\(\int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx\) [405]

Optimal result

Integrand size = 23, antiderivative size = 110 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=-\frac {\sqrt {c-a c x}}{4 x^4}-\frac {5 a \sqrt {c-a c x}}{8 x^3}-\frac {25 a^2 \sqrt {c-a c x}}{32 x^2}-\frac {75 a^3 \sqrt {c-a c x}}{64 x}-\frac {75}{64} a^4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Output:

-1/4*(-a*c*x+c)^(1/2)/x^4-5/8*a*(-a*c*x+c)^(1/2)/x^3-25/32*a^2*(-a*c*x+c)^ 
(1/2)/x^2-75/64*a^3*(-a*c*x+c)^(1/2)/x-75/64*a^4*c^(1/2)*arctanh((-a*c*x+c 
)^(1/2)/c^(1/2))
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.65 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=-\frac {\sqrt {c-a c x} \left (16+40 a x+50 a^2 x^2+75 a^3 x^3\right )}{64 x^4}-\frac {75}{64} a^4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Input:

Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^5,x]
 

Output:

-1/64*(Sqrt[c - a*c*x]*(16 + 40*a*x + 50*a^2*x^2 + 75*a^3*x^3))/x^4 - (75* 
a^4*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/64
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6680, 35, 87, 52, 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.

\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx\)

\(\Big \downarrow \) 6680

\(\displaystyle \int \frac {(a x+1) \sqrt {c-a c x}}{x^5 (1-a x)}dx\)

\(\Big \downarrow \) 35

\(\displaystyle c \int \frac {a x+1}{x^5 \sqrt {c-a c x}}dx\)

\(\Big \downarrow \) 87

\(\displaystyle c \left (\frac {15}{8} a \int \frac {1}{x^4 \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{4 c x^4}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle c \left (\frac {15}{8} a \left (\frac {5}{6} a \int \frac {1}{x^3 \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{3 c x^3}\right )-\frac {\sqrt {c-a c x}}{4 c x^4}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle c \left (\frac {15}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \int \frac {1}{x^2 \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )-\frac {\sqrt {c-a c x}}{4 c x^4}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle c \left (\frac {15}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{x \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )-\frac {\sqrt {c-a c x}}{4 c x^4}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c \left (\frac {15}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \left (-\frac {\int \frac {1}{\frac {1}{a}-\frac {c-a c x}{a c}}d\sqrt {c-a c x}}{c}-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )-\frac {\sqrt {c-a c x}}{4 c x^4}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c \left (\frac {15}{8} a \left (\frac {5}{6} a \left (\frac {3}{4} a \left (-\frac {a \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a c x}}{c x}\right )-\frac {\sqrt {c-a c x}}{2 c x^2}\right )-\frac {\sqrt {c-a c x}}{3 c x^3}\right )-\frac {\sqrt {c-a c x}}{4 c x^4}\right )\)

Input:

Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a*c*x])/x^5,x]
 

Output:

c*(-1/4*Sqrt[c - a*c*x]/(c*x^4) + (15*a*(-1/3*Sqrt[c - a*c*x]/(c*x^3) + (5 
*a*(-1/2*Sqrt[c - a*c*x]/(c*x^2) + (3*a*(-(Sqrt[c - a*c*x]/(c*x)) - (a*Arc 
Tanh[Sqrt[c - a*c*x]/Sqrt[c]])/Sqrt[c]))/4))/6))/8)
 

Defintions of rubi rules used

rule 35

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])
 

rule 52

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]
 

rule 73

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]
 

rule 87

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]))))
 

rule 221

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]
 

rule 6680

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])
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {\left (75 a^{4} x^{4}-25 a^{3} x^{3}-10 a^{2} x^{2}-24 a x -16\right ) c}{64 x^{4} \sqrt {-c \left (a x -1\right )}}-\frac {75 a^{4} \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{64}\)	\(70\)
pseudoelliptic	\(-\frac {75 \left (\frac {\sqrt {-c \left (a x -1\right )}\, \left (75 a^{3} x^{3}+50 a^{2} x^{2}+40 a x +16\right ) \sqrt {c}}{75}+\operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}}{\sqrt {c}}\right ) c \,a^{4} x^{4}\right )}{64 \sqrt {c}\, x^{4}}\)	\(70\)
derivativedivides	\(-2 a^{4} c^{4} \left (\frac {-\frac {75 \left (-a c x +c \right )^{\frac {7}{2}}}{128 c^{3}}+\frac {275 \left (-a c x +c \right )^{\frac {5}{2}}}{128 c^{2}}-\frac {365 \left (-a c x +c \right )^{\frac {3}{2}}}{128 c}+\frac {181 \sqrt {-a c x +c}}{128}}{a^{4} c^{4} x^{4}}+\frac {75 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{128 c^{\frac {7}{2}}}\right )\)	\(93\)
default	\(-2 a^{4} c^{4} \left (\frac {-\frac {75 \left (-a c x +c \right )^{\frac {7}{2}}}{128 c^{3}}+\frac {275 \left (-a c x +c \right )^{\frac {5}{2}}}{128 c^{2}}-\frac {365 \left (-a c x +c \right )^{\frac {3}{2}}}{128 c}+\frac {181 \sqrt {-a c x +c}}{128}}{a^{4} c^{4} x^{4}}+\frac {75 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{128 c^{\frac {7}{2}}}\right )\)	\(93\)

