Integrand size = 16, antiderivative size = 123 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=\frac {7}{8} c^4 x \sqrt {1-a^2 x^2}+\frac {7 c^4 \left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {7 c^4 (1-a x) \left (1-a^2 x^2\right )^{3/2}}{20 a}+\frac {c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{3/2}}{5 a}+\frac {7 c^4 \arcsin (a x)}{8 a} \]
7/12*c^4*(-a^2*x^2+1)^(3/2)/a+7/20*c^4*(-a*x+1)*(-a^2*x^2+1)^(3/2)/a+1/5*c ^4*(-a*x+1)^2*(-a^2*x^2+1)^(3/2)/a+7/8*c^4*arcsin(a*x)/a+7/8*c^4*x*(-a^2*x ^2+1)^(1/2)
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.61 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=-\frac {c^4 \left (\sqrt {1-a^2 x^2} \left (-136-15 a x+112 a^2 x^2-90 a^3 x^3+24 a^4 x^4\right )+210 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{120 a} \]
-1/120*(c^4*(Sqrt[1 - a^2*x^2]*(-136 - 15*a*x + 112*a^2*x^2 - 90*a^3*x^3 + 24*a^4*x^4) + 210*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/a
Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6677, 27, 469, 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.
\(\displaystyle \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c \int c^3 (1-a x)^3 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c^4 \int (1-a x)^3 \sqrt {1-a^2 x^2}dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^4 \left (\frac {7}{5} \int (1-a x)^2 \sqrt {1-a^2 x^2}dx+\frac {\left (1-a^2 x^2\right )^{3/2} (1-a x)^2}{5 a}\right )\) |
\(\Big \downarrow \) 469 |
\(\displaystyle c^4 \left (\frac {7}{5} \left (\frac {5}{4} \int (1-a x) \sqrt {1-a^2 x^2}dx+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )+\frac {\left (1-a^2 x^2\right )^{3/2} (1-a x)^2}{5 a}\right )\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c^4 \left (\frac {7}{5} \left (\frac {5}{4} \left (\int \sqrt {1-a^2 x^2}dx+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )+\frac {\left (1-a^2 x^2\right )^{3/2} (1-a x)^2}{5 a}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle c^4 \left (\frac {7}{5} \left (\frac {5}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} x \sqrt {1-a^2 x^2}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )+\frac {\left (1-a^2 x^2\right )^{3/2} (1-a x)^2}{5 a}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle c^4 \left (\frac {7}{5} \left (\frac {5}{4} \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a}+\frac {1}{2} x \sqrt {1-a^2 x^2}+\frac {\arcsin (a x)}{2 a}\right )+\frac {(1-a x) \left (1-a^2 x^2\right )^{3/2}}{4 a}\right )+\frac {\left (1-a^2 x^2\right )^{3/2} (1-a x)^2}{5 a}\right )\) |
c^4*(((1 - a*x)^2*(1 - a^2*x^2)^(3/2))/(5*a) + (7*(((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))/5)
3.4.19.3.1 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[((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 = 0.00 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {\left (24 a^{4} x^{4}-90 a^{3} x^{3}+112 a^{2} x^{2}-15 a x -136\right ) \left (a^{2} x^{2}-1\right ) c^{4}}{120 a \sqrt {-a^{2} x^{2}+1}}+\frac {7 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{4}}{8 \sqrt {a^{2}}}\) | \(91\) |
meijerg | \(-\frac {c^{4} \left (-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}+8 a^{2} x^{2}+16\right ) \sqrt {-a^{2} x^{2}+1}}{15}\right )}{2 a \sqrt {\pi }}-\frac {3 c^{4} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}\right )}{2 \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {c^{4} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{a \sqrt {\pi }}-\frac {c^{4} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {3 c^{4} \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{2 a \sqrt {\pi }}+\frac {c^{4} \arcsin \left (a x \right )}{a}\) | \(277\) |
