Integrand size = 17, antiderivative size = 83 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=-\frac {c x \sqrt {1-a^2 x^2}}{16 a^4}-\frac {c x^3 \sqrt {1-a^2 x^2}}{24 a^2}+\frac {1}{6} c x^5 \sqrt {1-a^2 x^2}+\frac {c \arcsin (a x)}{16 a^5} \] Output:
-1/16*c*x*(-a^2*x^2+1)^(1/2)/a^4-1/24*c*x^3*(-a^2*x^2+1)^(1/2)/a^2+1/6*c*x ^5*(-a^2*x^2+1)^(1/2)+1/16*c*arcsin(a*x)/a^5
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.60 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=\frac {c \left (a x \sqrt {1-a^2 x^2} \left (-3-2 a^2 x^2+8 a^4 x^4\right )+3 \arcsin (a x)\right )}{48 a^5} \] Input:
Integrate[E^ArcTanh[a*x]*x^4*(c - a*c*x),x]
Output:
(c*(a*x*Sqrt[1 - a^2*x^2]*(-3 - 2*a^2*x^2 + 8*a^4*x^4) + 3*ArcSin[a*x]))/( 48*a^5)
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6678, 248, 262, 262, 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.
\(\displaystyle \int x^4 e^{\text {arctanh}(a x)} (c-a c x) \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int x^4 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle c \left (\frac {1}{6} \int \frac {x^4}{\sqrt {1-a^2 x^2}}dx+\frac {1}{6} x^5 \sqrt {1-a^2 x^2}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle c \left (\frac {1}{6} \left (\frac {3 \int \frac {x^2}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}\right )+\frac {1}{6} x^5 \sqrt {1-a^2 x^2}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle c \left (\frac {1}{6} \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}\right )}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}\right )+\frac {1}{6} x^5 \sqrt {1-a^2 x^2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c \left (\frac {1}{6} x^5 \sqrt {1-a^2 x^2}+\frac {1}{6} \left (\frac {3 \left (\frac {\arcsin (a x)}{2 a^3}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}\right )}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}\right )\right )\) |
Input:
Int[E^ArcTanh[a*x]*x^4*(c - a*c*x),x]
Output:
c*((x^5*Sqrt[1 - a^2*x^2])/6 + (-1/4*(x^3*Sqrt[1 - a^2*x^2])/a^2 + (3*(-1/ 2*(x*Sqrt[1 - a^2*x^2])/a^2 + ArcSin[a*x]/(2*a^3)))/(4*a^2))/6)
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_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(\frac {c \left (-3 \arctan \left (\frac {\sqrt {-a^{2} x^{2}+1}}{x a}\right )+\left (8 a^{5} x^{5}-2 a^{3} x^{3}-3 a x \right ) \sqrt {-a^{2} x^{2}+1}\right )}{48 a^{5}}\) | \(64\) |
risch | \(-\frac {x \left (8 a^{4} x^{4}-2 a^{2} x^{2}-3\right ) \left (a^{2} x^{2}-1\right ) c}{48 a^{4} \sqrt {-a^{2} x^{2}+1}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c}{16 a^{4} \sqrt {a^{2}}}\) | \(79\) |
meijerg | \(\frac {c \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {7}{2}} \left (56 a^{4} x^{4}+70 a^{2} x^{2}+105\right ) \sqrt {-a^{2} x^{2}+1}}{168 a^{6}}+\frac {5 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {7}{2}} \arcsin \left (a x \right )}{8 a^{7}}\right )}{2 a^{4} \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{2 a^{4} \sqrt {\pi }\, \sqrt {-a^{2}}}\) | \(158\) |
default | \(-c \left (a^{2} \left (-\frac {x^{5} \sqrt {-a^{2} x^{2}+1}}{6 a^{2}}+\frac {-\frac {5 x^{3} \sqrt {-a^{2} x^{2}+1}}{24 a^{2}}+\frac {5 \left (-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}\right )}{6 a^{2}}}{a^{2}}\right )+\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}-\frac {3 \left (-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )}{4 a^{2}}\right )\) | \(184\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x,method=_RETURNVERBOSE)
