Integrand size = 25, antiderivative size = 128 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{7 x^3 \sqrt {1-a x}}+\frac {8 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{35 x^2 \sqrt {1-a x}}-\frac {16 a^2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{105 x \sqrt {1-a x}} \] Output:
-2/7*(c-c/a/x)^(1/2)*(a*x+1)^(3/2)/x^3/(-a*x+1)^(1/2)+8/35*a*(c-c/a/x)^(1/ 2)*(a*x+1)^(3/2)/x^2/(-a*x+1)^(1/2)-16/105*a^2*(c-c/a/x)^(1/2)*(a*x+1)^(3/ 2)/x/(-a*x+1)^(1/2)
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.43 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2} \left (15-12 a x+8 a^2 x^2\right )}{105 x^3 \sqrt {1-a x}} \] Input:
Integrate[(E^ArcTanh[a*x]*Sqrt[c - c/(a*x)])/x^4,x]
Output:
(-2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2)*(15 - 12*a*x + 8*a^2*x^2))/(105*x^3* Sqrt[1 - a*x])
Time = 0.49 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6684, 6678, 516, 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 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {\sqrt {1-a^2 x^2}}{x^{9/2} \sqrt {1-a x}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 516 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {\sqrt {a x+1}}{x^{9/2}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {4}{7} a \int \frac {\sqrt {a x+1}}{x^{7/2}}dx-\frac {2 (a x+1)^{3/2}}{7 x^{7/2}}\right )}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {4}{7} a \left (-\frac {2}{5} a \int \frac {\sqrt {a x+1}}{x^{5/2}}dx-\frac {2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2}}{7 x^{7/2}}\right )}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {x} \left (-\frac {2 (a x+1)^{3/2}}{7 x^{7/2}}-\frac {4}{7} a \left (\frac {4 a (a x+1)^{3/2}}{15 x^{3/2}}-\frac {2 (a x+1)^{3/2}}{5 x^{5/2}}\right )\right ) \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}\) |
Input:
Int[(E^ArcTanh[a*x]*Sqrt[c - c/(a*x)])/x^4,x]
Output:
(Sqrt[c - c/(a*x)]*Sqrt[x]*((-2*(1 + a*x)^(3/2))/(7*x^(7/2)) - (4*a*((-2*( 1 + a*x)^(3/2))/(5*x^(5/2)) + (4*a*(1 + a*x)^(3/2))/(15*x^(3/2))))/7))/Sqr t[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[((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 = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.41
method | result | size |
orering | \(-\frac {2 \left (8 a^{2} x^{2}-12 a x +15\right ) \left (a x +1\right )^{2} \sqrt {c -\frac {c}{a x}}}{105 x^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(52\) |
gosper | \(-\frac {2 \left (a x +1\right )^{2} \left (8 a^{2} x^{2}-12 a x +15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{105 x^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(54\) |
default | \(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (8 a^{2} x^{2}-12 a x +15\right )}{105 x^{3} \left (a x -1\right )}\) | \(59\) |
risch | \(-\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (8 a^{4} x^{4}+4 a^{3} x^{3}-a^{2} x^{2}+18 a x +15\right )}{105 \sqrt {-a^{2} x^{2}+1}\, x^{3} \sqrt {-\left (a x +1\right ) a c x}}\) | \(98\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^4,x,method=_RETURNVERBOSE )
Output:
-2/105*(8*a^2*x^2-12*a*x+15)/x^3*(a*x+1)^2/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1 /2)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \, {\left (8 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + 3 \, a x + 15\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^4,x, algorithm="fri cas")
Output:
2/105*(8*a^3*x^3 - 4*a^2*x^2 + 3*a*x + 15)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x))/(a*x^4 - x^3)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{4} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(c-c/a/x)**(1/2)/x**4,x)
Output:
Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**4*sqrt(-(a*x - 1)*(a*x + 1) )), x)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{4}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^4,x, algorithm="max ima")
Output:
integrate((a*x + 1)*sqrt(c - c/(a*x))/(sqrt(-a^2*x^2 + 1)*x^4), x)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{4}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^4,x, algorithm="gia c")
Output:
integrate((a*x + 1)*sqrt(c - c/(a*x))/(sqrt(-a^2*x^2 + 1)*x^4), x)
Time = 14.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.47 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=-\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (\frac {16\,a^4\,x^4}{105}+\frac {8\,a^3\,x^3}{105}-\frac {2\,a^2\,x^2}{105}+\frac {12\,a\,x}{35}+\frac {2}{7}\right )}{x^3\,\sqrt {1-a^2\,x^2}} \] Input:
int(((c - c/(a*x))^(1/2)*(a*x + 1))/(x^4*(1 - a^2*x^2)^(1/2)),x)
Output:
-((c - c/(a*x))^(1/2)*((12*a*x)/35 - (2*a^2*x^2)/105 + (8*a^3*x^3)/105 + ( 16*a^4*x^4)/105 + 2/7))/(x^3*(1 - a^2*x^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.34 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \sqrt {x}\, \sqrt {c}\, \sqrt {a}\, \sqrt {a x +1}\, i \left (8 a^{3} x^{3}-4 a^{2} x^{2}+3 a x +15\right )}{105 a \,x^{4}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^4,x)
Output:
(2*sqrt(x)*sqrt(c)*sqrt(a)*sqrt(a*x + 1)*i*(8*a**3*x**3 - 4*a**2*x**2 + 3* a*x + 15))/(105*a*x**4)