Integrand size = 14, antiderivative size = 139 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {(1-a x)^{3/4} \sqrt [4]{1+a x}}{3 x^3}-\frac {5 a (1-a x)^{3/4} \sqrt [4]{1+a x}}{12 x^2}-\frac {11 a^2 (1-a x)^{3/4} \sqrt [4]{1+a x}}{24 x}-\frac {3}{8} a^3 \arctan \left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac {3}{8} a^3 \text {arctanh}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \]
-1/3*(-a*x+1)^(3/4)*(a*x+1)^(1/4)/x^3-5/12*a*(-a*x+1)^(3/4)*(a*x+1)^(1/4)/ x^2-11/24*a^2*(-a*x+1)^(3/4)*(a*x+1)^(1/4)/x-3/8*a^3*arctan((a*x+1)^(1/4)/ (-a*x+1)^(1/4))-3/8*a^3*arctanh((a*x+1)^(1/4)/(-a*x+1)^(1/4))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {(1-a x)^{3/4} \left (8+18 a x+21 a^2 x^2+11 a^3 x^3+6 a^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {1-a x}{1+a x}\right )\right )}{24 x^3 (1+a x)^{3/4}} \]
-1/24*((1 - a*x)^(3/4)*(8 + 18*a*x + 21*a^2*x^2 + 11*a^3*x^3 + 6*a^3*x^3*H ypergeometric2F1[3/4, 1, 7/4, (1 - a*x)/(1 + a*x)]))/(x^3*(1 + a*x)^(3/4))
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6676, 110, 27, 168, 27, 168, 27, 104, 756, 216, 219}
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.
\(\displaystyle \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 6676 |
\(\displaystyle \int \frac {\sqrt [4]{a x+1}}{x^4 \sqrt [4]{1-a x}}dx\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \frac {1}{3} \int \frac {a (4 a x+5)}{2 x^3 \sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} a \int \frac {4 a x+5}{x^3 \sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{2} \int -\frac {a (10 a x+11)}{2 x^2 \sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \int \frac {10 a x+11}{x^2 \sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \left (-\int -\frac {9 a}{2 x \sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {11 (1-a x)^{3/4} \sqrt [4]{a x+1}}{x}\right )-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \left (\frac {9}{2} a \int \frac {1}{x \sqrt [4]{1-a x} (a x+1)^{3/4}}dx-\frac {11 (1-a x)^{3/4} \sqrt [4]{a x+1}}{x}\right )-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \left (18 a \int \frac {1}{\frac {a x+1}{1-a x}-1}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {11 (1-a x)^{3/4} \sqrt [4]{a x+1}}{x}\right )-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \left (18 a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt {a x+1}}{\sqrt {1-a x}}+1}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {11 (1-a x)^{3/4} \sqrt [4]{a x+1}}{x}\right )-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \left (18 a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )-\frac {11 (1-a x)^{3/4} \sqrt [4]{a x+1}}{x}\right )-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \left (18 a \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )-\frac {11 (1-a x)^{3/4} \sqrt [4]{a x+1}}{x}\right )-\frac {5 (1-a x)^{3/4} \sqrt [4]{a x+1}}{2 x^2}\right )-\frac {(1-a x)^{3/4} \sqrt [4]{a x+1}}{3 x^3}\) |
-1/3*((1 - a*x)^(3/4)*(1 + a*x)^(1/4))/x^3 + (a*((-5*(1 - a*x)^(3/4)*(1 + a*x)^(1/4))/(2*x^2) + (a*((-11*(1 - a*x)^(3/4)*(1 + a*x)^(1/4))/x + 18*a*( -1/2*ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)] - ArcTanh[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2)))/4))/6
3.1.67.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f)) Int[(a + b*x)^(m + 1) *(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x) ^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c, m, n}, x] && !Int egerQ[(n - 1)/2]
\[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{x^{4}}d x\]
Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {18 \, a^{3} x^{3} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) + 9 \, a^{3} x^{3} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) - 9 \, a^{3} x^{3} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 2 \, {\left (11 \, a^{3} x^{3} - a^{2} x^{2} - 2 \, a x - 8\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{48 \, x^{3}} \]
-1/48*(18*a^3*x^3*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) + 9*a^3*x^3* log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) - 9*a^3*x^3*log(sqrt(-sqrt(-a ^2*x^2 + 1)/(a*x - 1)) - 1) - 2*(11*a^3*x^3 - a^2*x^2 - 2*a*x - 8)*sqrt(-s qrt(-a^2*x^2 + 1)/(a*x - 1)))/x^3
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int \frac {\sqrt {\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}}}{x^{4}}\, dx \]
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int { \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{x^{4}} \,d x } \]
\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int { \frac {\sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int \frac {\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{x^4} \,d x \]