3.2.87 \(\int x^3 (a+b \text {arctanh}(c \sqrt {x})) \, dx\) [187]

3.2.87.1 Optimal result

Integrand size = 16, antiderivative size = 88 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\frac {b \sqrt {x}}{4 c^7}+\frac {b x^{3/2}}{12 c^5}+\frac {b x^{5/2}}{20 c^3}+\frac {b x^{7/2}}{28 c}-\frac {b \text {arctanh}\left (c \sqrt {x}\right )}{4 c^8}+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \]

output

1/12*b*x^(3/2)/c^5+1/20*b*x^(5/2)/c^3+1/28*b*x^(7/2)/c-1/4*b*arctanh(c*x^( 
1/2))/c^8+1/4*x^4*(a+b*arctanh(c*x^(1/2)))+1/4*b*x^(1/2)/c^7
 

3.2.87.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\frac {b \sqrt {x}}{4 c^7}+\frac {b x^{3/2}}{12 c^5}+\frac {b x^{5/2}}{20 c^3}+\frac {b x^{7/2}}{28 c}+\frac {a x^4}{4}+\frac {1}{4} b x^4 \text {arctanh}\left (c \sqrt {x}\right )+\frac {b \log \left (1-c \sqrt {x}\right )}{8 c^8}-\frac {b \log \left (1+c \sqrt {x}\right )}{8 c^8} \]

input

Integrate[x^3*(a + b*ArcTanh[c*Sqrt[x]]),x]
 

output

(b*Sqrt[x])/(4*c^7) + (b*x^(3/2))/(12*c^5) + (b*x^(5/2))/(20*c^3) + (b*x^( 
7/2))/(28*c) + (a*x^4)/4 + (b*x^4*ArcTanh[c*Sqrt[x]])/4 + (b*Log[1 - c*Sqr 
t[x]])/(8*c^8) - (b*Log[1 + c*Sqrt[x]])/(8*c^8)
 

3.2.87.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6452, 60, 60, 60, 60, 73, 219}

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.

\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \int \frac {x^{7/2}}{1-c^2 x}dx\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \left (\frac {\int \frac {x^{5/2}}{1-c^2 x}dx}{c^2}-\frac {2 x^{7/2}}{7 c^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \left (\frac {\frac {\int \frac {x^{3/2}}{1-c^2 x}dx}{c^2}-\frac {2 x^{5/2}}{5 c^2}}{c^2}-\frac {2 x^{7/2}}{7 c^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \left (\frac {\frac {\frac {\int \frac {\sqrt {x}}{1-c^2 x}dx}{c^2}-\frac {2 x^{3/2}}{3 c^2}}{c^2}-\frac {2 x^{5/2}}{5 c^2}}{c^2}-\frac {2 x^{7/2}}{7 c^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \left (\frac {\frac {\frac {\frac {\int \frac {1}{\sqrt {x} \left (1-c^2 x\right )}dx}{c^2}-\frac {2 \sqrt {x}}{c^2}}{c^2}-\frac {2 x^{3/2}}{3 c^2}}{c^2}-\frac {2 x^{5/2}}{5 c^2}}{c^2}-\frac {2 x^{7/2}}{7 c^2}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \left (\frac {\frac {\frac {\frac {2 \int \frac {1}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {2 \sqrt {x}}{c^2}}{c^2}-\frac {2 x^{3/2}}{3 c^2}}{c^2}-\frac {2 x^{5/2}}{5 c^2}}{c^2}-\frac {2 x^{7/2}}{7 c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{8} b c \left (\frac {\frac {\frac {\frac {2 \text {arctanh}\left (c \sqrt {x}\right )}{c^3}-\frac {2 \sqrt {x}}{c^2}}{c^2}-\frac {2 x^{3/2}}{3 c^2}}{c^2}-\frac {2 x^{5/2}}{5 c^2}}{c^2}-\frac {2 x^{7/2}}{7 c^2}\right )\)

input

Int[x^3*(a + b*ArcTanh[c*Sqrt[x]]),x]
 

output

(x^4*(a + b*ArcTanh[c*Sqrt[x]]))/4 - (b*c*((-2*x^(7/2))/(7*c^2) + ((-2*x^( 
5/2))/(5*c^2) + ((-2*x^(3/2))/(3*c^2) + ((-2*Sqrt[x])/c^2 + (2*ArcTanh[c*S 
qrt[x]])/c^3)/c^2)/c^2)/c^2))/8
 

