3.2.11 \(\int x^7 (a+b \tanh ^{-1}(c x^3)) \, dx\) [111]

Optimal. Leaf size=176 \[ \frac {3 b x^5}{40 c}-\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}} \]

[Out]

3/40*b*x^5/c-1/8*b*arctanh(c^(1/3)*x)/c^(8/3)+1/8*x^8*(a+b*arctanh(c*x^3))+1/32*b*ln(1-c^(1/3)*x+c^(2/3)*x^2)/
c^(8/3)-1/32*b*ln(1+c^(1/3)*x+c^(2/3)*x^2)/c^(8/3)+1/16*b*arctan(-1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)/c
^(8/3)+1/16*b*arctan(1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)/c^(8/3)

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Rubi [A]
time = 0.20, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6037, 327, 302, 648, 632, 210, 642, 212} \begin {gather*} \frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{16 c^{8/3}}+\frac {b \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac {b \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {3 b x^5}{40 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcTanh[c*x^3]),x]

[Out]

(3*b*x^5)/(40*c) - (Sqrt[3]*b*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (Sqrt[3]*b*ArcTan[1/Sq
rt[3] + (2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) - (b*ArcTanh[c^(1/3)*x])/(8*c^(8/3)) + (x^8*(a + b*ArcTanh[c*x^3]
))/8 + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3)) - (b*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 302

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-a/b, n]], s = Denominator[Rt[-
a/b, n]], k, u}, Simp[u = Int[(r*Cos[2*k*m*(Pi/n)] - s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[2*k*(Pi/n)]
*x + s^2*x^2), x] + Int[(r*Cos[2*k*m*(Pi/n)] + s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[2*k*(Pi/n)]*x + s
^2*x^2), x]; 2*(r^(m + 2)/(a*n*s^m))*Int[1/(r^2 - s^2*x^2), x] + Dist[2*(r^(m + 1)/(a*n*s^m)), Sum[u, {k, 1, (
n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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^7 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {1}{8} (3 b c) \int \frac {x^{10}}{1-c^2 x^6} \, dx\\ &=\frac {3 b x^5}{40 c}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {(3 b) \int \frac {x^4}{1-c^2 x^6} \, dx}{8 c}\\ &=\frac {3 b x^5}{40 c}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {b \int \frac {1}{1-c^{2/3} x^2} \, dx}{8 c^{7/3}}-\frac {b \int \frac {-\frac {1}{2}-\frac {\sqrt [3]{c} x}{2}}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{7/3}}-\frac {b \int \frac {-\frac {1}{2}+\frac {\sqrt [3]{c} x}{2}}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{7/3}}\\ &=\frac {3 b x^5}{40 c}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \int \frac {-\sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{8/3}}-\frac {b \int \frac {\sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{8/3}}+\frac {(3 b) \int \frac {1}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{7/3}}+\frac {(3 b) \int \frac {1}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{7/3}}\\ &=\frac {3 b x^5}{40 c}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{c} x\right )}{16 c^{8/3}}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{c} x\right )}{16 c^{8/3}}\\ &=\frac {3 b x^5}{40 c}-\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \tan ^{-1}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}-\frac {b \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 198, normalized size = 1.12 \begin {gather*} \frac {3 b x^5}{40 c}+\frac {a x^8}{8}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {-1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 c^{8/3}}+\frac {1}{8} b x^8 \tanh ^{-1}\left (c x^3\right )+\frac {b \log \left (1-\sqrt [3]{c} x\right )}{16 c^{8/3}}-\frac {b \log \left (1+\sqrt [3]{c} x\right )}{16 c^{8/3}}+\frac {b \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {b \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*ArcTanh[c*x^3]),x]

[Out]

(3*b*x^5)/(40*c) + (a*x^8)/8 + (Sqrt[3]*b*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (Sqrt[3]*b*ArcTan
[(1 + 2*c^(1/3)*x)/Sqrt[3]])/(16*c^(8/3)) + (b*x^8*ArcTanh[c*x^3])/8 + (b*Log[1 - c^(1/3)*x])/(16*c^(8/3)) - (
b*Log[1 + c^(1/3)*x])/(16*c^(8/3)) + (b*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(32*c^(8/3)) - (b*Log[1 + c^(1/3)*x
+ c^(2/3)*x^2])/(32*c^(8/3))

