Integrand size = 16, antiderivative size = 125 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=-\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {2 b c^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}+\frac {2 b c^{7/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}} \]
-4/35*b*c/d^2/(d*x)^(5/2)-2/7*b*c^(7/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2) )/d^(9/2)-2/7*(a+b*arctanh(c*x))/d/(d*x)^(7/2)+2/7*b*c^(7/2)*arctanh(c^(1/ 2)*(d*x)^(1/2)/d^(1/2))/d^(9/2)-4/7*b*c^3/d^4/(d*x)^(1/2)
Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=-\frac {\sqrt {d x} \left (10 a+4 b c x+20 b c^3 x^3+10 b c^{7/2} x^{7/2} \arctan \left (\sqrt {c} \sqrt {x}\right )+10 b \text {arctanh}(c x)+5 b c^{7/2} x^{7/2} \log \left (1-\sqrt {c} \sqrt {x}\right )-5 b c^{7/2} x^{7/2} \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{35 d^5 x^4} \]
-1/35*(Sqrt[d*x]*(10*a + 4*b*c*x + 20*b*c^3*x^3 + 10*b*c^(7/2)*x^(7/2)*Arc Tan[Sqrt[c]*Sqrt[x]] + 10*b*ArcTanh[c*x] + 5*b*c^(7/2)*x^(7/2)*Log[1 - Sqr t[c]*Sqrt[x]] - 5*b*c^(7/2)*x^(7/2)*Log[1 + Sqrt[c]*Sqrt[x]]))/(d^5*x^4)
Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6464, 264, 264, 266, 27, 827, 218, 221}
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.
| \(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx\) |
| \(\Big \downarrow \) 6464 |
| \(\displaystyle \frac {2 b c \int \frac {1}{(d x)^{7/2} \left (1-c^2 x^2\right )}dx}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \int \frac {1}{(d x)^{3/2} \left (1-c^2 x^2\right )}dx}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \left (\frac {c^2 \int \frac {\sqrt {d x}}{1-c^2 x^2}dx}{d^2}-\frac {2}{d \sqrt {d x}}\right )}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \left (\frac {2 c^2 \int \frac {d^3 x}{d^2-c^2 d^2 x^2}d\sqrt {d x}}{d^3}-\frac {2}{d \sqrt {d x}}\right )}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \left (\frac {2 c^2 \int \frac {d x}{d^2-c^2 d^2 x^2}d\sqrt {d x}}{d}-\frac {2}{d \sqrt {d x}}\right )}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \left (\frac {2 c^2 \left (\frac {\int \frac {1}{d-c d x}d\sqrt {d x}}{2 c}-\frac {\int \frac {1}{c x d+d}d\sqrt {d x}}{2 c}\right )}{d}-\frac {2}{d \sqrt {d x}}\right )}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \left (\frac {2 c^2 \left (\frac {\int \frac {1}{d-c d x}d\sqrt {d x}}{2 c}-\frac {\arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}\right )}{d}-\frac {2}{d \sqrt {d x}}\right )}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 b c \left (\frac {c^2 \left (\frac {2 c^2 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}-\frac {\arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}\right )}{d}-\frac {2}{d \sqrt {d x}}\right )}{d^2}-\frac {2}{5 d (d x)^{5/2}}\right )}{7 d}-\frac {2 (a+b \text {arctanh}(c x))}{7 d (d x)^{7/2}}\) |
(-2*(a + b*ArcTanh[c*x]))/(7*d*(d*x)^(7/2)) + (2*b*c*(-2/(5*d*(d*x)^(5/2)) + (c^2*(-2/(d*Sqrt[d*x]) + (2*c^2*(-1/2*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d ]]/(c^(3/2)*Sqrt[d]) + ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]/(2*c^(3/2)*Sqr t[d])))/d))/d^2))/(7*d)
3.1.42.3.1 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)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))*((d_)*(x_))^(m_), x_Symbol] : > Simp[(d*x)^(m + 1)*((a + b*ArcTanh[c*x^n])/(d*(m + 1))), x] - Simp[b*c*(n /(d^n*(m + 1))) Int[(d*x)^(m + n)/(1 - c^2*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegerQ[n] && NeQ[m, -1]
Time = 0.59 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {4 b c}{35 d \left (d x \right )^{\frac {5}{2}}}-\frac {4 b \,c^{3}}{7 d^{3} \sqrt {d x}}+\frac {2 b \,c^{4} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 d^{3} \sqrt {c d}}-\frac {2 b \,c^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 d^{3} \sqrt {c d}}}{d}\) | \(107\) |
| default | \(\frac {-\frac {2 a}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {4 b c}{35 d \left (d x \right )^{\frac {5}{2}}}-\frac {4 b \,c^{3}}{7 d^{3} \sqrt {d x}}+\frac {2 b \,c^{4} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 d^{3} \sqrt {c d}}-\frac {2 b \,c^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 d^{3} \sqrt {c d}}}{d}\) | \(107\) |
