Integrand size = 16, antiderivative size = 83 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {5}{8} c^3 x \sqrt {1-a^2 x^2}+\frac {2 c^3 \left (1-a^2 x^2\right )^{3/2}}{3 a}-\frac {1}{4} c^3 x \left (1-a^2 x^2\right )^{3/2}+\frac {5 c^3 \arcsin (a x)}{8 a} \] Output:
5/8*c^3*x*(-a^2*x^2+1)^(1/2)+2/3*c^3*(-a^2*x^2+1)^(3/2)/a-1/4*c^3*x*(-a^2* x^2+1)^(3/2)+5/8*c^3*arcsin(a*x)/a
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {c^3 \left (\sqrt {1-a^2 x^2} \left (16+9 a x-16 a^2 x^2+6 a^3 x^3\right )-30 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{24 a} \] Input:
Integrate[E^ArcTanh[a*x]*(c - a*c*x)^3,x]
Output:
(c^3*(Sqrt[1 - a^2*x^2]*(16 + 9*a*x - 16*a^2*x^2 + 6*a^3*x^3) - 30*ArcSin[ Sqrt[1 - a*x]/Sqrt[2]]))/(24*a)
Time = 0.44 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6677, 27, 469, 455, 211, 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 |
\(\displaystyle c \int c^2 (1-a x)^2 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^3 \int (1-a x)^2 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^3 \left (\frac {5}{4} \int (1-a x) \sqrt {1-a^2 x^2}dx+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c^3 \left (\frac {5}{4} \left (\int \sqrt {1-a^2 x^2}dx+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle c^3 \left (\frac {5}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^3 \left (\frac {5}{4} \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )\) |
Input:
Int[E^ArcTanh[a*x]*(c - a*c*x)^3,x]
Output:
c^3*(((1 - a*x)*(1 - a^2*x^2)^(3/2))/(4*a) + (5*((x*Sqrt[1 - a^2*x^2])/2 + (1 - a^2*x^2)^(3/2)/(3*a) + ArcSin[a*x]/(2*a)))/4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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 = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {\left (6 a^{3} x^{3}-16 a^{2} x^{2}+9 a x +16\right ) \left (a^{2} x^{2}-1\right ) c^{3}}{24 a \sqrt {-a^{2} x^{2}+1}}+\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{8 \sqrt {a^{2}}}\) | \(83\) |
meijerg | \(-\frac {c^{3} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{2 \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{3} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{a \sqrt {\pi }}+\frac {c^{3} \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{a \sqrt {\pi }}+\frac {c^{3} \arcsin \left (a x \right )}{a}\) | \(162\) |
default | \(-c^{3} \left (a^{4} \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{a}-2 a^{3} \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )\right )\) | \(173\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-1/24*(6*a^3*x^3-16*a^2*x^2+9*a*x+16)*(a^2*x^2-1)/a/(-a^2*x^2+1)^(1/2)*c^3 +5/8/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c^3
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=-\frac {30 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (6 \, a^{3} c^{3} x^{3} - 16 \, a^{2} c^{3} x^{2} + 9 \, a c^{3} x + 16 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{24 \, a} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3,x, algorithm="fricas")
Output:
-1/24*(30*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (6*a^3*c^3*x^3 - 16 *a^2*c^3*x^2 + 9*a*c^3*x + 16*c^3)*sqrt(-a^2*x^2 + 1))/a
Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (71) = 142\).
Time = 0.92 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.76 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\begin {cases} \frac {5 c^{3} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{8 \sqrt {- a^{2}}} + \sqrt {- a^{2} x^{2} + 1} \left (\frac {a^{2} c^{3} x^{3}}{4} - \frac {2 a c^{3} x^{2}}{3} + \frac {3 c^{3} x}{8} + \frac {2 c^{3}}{3 a}\right ) & \text {for}\: a^{2} \neq 0 \\\begin {cases} c^{3} x & \text {for}\: a = 0 \\\frac {- \frac {a^{5} c^{3} x^{5}}{5} + \frac {a^{4} c^{3} x^{4}}{2} - a^{2} c^{3} x^{2} + a c^{3} x}{a} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**3,x)
Output:
Piecewise((5*c**3*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(8*s qrt(-a**2)) + sqrt(-a**2*x**2 + 1)*(a**2*c**3*x**3/4 - 2*a*c**3*x**2/3 + 3 *c**3*x/8 + 2*c**3/(3*a)), Ne(a**2, 0)), (Piecewise((c**3*x, Eq(a, 0)), (( -a**5*c**3*x**5/5 + a**4*c**3*x**4/2 - a**2*c**3*x**2 + a*c**3*x)/a, True) ), True))
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {1}{4} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{3} - \frac {2}{3} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{2} + \frac {3}{8} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x + \frac {5 \, c^{3} \arcsin \left (a x\right )}{8 \, a} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{3 \, a} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3,x, algorithm="maxima")
Output:
1/4*sqrt(-a^2*x^2 + 1)*a^2*c^3*x^3 - 2/3*sqrt(-a^2*x^2 + 1)*a*c^3*x^2 + 3/ 8*sqrt(-a^2*x^2 + 1)*c^3*x + 5/8*c^3*arcsin(a*x)/a + 2/3*sqrt(-a^2*x^2 + 1 )*c^3/a
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {5 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, {\left | a \right |}} + \frac {1}{24} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {16 \, c^{3}}{a} + {\left (9 \, c^{3} + 2 \, {\left (3 \, a^{2} c^{3} x - 8 \, a c^{3}\right )} x\right )} x\right )} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3,x, algorithm="giac")
Output:
5/8*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1/24*sqrt(-a^2*x^2 + 1)*(16*c^3/a + (9 *c^3 + 2*(3*a^2*c^3*x - 8*a*c^3)*x)*x)
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {3\,c^3\,x\,\sqrt {1-a^2\,x^2}}{8}+\frac {5\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,\sqrt {-a^2}}+\frac {2\,c^3\,\sqrt {1-a^2\,x^2}}{3\,a}-\frac {2\,a\,c^3\,x^2\,\sqrt {1-a^2\,x^2}}{3}+\frac {a^2\,c^3\,x^3\,\sqrt {1-a^2\,x^2}}{4} \] Input:
int(((c - a*c*x)^3*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
(3*c^3*x*(1 - a^2*x^2)^(1/2))/8 + (5*c^3*asinh(x*(-a^2)^(1/2)))/(8*(-a^2)^ (1/2)) + (2*c^3*(1 - a^2*x^2)^(1/2))/(3*a) - (2*a*c^3*x^2*(1 - a^2*x^2)^(1 /2))/3 + (a^2*c^3*x^3*(1 - a^2*x^2)^(1/2))/4
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^3 \, dx=\frac {c^{3} \left (15 \mathit {asin} \left (a x \right )+6 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+9 \sqrt {-a^{2} x^{2}+1}\, a x +16 \sqrt {-a^{2} x^{2}+1}-16\right )}{24 a} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3,x)
Output:
(c**3*(15*asin(a*x) + 6*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 16*sqrt( - a**2 *x**2 + 1)*a**2*x**2 + 9*sqrt( - a**2*x**2 + 1)*a*x + 16*sqrt( - a**2*x**2 + 1) - 16))/(24*a)