Integrand size = 21, antiderivative size = 115 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=-\frac {b \sqrt {1+(c+d x)^2}}{20 d e^6 (c+d x)^4}+\frac {3 b \sqrt {1+(c+d x)^2}}{40 d e^6 (c+d x)^2}-\frac {a+b \text {arcsinh}(c+d x)}{5 d e^6 (c+d x)^5}-\frac {3 b \text {arctanh}\left (\sqrt {1+(c+d x)^2}\right )}{40 d e^6} \]
1/5*(-a-b*arcsinh(d*x+c))/d/e^6/(d*x+c)^5-3/40*b*arctanh((1+(d*x+c)^2)^(1/ 2))/d/e^6-1/20*b*(1+(d*x+c)^2)^(1/2)/d/e^6/(d*x+c)^4+3/40*b*(1+(d*x+c)^2)^ (1/2)/d/e^6/(d*x+c)^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.53 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=-\frac {\frac {a+b \text {arcsinh}(c+d x)}{(c+d x)^5}+b \sqrt {1+(c+d x)^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1+(c+d x)^2\right )}{5 d e^6} \]
-1/5*((a + b*ArcSinh[c + d*x])/(c + d*x)^5 + b*Sqrt[1 + (c + d*x)^2]*Hyper geometric2F1[1/2, 3, 3/2, 1 + (c + d*x)^2])/(d*e^6)
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6274, 27, 6191, 243, 52, 52, 73, 220}
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.
| \(\displaystyle \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx\) |
| \(\Big \downarrow \) 6274 |
| \(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c+d x)}{e^6 (c+d x)^6}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c+d x)}{(c+d x)^6}d(c+d x)}{d e^6}\) |
| \(\Big \downarrow \) 6191 |
| \(\displaystyle \frac {\frac {1}{5} b \int \frac {1}{(c+d x)^5 \sqrt {(c+d x)^2+1}}d(c+d x)-\frac {a+b \text {arcsinh}(c+d x)}{5 (c+d x)^5}}{d e^6}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {1}{10} b \int \frac {1}{(c+d x)^6 \sqrt {(c+d x)^2+1}}d(c+d x)^2-\frac {a+b \text {arcsinh}(c+d x)}{5 (c+d x)^5}}{d e^6}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {\frac {1}{10} b \left (-\frac {3}{4} \int \frac {1}{(c+d x)^4 \sqrt {(c+d x)^2+1}}d(c+d x)^2-\frac {\sqrt {(c+d x)^2+1}}{2 (c+d x)^4}\right )-\frac {a+b \text {arcsinh}(c+d x)}{5 (c+d x)^5}}{d e^6}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {\frac {1}{10} b \left (-\frac {3}{4} \left (-\frac {1}{2} \int \frac {1}{(c+d x)^2 \sqrt {(c+d x)^2+1}}d(c+d x)^2-\frac {\sqrt {(c+d x)^2+1}}{(c+d x)^2}\right )-\frac {\sqrt {(c+d x)^2+1}}{2 (c+d x)^4}\right )-\frac {a+b \text {arcsinh}(c+d x)}{5 (c+d x)^5}}{d e^6}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {1}{10} b \left (-\frac {3}{4} \left (-\int \frac {1}{(c+d x)^4-1}d\sqrt {(c+d x)^2+1}-\frac {\sqrt {(c+d x)^2+1}}{(c+d x)^2}\right )-\frac {\sqrt {(c+d x)^2+1}}{2 (c+d x)^4}\right )-\frac {a+b \text {arcsinh}(c+d x)}{5 (c+d x)^5}}{d e^6}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\frac {1}{10} b \left (-\frac {3}{4} \left (\text {arctanh}\left (\sqrt {(c+d x)^2+1}\right )-\frac {\sqrt {(c+d x)^2+1}}{(c+d x)^2}\right )-\frac {\sqrt {(c+d x)^2+1}}{2 (c+d x)^4}\right )-\frac {a+b \text {arcsinh}(c+d x)}{5 (c+d x)^5}}{d e^6}\) |
(-1/5*(a + b*ArcSinh[c + d*x])/(c + d*x)^5 + (b*(-1/2*Sqrt[1 + (c + d*x)^2 ]/(c + d*x)^4 - (3*(-(Sqrt[1 + (c + d*x)^2]/(c + d*x)^2) + ArcTanh[Sqrt[1 + (c + d*x)^2]]))/4))/10)/(d*e^6)