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Maple [A]
time = 0.06, size = 186, normalized size = 1.06

method	result	size
default	\(\frac {x^{8} a}{8}+\frac {x^{8} b \arctanh \left (c \,x^{3}\right )}{8}+\frac {3 b \,x^{5}}{40 c}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{32 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{32 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}\)	\(186\)
risch	\(\frac {x^{8} b \ln \left (c \,x^{3}+1\right )}{16}+\frac {x^{8} a}{8}-\frac {b \,x^{8} \ln \left (-c \,x^{3}+1\right )}{16}+\frac {3 b \,x^{5}}{40 c}+\frac {b \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{32 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{32 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{16 c^{3} \left (\frac {1}{c}\right )^{\frac {1}{3}}}\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a+b*arctanh(c*x^3)),x,method=_RETURNVERBOSE)

[Out]

1/8*x^8*a+1/8*x^8*b*arctanh(c*x^3)+3/40*b*x^5/c-1/16*b/c^3/(1/c)^(1/3)*ln(x+(1/c)^(1/3))+1/32*b/c^3/(1/c)^(1/3
)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+1/16*b/c^3*3^(1/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x-1))+1/1
6*b/c^3/(1/c)^(1/3)*ln(x-(1/c)^(1/3))-1/32*b/c^3/(1/c)^(1/3)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))+1/16*b/c^3*3^(1
/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))

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Maxima [A]
time = 0.47, size = 164, normalized size = 0.93 \begin {gather*} \frac {1}{8} \, a x^{8} + \frac {1}{160} \, {\left (20 \, x^{8} \operatorname {artanh}\left (c x^{3}\right ) + {\left (\frac {12 \, x^{5}}{c^{2}} + \frac {10 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {11}{3}}} + \frac {10 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {11}{3}}} - \frac {5 \, \log \left (c^{\frac {2}{3}} x^{2} + c^{\frac {1}{3}} x + 1\right )}{c^{\frac {11}{3}}} + \frac {5 \, \log \left (c^{\frac {2}{3}} x^{2} - c^{\frac {1}{3}} x + 1\right )}{c^{\frac {11}{3}}} - \frac {10 \, \log \left (\frac {c^{\frac {1}{3}} x + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {11}{3}}} + \frac {10 \, \log \left (\frac {c^{\frac {1}{3}} x - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {11}{3}}}\right )} c\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

1/8*a*x^8 + 1/160*(20*x^8*arctanh(c*x^3) + (12*x^5/c^2 + 10*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*x + c^(1/3))
/c^(1/3))/c^(11/3) + 10*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*x - c^(1/3))/c^(1/3))/c^(11/3) - 5*log(c^(2/3)*x
^2 + c^(1/3)*x + 1)/c^(11/3) + 5*log(c^(2/3)*x^2 - c^(1/3)*x + 1)/c^(11/3) - 10*log((c^(1/3)*x + 1)/c^(1/3))/c
^(11/3) + 10*log((c^(1/3)*x - 1)/c^(1/3))/c^(11/3))*c)*b

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Fricas [A]
time = 0.36, size = 248, normalized size = 1.41 \begin {gather*} \frac {10 \, b c^{4} x^{8} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 20 \, a c^{4} x^{8} + 12 \, b c^{3} x^{5} + 10 \, \sqrt {3} b c \sqrt {-\left (-c^{2}\right )^{\frac {1}{3}}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c x + \left (-c^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\left (-c^{2}\right )^{\frac {1}{3}}}}{3 \, c}\right ) + 10 \, \sqrt {3} b {\left (c^{2}\right )}^{\frac {1}{6}} c \arctan \left (\frac {\sqrt {3} {\left (c^{2}\right )}^{\frac {1}{6}} {\left (2 \, c x + {\left (c^{2}\right )}^{\frac {1}{3}}\right )}}{3 \, c}\right ) + 5 \, \left (-c^{2}\right )^{\frac {2}{3}} b \log \left (c^{2} x^{2} + \left (-c^{2}\right )^{\frac {1}{3}} c x + \left (-c^{2}\right )^{\frac {2}{3}}\right ) - 5 \, b {\left (c^{2}\right )}^{\frac {2}{3}} \log \left (c^{2} x^{2} + {\left (c^{2}\right )}^{\frac {1}{3}} c x + {\left (c^{2}\right )}^{\frac {2}{3}}\right ) - 10 \, \left (-c^{2}\right )^{\frac {2}{3}} b \log \left (c x - \left (-c^{2}\right )^{\frac {1}{3}}\right ) + 10 \, b {\left (c^{2}\right )}^{\frac {2}{3}} \log \left (c x - {\left (c^{2}\right )}^{\frac {1}{3}}\right )}{160 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