Input:

int((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/64*(75*a^4*x^4-25*a^3*x^3-10*a^2*x^2-24*a*x-16)/x^4/(-c*(a*x-1))^(1/2)*c 
-75/64*a^4*c^(1/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.41 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=\left [\frac {75 \, a^{4} \sqrt {c} x^{4} \log \left (\frac {a c x + 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, {\left (75 \, a^{3} x^{3} + 50 \, a^{2} x^{2} + 40 \, a x + 16\right )} \sqrt {-a c x + c}}{128 \, x^{4}}, -\frac {75 \, a^{4} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + {\left (75 \, a^{3} x^{3} + 50 \, a^{2} x^{2} + 40 \, a x + 16\right )} \sqrt {-a c x + c}}{64 \, x^{4}}\right ] \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^5,x, algorithm="fricas 
")
 

Output:

[1/128*(75*a^4*sqrt(c)*x^4*log((a*c*x + 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/ 
x) - 2*(75*a^3*x^3 + 50*a^2*x^2 + 40*a*x + 16)*sqrt(-a*c*x + c))/x^4, -1/6 
4*(75*a^4*sqrt(-c)*x^4*arctan(sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + (75 
*a^3*x^3 + 50*a^2*x^2 + 40*a*x + 16)*sqrt(-a*c*x + c))/x^4]
 

Sympy [F]

\[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=- \int \frac {\sqrt {- a c x + c}}{a x^{6} - x^{5}}\, dx - \int \frac {a x \sqrt {- a c x + c}}{a x^{6} - x^{5}}\, dx \] Input:

integrate((a*x+1)**2/(-a**2*x**2+1)*(-a*c*x+c)**(1/2)/x**5,x)
 

Output:

-Integral(sqrt(-a*c*x + c)/(a*x**6 - x**5), x) - Integral(a*x*sqrt(-a*c*x 
+ c)/(a*x**6 - x**5), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.48 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=\frac {1}{128} \, a^{4} c^{4} {\left (\frac {2 \, {\left (75 \, {\left (-a c x + c\right )}^{\frac {7}{2}} - 275 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c + 365 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} - 181 \, \sqrt {-a c x + c} c^{3}\right )}}{{\left (a c x - c\right )}^{4} c^{3} + 4 \, {\left (a c x - c\right )}^{3} c^{4} + 6 \, {\left (a c x - c\right )}^{2} c^{5} + 4 \, {\left (a c x - c\right )} c^{6} + c^{7}} + \frac {75 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{c^{\frac {7}{2}}}\right )} \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^5,x, algorithm="maxima 
")
 

Output:

1/128*a^4*c^4*(2*(75*(-a*c*x + c)^(7/2) - 275*(-a*c*x + c)^(5/2)*c + 365*( 
-a*c*x + c)^(3/2)*c^2 - 181*sqrt(-a*c*x + c)*c^3)/((a*c*x - c)^4*c^3 + 4*( 
a*c*x - c)^3*c^4 + 6*(a*c*x - c)^2*c^5 + 4*(a*c*x - c)*c^6 + c^7) + 75*log 
((sqrt(-a*c*x + c) - sqrt(c))/(sqrt(-a*c*x + c) + sqrt(c)))/c^(7/2))
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=\frac {\frac {75 \, a^{5} c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {75 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} a^{5} c + 275 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{5} c^{2} - 365 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{5} c^{3} + 181 \, \sqrt {-a c x + c} a^{5} c^{4}}{a^{4} c^{4} x^{4}}}{64 \, a} \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^5,x, algorithm="giac")
 

Output:

1/64*(75*a^5*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - (75*(a*c*x - c 
)^3*sqrt(-a*c*x + c)*a^5*c + 275*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^5*c^2 - 
365*(-a*c*x + c)^(3/2)*a^5*c^3 + 181*sqrt(-a*c*x + c)*a^5*c^4)/(a^4*c^4*x^ 
4))/a
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=\frac {365\,{\left (c-a\,c\,x\right )}^{3/2}}{64\,c\,x^4}-\frac {181\,\sqrt {c-a\,c\,x}}{64\,x^4}-\frac {275\,{\left (c-a\,c\,x\right )}^{5/2}}{64\,c^2\,x^4}+\frac {75\,{\left (c-a\,c\,x\right )}^{7/2}}{64\,c^3\,x^4}+\frac {a^4\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,75{}\mathrm {i}}{64} \] Input:

int(-((c - a*c*x)^(1/2)*(a*x + 1)^2)/(x^5*(a^2*x^2 - 1)),x)
 

Output:

(a^4*c^(1/2)*atan(((c - a*c*x)^(1/2)*1i)/c^(1/2))*75i)/64 - (181*(c - a*c* 
x)^(1/2))/(64*x^4) + (365*(c - a*c*x)^(3/2))/(64*c*x^4) - (275*(c - a*c*x) 
^(5/2))/(64*c^2*x^4) + (75*(c - a*c*x)^(7/2))/(64*c^3*x^4)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^5} \, dx=\frac {\sqrt {c}\, \left (-150 \sqrt {-a x +1}\, a^{3} x^{3}-100 \sqrt {-a x +1}\, a^{2} x^{2}-80 \sqrt {-a x +1}\, a x -32 \sqrt {-a x +1}+75 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{4} x^{4}-75 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{4} x^{4}\right )}{128 x^{4}} \] Input:

int((a*x+1)^2/(-a^2*x^2+1)*(-a*c*x+c)^(1/2)/x^5,x)
 

Output:

(sqrt(c)*( - 150*sqrt( - a*x + 1)*a**3*x**3 - 100*sqrt( - a*x + 1)*a**2*x* 
*2 - 80*sqrt( - a*x + 1)*a*x - 32*sqrt( - a*x + 1) + 75*log(sqrt( - a*x + 
1) - 1)*a**4*x**4 - 75*log(sqrt( - a*x + 1) + 1)*a**4*x**4))/(128*x**4)