3.2.25.3.1 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_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^n/(d*(m + 1))), x] - Simp[b*c* (n/(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b* ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.40 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {-\frac {a}{5 e^{6} \left (d x +c \right )^{5}}+\frac {b \left (-\frac {\operatorname {arcsinh}\left (d x +c \right )}{5 \left (d x +c \right )^{5}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{20 \left (d x +c \right )^{4}}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}}{40 \left (d x +c \right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{40}\right )}{e^{6}}}{d}\) | \(94\) |
| default | \(\frac {-\frac {a}{5 e^{6} \left (d x +c \right )^{5}}+\frac {b \left (-\frac {\operatorname {arcsinh}\left (d x +c \right )}{5 \left (d x +c \right )^{5}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{20 \left (d x +c \right )^{4}}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}}{40 \left (d x +c \right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{40}\right )}{e^{6}}}{d}\) | \(94\) |
| parts | \(-\frac {a}{5 e^{6} \left (d x +c \right )^{5} d}+\frac {b \left (-\frac {\operatorname {arcsinh}\left (d x +c \right )}{5 \left (d x +c \right )^{5}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{20 \left (d x +c \right )^{4}}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}}{40 \left (d x +c \right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{40}\right )}{e^{6} d}\) | \(96\) |
1/d*(-1/5*a/e^6/(d*x+c)^5+b/e^6*(-1/5/(d*x+c)^5*arcsinh(d*x+c)-1/20/(d*x+c )^4*(1+(d*x+c)^2)^(1/2)+3/40/(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-3/40*arctanh(1/ (1+(d*x+c)^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 509 vs. \(2 (101) = 202\).
Time = 0.31 (sec) , antiderivative size = 509, normalized size of antiderivative = 4.43 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=-\frac {8 \, a c^{5} - 8 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + 10 \, b c^{3} d^{2} x^{2} + 5 \, b c^{4} d x\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 3 \, {\left (b c^{5} d^{5} x^{5} + 5 \, b c^{6} d^{4} x^{4} + 10 \, b c^{7} d^{3} x^{3} + 10 \, b c^{8} d^{2} x^{2} + 5 \, b c^{9} d x + b c^{10}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} + 1\right ) - 8 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + 10 \, b c^{3} d^{2} x^{2} + 5 \, b c^{4} d x + b c^{5}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 3 \, {\left (b c^{5} d^{5} x^{5} + 5 \, b c^{6} d^{4} x^{4} + 10 \, b c^{7} d^{3} x^{3} + 10 \, b c^{8} d^{2} x^{2} + 5 \, b c^{9} d x + b c^{10}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} - 1\right ) - {\left (3 \, b c^{5} d^{3} x^{3} + 9 \, b c^{6} d^{2} x^{2} + 3 \, b c^{8} - 2 \, b c^{6} + {\left (9 \, b c^{7} - 2 \, b c^{5}\right )} d x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{40 \, {\left (c^{5} d^{6} e^{6} x^{5} + 5 \, c^{6} d^{5} e^{6} x^{4} + 10 \, c^{7} d^{4} e^{6} x^{3} + 10 \, c^{8} d^{3} e^{6} x^{2} + 5 \, c^{9} d^{2} e^{6} x + c^{10} d e^{6}\right )}} \]
