Integrand size = 22, antiderivative size = 94 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=-\frac {4 (1-a x)}{a c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c (1+a x)^3}-\frac {3 \arcsin (a x)}{a c} \] Output:
(4*a*x-4)/a/c/(-a^2*x^2+1)^(1/2)-(-a^2*x^2+1)^(1/2)/a/c+1/3*(-a^2*x^2+1)^( 3/2)/a/c/(a*x+1)^3-3*arcsin(a*x)/a/c
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {-14-5 a x+16 a^2 x^2+3 a^3 x^3-9 (1+a x) \sqrt {1-a^2 x^2} \arcsin (a x)}{3 a c (1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))),x]
Output:
(-14 - 5*a*x + 16*a^2*x^2 + 3*a^3*x^3 - 9*(1 + a*x)*Sqrt[1 - a^2*x^2]*ArcS in[a*x])/(3*a*c*(1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6707, 6699, 529, 27, 462, 455, 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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle -\frac {a^2 \int \frac {e^{-3 \text {arctanh}(a x)} x^2}{1-a^2 x^2}dx}{c}\) |
\(\Big \downarrow \) 6699 |
\(\displaystyle -\frac {a^2 \int \frac {x^2 (1-a x)^3}{\left (1-a^2 x^2\right )^{5/2}}dx}{c}\) |
\(\Big \downarrow \) 529 |
\(\displaystyle -\frac {a^2 \left (-\frac {1}{3} \int \frac {3 (1-a x)^3}{a^2 \left (1-a^2 x^2\right )^{3/2}}dx-\frac {(1-a x)^3}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^2 \left (-\frac {\int \frac {(1-a x)^3}{\left (1-a^2 x^2\right )^{3/2}}dx}{a^2}-\frac {(1-a x)^3}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\) |
\(\Big \downarrow \) 462 |
\(\displaystyle -\frac {a^2 \left (-\frac {-\int \frac {3-a x}{\sqrt {1-a^2 x^2}}dx-\frac {4 (1-a x)}{a \sqrt {1-a^2 x^2}}}{a^2}-\frac {(1-a x)^3}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle -\frac {a^2 \left (-\frac {-3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {4 (1-a x)}{a \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a}}{a^2}-\frac {(1-a x)^3}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {a^2 \left (-\frac {-\frac {4 (1-a x)}{a \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a}-\frac {3 \arcsin (a x)}{a}}{a^2}-\frac {(1-a x)^3}{3 a^3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\) |
Input:
Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))),x]
Output:
-((a^2*(-1/3*(1 - a*x)^3/(a^3*(1 - a^2*x^2)^(3/2)) - ((-4*(1 - a*x))/(a*Sq rt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/a - (3*ArcSin[a*x])/a)/a^2))/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_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp [(-2^(n - 1))*d*c^(n - 2)*((c + d*x)/(b*Sqrt[a + b*x^2])), x] + Simp[d^2/b 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] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[n, 2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^p Int[x^m*((1 - a^2*x^2)^(p + n/2)/(1 - a*x)^n), x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c , 0]) && ILtQ[(n - 1)/2, 0] && !IntegerQ[p - n/2]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.48
method | result | size |
risch | \(\frac {a^{2} x^{2}-1}{a c \sqrt {-a^{2} x^{2}+1}}-\frac {\left (\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}-\frac {2 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 a^{5} \left (x +\frac {1}{a}\right )^{2}}+\frac {13 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 a^{4} \left (x +\frac {1}{a}\right )}\right ) a^{2}}{c}\) | \(139\) |
default | \(\text {Expression too large to display}\) | \(825\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2),x,method=_RETURNVERBOSE)
Output:
1/a/c*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)-(3/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2) *x/(-a^2*x^2+1)^(1/2))-2/3/a^5/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2 )+13/3/a^4/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))/c*a^2
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=-\frac {14 \, a^{2} x^{2} + 28 \, a x - 18 \, {\left (a^{2} x^{2} + 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{2} x^{2} + 19 \, a x + 14\right )} \sqrt {-a^{2} x^{2} + 1} + 14}{3 \, {\left (a^{3} c x^{2} + 2 \, a^{2} c x + a c\right )}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2),x, algorithm="frica s")
Output:
-1/3*(14*a^2*x^2 + 28*a*x - 18*(a^2*x^2 + 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^2*x^2 + 19*a*x + 14)*sqrt(-a^2*x^2 + 1) + 14)/(a^ 3*c*x^2 + 2*a^2*c*x + a*c)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {a^{2} \left (\int \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + 3 a^{4} x^{4} + 2 a^{3} x^{3} - 2 a^{2} x^{2} - 3 a x - 1}\, dx + \int \left (- \frac {a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + 3 a^{4} x^{4} + 2 a^{3} x^{3} - 2 a^{2} x^{2} - 3 a x - 1}\right )\, dx\right )}{c} \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(c-c/a**2/x**2),x)
Output:
a**2*(Integral(x**2*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + 3*a**4*x**4 + 2*a**3 *x**3 - 2*a**2*x**2 - 3*a*x - 1), x) + Integral(-a**2*x**4*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + 3*a**4*x**4 + 2*a**3*x**3 - 2*a**2*x**2 - 3*a*x - 1), x ))/c
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2),x, algorithm="maxim a")
Output:
integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))), x)
Exception generated. \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2),x, algorithm="giac" )
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {2\,a\,\sqrt {1-a^2\,x^2}}{3\,\left (c\,a^4\,x^2+2\,c\,a^3\,x+c\,a^2\right )}+\frac {13\,\sqrt {1-a^2\,x^2}}{3\,\left (\frac {c\,\sqrt {-a^2}}{a}+c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c} \] Input:
int((1 - a^2*x^2)^(3/2)/((c - c/(a^2*x^2))*(a*x + 1)^3),x)
Output:
(2*a*(1 - a^2*x^2)^(1/2))/(3*(a^2*c + a^4*c*x^2 + 2*a^3*c*x)) + (13*(1 - a ^2*x^2)^(1/2))/(3*((c*(-a^2)^(1/2))/a + c*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (3*asinh(x*(-a^2)^(1/2)))/(c*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/(a*c)
Time = 0.15 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.91 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx=\frac {-9 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -9 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )+9 \mathit {asin} \left (a x \right ) a^{2} x^{2}+18 \mathit {asin} \left (a x \right ) a x +9 \mathit {asin} \left (a x \right )+3 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+11 \sqrt {-a^{2} x^{2}+1}\, a x +6 \sqrt {-a^{2} x^{2}+1}+3 a^{3} x^{3}+24 a^{2} x^{2}+11 a x -6}{3 a c \left (\sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}-a^{2} x^{2}-2 a x -1\right )} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2),x)
Output:
( - 9*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x - 9*sqrt( - a**2*x**2 + 1)*asin (a*x) + 9*asin(a*x)*a**2*x**2 + 18*asin(a*x)*a*x + 9*asin(a*x) + 3*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 11*sqrt( - a**2*x**2 + 1)*a*x + 6*sqrt( - a**2 *x**2 + 1) + 3*a**3*x**3 + 24*a**2*x**2 + 11*a*x - 6)/(3*a*c*(sqrt( - a**2 *x**2 + 1)*a*x + sqrt( - a**2*x**2 + 1) - a**2*x**2 - 2*a*x - 1))