Integrand size = 20, antiderivative size = 185 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=-\frac {b c}{6 d x^2}+\frac {b c^2}{2 d x}-\frac {b c^3 \text {arctanh}(c x)}{2 d}-\frac {a+b \text {arctanh}(c x)}{3 d x^3}+\frac {c (a+b \text {arctanh}(c x))}{2 d x^2}-\frac {c^2 (a+b \text {arctanh}(c x))}{d x}+\frac {4 b c^3 \log (x)}{3 d}-\frac {2 b c^3 \log \left (1-c^2 x^2\right )}{3 d}-\frac {c^3 (a+b \text {arctanh}(c x)) \log \left (2-\frac {2}{1+c x}\right )}{d}+\frac {b c^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{2 d} \]
-1/6*b*c/d/x^2+1/2*b*c^2/d/x-1/2*b*c^3*arctanh(c*x)/d+1/3*(-a-b*arctanh(c* x))/d/x^3+1/2*c*(a+b*arctanh(c*x))/d/x^2-c^2*(a+b*arctanh(c*x))/d/x+4/3*b* c^3*ln(x)/d-2/3*b*c^3*ln(-c^2*x^2+1)/d-c^3*(a+b*arctanh(c*x))*ln(2-2/(c*x+ 1))/d+1/2*b*c^3*polylog(2,-1+2/(c*x+1))/d
Time = 0.43 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=\frac {-2 a+3 a c x-b c x-6 a c^2 x^2+3 b c^2 x^2+b c^3 x^3-b \text {arctanh}(c x) \left (2-3 c x+6 c^2 x^2+3 c^3 x^3+6 c^3 x^3 \log \left (1-e^{-2 \text {arctanh}(c x)}\right )\right )-6 a c^3 x^3 \log (x)+8 b c^3 x^3 \log (c x)+6 a c^3 x^3 \log (1+c x)-4 b c^3 x^3 \log \left (1-c^2 x^2\right )+3 b c^3 x^3 \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}(c x)}\right )}{6 d x^3} \]
(-2*a + 3*a*c*x - b*c*x - 6*a*c^2*x^2 + 3*b*c^2*x^2 + b*c^3*x^3 - b*ArcTan h[c*x]*(2 - 3*c*x + 6*c^2*x^2 + 3*c^3*x^3 + 6*c^3*x^3*Log[1 - E^(-2*ArcTan h[c*x])]) - 6*a*c^3*x^3*Log[x] + 8*b*c^3*x^3*Log[c*x] + 6*a*c^3*x^3*Log[1 + c*x] - 4*b*c^3*x^3*Log[1 - c^2*x^2] + 3*b*c^3*x^3*PolyLog[2, E^(-2*ArcTa nh[c*x])])/(6*d*x^3)
Time = 1.23 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6496, 27, 6452, 243, 54, 2009, 6496, 6452, 264, 219, 6496, 6452, 243, 47, 14, 16, 6494, 2897}
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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{x^4 (c d x+d)} \, dx\) |
\(\Big \downarrow \) 6496 |
\(\displaystyle \frac {\int \frac {a+b \text {arctanh}(c x)}{x^4}dx}{d}-c \int \frac {a+b \text {arctanh}(c x)}{d x^3 (c x+1)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a+b \text {arctanh}(c x)}{x^4}dx}{d}-\frac {c \int \frac {a+b \text {arctanh}(c x)}{x^3 (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {\frac {1}{3} b c \int \frac {1}{x^3 \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \int \frac {a+b \text {arctanh}(c x)}{x^3 (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{6} b c \int \frac {1}{x^4 \left (1-c^2 x^2\right )}dx^2-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \int \frac {a+b \text {arctanh}(c x)}{x^3 (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\frac {1}{6} b c \int \left (-\frac {c^4}{c^2 x^2-1}+\frac {c^2}{x^2}+\frac {1}{x^4}\right )dx^2-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \int \frac {a+b \text {arctanh}(c x)}{x^3 (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \int \frac {a+b \text {arctanh}(c x)}{x^3 (c x+1)}dx}{d}\) |
\(\Big \downarrow \) 6496 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (\int \frac {a+b \text {arctanh}(c x)}{x^3}dx-c \int \frac {a+b \text {arctanh}(c x)}{x^2 (c x+1)}dx\right )}{d}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x^2 (c x+1)}dx+\frac {1}{2} b c \int \frac {1}{x^2 \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{2 x^2}\right )}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x^2 (c x+1)}dx+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-c^2 x^2}dx-\frac {1}{x}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x^2 (c x+1)}dx-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 6496 