3.2.87.3.1 Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

3.2.87.4 Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90

method	result	size
parts	\(\frac {a \,x^{4}}{4}+\frac {2 b \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )}{8}+\frac {c^{7} x^{\frac {7}{2}}}{56}+\frac {c^{5} x^{\frac {5}{2}}}{40}+\frac {c^{3} x^{\frac {3}{2}}}{24}+\frac {c \sqrt {x}}{8}+\frac {\ln \left (c \sqrt {x}-1\right )}{16}-\frac {\ln \left (1+c \sqrt {x}\right )}{16}\right )}{c^{8}}\)	\(79\)
derivativedivides	\(\frac {\frac {a \,c^{8} x^{4}}{4}+2 b \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )}{8}+\frac {c^{7} x^{\frac {7}{2}}}{56}+\frac {c^{5} x^{\frac {5}{2}}}{40}+\frac {c^{3} x^{\frac {3}{2}}}{24}+\frac {c \sqrt {x}}{8}+\frac {\ln \left (c \sqrt {x}-1\right )}{16}-\frac {\ln \left (1+c \sqrt {x}\right )}{16}\right )}{c^{8}}\)	\(83\)
default	\(\frac {\frac {a \,c^{8} x^{4}}{4}+2 b \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )}{8}+\frac {c^{7} x^{\frac {7}{2}}}{56}+\frac {c^{5} x^{\frac {5}{2}}}{40}+\frac {c^{3} x^{\frac {3}{2}}}{24}+\frac {c \sqrt {x}}{8}+\frac {\ln \left (c \sqrt {x}-1\right )}{16}-\frac {\ln \left (1+c \sqrt {x}\right )}{16}\right )}{c^{8}}\)	\(83\)

input

int(x^3*(a+b*arctanh(c*x^(1/2))),x,method=_RETURNVERBOSE)
 

output

1/4*a*x^4+2*b/c^8*(1/8*c^8*x^4*arctanh(c*x^(1/2))+1/56*c^7*x^(7/2)+1/40*c^ 
5*x^(5/2)+1/24*c^3*x^(3/2)+1/8*c*x^(1/2)+1/16*ln(c*x^(1/2)-1)-1/16*ln(1+c* 
x^(1/2)))
 

3.2.87.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.01 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\frac {210 \, a c^{8} x^{4} + 105 \, {\left (b c^{8} x^{4} - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 2 \, {\left (15 \, b c^{7} x^{3} + 21 \, b c^{5} x^{2} + 35 \, b c^{3} x + 105 \, b c\right )} \sqrt {x}}{840 \, c^{8}} \]

input

integrate(x^3*(a+b*arctanh(c*x^(1/2))),x, algorithm="fricas")
 

output

1/840*(210*a*c^8*x^4 + 105*(b*c^8*x^4 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/ 
(c^2*x - 1)) + 2*(15*b*c^7*x^3 + 21*b*c^5*x^2 + 35*b*c^3*x + 105*b*c)*sqrt 
(x))/c^8
 

3.2.87.6 Sympy [F]

\[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\int x^{3} \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )\, dx \]

input

integrate(x**3*(a+b*atanh(c*x**(1/2))),x)
 

output

Integral(x**3*(a + b*atanh(c*sqrt(x))), x)
 

3.2.87.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{840} \, {\left (210 \, x^{4} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (15 \, c^{6} x^{\frac {7}{2}} + 21 \, c^{4} x^{\frac {5}{2}} + 35 \, c^{2} x^{\frac {3}{2}} + 105 \, \sqrt {x}\right )}}{c^{8}} - \frac {105 \, \log \left (c \sqrt {x} + 1\right )}{c^{9}} + \frac {105 \, \log \left (c \sqrt {x} - 1\right )}{c^{9}}\right )}\right )} b \]

input

integrate(x^3*(a+b*arctanh(c*x^(1/2))),x, algorithm="maxima")
 

output

1/4*a*x^4 + 1/840*(210*x^4*arctanh(c*sqrt(x)) + c*(2*(15*c^6*x^(7/2) + 21* 
c^4*x^(5/2) + 35*c^2*x^(3/2) + 105*sqrt(x))/c^8 - 105*log(c*sqrt(x) + 1)/c 
^9 + 105*log(c*sqrt(x) - 1)/c^9))*b
 

3.2.87.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (64) = 128\).

