Integrand size = 27, antiderivative size = 169 \[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {208 a^3 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{105 \sqrt {1-a x}}-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{7 x^3 \sqrt {1-a x}}+\frac {26 a \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{35 x^2 \sqrt {1-a x}}-\frac {104 a^2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{105 x \sqrt {1-a x}} \] Output:
208/105*a^3*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/(-a*x+1)^(1/2)-2/7*(c-c/a/x)^(1/ 2)*(a*x+1)^(1/2)/x^3/(-a*x+1)^(1/2)+26/35*a*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/ x^2/(-a*x+1)^(1/2)-104/105*a^2*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/x/(-a*x+1)^(1 /2)
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.37 \[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x} \left (-15+39 a x-52 a^2 x^2+104 a^3 x^3\right )}{105 x^3 \sqrt {1-a x}} \] Input:
Integrate[Sqrt[c - c/(a*x)]/(E^ArcTanh[a*x]*x^4),x]
Output:
(2*Sqrt[c - c/(a*x)]*Sqrt[1 + a*x]*(-15 + 39*a*x - 52*a^2*x^2 + 104*a^3*x^ 3))/(105*x^3*Sqrt[1 - a*x])
Time = 0.50 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.69, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6684, 6678, 516, 87, 55, 55, 48}
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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx\) |
\(\Big \downarrow \) 6684 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {e^{-\text {arctanh}(a x)} \sqrt {1-a x}}{x^{9/2}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {(1-a x)^{3/2}}{x^{9/2} \sqrt {1-a^2 x^2}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 516 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {1-a x}{x^{9/2} \sqrt {a x+1}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {13}{7} a \int \frac {1}{x^{7/2} \sqrt {a x+1}}dx-\frac {2 \sqrt {a x+1}}{7 x^{7/2}}\right )}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {13}{7} a \left (-\frac {4}{5} a \int \frac {1}{x^{5/2} \sqrt {a x+1}}dx-\frac {2 \sqrt {a x+1}}{5 x^{5/2}}\right )-\frac {2 \sqrt {a x+1}}{7 x^{7/2}}\right )}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {13}{7} a \left (-\frac {4}{5} a \left (-\frac {2}{3} a \int \frac {1}{x^{3/2} \sqrt {a x+1}}dx-\frac {2 \sqrt {a x+1}}{3 x^{3/2}}\right )-\frac {2 \sqrt {a x+1}}{5 x^{5/2}}\right )-\frac {2 \sqrt {a x+1}}{7 x^{7/2}}\right )}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {x} \left (-\frac {13}{7} a \left (-\frac {4}{5} a \left (\frac {4 a \sqrt {a x+1}}{3 \sqrt {x}}-\frac {2 \sqrt {a x+1}}{3 x^{3/2}}\right )-\frac {2 \sqrt {a x+1}}{5 x^{5/2}}\right )-\frac {2 \sqrt {a x+1}}{7 x^{7/2}}\right ) \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}\) |
Input:
Int[Sqrt[c - c/(a*x)]/(E^ArcTanh[a*x]*x^4),x]
Output:
(Sqrt[c - c/(a*x)]*Sqrt[x]*((-2*Sqrt[1 + a*x])/(7*x^(7/2)) - (13*a*((-2*Sq rt[1 + a*x])/(5*x^(5/2)) - (4*a*((-2*Sqrt[1 + a*x])/(3*x^(3/2)) + (4*a*Sqr t[1 + a*x])/(3*Sqrt[x])))/5))/7))/Sqrt[1 - a*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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; Free Q[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Simp[x^p*((c + d/x)^p/(1 + c*(x/d))^p) Int[u*(1 + c*(x/d))^p*(E^(n*Ar cTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*d^2, 0] && !IntegerQ[p]
Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.36
method | result | size |