default | \(c^{4} \left (\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+a^{5} \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )-3 a^{4} \left (-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )+2 a^{3} \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )+2 a^{2} \left (-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )+\frac {3 \sqrt {-a^{2} x^{2}+1}}{a}\right )\) | \(294\) |
1/120*(24*a^4*x^4-90*a^3*x^3+112*a^2*x^2-15*a*x-136)*(a^2*x^2-1)/a/(-a^2*x ^2+1)^(1/2)*c^4+7/8/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c ^4
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.75 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=-\frac {210 \, c^{4} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (24 \, a^{4} c^{4} x^{4} - 90 \, a^{3} c^{4} x^{3} + 112 \, a^{2} c^{4} x^{2} - 15 \, a c^{4} x - 136 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{120 \, a} \]
-1/120*(210*c^4*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (24*a^4*c^4*x^4 - 90*a^3*c^4*x^3 + 112*a^2*c^4*x^2 - 15*a*c^4*x - 136*c^4)*sqrt(-a^2*x^2 + 1))/a
Time = 0.86 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.54 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=\begin {cases} \frac {7 c^{4} \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{8 \sqrt {- a^{2}}} + \sqrt {- a^{2} x^{2} + 1} \left (- \frac {a^{3} c^{4} x^{4}}{5} + \frac {3 a^{2} c^{4} x^{3}}{4} - \frac {14 a c^{4} x^{2}}{15} + \frac {c^{4} x}{8} + \frac {17 c^{4}}{15 a}\right ) & \text {for}\: a^{2} \neq 0 \\\begin {cases} c^{4} x & \text {for}\: a = 0 \\\frac {\frac {a^{6} c^{4} x^{6}}{6} - \frac {3 a^{5} c^{4} x^{5}}{5} + \frac {a^{4} c^{4} x^{4}}{2} + \frac {2 a^{3} c^{4} x^{3}}{3} - \frac {3 a^{2} c^{4} x^{2}}{2} + a c^{4} x}{a} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]
Piecewise((7*c**4*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**3*c**4*x**4/5 + 3*a**2*c**4*x**3/4 - 14*a*c**4*x**2/15 + c**4*x/8 + 17*c**4/(15*a)), Ne(a**2, 0)), (Piecewis e((c**4*x, Eq(a, 0)), ((a**6*c**4*x**6/6 - 3*a**5*c**4*x**5/5 + a**4*c**4* x**4/2 + 2*a**3*c**4*x**3/3 - 3*a**2*c**4*x**2/2 + a*c**4*x)/a, True)), Tr ue))
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.96 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=-\frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{4} x^{4} + \frac {3}{4} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{4} x^{3} - \frac {14}{15} \, \sqrt {-a^{2} x^{2} + 1} a c^{4} x^{2} + \frac {1}{8} \, \sqrt {-a^{2} x^{2} + 1} c^{4} x + \frac {7 \, c^{4} \arcsin \left (a x\right )}{8 \, a} + \frac {17 \, \sqrt {-a^{2} x^{2} + 1} c^{4}}{15 \, a} \]
-1/5*sqrt(-a^2*x^2 + 1)*a^3*c^4*x^4 + 3/4*sqrt(-a^2*x^2 + 1)*a^2*c^4*x^3 - 14/15*sqrt(-a^2*x^2 + 1)*a*c^4*x^2 + 1/8*sqrt(-a^2*x^2 + 1)*c^4*x + 7/8*c ^4*arcsin(a*x)/a + 17/15*sqrt(-a^2*x^2 + 1)*c^4/a
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=\frac {7 \, c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, {\left | a \right |}} + \frac {1}{120} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {136 \, c^{4}}{a} + {\left (15 \, c^{4} - 2 \, {\left (56 \, a c^{4} + 3 \, {\left (4 \, a^{3} c^{4} x - 15 \, a^{2} c^{4}\right )} x\right )} x\right )} x\right )} \]
7/8*c^4*arcsin(a*x)*sgn(a)/abs(a) + 1/120*sqrt(-a^2*x^2 + 1)*(136*c^4/a + (15*c^4 - 2*(56*a*c^4 + 3*(4*a^3*c^4*x - 15*a^2*c^4)*x)*x)*x)
Time = 0.00 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04 \[ \int e^{\text {arctanh}(a x)} (c-a c x)^4 \, dx=\frac {c^4\,x\,\sqrt {1-a^2\,x^2}}{8}+\frac {7\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,\sqrt {-a^2}}+\frac {17\,c^4\,\sqrt {1-a^2\,x^2}}{15\,a}-\frac {14\,a\,c^4\,x^2\,\sqrt {1-a^2\,x^2}}{15}+\frac {3\,a^2\,c^4\,x^3\,\sqrt {1-a^2\,x^2}}{4}-\frac {a^3\,c^4\,x^4\,\sqrt {1-a^2\,x^2}}{5} \]