3.1.58 \(\int x^4 (a+b \tanh ^{-1}(c x^2)) \, dx\) [58]

Optimal. Leaf size=65 \[ \frac {2 b x^3}{15 c}+\frac {b \text {ArcTan}\left (\sqrt {c} x\right )}{5 c^{5/2}}-\frac {b \tanh ^{-1}\left (\sqrt {c} x\right )}{5 c^{5/2}}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6037, 327, 304, 209, 212} \begin {gather*} \frac {1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {b \text {ArcTan}\left (\sqrt {c} x\right )}{5 c^{5/2}}-\frac {b \tanh ^{-1}\left (\sqrt {c} x\right )}{5 c^{5/2}}+\frac {2 b x^3}{15 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x^2]),x]

[Out]

(2*b*x^3)/(15*c) + (b*ArcTan[Sqrt[c]*x])/(5*c^(5/2)) - (b*ArcTanh[Sqrt[c]*x])/(5*c^(5/2)) + (x^5*(a + b*ArcTan
h[c*x^2]))/5

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 327

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

Rule 6037

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] - Dist[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
]

Rubi steps

\begin {align*} \int x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {1}{5} (2 b c) \int \frac {x^6}{1-c^2 x^4} \, dx\\ &=\frac {2 b x^3}{15 c}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {(2 b) \int \frac {x^2}{1-c^2 x^4} \, dx}{5 c}\\ &=\frac {2 b x^3}{15 c}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {b \int \frac {1}{1-c x^2} \, dx}{5 c^2}+\frac {b \int \frac {1}{1+c x^2} \, dx}{5 c^2}\\ &=\frac {2 b x^3}{15 c}+\frac {b \tan ^{-1}\left (\sqrt {c} x\right )}{5 c^{5/2}}-\frac {b \tanh ^{-1}\left (\sqrt {c} x\right )}{5 c^{5/2}}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 93, normalized size = 1.43 \begin {gather*} \frac {2 b x^3}{15 c}+\frac {a x^5}{5}+\frac {b \text {ArcTan}\left (\sqrt {c} x\right )}{5 c^{5/2}}+\frac {1}{5} b x^5 \tanh ^{-1}\left (c x^2\right )+\frac {b \log \left (1-\sqrt {c} x\right )}{10 c^{5/2}}-\frac {b \log \left (1+\sqrt {c} x\right )}{10 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcTanh[c*x^2]),x]

[Out]

(2*b*x^3)/(15*c) + (a*x^5)/5 + (b*ArcTan[Sqrt[c]*x])/(5*c^(5/2)) + (b*x^5*ArcTanh[c*x^2])/5 + (b*Log[1 - Sqrt[
c]*x])/(10*c^(5/2)) - (b*Log[1 + Sqrt[c]*x])/(10*c^(5/2))

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 53, normalized size = 0.82

method	result	size
default	\(\frac {a \,x^{5}}{5}+\frac {x^{5} b \arctanh \left (c \,x^{2}\right )}{5}+\frac {2 b \,x^{3}}{15 c}-\frac {b \arctanh \left (x \sqrt {c}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctan \left (x \sqrt {c}\right )}{5 c^{\frac {5}{2}}}\)	\(53\)
risch	\(\frac {x^{5} b \ln \left (c \,x^{2}+1\right )}{10}-\frac {b \,x^{5} \ln \left (-c \,x^{2}+1\right )}{10}+\frac {a \,x^{5}}{5}+\frac {2 b \,x^{3}}{15 c}+\frac {b \ln \left (1-x \sqrt {c}\right )}{10 c^{\frac {5}{2}}}-\frac {b \ln \left (1+x \sqrt {c}\right )}{10 c^{\frac {5}{2}}}+\frac {\sqrt {-c}\, \ln \left (-\sqrt {-c}\, c -x \,c^{2}\right ) b}{10 c^{3}}-\frac {\sqrt {-c}\, \ln \left (-\sqrt {-c}\, c +x \,c^{2}\right ) b}{10 c^{3}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x^2)),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.46, size = 69, normalized size = 1.06 \begin {gather*} \frac {1}{5} \, a x^{5} + \frac {1}{30} \, {\left (6 \, x^{5} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {4 \, x^{3}}{c^{2}} + \frac {6 \, \arctan \left (\sqrt {c} x\right )}{c^{\frac {7}{2}}} + \frac {3 \, \log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{c^{\frac {7}{2}}}\right )}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x^2)),x, algorithm="maxima")

[Out]

1/5*a*x^5 + 1/30*(6*x^5*arctanh(c*x^2) + c*(4*x^3/c^2 + 6*arctan(sqrt(c)*x)/c^(7/2) + 3*log((c*x - sqrt(c))/(c
*x + sqrt(c)))/c^(7/2)))*b