Time = 0.29 (sec) , antiderivative size = 359, normalized size of antiderivative = 4.08 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {2}{105} \, b c {\left (\frac {\frac {105 \, {\left (c \sqrt {x} + 1\right )}^{6}}{{\left (c \sqrt {x} - 1\right )}^{6}} - \frac {315 \, {\left (c \sqrt {x} + 1\right )}^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {770 \, {\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} - \frac {770 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {609 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} - \frac {203 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 44}{c^{9} {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} - 1\right )}^{7}} + \frac {105 \, {\left (\frac {{\left (c \sqrt {x} + 1\right )}^{7}}{{\left (c \sqrt {x} - 1\right )}^{7}} + \frac {7 \, {\left (c \sqrt {x} + 1\right )}^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {7 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )} \log \left (-\frac {\frac {c {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} + 1\right )}}{\frac {{\left (c \sqrt {x} + 1\right )} c}{c \sqrt {x} - 1} - c} + 1}{\frac {c {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} + 1\right )}}{\frac {{\left (c \sqrt {x} + 1\right )} c}{c \sqrt {x} - 1} - c} - 1}\right )}{c^{9} {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} - 1\right )}^{8}}\right )} \]

input

integrate(x^3*(a+b*arctanh(c*x^(1/2))),x, algorithm="giac")
 

output

1/4*a*x^4 + 2/105*b*c*((105*(c*sqrt(x) + 1)^6/(c*sqrt(x) - 1)^6 - 315*(c*s 
qrt(x) + 1)^5/(c*sqrt(x) - 1)^5 + 770*(c*sqrt(x) + 1)^4/(c*sqrt(x) - 1)^4 
- 770*(c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 + 609*(c*sqrt(x) + 1)^2/(c*sqrt( 
x) - 1)^2 - 203*(c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 44)/(c^9*((c*sqrt(x) + 1 
)/(c*sqrt(x) - 1) - 1)^7) + 105*((c*sqrt(x) + 1)^7/(c*sqrt(x) - 1)^7 + 7*( 
c*sqrt(x) + 1)^5/(c*sqrt(x) - 1)^5 + 7*(c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 
 + (c*sqrt(x) + 1)/(c*sqrt(x) - 1))*log(-(c*((c*sqrt(x) + 1)/(c*sqrt(x) - 
1) + 1)/((c*sqrt(x) + 1)*c/(c*sqrt(x) - 1) - c) + 1)/(c*((c*sqrt(x) + 1)/( 
c*sqrt(x) - 1) + 1)/((c*sqrt(x) + 1)*c/(c*sqrt(x) - 1) - c) - 1))/(c^9*((c 
*sqrt(x) + 1)/(c*sqrt(x) - 1) - 1)^8))
 

3.2.87.9 Mupad [B] (verification not implemented)

Time = 3.91 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \, dx=\frac {\frac {b\,c^3\,x^{3/2}}{12}-\frac {b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{4}+\frac {b\,c^5\,x^{5/2}}{20}+\frac {b\,c^7\,x^{7/2}}{28}+\frac {b\,c\,\sqrt {x}}{4}}{c^8}+\frac {b\,\left (105\,x^4\,\ln \left (c\,\sqrt {x}+1\right )-105\,x^4\,\ln \left (1-c\,\sqrt {x}\right )\right )}{840}+\frac {a\,x^4}{4} \]

input

int(x^3*(a + b*atanh(c*x^(1/2))),x)
 

output

((b*c^3*x^(3/2))/12 - (b*atanh(c*x^(1/2)))/4 + (b*c^5*x^(5/2))/20 + (b*c^7 
*x^(7/2))/28 + (b*c*x^(1/2))/4)/c^8 + (b*(105*x^4*log(c*x^(1/2) + 1) - 105 
*x^4*log(1 - c*x^(1/2))))/840 + (a*x^4)/4