Integrand size = 16, antiderivative size = 124 \[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\frac {4 b d^2 \sqrt {d x}}{7 c^3}+\frac {4 b (d x)^{5/2}}{35 c}-\frac {2 b d^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}+\frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b d^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}} \]
4/35*b*(d*x)^(5/2)/c-2/7*b*d^(5/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^( 7/2)+2/7*(d*x)^(7/2)*(a+b*arctanh(c*x))/d-2/7*b*d^(5/2)*arctanh(c^(1/2)*(d *x)^(1/2)/d^(1/2))/c^(7/2)+4/7*b*d^2*(d*x)^(1/2)/c^3
Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\frac {(d x)^{5/2} \left (20 b \sqrt {c} \sqrt {x}+4 b c^{5/2} x^{5/2}+10 a c^{7/2} x^{7/2}-10 b \arctan \left (\sqrt {c} \sqrt {x}\right )+10 b c^{7/2} x^{7/2} \text {arctanh}(c x)+5 b \log \left (1-\sqrt {c} \sqrt {x}\right )-5 b \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{35 c^{7/2} x^{5/2}} \]
((d*x)^(5/2)*(20*b*Sqrt[c]*Sqrt[x] + 4*b*c^(5/2)*x^(5/2) + 10*a*c^(7/2)*x^ (7/2) - 10*b*ArcTan[Sqrt[c]*Sqrt[x]] + 10*b*c^(7/2)*x^(7/2)*ArcTanh[c*x] + 5*b*Log[1 - Sqrt[c]*Sqrt[x]] - 5*b*Log[1 + Sqrt[c]*Sqrt[x]]))/(35*c^(7/2) *x^(5/2))
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6464, 262, 262, 266, 756, 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.
\(\displaystyle \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6464 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \int \frac {(d x)^{7/2}}{1-c^2 x^2}dx}{7 d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \left (\frac {d^2 \int \frac {(d x)^{3/2}}{1-c^2 x^2}dx}{c^2}-\frac {2 d (d x)^{5/2}}{5 c^2}\right )}{7 d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \left (\frac {d^2 \left (\frac {d^2 \int \frac {1}{\sqrt {d x} \left (1-c^2 x^2\right )}dx}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{c^2}-\frac {2 d (d x)^{5/2}}{5 c^2}\right )}{7 d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \left (\frac {d^2 \left (\frac {2 d \int \frac {1}{1-c^2 x^2}d\sqrt {d x}}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{c^2}-\frac {2 d (d x)^{5/2}}{5 c^2}\right )}{7 d}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \left (\frac {d^2 \left (\frac {2 d \left (\frac {1}{2} d \int \frac {1}{d-c d x}d\sqrt {d x}+\frac {1}{2} d \int \frac {1}{c x d+d}d\sqrt {d x}\right )}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{c^2}-\frac {2 d (d x)^{5/2}}{5 c^2}\right )}{7 d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \left (\frac {d^2 \left (\frac {2 d \left (\frac {1}{2} d \int \frac {1}{d-c d x}d\sqrt {d x}+\frac {\sqrt {d} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt {c}}\right )}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{c^2}-\frac {2 d (d x)^{5/2}}{5 c^2}\right )}{7 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (d x)^{7/2} (a+b \text {arctanh}(c x))}{7 d}-\frac {2 b c \left (\frac {d^2 \left (\frac {2 d \left (\frac {\sqrt {d} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt {c}}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt {c}}\right )}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{c^2}-\frac {2 d (d x)^{5/2}}{5 c^2}\right )}{7 d}\) |
(2*(d*x)^(7/2)*(a + b*ArcTanh[c*x]))/(7*d) - (2*b*c*((-2*d*(d*x)^(5/2))/(5 *c^2) + (d^2*((-2*d*Sqrt[d*x])/c^2 + (2*d*((Sqrt[d]*ArcTan[(Sqrt[c]*Sqrt[d *x])/Sqrt[d]])/(2*Sqrt[c]) + (Sqrt[d]*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]] )/(2*Sqrt[c])))/c^2))/c^2))/(7*d)
3.1.35.3.1 Defintions of rubi rules used
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*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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[((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[((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 = 1.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {\frac {2 a \left (d x \right )^{\frac {7}{2}}}{7}+\frac {2 b \left (d x \right )^{\frac {7}{2}} \operatorname {arctanh}\left (c x \right )}{7}+\frac {4 b d \left (d x \right )^{\frac {5}{2}}}{35 c}+\frac {4 b \,d^{3} \sqrt {d x}}{7 c^{3}}-\frac {2 b \,d^{4} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}-\frac {2 b \,d^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}}{d}\) | \(107\) |
