Integrand size = 18, antiderivative size = 74 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=-\frac {2 \sqrt {1-a^2 x^2}}{a c (1-a x)}+\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a c (1-a x)^3}+\frac {\arcsin (a x)}{a c} \] Output:
-2*(-a^2*x^2+1)^(1/2)/a/c/(-a*x+1)+2/3*(-a^2*x^2+1)^(3/2)/a/c/(-a*x+1)^3+a rcsin(a*x)/a/c
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.61 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=\frac {4 \sqrt {2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1-a x)\right )}{3 a c (1-a x)^{3/2}} \] Input:
Integrate[E^(3*ArcTanh[a*x])/(c - a*c*x),x]
Output:
(4*Sqrt[2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - a*x)/2])/(3*a*c*(1 - a *x)^(3/2))
Time = 0.42 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6677, 27, 465, 463, 25, 223}
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.
\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{c^4 (1-a x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (1-a^2 x^2\right )^{3/2}}{(1-a x)^4}dx}{c}\) |
\(\Big \downarrow \) 465 |
\(\displaystyle \frac {\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a (1-a x)^3}-\int \frac {\sqrt {1-a^2 x^2}}{(1-a x)^2}dx}{c}\) |
\(\Big \downarrow \) 463 |
\(\displaystyle \frac {-\int -\frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a (1-a x)^3}-\frac {2 \sqrt {1-a^2 x^2}}{a (1-a x)}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a (1-a x)^3}-\frac {2 \sqrt {1-a^2 x^2}}{a (1-a x)}}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a (1-a x)^3}-\frac {2 \sqrt {1-a^2 x^2}}{a (1-a x)}+\frac {\arcsin (a x)}{a}}{c}\) |
Input:
Int[E^(3*ArcTanh[a*x])/(c - a*c*x),x]
Output:
((-2*Sqrt[1 - a^2*x^2])/(a*(1 - a*x)) + (2*(1 - a^2*x^2)^(3/2))/(3*a*(1 - a*x)^3) + ArcSin[a*x]/a)/c
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x ))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ (2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(68)=136\).
Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.51
method | result | size |
default | \(-\frac {a^{2} \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )+\frac {7 x}{\sqrt {-a^{2} x^{2}+1}}+\frac {\frac {8}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {8 \left (-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a \right )}{3 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}}{a}+\frac {4}{a \sqrt {-a^{2} x^{2}+1}}}{c}\) | \(186\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c),x,method=_RETURNVERBOSE)
Output:
-1/c*(a^2*(x/a^2/(-a^2*x^2+1)^(1/2)-1/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x /(-a^2*x^2+1)^(1/2)))+7*x/(-a^2*x^2+1)^(1/2)+8/a*(1/3/a/(x-1/a)/(-(x-1/a)^ 2*a^2-2*a*(x-1/a))^(1/2)+1/3/a*(-2*(x-1/a)*a^2-2*a)/(-(x-1/a)^2*a^2-2*a*(x -1/a))^(1/2))+4/a/(-a^2*x^2+1)^(1/2))
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=-\frac {2 \, {\left (2 \, a^{2} x^{2} - 4 \, a x + 3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x - 1\right )} + 2\right )}}{3 \, {\left (a^{3} c x^{2} - 2 \, a^{2} c x + a c\right )}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c),x, algorithm="fricas")
Output:
-2/3*(2*a^2*x^2 - 4*a*x + 3*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 2*sqrt(-a^2*x^2 + 1)*(2*a*x - 1) + 2)/(a^3*c*x^2 - 2*a^2* c*x + a*c)
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=- \frac {\int \frac {3 a x}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {3 a^{2} x^{2}}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a^{3} x^{3}}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(-a*c*x+c),x)
Output:
-(Integral(3*a*x/(-a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x **2 + 1) + a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral (3*a**2*x**2/(-a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(a** 3*x**3/(-a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(1/(-a**3* x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + a*x*sqrt(-a** 2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c
Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=-\frac {8 \, x}{3 \, \sqrt {-a^{2} x^{2} + 1} c} - \frac {8}{3 \, {\left (\sqrt {-a^{2} x^{2} + 1} a^{2} c x - \sqrt {-a^{2} x^{2} + 1} a c\right )}} + \frac {\arcsin \left (a x\right )}{a c} - \frac {4}{\sqrt {-a^{2} x^{2} + 1} a c} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c),x, algorithm="maxima")
Output:
-8/3*x/(sqrt(-a^2*x^2 + 1)*c) - 8/3/(sqrt(-a^2*x^2 + 1)*a^2*c*x - sqrt(-a^ 2*x^2 + 1)*a*c) + arcsin(a*x)/(a*c) - 4/(sqrt(-a^2*x^2 + 1)*a*c)
Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=\frac {\arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{c {\left | a \right |}} + \frac {8 \, {\left (\frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - 1\right )}}{3 \, c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{3} {\left | a \right |}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c),x, algorithm="giac")
Output:
arcsin(a*x)*sgn(a)/(c*abs(a)) + 8/3*(3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^ 2*x) - 1)/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^3*abs(a))
Time = 22.55 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=-\frac {4\,a\,\sqrt {1-a^2\,x^2}+3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}-8\,a^2\,x\,\sqrt {1-a^2\,x^2}+3\,a^2\,x^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}-6\,a\,x\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}{3\,a^2\,c\,{\left (a\,x-1\right )}^2} \] Input:
int((a*x + 1)^3/((1 - a^2*x^2)^(3/2)*(c - a*c*x)),x)
Output:
-(4*a*(1 - a^2*x^2)^(1/2) + 3*asinh(x*(-a^2)^(1/2))*(-a^2)^(1/2) - 8*a^2*x *(1 - a^2*x^2)^(1/2) + 3*a^2*x^2*asinh(x*(-a^2)^(1/2))*(-a^2)^(1/2) - 6*a* x*asinh(x*(-a^2)^(1/2))*(-a^2)^(1/2))/(3*a^2*c*(a*x - 1)^2)
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-a c x} \, dx=\frac {3 \mathit {asin} \left (a x \right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-9 \mathit {asin} \left (a x \right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+9 \mathit {asin} \left (a x \right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-3 \mathit {asin} \left (a x \right )-24 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )+8}{3 a c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c),x)
Output:
(3*asin(a*x)*tan(asin(a*x)/2)**3 - 9*asin(a*x)*tan(asin(a*x)/2)**2 + 9*asi n(a*x)*tan(asin(a*x)/2) - 3*asin(a*x) - 24*tan(asin(a*x)/2) + 8)/(3*a*c*(t an(asin(a*x)/2)**3 - 3*tan(asin(a*x)/2)**2 + 3*tan(asin(a*x)/2) - 1))