Output:
1/48*c*(-3*arctan((-a^2*x^2+1)^(1/2)/x/a)+(8*a^5*x^5-2*a^3*x^3-3*a*x)*(-a^ 2*x^2+1)^(1/2))/a^5
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=-\frac {6 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (8 \, a^{5} c x^{5} - 2 \, a^{3} c x^{3} - 3 \, a c x\right )} \sqrt {-a^{2} x^{2} + 1}}{48 \, a^{5}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x, algorithm="fricas")
Output:
-1/48*(6*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (8*a^5*c*x^5 - 2*a^3*c *x^3 - 3*a*c*x)*sqrt(-a^2*x^2 + 1))/a^5
Time = 0.72 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.20 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (\frac {c x^{5}}{6} - \frac {c x^{3}}{24 a^{2}} - \frac {c x}{16 a^{4}}\right ) + \frac {c \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{16 a^{4} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\- \frac {a^{2} c x^{7}}{7} + \frac {c x^{5}}{5} & \text {otherwise} \end {cases} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**4*(-a*c*x+c),x)
Output:
Piecewise((sqrt(-a**2*x**2 + 1)*(c*x**5/6 - c*x**3/(24*a**2) - c*x/(16*a** 4)) + c*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(16*a**4*sqrt( -a**2)), Ne(a**2, 0)), (-a**2*c*x**7/7 + c*x**5/5, True))
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=\frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} c x^{5} - \frac {\sqrt {-a^{2} x^{2} + 1} c x^{3}}{24 \, a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} c x}{16 \, a^{4}} + \frac {c \arcsin \left (a x\right )}{16 \, a^{5}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x, algorithm="maxima")
Output:
1/6*sqrt(-a^2*x^2 + 1)*c*x^5 - 1/24*sqrt(-a^2*x^2 + 1)*c*x^3/a^2 - 1/16*sq rt(-a^2*x^2 + 1)*c*x/a^4 + 1/16*c*arcsin(a*x)/a^5
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.69 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=\frac {1}{48} \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, {\left (4 \, c x^{2} - \frac {c}{a^{2}}\right )} x^{2} - \frac {3 \, c}{a^{4}}\right )} x + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{16 \, a^{4} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x, algorithm="giac")
Output:
1/48*sqrt(-a^2*x^2 + 1)*(2*(4*c*x^2 - c/a^2)*x^2 - 3*c/a^4)*x + 1/16*c*arc sin(a*x)*sgn(a)/(a^4*abs(a))
Time = 14.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=\frac {c\,x^5\,\sqrt {1-a^2\,x^2}}{6}-\frac {c\,x^3\,\sqrt {1-a^2\,x^2}}{24\,a^2}-\frac {c\,x\,\sqrt {1-a^2\,x^2}}{16\,a^4}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{16\,a^4\,\sqrt {-a^2}} \] Input:
int((x^4*(c - a*c*x)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
(c*x^5*(1 - a^2*x^2)^(1/2))/6 - (c*x^3*(1 - a^2*x^2)^(1/2))/(24*a^2) - (c* x*(1 - a^2*x^2)^(1/2))/(16*a^4) + (c*asinh(x*(-a^2)^(1/2)))/(16*a^4*(-a^2) ^(1/2))
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int e^{\text {arctanh}(a x)} x^4 (c-a c x) \, dx=\frac {c \left (3 \mathit {asin} \left (a x \right )+8 \sqrt {-a^{2} x^{2}+1}\, a^{5} x^{5}-2 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-3 \sqrt {-a^{2} x^{2}+1}\, a x \right )}{48 a^{5}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x)
Output:
(c*(3*asin(a*x) + 8*sqrt( - a**2*x**2 + 1)*a**5*x**5 - 2*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 3*sqrt( - a**2*x**2 + 1)*a*x))/(48*a**5)