orering | \(-\frac {2 \sqrt {c -\frac {c}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, \left (104 a^{3} x^{3}-52 a^{2} x^{2}+39 a x -15\right )}{105 x^{3} \left (a x -1\right )}\) | \(60\) |
gosper | \(-\frac {2 \left (104 a^{3} x^{3}-52 a^{2} x^{2}+39 a x -15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}}{105 x^{3} \left (a x -1\right )}\) | \(62\) |
default | \(-\frac {2 \left (104 a^{3} x^{3}-52 a^{2} x^{2}+39 a x -15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}}{105 x^{3} \left (a x -1\right )}\) | \(62\) |
risch | \(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (104 a^{4} x^{4}+52 a^{3} x^{3}-13 a^{2} x^{2}+24 a x -15\right )}{105 \sqrt {-a^{2} x^{2}+1}\, x^{3} \sqrt {-\left (a x +1\right ) a c x}}\) | \(98\) |
Input:
int((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x,method=_RETURNVERBOSE )
Output:
-2/105*(c-c/a/x)^(1/2)/x^3*(-a^2*x^2+1)^(1/2)*(104*a^3*x^3-52*a^2*x^2+39*a *x-15)/(a*x-1)
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.39 \[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=-\frac {2 \, {\left (104 \, a^{3} x^{3} - 52 \, a^{2} x^{2} + 39 \, a x - 15\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \] Input:
integrate((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="fri cas")
Output:
-2/105*(104*a^3*x^3 - 52*a^2*x^2 + 39*a*x - 15)*sqrt(-a^2*x^2 + 1)*sqrt((a *c*x - c)/(a*x))/(a*x^4 - x^3)
\[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{4} \left (a x + 1\right )}\, dx \] Input:
integrate((c-c/a/x)**(1/2)/(a*x+1)*(-a**2*x**2+1)**(1/2)/x**4,x)
Output:
Integral(sqrt(-c*(-1 + 1/(a*x)))*sqrt(-(a*x - 1)*(a*x + 1))/(x**4*(a*x + 1 )), x)
\[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a x}}}{{\left (a x + 1\right )} x^{4}} \,d x } \] Input:
integrate((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="max ima")
Output:
integrate(sqrt(-a^2*x^2 + 1)*sqrt(c - c/(a*x))/((a*x + 1)*x^4), x)
\[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a x}}}{{\left (a x + 1\right )} x^{4}} \,d x } \] Input:
integrate((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="gia c")
Output:
integrate(sqrt(-a^2*x^2 + 1)*sqrt(c - c/(a*x))/((a*x + 1)*x^4), x)
Time = 14.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=-\frac {152\,\sqrt {1-a^2\,x^2}\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3\,\left (a\,x-1\right )}-\frac {26\,\sqrt {1-a^2\,x^2}\,\left (8\,a^2\,x^2+4\,a\,x+7\right )\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3} \] Input:
int(((c - c/(a*x))^(1/2)*(1 - a^2*x^2)^(1/2))/(x^4*(a*x + 1)),x)
Output:
- (152*(1 - a^2*x^2)^(1/2)*((c*(a*x - 1))/(a*x))^(1/2))/(105*x^3*(a*x - 1) ) - (26*(1 - a^2*x^2)^(1/2)*(4*a*x + 8*a^2*x^2 + 7)*((c*(a*x - 1))/(a*x))^ (1/2))/(105*x^3)
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.49 \[ \int \frac {e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \sqrt {c}\, i \left (-104 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{3} x^{3}+52 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{2} x^{2}-39 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a x +15 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}+104 a^{4} x^{4}\right )}{105 a \,x^{4}} \] Input:
int((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x)
Output:
(2*sqrt(c)*i*( - 104*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a**3*x**3 + 52*sqrt(x)* sqrt(a)*sqrt(a*x + 1)*a**2*x**2 - 39*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a*x + 1 5*sqrt(x)*sqrt(a)*sqrt(a*x + 1) + 104*a**4*x**4))/(105*a*x**4)