3.1.37 \(\int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx\) [37]

3.1.37.1 Optimal result

Integrand size = 16, antiderivative size = 106 \[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\frac {4 b \sqrt {d x}}{3 c}-\frac {2 b \sqrt {d} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 c^{3/2}}+\frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 c^{3/2}} \]

output

2/3*(d*x)^(3/2)*(a+b*arctanh(c*x))/d-2/3*b*arctan(c^(1/2)*(d*x)^(1/2)/d^(1 
/2))*d^(1/2)/c^(3/2)-2/3*b*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))*d^(1/2)/c^ 
(3/2)+4/3*b*(d*x)^(1/2)/c
 

3.1.37.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.08 \[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\frac {\sqrt {d x} \left (4 b \sqrt {c} \sqrt {x}+2 a c^{3/2} x^{3/2}-2 b \arctan \left (\sqrt {c} \sqrt {x}\right )+2 b c^{3/2} x^{3/2} \text {arctanh}(c x)+b \log \left (1-\sqrt {c} \sqrt {x}\right )-b \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{3 c^{3/2} \sqrt {x}} \]

input

Integrate[Sqrt[d*x]*(a + b*ArcTanh[c*x]),x]
 

output

(Sqrt[d*x]*(4*b*Sqrt[c]*Sqrt[x] + 2*a*c^(3/2)*x^(3/2) - 2*b*ArcTan[Sqrt[c] 
*Sqrt[x]] + 2*b*c^(3/2)*x^(3/2)*ArcTanh[c*x] + b*Log[1 - Sqrt[c]*Sqrt[x]] 
- b*Log[1 + Sqrt[c]*Sqrt[x]]))/(3*c^(3/2)*Sqrt[x])
 

3.1.37.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6464, 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.

\(\displaystyle \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx\)

\(\Big \downarrow \) 6464

\(\displaystyle \frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b c \int \frac {(d x)^{3/2}}{1-c^2 x^2}dx}{3 d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b c \left (\frac {d^2 \int \frac {1}{\sqrt {d x} \left (1-c^2 x^2\right )}dx}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{3 d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b c \left (\frac {2 d \int \frac {1}{1-c^2 x^2}d\sqrt {d x}}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{3 d}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b c \left (\frac {2 d \left (\frac {1}{2} d \int \frac {1}{d-c d x}d\sqrt {d x}+\frac {1}{2} d \int \frac {1}{c x d+d}d\sqrt {d x}\right )}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{3 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b c \left (\frac {2 d \left (\frac {1}{2} d \int \frac {1}{d-c d x}d\sqrt {d x}+\frac {\sqrt {d} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt {c}}\right )}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{3 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (d x)^{3/2} (a+b \text {arctanh}(c x))}{3 d}-\frac {2 b c \left (\frac {2 d \left (\frac {\sqrt {d} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt {c}}+\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 \sqrt {c}}\right )}{c^2}-\frac {2 d \sqrt {d x}}{c^2}\right )}{3 d}\)

input

Int[Sqrt[d*x]*(a + b*ArcTanh[c*x]),x]
 

output

(2*(d*x)^(3/2)*(a + b*ArcTanh[c*x]))/(3*d) - (2*b*c*((-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))/(3*d)
 

3.1.37.3.1 Defintions of rubi rules used

rule 218

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]
 

rule 221

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]
 

rule 262

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]
 

rule 266

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]
 

rule 756

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]
 

rule 6464

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]
 

3.1.37.4 Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.84

method	result	size
parts	\(\frac {2 a \left (d x \right )^{\frac {3}{2}}}{3 d}+\frac {2 b \left (d x \right )^{\frac {3}{2}} \operatorname {arctanh}\left (c x \right )}{3 d}+\frac {4 b \sqrt {d x}}{3 c}-\frac {2 b d \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}-\frac {2 b d \,\operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}\)	\(89\)
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {3}{2}} a}{3}+\frac {2 \left (d x \right )^{\frac {3}{2}} b \,\operatorname {arctanh}\left (c x \right )}{3}+\frac {4 b d \sqrt {d x}}{3 c}-\frac {2 b \,d^{2} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}-\frac {2 b \,d^{2} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}}{d}\)	\(93\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {3}{2}} a}{3}+\frac {2 \left (d x \right )^{\frac {3}{2}} b \,\operatorname {arctanh}\left (c x \right )}{3}+\frac {4 b d \sqrt {d x}}{3 c}-\frac {2 b \,d^{2} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}-\frac {2 b \,d^{2} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}}{d}\)	\(93\)