1/160*(10*b*c^4*x^8*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 20*a*c^4*x^8 + 12*b*c^3*x^5 + 10*sqrt(3)*b*c*sqrt(-(-c^2)^
(1/3))*arctan(1/3*sqrt(3)*(2*c*x + (-c^2)^(1/3))*sqrt(-(-c^2)^(1/3))/c) + 10*sqrt(3)*b*(c^2)^(1/6)*c*arctan(1/
3*sqrt(3)*(c^2)^(1/6)*(2*c*x + (c^2)^(1/3))/c) + 5*(-c^2)^(2/3)*b*log(c^2*x^2 + (-c^2)^(1/3)*c*x + (-c^2)^(2/3
)) - 5*b*(c^2)^(2/3)*log(c^2*x^2 + (c^2)^(1/3)*c*x + (c^2)^(2/3)) - 10*(-c^2)^(2/3)*b*log(c*x - (-c^2)^(1/3))
+ 10*b*(c^2)^(2/3)*log(c*x - (c^2)^(1/3)))/c^4

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(a+b*atanh(c*x**3)),x)

[Out]

Timed out

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Giac [A]
time = 0.48, size = 208, normalized size = 1.18 \begin {gather*} \frac {1}{16} \, b x^{8} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + \frac {1}{8} \, a x^{8} + \frac {3 \, b x^{5}}{40 \, c} - \frac {b \left (-\frac {1}{c}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {1}{c}\right )^{\frac {1}{3}} \right |}\right )}{16 \, c^{2}} + \frac {\sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {1}{c}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {1}{c}\right )^{\frac {1}{3}}}\right )}{16 \, \left (-c^{2}\right )^{\frac {1}{3}} c^{2}} + \frac {\sqrt {3} b {\left | c \right |}^{\frac {4}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} c^{\frac {1}{3}} {\left (2 \, x + \frac {1}{c^{\frac {1}{3}}}\right )}\right )}{16 \, c^{4}} - \frac {b \log \left (x^{2} + x \left (-\frac {1}{c}\right )^{\frac {1}{3}} + \left (-\frac {1}{c}\right )^{\frac {2}{3}}\right )}{32 \, \left (-c^{2}\right )^{\frac {1}{3}} c^{2}} - \frac {b \log \left (x^{2} + \frac {x}{c^{\frac {1}{3}}} + \frac {1}{c^{\frac {2}{3}}}\right )}{32 \, c^{2} {\left | c \right |}^{\frac {2}{3}}} + \frac {b \log \left ({\left | x - \frac {1}{c^{\frac {1}{3}}} \right |}\right )}{16 \, c^{\frac {8}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

1/16*b*x^8*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/8*a*x^8 + 3/40*b*x^5/c - 1/16*b*(-1/c)^(2/3)*log(abs(x - (-1/c)^(
1/3)))/c^2 + 1/16*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*x + (-1/c)^(1/3))/(-1/c)^(1/3))/((-c^2)^(1/3)*c^2) + 1/16*sq
rt(3)*b*abs(c)^(4/3)*arctan(1/3*sqrt(3)*c^(1/3)*(2*x + 1/c^(1/3)))/c^4 - 1/32*b*log(x^2 + x*(-1/c)^(1/3) + (-1
/c)^(2/3))/((-c^2)^(1/3)*c^2) - 1/32*b*log(x^2 + x/c^(1/3) + 1/c^(2/3))/(c^2*abs(c)^(2/3)) + 1/16*b*log(abs(x
- 1/c^(1/3)))/c^(8/3)

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Mupad [B]
time = 1.26, size = 127, normalized size = 0.72 \begin {gather*} \frac {a\,x^8}{8}+\frac {b\,\left (-\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )}{2}+\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )}{2}+\mathrm {atan}\left (c^{1/3}\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{8\,c^{8/3}}+\frac {3\,b\,x^5}{40\,c}+\frac {b\,x^8\,\ln \left (c\,x^3+1\right )}{16}-\frac {b\,x^8\,\ln \left (1-c\,x^3\right )}{16}+\frac {\sqrt {3}\,b\,\left (\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )+\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )\right )}{16\,c^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*atanh(c*x^3)),x)

[Out]

(a*x^8)/8 + (b*(atan((c^(1/3)*x*(3^(1/2) + 1i))/2)/2 - atan((c^(1/3)*x*(3^(1/2) - 1i))/2)/2 + atan(c^(1/3)*x*1
i))*1i)/(8*c^(8/3)) + (3*b*x^5)/(40*c) + (b*x^8*log(c*x^3 + 1))/16 - (b*x^8*log(1 - c*x^3))/16 + (3^(1/2)*b*(a
tan((c^(1/3)*x*(3^(1/2) - 1i))/2) + atan((c^(1/3)*x*(3^(1/2) + 1i))/2)))/(16*c^(8/3))

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