________________________________________________________________________________________

Fricas [A] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (49) = 98\).
time = 0.36, size = 197, normalized size = 3.03 \begin {gather*} \left [\frac {3 \, b c^{3} x^{5} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c^{3} x^{5} + 4 \, b c^{2} x^{3} + 6 \, b \sqrt {c} \arctan \left (\sqrt {c} x\right ) + 3 \, b \sqrt {c} \log \left (\frac {c x^{2} - 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right )}{30 \, c^{3}}, \frac {3 \, b c^{3} x^{5} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c^{3} x^{5} + 4 \, b c^{2} x^{3} + 6 \, b \sqrt {-c} \arctan \left (\sqrt {-c} x\right ) - 3 \, b \sqrt {-c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right )}{30 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x^2)),x, algorithm="fricas")

[Out]

[1/30*(3*b*c^3*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*a*c^3*x^5 + 4*b*c^2*x^3 + 6*b*sqrt(c)*arctan(sqrt(c)*x) +
3*b*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)))/c^3, 1/30*(3*b*c^3*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1))
+ 6*a*c^3*x^5 + 4*b*c^2*x^3 + 6*b*sqrt(-c)*arctan(sqrt(-c)*x) - 3*b*sqrt(-c)*log((c*x^2 - 2*sqrt(-c)*x - 1)/(c
*x^2 + 1)))/c^3]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (58) = 116\).
time = 5.80, size = 185, normalized size = 2.85 \begin {gather*} \begin {cases} \frac {a x^{5}}{5} + \frac {b x^{5} \operatorname {atanh}{\left (c x^{2} \right )}}{5} + \frac {2 b x^{3}}{15 c} - \frac {b \sqrt {- \frac {1}{c}} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{10 c^{2}} + \frac {b \sqrt {- \frac {1}{c}} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{10 c^{2}} - \frac {b \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{10 c^{3} \sqrt {\frac {1}{c}}} - \frac {b \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{10 c^{3} \sqrt {\frac {1}{c}}} + \frac {b \log {\left (x - \sqrt {\frac {1}{c}} \right )}}{5 c^{3} \sqrt {\frac {1}{c}}} + \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{5 c^{3} \sqrt {\frac {1}{c}}} & \text {for}\: c \neq 0 \\\frac {a x^{5}}{5} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x**2)),x)

[Out]

Piecewise((a*x**5/5 + b*x**5*atanh(c*x**2)/5 + 2*b*x**3/(15*c) - b*sqrt(-1/c)*log(x - sqrt(-1/c))/(10*c**2) +
b*sqrt(-1/c)*log(x + sqrt(-1/c))/(10*c**2) - b*log(x - sqrt(-1/c))/(10*c**3*sqrt(1/c)) - b*log(x + sqrt(-1/c))
/(10*c**3*sqrt(1/c)) + b*log(x - sqrt(1/c))/(5*c**3*sqrt(1/c)) + b*atanh(c*x**2)/(5*c**3*sqrt(1/c)), Ne(c, 0))
, (a*x**5/5, True))

________________________________________________________________________________________

Giac [A]
time = 0.48, size = 73, normalized size = 1.12 \begin {gather*} \frac {1}{10} \, b x^{5} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {1}{5} \, a x^{5} + \frac {2 \, b x^{3}}{15 \, c} + \frac {b \arctan \left (\sqrt {c} x\right )}{5 \, c^{\frac {5}{2}}} + \frac {b \arctan \left (\frac {c x}{\sqrt {-c}}\right )}{5 \, \sqrt {-c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x^2)),x, algorithm="giac")

[Out]

1/10*b*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 1/5*a*x^5 + 2/15*b*x^3/c + 1/5*b*arctan(sqrt(c)*x)/c^(5/2) + 1/5*b*
arctan(c*x/sqrt(-c))/(sqrt(-c)*c^2)

________________________________________________________________________________________

Mupad [B]
time = 0.98, size = 72, normalized size = 1.11 \begin {gather*} \frac {a\,x^5}{5}+\frac {2\,b\,x^3}{15\,c}+\frac {b\,\mathrm {atan}\left (\sqrt {c}\,x\right )}{5\,c^{5/2}}+\frac {b\,x^5\,\ln \left (c\,x^2+1\right )}{10}-\frac {b\,x^5\,\ln \left (1-c\,x^2\right )}{10}+\frac {b\,\mathrm {atan}\left (\sqrt {c}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{5\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*atanh(c*x^2)),x)

[Out]

(a*x^5)/5 + (2*b*x^3)/(15*c) + (b*atan(c^(1/2)*x))/(5*c^(5/2)) + (b*atan(c^(1/2)*x*1i)*1i)/(5*c^(5/2)) + (b*x^
5*log(c*x^2 + 1))/10 - (b*x^5*log(1 - c*x^2))/10

________________________________________________________________________________________