| parts | \(-\frac {2 a}{7 \left (d x \right )^{\frac {7}{2}} d}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{7 d \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \,c^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 d^{4} \sqrt {c d}}-\frac {4 b c}{35 d^{2} \left (d x \right )^{\frac {5}{2}}}-\frac {4 b \,c^{3}}{7 d^{4} \sqrt {d x}}+\frac {2 b \,c^{4} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 d^{4} \sqrt {c d}}\) | \(108\) |
2/d*(-1/7*a/(d*x)^(7/2)-1/7*b/(d*x)^(7/2)*arctanh(c*x)-2/35*b/d*c/(d*x)^(5 /2)-2/7*b/d^3*c^3/(d*x)^(1/2)+1/7*b/d^3*c^4/(c*d)^(1/2)*arctanh(c*(d*x)^(1 /2)/(c*d)^(1/2))-1/7*b/d^3*c^4/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2 )))
Time = 0.28 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.18 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=\left [\frac {10 \, b c^{3} d x^{4} \sqrt {\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {\frac {c}{d}}}{c x}\right ) + 5 \, b c^{3} d x^{4} \sqrt {\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {\frac {c}{d}} + 1}{c x - 1}\right ) - {\left (20 \, b c^{3} x^{3} + 4 \, b c x + 5 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a\right )} \sqrt {d x}}{35 \, d^{5} x^{4}}, -\frac {10 \, b c^{3} d x^{4} \sqrt {-\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {c}{d}}}{c x}\right ) - 5 \, b c^{3} d x^{4} \sqrt {-\frac {c}{d}} \log \left (\frac {c x - 2 \, \sqrt {d x} \sqrt {-\frac {c}{d}} - 1}{c x + 1}\right ) + {\left (20 \, b c^{3} x^{3} + 4 \, b c x + 5 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a\right )} \sqrt {d x}}{35 \, d^{5} x^{4}}\right ] \]
[1/35*(10*b*c^3*d*x^4*sqrt(c/d)*arctan(sqrt(d*x)*sqrt(c/d)/(c*x)) + 5*b*c^ 3*d*x^4*sqrt(c/d)*log((c*x + 2*sqrt(d*x)*sqrt(c/d) + 1)/(c*x - 1)) - (20*b *c^3*x^3 + 4*b*c*x + 5*b*log(-(c*x + 1)/(c*x - 1)) + 10*a)*sqrt(d*x))/(d^5 *x^4), -1/35*(10*b*c^3*d*x^4*sqrt(-c/d)*arctan(sqrt(d*x)*sqrt(-c/d)/(c*x)) - 5*b*c^3*d*x^4*sqrt(-c/d)*log((c*x - 2*sqrt(d*x)*sqrt(-c/d) - 1)/(c*x + 1)) + (20*b*c^3*x^3 + 4*b*c*x + 5*b*log(-(c*x + 1)/(c*x - 1)) + 10*a)*sqrt (d*x))/(d^5*x^4)]
\[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=\int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\left (d x\right )^{\frac {9}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=-\frac {b {\left (\frac {{\left (\frac {10 \, c^{3} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d^{2}} + \frac {5 \, c^{3} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} d^{2}} + \frac {4 \, {\left (5 \, c^{2} d^{2} x^{2} + d^{2}\right )}}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )} c}{d} + \frac {10 \, \operatorname {artanh}\left (c x\right )}{\left (d x\right )^{\frac {7}{2}}}\right )} + \frac {10 \, a}{\left (d x\right )^{\frac {7}{2}}}}{35 \, d} \]
-1/35*(b*((10*c^3*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*d^2) + 5*c^3*lo g((sqrt(d*x)*c - sqrt(c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*d^2) + 4 *(5*c^2*d^2*x^2 + d^2)/((d*x)^(5/2)*d^2))*c/d + 10*arctanh(c*x)/(d*x)^(7/2 )) + 10*a/(d*x)^(7/2))/d
Time = 0.34 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=-\frac {\frac {10 \, b c^{4} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d^{3}} + \frac {10 \, b c^{4} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} d^{3}} + \frac {5 \, b \log \left (-\frac {c d x + d}{c d x - d}\right )}{\sqrt {d x} d^{3} x^{3}} + \frac {2 \, {\left (10 \, b c^{3} d^{3} x^{3} + 2 \, b c d^{3} x + 5 \, a d^{3}\right )}}{\sqrt {d x} d^{6} x^{3}}}{35 \, d} \]
-1/35*(10*b*c^4*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*d^3) + 10*b*c^4*a rctan(sqrt(d*x)*c/sqrt(-c*d))/(sqrt(-c*d)*d^3) + 5*b*log(-(c*d*x + d)/(c*d *x - d))/(sqrt(d*x)*d^3*x^3) + 2*(10*b*c^3*d^3*x^3 + 2*b*c*d^3*x + 5*a*d^3 )/(sqrt(d*x)*d^6*x^3))/d
Timed out. \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{9/2}} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{{\left (d\,x\right )}^{9/2}} \,d x \]