Integrand size = 20, antiderivative size = 115 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {3 \sqrt {1-a^2 x^2}}{a \sqrt {c-a c x}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{a (c-a c x)^{5/2}}-\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a \sqrt {c}} \] Output:
3*(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(1/2)+c^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c )^(5/2)-3*2^(1/2)*arctanh(1/2*c^(1/2)*(-a^2*x^2+1)^(1/2)*2^(1/2)/(-a*c*x+c )^(1/2))/a/c^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.50 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {(1+a x)^{5/2} (c-a c x)^{3/2} \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},\frac {1}{2} (1+a x)\right )}{10 a c^2 (1-a x)^{3/2}} \] Input:
Integrate[E^(3*ArcTanh[a*x])/Sqrt[c - a*c*x],x]
Output:
((1 + a*x)^(5/2)*(c - a*c*x)^(3/2)*Hypergeometric2F1[2, 5/2, 7/2, (1 + a*x )/2])/(10*a*c^2*(1 - a*x)^(3/2))
Time = 0.32 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6677, 465, 466, 471, 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.
| \(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx\) |
| \(\Big \downarrow \) 6677 |
| \(\displaystyle c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{7/2}}dx\) |
| \(\Big \downarrow \) 465 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{a c (c-a c x)^{5/2}}-\frac {3 \int \frac {\sqrt {1-a^2 x^2}}{(c-a c x)^{3/2}}dx}{2 c^2}\right )\) |
| \(\Big \downarrow \) 466 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{a c (c-a c x)^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-a c x} \sqrt {1-a^2 x^2}}dx}{c}-\frac {2 \sqrt {1-a^2 x^2}}{a c \sqrt {c-a c x}}\right )}{2 c^2}\right )\) |
| \(\Big \downarrow \) 471 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{a c (c-a c x)^{5/2}}-\frac {3 \left (-4 a \int \frac {1}{\frac {a^2 c^2 \left (1-a^2 x^2\right )}{c-a c x}-2 a^2 c}d\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}-\frac {2 \sqrt {1-a^2 x^2}}{a c \sqrt {c-a c x}}\right )}{2 c^2}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{a c (c-a c x)^{5/2}}-\frac {3 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{a c^{3/2}}-\frac {2 \sqrt {1-a^2 x^2}}{a c \sqrt {c-a c x}}\right )}{2 c^2}\right )\) |
Input:
Int[E^(3*ArcTanh[a*x])/Sqrt[c - a*c*x],x]
Output:
c^3*((1 - a^2*x^2)^(3/2)/(a*c*(c - a*c*x)^(5/2)) - (3*((-2*Sqrt[1 - a^2*x^ 2])/(a*c*Sqrt[c - a*c*x]) + (2*Sqrt[2]*ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2]) /(Sqrt[2]*Sqrt[c - a*c*x])])/(a*c^(3/2))))/(2*c^2))
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + p + 1))) Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n + 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 2*(n + 2*p + 1))) Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 ] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10
| method | result | size |
| default | \(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x -2 a x \sqrt {c \left (a x +1\right )}\, \sqrt {c}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +4 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{c^{\frac {3}{2}} \left (a x -1\right )^{2} \sqrt {c \left (a x +1\right )}\, a}\) | \(127\) |
| risch | \(-\frac {2 \left (a x +1\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{a \sqrt {c \left (a x +1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {4 \left (-\frac {\sqrt {a c x +c}}{2 \left (a c x -c \right )}-\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{a \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\) | \(170\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(3*2^(1/2)*arctanh(1/2*(c*(a*x+1))^( 1/2)*2^(1/2)/c^(1/2))*a*c*x-2*a*x*(c*(a*x+1))^(1/2)*c^(1/2)-3*2^(1/2)*arct anh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*c+4*(c*(a*x+1))^(1/2)*c^(1/2))/ c^(3/2)/(a*x-1)^2/(c*(a*x+1))^(1/2)/a
Time = 0.09 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.23 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\left [-\frac {4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x - 2\right )} - \frac {3 \, \sqrt {2} {\left (a^{2} c x^{2} - 2 \, a c x + c\right )} \log \left (-\frac {a^{2} x^{2} + 2 \, a x + \frac {2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{\sqrt {c}} - 3}{a^{2} x^{2} - 2 \, a x + 1}\right )}{\sqrt {c}}}{2 \, {\left (a^{3} c x^{2} - 2 \, a^{2} c x + a c\right )}}, -\frac {3 \, \sqrt {2} {\left (a^{2} c x^{2} - 2 \, a c x + c\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-\frac {1}{c}}}{2 \, {\left (a x - 1\right )}}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x - 2\right )}}{a^{3} c x^{2} - 2 \, a^{2} c x + a c}\right ] \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x, algorithm="fric as")
Output:
[-1/2*(4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x - 2) - 3*sqrt(2)*(a^2*c* x^2 - 2*a*c*x + c)*log(-(a^2*x^2 + 2*a*x + 2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sq rt(-a*c*x + c)/sqrt(c) - 3)/(a^2*x^2 - 2*a*x + 1))/sqrt(c))/(a^3*c*x^2 - 2 *a^2*c*x + a*c), -(3*sqrt(2)*(a^2*c*x^2 - 2*a*c*x + c)*sqrt(-1/c)*arctan(1 /2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-1/c)/(a*x - 1)) + 2*s qrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x - 2))/(a^3*c*x^2 - 2*a^2*c*x + a*c )]
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\int \frac {\left (a x + 1\right )^{3}}{\sqrt {- c \left (a x - 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**(1/2),x)
Output:
Integral((a*x + 1)**3/(sqrt(-c*(a*x - 1))*(-(a*x - 1)*(a*x + 1))**(3/2)), x)
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\int { \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {-a c x + c}} \,d x } \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x, algorithm="maxi ma")
Output:
integrate((a*x + 1)^3/((-a^2*x^2 + 1)^(3/2)*sqrt(-a*c*x + c)), x)
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {\frac {3 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + 2 \, \sqrt {a c x + c} - \frac {2 \, \sqrt {a c x + c} c}{a c x - c}}{a {\left | c \right |}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x, algorithm="giac ")
Output:
(3*sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) + 2*sqr t(a*c*x + c) - 2*sqrt(a*c*x + c)*c/(a*c*x - c))/(a*abs(c))
Timed out. \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\int \frac {{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}\,\sqrt {c-a\,c\,x}} \,d x \] Input:
int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(1/2)),x)
Output:
int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {\sqrt {c}\, \left (8 \sqrt {a x +1}\, a x -16 \sqrt {a x +1}+12 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x -12 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )-9 \sqrt {2}\, a x +9 \sqrt {2}\right )}{4 a c \left (a x -1\right )} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(1/2),x)
Output:
(sqrt(c)*(8*sqrt(a*x + 1)*a*x - 16*sqrt(a*x + 1) + 12*sqrt(2)*log(tan(asin (sqrt( - a*x + 1)/sqrt(2))/2))*a*x - 12*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2)) - 9*sqrt(2)*a*x + 9*sqrt(2)))/(4*a*c*(a*x - 1))