\(\int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx\) [41]

Optimal result

Integrand size = 16, antiderivative size = 107 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=-\frac {4 b c}{15 d^2 (d x)^{3/2}}+\frac {2 b c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}-\frac {2 (a+b \text {arctanh}(c x))}{5 d (d x)^{5/2}}+\frac {2 b c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}} \] Output:

-4/15*b*c/d^2/(d*x)^(3/2)+2/5*b*c^(5/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2) 
)/d^(7/2)-2/5*(a+b*arctanh(c*x))/d/(d*x)^(5/2)+2/5*b*c^(5/2)*arctanh(c^(1/ 
2)*(d*x)^(1/2)/d^(1/2))/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\frac {x \left (-6 a-4 b c x+6 b c^{5/2} x^{5/2} \arctan \left (\sqrt {c} \sqrt {x}\right )-6 b \text {arctanh}(c x)-3 b c^{5/2} x^{5/2} \log \left (1-\sqrt {c} \sqrt {x}\right )+3 b c^{5/2} x^{5/2} \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{15 (d x)^{7/2}} \] Input:

Integrate[(a + b*ArcTanh[c*x])/(d*x)^(7/2),x]
 

Output:

(x*(-6*a - 4*b*c*x + 6*b*c^(5/2)*x^(5/2)*ArcTan[Sqrt[c]*Sqrt[x]] - 6*b*Arc 
Tanh[c*x] - 3*b*c^(5/2)*x^(5/2)*Log[1 - Sqrt[c]*Sqrt[x]] + 3*b*c^(5/2)*x^( 
5/2)*Log[1 + Sqrt[c]*Sqrt[x]]))/(15*(d*x)^(7/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6464, 264, 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.

\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx\)

\(\Big \downarrow \) 6464

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[(a + b*ArcTanh[c*x])/(d*x)^(7/2),x]
 

Output:

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

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 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 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]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{5 \left (d x \right )^{\frac {5}{2}}}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {2 b \,c^{3} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{2} \sqrt {c d}}-\frac {4 b c}{15 d \left (d x \right )^{\frac {3}{2}}}+\frac {2 b \,c^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{2} \sqrt {c d}}}{d}\)	\(93\)
default	\(\frac {-\frac {2 a}{5 \left (d x \right )^{\frac {5}{2}}}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{5 \left (d x \right )^{\frac {5}{2}}}+\frac {2 b \,c^{3} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{2} \sqrt {c d}}-\frac {4 b c}{15 d \left (d x \right )^{\frac {3}{2}}}+\frac {2 b \,c^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{2} \sqrt {c d}}}{d}\)	\(93\)
parts	\(-\frac {2 a}{5 \left (d x \right )^{\frac {5}{2}} d}-\frac {2 b \,\operatorname {arctanh}\left (c x \right )}{5 d \left (d x \right )^{\frac {5}{2}}}+\frac {2 b \,c^{3} \operatorname {arctanh}\left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{3} \sqrt {c d}}+\frac {2 b \,c^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{3} \sqrt {c d}}-\frac {4 b c}{15 d^{2} \left (d x \right )^{\frac {3}{2}}}\)	\(94\)

Input:

int((a+b*arctanh(c*x))/(d*x)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/d*(-1/5*a/(d*x)^(5/2)-1/5*b/(d*x)^(5/2)*arctanh(c*x)+1/5*b/d^2*c^3/(c*d) 
^(1/2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))-2/15*b/d*c/(d*x)^(3/2)+1/5*b/d^2 
*c^3/(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 = 242, normalized size of antiderivative = 2.26 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\left [\frac {6 \, b c^{2} d x^{3} \sqrt {\frac {c}{d}} \arctan \left (\sqrt {d x} \sqrt {\frac {c}{d}}\right ) + 3 \, b c^{2} d x^{3} \sqrt {\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {\frac {c}{d}} + 1}{c x - 1}\right ) - {\left (4 \, b c x + 3 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a\right )} \sqrt {d x}}{15 \, d^{4} x^{3}}, -\frac {6 \, b c^{2} d x^{3} \sqrt {-\frac {c}{d}} \arctan \left (\sqrt {d x} \sqrt {-\frac {c}{d}}\right ) - 3 \, b c^{2} d x^{3} \sqrt {-\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {-\frac {c}{d}} - 1}{c x + 1}\right ) + {\left (4 \, b c x + 3 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a\right )} \sqrt {d x}}{15 \, d^{4} x^{3}}\right ] \] Input:

integrate((a+b*arctanh(c*x))/(d*x)^(7/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\left (d x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((a+b*atanh(c*x))/(d*x)**(7/2),x)
 

Output:

Integral((a + b*atanh(c*x))/(d*x)**(7/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\frac {b {\left (\frac {{\left (\frac {6 \, c^{2} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d} - \frac {3 \, c^{2} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} d} - \frac {4}{\left (d x\right )^{\frac {3}{2}}}\right )} c}{d} - \frac {6 \, \operatorname {artanh}\left (c x\right )}{\left (d x\right )^{\frac {5}{2}}}\right )} - \frac {6 \, a}{\left (d x\right )^{\frac {5}{2}}}}{15 \, d} \] Input:

integrate((a+b*arctanh(c*x))/(d*x)^(7/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\frac {6 \, b c^{3} {\left (\frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d^{2}} - \frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} d^{2}}\right )} - \frac {3 \, b \log \left (-\frac {c d x + d}{c d x - d}\right )}{\sqrt {d x} d^{2} x^{2}} - \frac {2 \, {\left (2 \, b c d x + 3 \, a d\right )}}{\sqrt {d x} d^{3} x^{2}}}{15 \, d} \] Input:

integrate((a+b*arctanh(c*x))/(d*x)^(7/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{{\left (d\,x\right )}^{7/2}} \,d x \] Input:

int((a + b*atanh(c*x))/(d*x)^(7/2),x)
 

Output:

int((a + b*atanh(c*x))/(d*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \text {arctanh}(c x)}{(d x)^{7/2}} \, dx=\frac {\sqrt {d}\, \left (6 \sqrt {x}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}}\right ) b \,c^{2} x^{2}-6 \sqrt {x}\, \sqrt {c}\, \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}-6 \mathit {atanh} \left (c x \right ) b -6 \sqrt {x}\, \sqrt {c}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {c}-1\right ) b \,c^{2} x^{2}+3 \sqrt {x}\, \sqrt {c}\, \mathrm {log}\left (c x +1\right ) b \,c^{2} x^{2}-6 a -4 b c x \right )}{15 \sqrt {x}\, d^{4} x^{2}} \] Input:

int((a+b*atanh(c*x))/(d*x)^(7/2),x)
 

Output:

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