\(\int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx\) [36]

Optimal result

Integrand size = 16, antiderivative size = 106 \[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\frac {4 b (d x)^{3/2}}{15 c}+\frac {2 b d^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}}+\frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b d^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}} \] Output:

4/15*b*(d*x)^(3/2)/c+2/5*b*d^(3/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^( 
5/2)+2/5*(d*x)^(5/2)*(a+b*arctanh(c*x))/d-2/5*b*d^(3/2)*arctanh(c^(1/2)*(d 
*x)^(1/2)/d^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.08 \[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\frac {(d x)^{3/2} \left (4 b c^{3/2} x^{3/2}+6 a c^{5/2} x^{5/2}+6 b \arctan \left (\sqrt {c} \sqrt {x}\right )+6 b c^{5/2} x^{5/2} \text {arctanh}(c x)+3 b \log \left (1-\sqrt {c} \sqrt {x}\right )-3 b \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{15 c^{5/2} x^{3/2}} \] Input:

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

Output:

((d*x)^(3/2)*(4*b*c^(3/2)*x^(3/2) + 6*a*c^(5/2)*x^(5/2) + 6*b*ArcTan[Sqrt[ 
c]*Sqrt[x]] + 6*b*c^(5/2)*x^(5/2)*ArcTanh[c*x] + 3*b*Log[1 - Sqrt[c]*Sqrt[ 
x]] - 3*b*Log[1 + Sqrt[c]*Sqrt[x]]))/(15*c^(5/2)*x^(3/2))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6464, 262, 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.

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

\(\Big \downarrow \) 6464

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \int \frac {(d x)^{5/2}}{1-c^2 x^2}dx}{5 d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \left (\frac {d^2 \int \frac {\sqrt {d x}}{1-c^2 x^2}dx}{c^2}-\frac {2 d (d x)^{3/2}}{3 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \left (\frac {2 d \int \frac {d^3 x}{d^2-c^2 d^2 x^2}d\sqrt {d x}}{c^2}-\frac {2 d (d x)^{3/2}}{3 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \left (\frac {2 d^3 \int \frac {d x}{d^2-c^2 d^2 x^2}d\sqrt {d x}}{c^2}-\frac {2 d (d x)^{3/2}}{3 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \left (\frac {2 d^3 \left (\frac {\int \frac {1}{d-c d x}d\sqrt {d x}}{2 c}-\frac {\int \frac {1}{c x d+d}d\sqrt {d x}}{2 c}\right )}{c^2}-\frac {2 d (d x)^{3/2}}{3 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \left (\frac {2 d^3 \left (\frac {\int \frac {1}{d-c d x}d\sqrt {d x}}{2 c}-\frac {\arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}\right )}{c^2}-\frac {2 d (d x)^{3/2}}{3 c^2}\right )}{5 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 (d x)^{5/2} (a+b \text {arctanh}(c x))}{5 d}-\frac {2 b c \left (\frac {2 d^3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}-\frac {\arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{2 c^{3/2} \sqrt {d}}\right )}{c^2}-\frac {2 d (d x)^{3/2}}{3 c^2}\right )}{5 d}\)

Input:

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

Output:

(2*(d*x)^(5/2)*(a + b*ArcTanh[c*x]))/(5*d) - (2*b*c*((-2*d*(d*x)^(3/2))/(3 
*c^2) + (2*d^3*(-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])))/c^2))/(5*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 827

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]
 

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]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+\frac {2 \left (d x \right )^{\frac {5}{2}} b \,\operatorname {arctanh}\left (c x \right )}{5}+\frac {4 b d \left (d x \right )^{\frac {3}{2}}}{15 c}-\frac {2 b \,d^{3} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 c^{2} \sqrt {c d}}+\frac {2 b \,d^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 c^{2} \sqrt {c d}}}{d}\)	\(93\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+\frac {2 \left (d x \right )^{\frac {5}{2}} b \,\operatorname {arctanh}\left (c x \right )}{5}+\frac {4 b d \left (d x \right )^{\frac {3}{2}}}{15 c}-\frac {2 b \,d^{3} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 c^{2} \sqrt {c d}}+\frac {2 b \,d^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 c^{2} \sqrt {c d}}}{d}\)	\(93\)
parts	\(\frac {2 a \left (d x \right )^{\frac {5}{2}}}{5 d}+\frac {2 b \left (d x \right )^{\frac {5}{2}} \operatorname {arctanh}\left (c x \right )}{5 d}+\frac {4 b \left (d x \right )^{\frac {3}{2}}}{15 c}-\frac {2 b \,d^{2} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 c^{2} \sqrt {c d}}+\frac {2 b \,d^{2} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 c^{2} \sqrt {c d}}\)	\(93\)

Input:

int((d*x)^(3/2)*(a+b*arctanh(c*x)),x,method=_RETURNVERBOSE)
 

Output:

