Integrand size = 23, antiderivative size = 89 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=-\frac {\sqrt {c-a c x}}{3 x^3}-\frac {11 a \sqrt {c-a c x}}{12 x^2}-\frac {11 a^2 \sqrt {c-a c x}}{8 x}-\frac {11}{8} a^3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Output:
-1/3*(-a*c*x+c)^(1/2)/x^3-11/12*a*(-a*c*x+c)^(1/2)/x^2-11/8*a^2*(-a*c*x+c) ^(1/2)/x-11/8*a^3*c^(1/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/2))
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.71 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=-\frac {\sqrt {c-a c x} \left (8+22 a x+33 a^2 x^2\right )}{24 x^3}-\frac {11}{8} a^3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Input:
Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^4,x]
Output:
-1/24*(Sqrt[c - a*c*x]*(8 + 22*a*x + 33*a^2*x^2))/x^3 - (11*a^3*Sqrt[c]*Ar cTanh[Sqrt[c - a*c*x]/Sqrt[c]])/8
Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6680, 35, 87, 52, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle \int \frac {(a x+1) \sqrt {c-a c x}}{x^4 (1-a x)}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle c \int \frac {a x+1}{x^4 \sqrt {c-a c x}}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle c \left (\frac {11}{6} a \int \frac {1}{x^3 \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle c \left (\frac {11}{6} a \left (\frac {3}{4} a \int \frac {1}{x^2 \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle c \left (\frac {11}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{x \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle c \left (\frac {11}{6} a \left (\frac {3}{4} a \left (-\frac {\int \frac {1}{\frac {1}{a}-\frac {c-a c x}{a c}}d\sqrt {c-a c x}}{c}-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle c \left (\frac {11}{6} a \left (\frac {3}{4} a \left (-\frac {a \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^4,x]
Output:
c*(-1/3*Sqrt[c - a*c*x]/(c*x^3) + (11*a*(-1/2*Sqrt[c - a*c*x]/(c*x^2) + (3 *a*(-(Sqrt[c - a*c*x]/(c*x)) - (a*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/Sqrt[c ]))/4))/6)
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 = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {\left (33 a^{3} x^{3}-11 a^{2} x^{2}-14 a x -8\right ) c}{24 x^{3} \sqrt {-c \left (a x -1\right )}}-\frac {11 a^{3} \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{8}\) | \(62\) |
pseudoelliptic | \(-\frac {11 \left (\frac {\sqrt {-c \left (a x -1\right )}\, \left (33 a^{2} x^{2}+22 a x +8\right ) \sqrt {c}}{33}+\operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}}{\sqrt {c}}\right ) a^{3} c \,x^{3}\right )}{8 \sqrt {c}\, x^{3}}\) | \(62\) |
default | \(-2 a^{3} c^{3} \left (\frac {\frac {11 \left (-a c x +c \right )^{\frac {5}{2}}}{16 c^{2}}-\frac {11 \left (-a c x +c \right )^{\frac {3}{2}}}{6 c}+\frac {21 \sqrt {-a c x +c}}{16}}{a^{3} c^{3} x^{3}}+\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{16 c^{\frac {5}{2}}}\right )\) | \(79\) |
derivativedivides | \(2 a^{3} c^{3} \left (-\frac {\frac {11 \left (-a c x +c \right )^{\frac {5}{2}}}{16 c^{2}}-\frac {11 \left (-a c x +c \right )^{\frac {3}{2}}}{6 c}+\frac {21 \sqrt {-a c x +c}}{16}}{a^{3} c^{3} x^{3}}-\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{16 c^{\frac {5}{2}}}\right )\) | \(80\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
1/24*(33*a^3*x^3-11*a^2*x^2-14*a*x-8)/x^3/(-c*(a*x-1))^(1/2)*c-11/8*a^3*c^ (1/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/2))
Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.56 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=\left [\frac {33 \, a^{3} \sqrt {c} x^{3} \log \left (\frac {a c x + 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, {\left (33 \, a^{2} x^{2} + 22 \, a x + 8\right )} \sqrt {-a c x + c}}{48 \, x^{3}}, -\frac {33 \, a^{3} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + {\left (33 \, a^{2} x^{2} + 22 \, a x + 8\right )} \sqrt {-a c x + c}}{24 \, x^{3}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^4,x, algorithm="fricas ")
Output:
[1/48*(33*a^3*sqrt(c)*x^3*log((a*c*x + 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x ) - 2*(33*a^2*x^2 + 22*a*x + 8)*sqrt(-a*c*x + c))/x^3, -1/24*(33*a^3*sqrt( -c)*x^3*arctan(sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + (33*a^2*x^2 + 22*a *x + 8)*sqrt(-a*c*x + c))/x^3]
\[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=- \int \frac {\sqrt {- a c x + c}}{a x^{5} - x^{4}}\, dx - \int \frac {a x \sqrt {- a c x + c}}{a x^{5} - x^{4}}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a*c*x+c)**(1/2)/x**4,x)
Output:
-Integral(sqrt(-a*c*x + c)/(a*x**5 - x**4), x) - Integral(a*x*sqrt(-a*c*x + c)/(a*x**5 - x**4), x)
Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.51 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=-\frac {1}{48} \, a^{3} c^{3} {\left (\frac {2 \, {\left (33 \, {\left (-a c x + c\right )}^{\frac {5}{2}} - 88 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 63 \, \sqrt {-a c x + c} c^{2}\right )}}{{\left (a c x - c\right )}^{3} c^{2} + 3 \, {\left (a c x - c\right )}^{2} c^{3} + 3 \, {\left (a c x - c\right )} c^{4} + c^{5}} - \frac {33 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^4,x, algorithm="maxima ")
Output:
-1/48*a^3*c^3*(2*(33*(-a*c*x + c)^(5/2) - 88*(-a*c*x + c)^(3/2)*c + 63*sqr t(-a*c*x + c)*c^2)/((a*c*x - c)^3*c^2 + 3*(a*c*x - c)^2*c^3 + 3*(a*c*x - c )*c^4 + c^5) - 33*log((sqrt(-a*c*x + c) - sqrt(c))/(sqrt(-a*c*x + c) + sqr t(c)))/c^(5/2))
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=\frac {\frac {33 \, a^{4} c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {33 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{4} c - 88 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{4} c^{2} + 63 \, \sqrt {-a c x + c} a^{4} c^{3}}{a^{3} c^{3} x^{3}}}{24 \, a} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^4,x, algorithm="giac")
Output:
1/24*(33*a^4*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - (33*(a*c*x - c )^2*sqrt(-a*c*x + c)*a^4*c - 88*(-a*c*x + c)^(3/2)*a^4*c^2 + 63*sqrt(-a*c* x + c)*a^4*c^3)/(a^3*c^3*x^3))/a
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=\frac {11\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,c\,x^3}-\frac {21\,\sqrt {c-a\,c\,x}}{8\,x^3}-\frac {11\,{\left (c-a\,c\,x\right )}^{5/2}}{8\,c^2\,x^3}+\frac {a^3\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,11{}\mathrm {i}}{8} \] Input:
int(-((c - a*c*x)^(1/2)*(a*x + 1)^2)/(x^4*(a^2*x^2 - 1)),x)
Output:
(a^3*c^(1/2)*atan(((c - a*c*x)^(1/2)*1i)/c^(1/2))*11i)/8 - (21*(c - a*c*x) ^(1/2))/(8*x^3) + (11*(c - a*c*x)^(3/2))/(3*c*x^3) - (11*(c - a*c*x)^(5/2) )/(8*c^2*x^3)
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^4} \, dx=\frac {\sqrt {c}\, \left (-66 \sqrt {-a x +1}\, a^{2} x^{2}-44 \sqrt {-a x +1}\, a x -16 \sqrt {-a x +1}+33 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{3} x^{3}-33 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{3} x^{3}\right )}{48 x^{3}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^4,x)
Output:
(sqrt(c)*( - 66*sqrt( - a*x + 1)*a**2*x**2 - 44*sqrt( - a*x + 1)*a*x - 16* sqrt( - a*x + 1) + 33*log(sqrt( - a*x + 1) - 1)*a**3*x**3 - 33*log(sqrt( - a*x + 1) + 1)*a**3*x**3))/(48*x**3)