Integrand size = 16, antiderivative size = 85 \[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\frac {2 b \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}}+\frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}} \]
2*b*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(1/2)/d^(1/2)-2*b*arctanh(c^(1/2 )*(d*x)^(1/2)/d^(1/2))/c^(1/2)/d^(1/2)+2*(a+b*arctanh(c*x))*(d*x)^(1/2)/d
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\frac {\sqrt {x} \left (2 a \sqrt {c} \sqrt {x}+2 b \arctan \left (\sqrt {c} \sqrt {x}\right )+2 b \sqrt {c} \sqrt {x} \text {arctanh}(c x)+b \log \left (1-\sqrt {c} \sqrt {x}\right )-b \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{\sqrt {c} \sqrt {d x}} \]
(Sqrt[x]*(2*a*Sqrt[c]*Sqrt[x] + 2*b*ArcTan[Sqrt[c]*Sqrt[x]] + 2*b*Sqrt[c]* Sqrt[x]*ArcTanh[c*x] + b*Log[1 - Sqrt[c]*Sqrt[x]] - b*Log[1 + Sqrt[c]*Sqrt [x]]))/(Sqrt[c]*Sqrt[d*x])
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6464, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx\) |
\(\Big \downarrow \) 6464 |
\(\displaystyle \frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-\frac {2 b c \int \frac {\sqrt {d x}}{1-c^2 x^2}dx}{d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-\frac {4 b c \int \frac {d^3 x}{d^2-c^2 d^2 x^2}d\sqrt {d x}}{d^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-4 b c \int \frac {d x}{d^2-c^2 d^2 x^2}d\sqrt {d x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-4 b c \left (\frac {\int \frac {1}{d-c d x}d\sqrt {d x}}{2 c}-\frac {\int \frac {1}{c x d+d}d\sqrt {d x}}{2 c}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-4 b c \left (\frac {\int \frac {1}{d-c d x}d\sqrt {d x}}{2 c}-\frac {\arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {d x} (a+b \text {arctanh}(c x))}{d}-4 b c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}-\frac {\arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}\right )\) |
(2*Sqrt[d*x]*(a + b*ArcTanh[c*x]))/d - 4*b*c*(-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)*Sqrt[d]))
3.1.38.3.1 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] :> 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 = 0.57 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {2 \sqrt {d x}\, a +2 \sqrt {d x}\, b \,\operatorname {arctanh}\left (c x \right )-\frac {2 b d \,\operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}+\frac {2 b d \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}}{d}\) | \(68\) |
default | \(\frac {2 \sqrt {d x}\, a +2 \sqrt {d x}\, b \,\operatorname {arctanh}\left (c x \right )-\frac {2 b d \,\operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}+\frac {2 b d \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}}{d}\) | \(68\) |
parts | \(\frac {2 a \sqrt {d x}}{d}+\frac {2 b \sqrt {d x}\, \operatorname {arctanh}\left (c x \right )}{d}+\frac {2 b \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}-\frac {2 b \,\operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}\) | \(70\) |
2/d*((d*x)^(1/2)*a+(d*x)^(1/2)*b*arctanh(c*x)-b*d/(c*d)^(1/2)*arctanh(c*(d *x)^(1/2)/(c*d)^(1/2))+b*d/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.48 \[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\left [-\frac {2 \, \sqrt {c d} b \arctan \left (\frac {\sqrt {c d} \sqrt {d x}}{c d x}\right ) - \sqrt {c d} b \log \left (\frac {c d x - 2 \, \sqrt {c d} \sqrt {d x} + d}{c x - 1}\right ) - {\left (b c \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c\right )} \sqrt {d x}}{c d}, \frac {2 \, \sqrt {-c d} b \arctan \left (\frac {\sqrt {-c d} \sqrt {d x}}{c d x}\right ) - \sqrt {-c d} b \log \left (\frac {c d x - 2 \, \sqrt {-c d} \sqrt {d x} - d}{c x + 1}\right ) + {\left (b c \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c\right )} \sqrt {d x}}{c d}\right ] \]
[-(2*sqrt(c*d)*b*arctan(sqrt(c*d)*sqrt(d*x)/(c*d*x)) - sqrt(c*d)*b*log((c* d*x - 2*sqrt(c*d)*sqrt(d*x) + d)/(c*x - 1)) - (b*c*log(-(c*x + 1)/(c*x - 1 )) + 2*a*c)*sqrt(d*x))/(c*d), (2*sqrt(-c*d)*b*arctan(sqrt(-c*d)*sqrt(d*x)/ (c*d*x)) - sqrt(-c*d)*b*log((c*d*x - 2*sqrt(-c*d)*sqrt(d*x) - d)/(c*x + 1) ) + (b*c*log(-(c*x + 1)/(c*x - 1)) + 2*a*c)*sqrt(d*x))/(c*d)]
\[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\sqrt {d x}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.21 \[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\frac {{\left (2 \, \sqrt {d x} \operatorname {artanh}\left (c x\right ) + \frac {{\left (\frac {2 \, d^{2} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c} + \frac {d^{2} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c}\right )} c}{d}\right )} b + 2 \, \sqrt {d x} a}{d} \]
((2*sqrt(d*x)*arctanh(c*x) + (2*d^2*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c* d)*c) + d^2*log((sqrt(d*x)*c - sqrt(c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt (c*d)*c))*c/d)*b + 2*sqrt(d*x)*a)/d
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\frac {{\left (2 \, c d {\left (\frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c} + \frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} c}\right )} + \sqrt {d x} \log \left (-\frac {c x + 1}{c x - 1}\right )\right )} b + 2 \, \sqrt {d x} a}{d} \]
((2*c*d*(arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c) + arctan(sqrt(d*x)*c/ sqrt(-c*d))/(sqrt(-c*d)*c)) + sqrt(d*x)*log(-(c*x + 1)/(c*x - 1)))*b + 2*s qrt(d*x)*a)/d
Timed out. \[ \int \frac {a+b \text {arctanh}(c x)}{\sqrt {d x}} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{\sqrt {d\,x}} \,d x \]