Integrand size = 18, antiderivative size = 133 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {35 c^3 \sqrt {1-a^2 x^2}}{8 a}+\frac {35 c^3 (1-a x) \sqrt {1-a^2 x^2}}{24 a}+\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {35 c^3 \arcsin (a x)}{8 a} \]
35/8*c^3*arcsin(a*x)/a+35/8*c^3*(-a^2*x^2+1)^(1/2)/a+35/24*c^3*(-a*x+1)*(- a^2*x^2+1)^(1/2)/a+7/12*c^3*(-a*x+1)^2*(-a^2*x^2+1)^(1/2)/a+1/4*c^3*(-a*x+ 1)^3*(-a^2*x^2+1)^(1/2)/a
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {c^3 \left (\frac {\sqrt {1+a x} \left (160-241 a x+113 a^2 x^2-38 a^3 x^3+6 a^4 x^4\right )}{\sqrt {1-a x}}-210 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{24 a} \]
(c^3*((Sqrt[1 + a*x]*(160 - 241*a*x + 113*a^2*x^2 - 38*a^3*x^3 + 6*a^4*x^4 ))/Sqrt[1 - a*x] - 210*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(24*a)
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6677, 27, 469, 469, 469, 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.
\(\displaystyle \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {c^4 (1-a x)^4}{\sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^3 \int \frac {(1-a x)^4}{\sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^3 \left (\frac {7}{4} \int \frac {(1-a x)^3}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2} (1-a x)^3}{4 a}\right )\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^3 \left (\frac {7}{4} \left (\frac {5}{3} \int \frac {(1-a x)^2}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^3}{4 a}\right )\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^3 \left (\frac {7}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \frac {1-a x}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2} (1-a x)}{2 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^3}{4 a}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c^3 \left (\frac {7}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)}{2 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^3}{4 a}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^3 \left (\frac {7}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\frac {\sqrt {1-a^2 x^2}}{a}+\frac {\arcsin (a x)}{a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)}{2 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^2}{3 a}\right )+\frac {\sqrt {1-a^2 x^2} (1-a x)^3}{4 a}\right )\) |
c^3*(((1 - a*x)^3*Sqrt[1 - a^2*x^2])/(4*a) + (7*(((1 - a*x)^2*Sqrt[1 - a^2 *x^2])/(3*a) + (5*(((1 - a*x)*Sqrt[1 - a^2*x^2])/(2*a) + (3*(Sqrt[1 - a^2* x^2]/a + ArcSin[a*x]/a))/2))/3))/4)
3.2.98.3.1 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)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* 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]
Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.62
method | result | size |
risch | \(\frac {\left (6 a^{3} x^{3}-32 a^{2} x^{2}+81 a x -160\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{24 a \sqrt {-a^{2} x^{2}+1}}+\frac {35 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{8 \sqrt {a^{2}}}\) | \(83\) |
default | \(-c^{3} \left (\frac {7 \sqrt {-a^{2} x^{2}+1}\, x}{2}+\frac {7 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}+a^{2} \left (-\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 a^{2}}+\frac {\frac {\sqrt {-a^{2} x^{2}+1}\, x}{2}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}}{4 a^{2}}\right )+\frac {4 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a}-\frac {8 \left (\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{a}\right )\) | \(204\) |
1/24*(6*a^3*x^3-32*a^2*x^2+81*a*x-160)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)*c^ 3+35/8/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c^3
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=-\frac {210 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (6 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} + 81 \, a c^{3} x - 160 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{24 \, a} \]
-1/24*(210*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (6*a^3*c^3*x^3 - 3 2*a^2*c^3*x^2 + 81*a*c^3*x - 160*c^3)*sqrt(-a^2*x^2 + 1))/a
\[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=- c^{3} \left (\int \left (- \frac {\sqrt {- a^{2} x^{2} + 1}}{a x + 1}\right )\, dx + \int \frac {3 a x \sqrt {- a^{2} x^{2} + 1}}{a x + 1}\, dx + \int \left (- \frac {3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a x + 1}\right )\, dx + \int \frac {a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1}}{a x + 1}\, dx\right ) \]
-c**3*(Integral(-sqrt(-a**2*x**2 + 1)/(a*x + 1), x) + Integral(3*a*x*sqrt( -a**2*x**2 + 1)/(a*x + 1), x) + Integral(-3*a**2*x**2*sqrt(-a**2*x**2 + 1) /(a*x + 1), x) + Integral(a**3*x**3*sqrt(-a**2*x**2 + 1)/(a*x + 1), x))
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.67 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {1}{4} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c^{3} x - \frac {29}{8} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x - \frac {4 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c^{3}}{3 \, a} + \frac {35 \, c^{3} \arcsin \left (a x\right )}{8 \, a} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{a} \]
1/4*(-a^2*x^2 + 1)^(3/2)*c^3*x - 29/8*sqrt(-a^2*x^2 + 1)*c^3*x - 4/3*(-a^2 *x^2 + 1)^(3/2)*c^3/a + 35/8*c^3*arcsin(a*x)/a + 8*sqrt(-a^2*x^2 + 1)*c^3/ a
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.50 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {35 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, {\left | a \right |}} + \frac {1}{24} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {160 \, c^{3}}{a} - {\left (81 \, c^{3} + 2 \, {\left (3 \, a^{2} c^{3} x - 16 \, a c^{3}\right )} x\right )} x\right )} \]
35/8*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1/24*sqrt(-a^2*x^2 + 1)*(160*c^3/a - (81*c^3 + 2*(3*a^2*c^3*x - 16*a*c^3)*x)*x)
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int e^{-\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {35\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,\sqrt {-a^2}}-\frac {27\,c^3\,x\,\sqrt {1-a^2\,x^2}}{8}+\frac {20\,c^3\,\sqrt {1-a^2\,x^2}}{3\,a}+\frac {4\,a\,c^3\,x^2\,\sqrt {1-a^2\,x^2}}{3}-\frac {a^2\,c^3\,x^3\,\sqrt {1-a^2\,x^2}}{4} \]