}{a^{2} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left | a \right |}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} - \frac {7 \, \sqrt {c} \log \left ({\left | a \right |} {\left | c \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, a^{2}} + \frac {7 \, \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} \sqrt {c} {\left | a \right |} + a c \right |}\right )}{8 \, a^{2} \mathrm {sgn}\left (x\right )} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x,x, algorithm="giac")
Output:
-1/4*sqrt(a^2*c*x^2 - a*c*x)*(2*x*abs(a)/(a^2*sgn(x)) + 7*abs(a)/(a^3*sgn( x))) - 7/8*sqrt(c)*log(abs(a)*abs(c))*sgn(x)/a^2 + 7/8*sqrt(c)*log(abs(-2* (sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*sqrt(c)*abs(a) + a*c))/(a^2*sgn( x))
Timed out. \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\int -\frac {x\,\sqrt {c-\frac {c}{a\,x}}\,{\left (a\,x+1\right )}^2}{a^2\,x^2-1} \,d x \] Input:
int(-(x*(c - c/(a*x))^(1/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
Output:
int(-(x*(c - c/(a*x))^(1/2)*(a*x + 1)^2)/(a^2*x^2 - 1), x)
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.61 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\frac {\sqrt {c}\, \left (-2 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a x -7 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}-7 \,\mathrm {log}\left (\sqrt {a x -1}+\sqrt {x}\, \sqrt {a}\right )\right )}{4 a^{2}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(c-c/a/x)^(1/2)*x,x)
Output:
(sqrt(c)*( - 2*sqrt(x)*sqrt(a)*sqrt(a*x - 1)*a*x - 7*sqrt(x)*sqrt(a)*sqrt( a*x - 1) - 7*log(sqrt(a*x - 1) + sqrt(x)*sqrt(a))))/(4*a**2)