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (\int \frac {a+b \text {arctanh}(c x)}{x^2}dx-c \int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+b c \int \frac {1}{x \left (1-c^2 x^2\right )}dx-\frac {a+b \text {arctanh}(c x)}{x}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+\frac {1}{2} b c \int \frac {1}{x^2 \left (1-c^2 x^2\right )}dx^2-\frac {a+b \text {arctanh}(c x)}{x}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-c^2 x^2}dx^2+\int \frac {1}{x^2}dx^2\right )-\frac {a+b \text {arctanh}(c x)}{x}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx+\frac {1}{2} b c \left (c^2 \int \frac {1}{1-c^2 x^2}dx^2+\log \left (x^2\right )\right )-\frac {a+b \text {arctanh}(c x)}{x}\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \int \frac {a+b \text {arctanh}(c x)}{x (c x+1)}dx-\frac {a+b \text {arctanh}(c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (1-c^2 x^2\right )\right )\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-b c \int \frac {\log \left (2-\frac {2}{c x+1}\right )}{1-c^2 x^2}dx\right )-\frac {a+b \text {arctanh}(c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (1-c^2 x^2\right )\right )\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {\frac {1}{6} b c \left (c^2 \log \left (x^2\right )-c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )-\frac {a+b \text {arctanh}(c x)}{3 x^3}}{d}-\frac {c \left (-c \left (-c \left (\log \left (2-\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{c x+1}-1\right )\right )-\frac {a+b \text {arctanh}(c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (1-c^2 x^2\right )\right )\right )-\frac {a+b \text {arctanh}(c x)}{2 x^2}+\frac {1}{2} b c \left (c \text {arctanh}(c x)-\frac {1}{x}\right )\right )}{d}\) |
(-1/3*(a + b*ArcTanh[c*x])/x^3 + (b*c*(-x^(-2) + c^2*Log[x^2] - c^2*Log[1 - c^2*x^2]))/6)/d - (c*(-1/2*(a + b*ArcTanh[c*x])/x^2 + (b*c*(-x^(-1) + c* ArcTanh[c*x]))/2 - c*(-((a + b*ArcTanh[c*x])/x) + (b*c*(Log[x^2] - Log[1 - c^2*x^2]))/2 - c*((a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - (b*PolyLog[ 2, -1 + 2/(1 + c*x)])/2))))/d
3.1.50.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x _Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Simp[b*c*(p/d) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))] /(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c ^2*d^2 - e^2, 0]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Simp[e/(d*f) Int[(f*x)^(m + 1)*((a + b*ArcTanh[c*x])^p/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && LtQ[m, -1]
Time = 1.34 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.18
method | result | size |
parts | \(\frac {a \left (c^{3} \ln \left (c x +1\right )-\frac {1}{3 x^{3}}-\frac {c^{2}}{x}+\frac {c}{2 x^{2}}-c^{3} \ln \left (x \right )\right )}{d}+\frac {b \,c^{3} \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}+\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {11 \ln \left (c x +1\right )}{12}-\frac {5 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}+\frac {1}{2 c x}+\frac {4 \ln \left (c x \right )}{3}+\frac {\operatorname {dilog}\left (c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {\operatorname {dilog}\left (c x \right )}{2}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d}\) | \(218\) |