input

int((d*x)^(1/2)*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
 

output

2/3*a*(d*x)^(3/2)/d+2/3*b/d*(d*x)^(3/2)*arctanh(c*x)+4/3*b*(d*x)^(1/2)/c-2 
/3*b*d/c/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2))-2/3*b*d/c/(c*d)^(1/ 
2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))
 

3.1.37.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.10 \[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\left [-\frac {2 \, b \sqrt {\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {\frac {d}{c}}}{d}\right ) - b \sqrt {\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {\frac {d}{c}} + d}{c x - 1}\right ) - {\left (b c x \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c x + 4 \, b\right )} \sqrt {d x}}{3 \, c}, \frac {2 \, b \sqrt {-\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {-\frac {d}{c}}}{d}\right ) + b \sqrt {-\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {-\frac {d}{c}} - d}{c x + 1}\right ) + {\left (b c x \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c x + 4 \, b\right )} \sqrt {d x}}{3 \, c}\right ] \]

input

integrate((d*x)^(1/2)*(a+b*arctanh(c*x)),x, algorithm="fricas")
 

output

[-1/3*(2*b*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) - b*sqrt(d/c)*log((c* 
d*x - 2*sqrt(d*x)*c*sqrt(d/c) + d)/(c*x - 1)) - (b*c*x*log(-(c*x + 1)/(c*x 
 - 1)) + 2*a*c*x + 4*b)*sqrt(d*x))/c, 1/3*(2*b*sqrt(-d/c)*arctan(sqrt(d*x) 
*c*sqrt(-d/c)/d) + b*sqrt(-d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(-d/c) - d) 
/(c*x + 1)) + (b*c*x*log(-(c*x + 1)/(c*x - 1)) + 2*a*c*x + 4*b)*sqrt(d*x)) 
/c]
 

3.1.37.6 Sympy [F]

\[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\int \sqrt {d x} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )\, dx \]

input

integrate((d*x)**(1/2)*(a+b*atanh(c*x)),x)
 

output

Integral(sqrt(d*x)*(a + b*atanh(c*x)), x)
 

3.1.37.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12 \[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\frac {2 \, \left (d x\right )^{\frac {3}{2}} a + {\left (2 \, \left (d x\right )^{\frac {3}{2}} \operatorname {artanh}\left (c x\right ) - \frac {{\left (\frac {2 \, d^{3} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c^{2}} - \frac {d^{3} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c^{2}} - \frac {4 \, \sqrt {d x} d^{2}}{c^{2}}\right )} c}{d}\right )} b}{3 \, d} \]

input

integrate((d*x)^(1/2)*(a+b*arctanh(c*x)),x, algorithm="maxima")
 

output

1/3*(2*(d*x)^(3/2)*a + (2*(d*x)^(3/2)*arctanh(c*x) - (2*d^3*arctan(sqrt(d* 
x)*c/sqrt(c*d))/(sqrt(c*d)*c^2) - d^3*log((sqrt(d*x)*c - sqrt(c*d))/(sqrt( 
d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c^2) - 4*sqrt(d*x)*d^2/c^2)*c/d)*b)/d
 

3.1.37.8 Giac [F]

\[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\int { \sqrt {d x} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )} \,d x } \]

input

integrate((d*x)^(1/2)*(a+b*arctanh(c*x)),x, algorithm="giac")
 

output

integrate(sqrt(d*x)*(b*arctanh(c*x) + a), x)
 

3.1.37.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {d x} (a+b \text {arctanh}(c x)) \, dx=\int \left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\sqrt {d\,x} \,d x \]

input

int((a + b*atanh(c*x))*(d*x)^(1/2),x)
 

output

int((a + b*atanh(c*x))*(d*x)^(1/2), x)