-1/40*(8*a*c^5 - 8*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + 10*b*c^ 3*d^2*x^2 + 5*b*c^4*d*x)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + 3*(b*c^5*d^5*x^5 + 5*b*c^6*d^4*x^4 + 10*b*c^7*d^3*x^3 + 10*b*c^8*d^2*x^2 + 5*b*c^9*d*x + b*c^10)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) + 1) - 8*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + 10*b*c^3*d^2*x^2 + 5*b*c^4*d*x + b*c^5)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - 3*(b*c^5*d^5*x^5 + 5*b*c^6*d^4*x^4 + 10*b*c^7*d^3*x^3 + 10*b*c^8*d^2*x^2 + 5*b*c^9*d*x + b*c^10)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) - 1) - (3*b*c^5*d^3*x^3 + 9*b*c^6*d^2*x^2 + 3*b*c^8 - 2*b*c^6 + (9*b*c^7 - 2*b*c^5)*d*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(c^5*d^6*e^6*x^5 + 5*c^6* d^5*e^6*x^4 + 10*c^7*d^4*e^6*x^3 + 10*c^8*d^3*e^6*x^2 + 5*c^9*d^2*e^6*x + c^10*d*e^6)
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=\frac {\int \frac {a}{c^{6} + 6 c^{5} d x + 15 c^{4} d^{2} x^{2} + 20 c^{3} d^{3} x^{3} + 15 c^{2} d^{4} x^{4} + 6 c d^{5} x^{5} + d^{6} x^{6}}\, dx + \int \frac {b \operatorname {asinh}{\left (c + d x \right )}}{c^{6} + 6 c^{5} d x + 15 c^{4} d^{2} x^{2} + 20 c^{3} d^{3} x^{3} + 15 c^{2} d^{4} x^{4} + 6 c d^{5} x^{5} + d^{6} x^{6}}\, dx}{e^{6}} \]
(Integral(a/(c**6 + 6*c**5*d*x + 15*c**4*d**2*x**2 + 20*c**3*d**3*x**3 + 1 5*c**2*d**4*x**4 + 6*c*d**5*x**5 + d**6*x**6), x) + Integral(b*asinh(c + d *x)/(c**6 + 6*c**5*d*x + 15*c**4*d**2*x**2 + 20*c**3*d**3*x**3 + 15*c**2*d **4*x**4 + 6*c*d**5*x**5 + d**6*x**6), x))/e**6
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=\int { \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{6}} \,d x } \]
1/30*b*(2*(3*d^4*x^4 + 12*c*d^3*x^3 + 3*c^4 + (18*c^2*d^2 - d^2)*x^2 - c^2 + 2*(6*c^3*d - c*d)*x - 3*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) ))/(d^6*e^6*x^5 + 5*c*d^5*e^6*x^4 + 10*c^2*d^4*e^6*x^3 + 10*c^3*d^3*e^6*x^ 2 + 5*c^4*d^2*e^6*x + c^5*d*e^6) - 3*I*(log(I*(d^2*x + c*d)/d + 1) - log(- I*(d^2*x + c*d)/d + 1))/(d*e^6) + 30*integrate(1/5/(d^8*e^6*x^8 + 8*c*d^7* e^6*x^7 + c^8*e^6 + c^6*e^6 + (28*c^2*d^6*e^6 + d^6*e^6)*x^6 + 2*(28*c^3*d ^5*e^6 + 3*c*d^5*e^6)*x^5 + 5*(14*c^4*d^4*e^6 + 3*c^2*d^4*e^6)*x^4 + 4*(14 *c^5*d^3*e^6 + 5*c^3*d^3*e^6)*x^3 + (28*c^6*d^2*e^6 + 15*c^4*d^2*e^6)*x^2 + 2*(4*c^7*d*e^6 + 3*c^5*d*e^6)*x + (d^7*e^6*x^7 + 7*c*d^6*e^6*x^6 + c^7*e ^6 + c^5*e^6 + (21*c^2*d^5*e^6 + d^5*e^6)*x^5 + 5*(7*c^3*d^4*e^6 + c*d^4*e ^6)*x^4 + 5*(7*c^4*d^3*e^6 + 2*c^2*d^3*e^6)*x^3 + (21*c^5*d^2*e^6 + 10*c^3 *d^2*e^6)*x^2 + (7*c^6*d*e^6 + 5*c^4*d*e^6)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^ 2 + 1)), x)) - 1/5*a/(d^6*e^6*x^5 + 5*c*d^5*e^6*x^4 + 10*c^2*d^4*e^6*x^3 + 10*c^3*d^3*e^6*x^2 + 5*c^4*d^2*e^6*x + c^5*d*e^6)
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=\int { \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{6}} \,d x } \]
Timed out. \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^6} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^6} \,d x \]