\(\int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx\) [40]

Optimal result

Integrand size = 16, antiderivative size = 107 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=-\frac {4 b c}{3 d^2 \sqrt {d x}}-\frac {2 b c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {2 (a+b \text {arctanh}(c x))}{3 d (d x)^{3/2}}+\frac {2 b c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 d^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=-\frac {x \left (2 a+4 b c x+2 b c^{3/2} x^{3/2} \arctan \left (\sqrt {c} \sqrt {x}\right )+2 b \text {arctanh}(c x)+b c^{3/2} x^{3/2} \log \left (1-\sqrt {c} \sqrt {x}\right )-b c^{3/2} x^{3/2} \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{3 (d x)^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6464, 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.

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

\(\Big \downarrow \) 6464

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-2*(a + b*ArcTanh[c*x]))/(3*d*(d*x)^(3/2)) + (2*b*c*(-2/(d*Sqrt[d*x]) + ( 
2*c^2*(-1/2*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]/(c^(3/2)*Sqrt[d]) + ArcTan 
h[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]/(2*c^(3/2)*Sqrt[d])))/d))/(3*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 264

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]
 

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 = 0.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 b \,c^{2} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 d \sqrt {c d}}-\frac {2 b \,c^{2} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 d \sqrt {c d}}-\frac {4 b c}{3 d \sqrt {d x}}}{d}\)	\(93\)
default	\(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {2 b \,c^{2} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 d \sqrt {c d}}-\frac {2 b \,c^{2} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 d \sqrt {c d}}-\frac {4 b c}{3 d \sqrt {d x}}}{d}\)	\(93\)
parts	\(-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}} d}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{3 d \left (d x \right )^{\frac {3}{2}}}+\frac {2 b \,c^{2} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 d^{2} \sqrt {c d}}-\frac {2 b \,c^{2} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 d^{2} \sqrt {c d}}-\frac {4 b c}{3 d^{2} \sqrt {d x}}\)	\(94\)

Input:

int((a+b*arctanh(c*x))/(d*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.16 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=\left [-\frac {2 \, b c d x^{2} \sqrt {\frac {c}{d}} \arctan \left (\sqrt {d x} \sqrt {\frac {c}{d}}\right ) - b c d x^{2} \sqrt {\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {\frac {c}{d}} + 1}{c x - 1}\right ) + {\left (4 \, b c x + b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a\right )} \sqrt {d x}}{3 \, d^{3} x^{2}}, -\frac {2 \, b c d x^{2} \sqrt {-\frac {c}{d}} \arctan \left (\sqrt {d x} \sqrt {-\frac {c}{d}}\right ) - b c d x^{2} \sqrt {-\frac {c}{d}} \log \left (\frac {c x - 2 \, \sqrt {d x} \sqrt {-\frac {c}{d}} - 1}{c x + 1}\right ) + {\left (4 \, b c x + b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a\right )} \sqrt {d x}}{3 \, d^{3} x^{2}}\right ] \] Input:

integrate((a+b*arctanh(c*x))/(d*x)^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=\int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\left (d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+b*atanh(c*x))/(d*x)**(5/2),x)
 

Output:

Integral((a + b*atanh(c*x))/(d*x)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=-\frac {b {\left (\frac {{\left (\frac {2 \, c \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d}} + \frac {c \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d}} + \frac {4}{\sqrt {d x}}\right )} c}{d} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{\left (d x\right )^{\frac {3}{2}}}\right )} + \frac {2 \, a}{\left (d x\right )^{\frac {3}{2}}}}{3 \, d} \] Input:

integrate((a+b*arctanh(c*x))/(d*x)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=-\frac {\frac {2 \, b c^{2} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d} + \frac {2 \, b c^{2} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} d} + \frac {b \log \left (-\frac {c d x + d}{c d x - d}\right )}{\sqrt {d x} d x} + \frac {2 \, {\left (2 \, b c d x + a d\right )}}{\sqrt {d x} d^{2} x}}{3 \, d} \] Input:

integrate((a+b*arctanh(c*x))/(d*x)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{{\left (d\,x\right )}^{5/2}} \,d x \] Input:

int((a + b*atanh(c*x))/(d*x)^(5/2),x)
 

Output:

int((a + b*atanh(c*x))/(d*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{5/2}} \, dx=\frac {\sqrt {d}\, \left (-2 \sqrt {x}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}}\right ) b c x -2 \sqrt {x}\, \sqrt {c}\, \mathit {atanh} \left (c x \right ) b c x -2 \mathit {atanh} \left (c x \right ) b -2 \sqrt {x}\, \sqrt {c}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {c}-1\right ) b c x +\sqrt {x}\, \sqrt {c}\, \mathrm {log}\left (c x +1\right ) b c x -2 a -4 b c x \right )}{3 \sqrt {x}\, d^{3} x} \] Input:

int((a+b*atanh(c*x))/(d*x)^(5/2),x)
 

Output:

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