2/d*(1/5*(d*x)^(5/2)*a+1/5*(d*x)^(5/2)*b*arctanh(c*x)+2/15*b*d*(d*x)^(3/2) 
/c-1/5*b*d^3/c^2/(c*d)^(1/2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))+1/5*b*d^3/ 
c^2/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.41 \[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\left [\frac {6 \, b d \sqrt {\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {\frac {d}{c}}}{d}\right ) + 3 \, b d \sqrt {\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {\frac {d}{c}} + d}{c x - 1}\right ) + {\left (3 \, b c^{2} d x^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a c^{2} d x^{2} + 4 \, b c d x\right )} \sqrt {d x}}{15 \, c^{2}}, \frac {6 \, b d \sqrt {-\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {-\frac {d}{c}}}{d}\right ) + 3 \, b d \sqrt {-\frac {d}{c}} \log \left (\frac {c d x + 2 \, \sqrt {d x} c \sqrt {-\frac {d}{c}} - d}{c x + 1}\right ) + {\left (3 \, b c^{2} d x^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a c^{2} d x^{2} + 4 \, b c d x\right )} \sqrt {d x}}{15 \, c^{2}}\right ] \] Input:

integrate((d*x)^(3/2)*(a+b*arctanh(c*x)),x, algorithm="fricas")
 

Output:

[1/15*(6*b*d*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) + 3*b*d*sqrt(d/c)*l 
og((c*d*x - 2*sqrt(d*x)*c*sqrt(d/c) + d)/(c*x - 1)) + (3*b*c^2*d*x^2*log(- 
(c*x + 1)/(c*x - 1)) + 6*a*c^2*d*x^2 + 4*b*c*d*x)*sqrt(d*x))/c^2, 1/15*(6* 
b*d*sqrt(-d/c)*arctan(sqrt(d*x)*c*sqrt(-d/c)/d) + 3*b*d*sqrt(-d/c)*log((c* 
d*x + 2*sqrt(d*x)*c*sqrt(-d/c) - d)/(c*x + 1)) + (3*b*c^2*d*x^2*log(-(c*x 
+ 1)/(c*x - 1)) + 6*a*c^2*d*x^2 + 4*b*c*d*x)*sqrt(d*x))/c^2]
 

Sympy [F]

\[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\int \left (d x\right )^{\frac {3}{2}} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )\, dx \] Input:

integrate((d*x)**(3/2)*(a+b*atanh(c*x)),x)
 

Output:

Integral((d*x)**(3/2)*(a + b*atanh(c*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11 \[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\frac {6 \, \left (d x\right )^{\frac {5}{2}} a + {\left (6 \, \left (d x\right )^{\frac {5}{2}} \operatorname {artanh}\left (c x\right ) + \frac {{\left (\frac {4 \, \left (d x\right )^{\frac {3}{2}} d^{2}}{c^{2}} + \frac {6 \, d^{4} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c^{3}} + \frac {3 \, d^{4} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c^{3}}\right )} c}{d}\right )} b}{15 \, d} \] Input:

integrate((d*x)^(3/2)*(a+b*arctanh(c*x)),x, algorithm="maxima")
 

Output:

1/15*(6*(d*x)^(5/2)*a + (6*(d*x)^(5/2)*arctanh(c*x) + (4*(d*x)^(3/2)*d^2/c 
^2 + 6*d^4*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c^3) + 3*d^4*log((sqrt 
(d*x)*c - sqrt(c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c^3))*c/d)*b)/d
 

Giac [F]

\[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )} \,d x } \] Input:

integrate((d*x)^(3/2)*(a+b*arctanh(c*x)),x, algorithm="giac")
 

Output:

integrate((d*x)^(3/2)*(b*arctanh(c*x) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\int \left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{3/2} \,d x \] Input:

int((a + b*atanh(c*x))*(d*x)^(3/2),x)
 

Output:

int((a + b*atanh(c*x))*(d*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.86 \[ \int (d x)^{3/2} (a+b \text {arctanh}(c x)) \, dx=\frac {\sqrt {d}\, d \left (6 \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}}\right ) b +6 \sqrt {c}\, \mathit {atanh} \left (c x \right ) b +6 \sqrt {x}\, \mathit {atanh} \left (c x \right ) b \,c^{3} x^{2}+6 \sqrt {c}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {c}-1\right ) b -3 \sqrt {c}\, \mathrm {log}\left (c x +1\right ) b +6 \sqrt {x}\, a \,c^{3} x^{2}+4 \sqrt {x}\, b \,c^{2} x \right )}{15 c^{3}} \] Input:

int((d*x)^(3/2)*(a+b*atanh(c*x)),x)
 

Output:

(sqrt(d)*d*(6*sqrt(c)*atan((sqrt(x)*c)/sqrt(c))*b + 6*sqrt(c)*atanh(c*x)*b 
 + 6*sqrt(x)*atanh(c*x)*b*c**3*x**2 + 6*sqrt(c)*log(sqrt(x)*sqrt(c) - 1)*b 
 - 3*sqrt(c)*log(c*x + 1)*b + 6*sqrt(x)*a*c**3*x**2 + 4*sqrt(x)*b*c**2*x)) 
/(15*c**3)