Integrand size = 16, antiderivative size = 91 \[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=-\frac {2 c (1-a x)^3}{a \sqrt {1-a^2 x^2}}-\frac {15 c \sqrt {1-a^2 x^2}}{2 a}-\frac {5 c (1-a x) \sqrt {1-a^2 x^2}}{2 a}-\frac {15 c \arcsin (a x)}{2 a} \]
-15/2*c*arcsin(a*x)/a-2*c*(-a*x+1)^3/a/(-a^2*x^2+1)^(1/2)-15/2*c*(-a^2*x^2 +1)^(1/2)/a-5/2*c*(-a*x+1)*(-a^2*x^2+1)^(1/2)/a
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.47 \[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=-\frac {c (1-a x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} (1-a x)\right )}{7 \sqrt {2} a} \]
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6677, 27, 462, 2346, 25, 27, 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.
\(\displaystyle \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {c^4 (1-a x)^4}{\left (1-a^2 x^2\right )^{3/2}}dx}{c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \int \frac {(1-a x)^4}{\left (1-a^2 x^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 462 |
\(\displaystyle c \left (-\int \frac {a^2 x^2-4 a x+7}{\sqrt {1-a^2 x^2}}dx-\frac {8 (1-a x)}{a \sqrt {1-a^2 x^2}}\right )\) |
\(\Big \downarrow \) 2346 |
\(\displaystyle c \left (\frac {\int -\frac {a^2 (15-8 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}+\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {8 (1-a x)}{a \sqrt {1-a^2 x^2}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c \left (-\frac {\int \frac {a^2 (15-8 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}+\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {8 (1-a x)}{a \sqrt {1-a^2 x^2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (-\frac {1}{2} \int \frac {15-8 a x}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {8 (1-a x)}{a \sqrt {1-a^2 x^2}}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c \left (\frac {1}{2} \left (-15 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {8 \sqrt {1-a^2 x^2}}{a}\right )+\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {8 (1-a x)}{a \sqrt {1-a^2 x^2}}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c \left (\frac {1}{2} \left (-\frac {8 \sqrt {1-a^2 x^2}}{a}-\frac {15 \arcsin (a x)}{a}\right )+\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {8 (1-a x)}{a \sqrt {1-a^2 x^2}}\right )\) |
c*((-8*(1 - a*x))/(a*Sqrt[1 - a^2*x^2]) + (x*Sqrt[1 - a^2*x^2])/2 + ((-8*S qrt[1 - a^2*x^2])/a - (15*ArcSin[a*x])/a)/2)
3.3.19.3.1 Defintions of rubi rules used
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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
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 = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.09
method | result | size |
risch | \(-\frac {\left (a x -8\right ) \left (a^{2} x^{2}-1\right ) c}{2 a \sqrt {-a^{2} x^{2}+1}}+\left (-\frac {15 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}-\frac {8 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a^{2} \left (x +\frac {1}{a}\right )}\right ) c\) | \(99\) |
default | \(-c \left (\frac {\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )}{a^{2}}-\frac {2 \left (-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{3}}-2 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )\right )}{a^{3}}\right )\) | \(347\) |
-1/2*(a*x-8)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)*c+(-15/2/(a^2)^(1/2)*arctan( (a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-8/a^2/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a ))^(1/2))*c
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89 \[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=-\frac {24 \, a c x - 30 \, {\left (a c x + c\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (a^{2} c x^{2} - 7 \, a c x - 24 \, c\right )} \sqrt {-a^{2} x^{2} + 1} + 24 \, c}{2 \, {\left (a^{2} x + a\right )}} \]
-1/2*(24*a*c*x - 30*(a*c*x + c)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - ( a^2*c*x^2 - 7*a*c*x - 24*c)*sqrt(-a^2*x^2 + 1) + 24*c)/(a^2*x + a)
\[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=- c \left (\int \left (- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\right )\, dx + \int \frac {a x \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\, dx + \int \frac {a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\, dx + \int \left (- \frac {a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\right )\, dx\right ) \]
-c*(Integral(-sqrt(-a**2*x**2 + 1)/(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1), x) + Integral(a*x*sqrt(-a**2*x**2 + 1)/(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1), x) + Integral(a**2*x**2*sqrt(-a**2*x**2 + 1)/(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 1), x) + Integral(-a**3*x**3*sqrt(-a**2*x**2 + 1)/(a**3*x**3 + 3 *a**2*x**2 + 3*a*x + 1), x))
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.20 \[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=\frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{a^{3} x^{2} + 2 \, a^{2} x + a} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{2 \, {\left (a^{2} x + a\right )}} - \frac {15 \, c \arcsin \left (a x\right )}{2 \, a} - \frac {12 \, \sqrt {-a^{2} x^{2} + 1} c}{a^{2} x + a} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} c}{2 \, a} \]
2*(-a^2*x^2 + 1)^(3/2)*c/(a^3*x^2 + 2*a^2*x + a) - 1/2*(-a^2*x^2 + 1)^(3/2 )*c/(a^2*x + a) - 15/2*c*arcsin(a*x)/a - 12*sqrt(-a^2*x^2 + 1)*c/(a^2*x + a) - 3/2*sqrt(-a^2*x^2 + 1)*c/a
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.80 \[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=-\frac {15 \, c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, {\left | a \right |}} + \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} {\left (c x - \frac {8 \, c}{a}\right )} + \frac {16 \, c}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]
-15/2*c*arcsin(a*x)*sgn(a)/abs(a) + 1/2*sqrt(-a^2*x^2 + 1)*(c*x - 8*c/a) + 16*c/(((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))
Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05 \[ \int e^{-3 \text {arctanh}(a x)} (c-a c x) \, dx=\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {4\,a\,c}{\sqrt {-a^2}}+\frac {c\,x\,\sqrt {-a^2}}{2}\right )-\frac {15\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2}+\frac {8\,c\,\sqrt {1-a^2\,x^2}}{x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}}}{\sqrt {-a^2}} \]