derivativedivides | \(c^{3} \left (\frac {a \left (\ln \left (c x +1\right )-\frac {1}{3 c^{3} x^{3}}-\frac {1}{c x}+\frac {1}{2 c^{2} x^{2}}-\ln \left (c x \right )\right )}{d}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}+\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {11 \ln \left (c x +1\right )}{12}-\frac {5 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}+\frac {1}{2 c x}+\frac {4 \ln \left (c x \right )}{3}+\frac {\operatorname {dilog}\left (c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {\operatorname {dilog}\left (c x \right )}{2}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d}\right )\) | \(219\) |
default | \(c^{3} \left (\frac {a \left (\ln \left (c x +1\right )-\frac {1}{3 c^{3} x^{3}}-\frac {1}{c x}+\frac {1}{2 c^{2} x^{2}}-\ln \left (c x \right )\right )}{d}+\frac {b \left (\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x}+\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {11 \ln \left (c x +1\right )}{12}-\frac {5 \ln \left (c x -1\right )}{12}-\frac {1}{6 c^{2} x^{2}}+\frac {1}{2 c x}+\frac {4 \ln \left (c x \right )}{3}+\frac {\operatorname {dilog}\left (c x +1\right )}{2}+\frac {\ln \left (c x \right ) \ln \left (c x +1\right )}{2}+\frac {\operatorname {dilog}\left (c x \right )}{2}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c x +1\right )^{2}}{4}\right )}{d}\right )\) | \(219\) |
risch | \(-\frac {b c}{6 d \,x^{2}}+\frac {b \,c^{2}}{2 d x}-\frac {c b \ln \left (-c x +1\right )}{4 d \,x^{2}}-\frac {c^{3} b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d}+\frac {c^{3} b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d}+\frac {c^{2} b \ln \left (-c x +1\right )}{2 d x}+\frac {b c \ln \left (c x +1\right )}{4 d \,x^{2}}-\frac {b \,c^{2} \ln \left (c x +1\right )}{2 d x}-\frac {a}{3 d \,x^{3}}+\frac {c a}{2 d \,x^{2}}-\frac {c^{2} a}{d x}+\frac {5 c^{3} b \ln \left (-c x \right )}{12 d}-\frac {5 c^{3} b \ln \left (-c x +1\right )}{12 d}+\frac {b \ln \left (-c x +1\right )}{6 d \,x^{3}}+\frac {c^{3} b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d}-\frac {c^{3} \operatorname {dilog}\left (-c x +1\right ) b}{2 d}+\frac {c^{3} a \ln \left (-c x -1\right )}{d}-\frac {c^{3} \ln \left (-c x \right ) a}{d}+\frac {11 b \,c^{3} \ln \left (c x \right )}{12 d}-\frac {11 b \,c^{3} \ln \left (c x +1\right )}{12 d}-\frac {b \ln \left (c x +1\right )}{6 d \,x^{3}}+\frac {b \,c^{3} \operatorname {dilog}\left (c x +1\right )}{2 d}+\frac {b \,c^{3} \ln \left (c x +1\right )^{2}}{4 d}\) | \(353\) |
a/d*(c^3*ln(c*x+1)-1/3/x^3-c^2/x+1/2*c/x^2-c^3*ln(x))+b/d*c^3*(arctanh(c*x )*ln(c*x+1)-1/3/c^3/x^3*arctanh(c*x)-1/c/x*arctanh(c*x)+1/2/c^2/x^2*arctan h(c*x)-ln(c*x)*arctanh(c*x)-11/12*ln(c*x+1)-5/12*ln(c*x-1)-1/6/c^2/x^2+1/2 /c/x+4/3*ln(c*x)+1/2*dilog(c*x+1)+1/2*ln(c*x)*ln(c*x+1)+1/2*dilog(c*x)+1/2 *(ln(c*x+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)-1/2*dilog(1/2*c*x+1/2)-1/4*l n(c*x+1)^2)
\[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=\int { \frac {b \operatorname {artanh}\left (c x\right ) + a}{{\left (c d x + d\right )} x^{4}} \,d x } \]
\[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=\frac {\int \frac {a}{c x^{5} + x^{4}}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c x^{5} + x^{4}}\, dx}{d} \]
\[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=\int { \frac {b \operatorname {artanh}\left (c x\right ) + a}{{\left (c d x + d\right )} x^{4}} \,d x } \]
1/6*(6*c^3*log(c*x + 1)/d - 6*c^3*log(x)/d - (6*c^2*x^2 - 3*c*x + 2)/(d*x^ 3))*a + 1/2*b*integrate((log(c*x + 1) - log(-c*x + 1))/(c*d*x^5 + d*x^4), x)
\[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=\int { \frac {b \operatorname {artanh}\left (c x\right ) + a}{{\left (c d x + d\right )} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \text {arctanh}(c x)}{x^4 (d+c d x)} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x^4\,\left (d+c\,d\,x\right )} \,d x \]