default | \(\frac {\frac {2 a \left (d x \right )^{\frac {7}{2}}}{7}+\frac {2 b \left (d x \right )^{\frac {7}{2}} \operatorname {arctanh}\left (c x \right )}{7}+\frac {4 b d \left (d x \right )^{\frac {5}{2}}}{35 c}+\frac {4 b \,d^{3} \sqrt {d x}}{7 c^{3}}-\frac {2 b \,d^{4} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}-\frac {2 b \,d^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}}{d}\) | \(107\) |
parts | \(\frac {2 a \left (d x \right )^{\frac {7}{2}}}{7 d}+\frac {2 b \left (d x \right )^{\frac {7}{2}} \operatorname {arctanh}\left (c x \right )}{7 d}+\frac {4 b \left (d x \right )^{\frac {5}{2}}}{35 c}+\frac {4 b \,d^{2} \sqrt {d x}}{7 c^{3}}-\frac {2 b \,d^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}-\frac {2 b \,d^{3} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}\) | \(107\) |
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^4/c^3/(c*d)^(1/2)*arctanh(c*(d*x)^(1/ 2)/(c*d)^(1/2))-1/7*b*d^4/c^3/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2) ))
Time = 0.28 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.39 \[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\left [-\frac {10 \, b d^{2} \sqrt {\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {\frac {d}{c}}}{d}\right ) - 5 \, b d^{2} \sqrt {\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {\frac {d}{c}} + d}{c x - 1}\right ) - {\left (5 \, b c^{3} d^{2} x^{3} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a c^{3} d^{2} x^{3} + 4 \, b c^{2} d^{2} x^{2} + 20 \, b d^{2}\right )} \sqrt {d x}}{35 \, c^{3}}, \frac {10 \, b d^{2} \sqrt {-\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {-\frac {d}{c}}}{d}\right ) + 5 \, b d^{2} \sqrt {-\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {-\frac {d}{c}} - d}{c x + 1}\right ) + {\left (5 \, b c^{3} d^{2} x^{3} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a c^{3} d^{2} x^{3} + 4 \, b c^{2} d^{2} x^{2} + 20 \, b d^{2}\right )} \sqrt {d x}}{35 \, c^{3}}\right ] \]
[-1/35*(10*b*d^2*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) - 5*b*d^2*sqrt( d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(d/c) + d)/(c*x - 1)) - (5*b*c^3*d^2*x ^3*log(-(c*x + 1)/(c*x - 1)) + 10*a*c^3*d^2*x^3 + 4*b*c^2*d^2*x^2 + 20*b*d ^2)*sqrt(d*x))/c^3, 1/35*(10*b*d^2*sqrt(-d/c)*arctan(sqrt(d*x)*c*sqrt(-d/c )/d) + 5*b*d^2*sqrt(-d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(-d/c) - d)/(c*x + 1)) + (5*b*c^3*d^2*x^3*log(-(c*x + 1)/(c*x - 1)) + 10*a*c^3*d^2*x^3 + 4* b*c^2*d^2*x^2 + 20*b*d^2)*sqrt(d*x))/c^3]
\[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\int \left (d x\right )^{\frac {5}{2}} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )\, dx \]
Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\frac {10 \, \left (d x\right )^{\frac {7}{2}} a + {\left (10 \, \left (d x\right )^{\frac {7}{2}} \operatorname {artanh}\left (c x\right ) - \frac {{\left (\frac {10 \, d^{5} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c^{4}} - \frac {5 \, d^{5} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c^{4}} - \frac {4 \, {\left (\left (d x\right )^{\frac {5}{2}} c^{2} d^{2} + 5 \, \sqrt {d x} d^{4}\right )}}{c^{4}}\right )} c}{d}\right )} b}{35 \, d} \]
1/35*(10*(d*x)^(7/2)*a + (10*(d*x)^(7/2)*arctanh(c*x) - (10*d^5*arctan(sqr t(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c^4) - 5*d^5*log((sqrt(d*x)*c - sqrt(c*d))/ (sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c^4) - 4*((d*x)^(5/2)*c^2*d^2 + 5*sq rt(d*x)*d^4)/c^4)*c/d)*b)/d
\[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )} \,d x } \]
Timed out. \[ \int (d x)^{5/2} (a+b \text {arctanh}(c x)) \, dx=\int \left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{5/2} \,d x \]