3.2.8 \(\int x^7 (a+b \text {ArcTan}(c x^3)) \, dx\) [108]

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

[Out]

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

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Rubi [A]
time = 0.31, 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 = {4946, 327, 301, 648, 632, 210, 642, 209} \begin {gather*} \frac {1}{8} x^8 \left (a+b \text {ArcTan}\left (c x^3\right )\right )+\frac {b \text {ArcTan}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}-\frac {b \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{16 c^{8/3}}+\frac {b \text {ArcTan}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac {\sqrt {3} b \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{32 c^{8/3}}-\frac {3 b x^5}{40 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcTan[c*x^3]),x]

[Out]

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

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 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 301

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 - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(
2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 +
2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(-1)^(m/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] && PosQ[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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[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 \tan ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {1}{8} x^8 \left (a+b \tan ^{-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 \tan ^{-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 \tan ^{-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 {1}{2} \sqrt {3} \sqrt [3]{c} x}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{7/3}}+\frac {b \int \frac {-\frac {1}{2}-\frac {1}{2} \sqrt {3} \sqrt [3]{c} x}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{7/3}}\\ &=-\frac {3 b x^5}{40 c}+\frac {b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tan ^{-1}\left (c x^3\right )\right )+\frac {\left (\sqrt {3} b\right ) \int \frac {-\sqrt {3} \sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{8/3}}-\frac {\left (\sqrt {3} b\right ) \int \frac {\sqrt {3} \sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{8/3}}+\frac {b \int \frac {1}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{7/3}}+\frac {b \int \frac {1}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{32 c^{7/3}}\\ &=-\frac {3 b x^5}{40 c}+\frac {b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tan ^{-1}\left (c x^3\right )\right )+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}+\frac {b \text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 \sqrt {3} c^{8/3}}-\frac {b \text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1+\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{16 \sqrt {3} c^{8/3}}\\ &=-\frac {3 b x^5}{40 c}+\frac {b \tan ^{-1}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} x^8 \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{16 c^{8/3}}+\frac {b \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{c} x\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 181, normalized size = 1.03 \begin {gather*} -\frac {3 b x^5}{40 c}+\frac {a x^8}{8}+\frac {b \text {ArcTan}\left (\sqrt [3]{c} x\right )}{8 c^{8/3}}+\frac {1}{8} b x^8 \text {ArcTan}\left (c x^3\right )-\frac {b \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{16 c^{8/3}}+\frac {b \text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{c} x\right )}{16 c^{8/3}}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{32 c^{8/3}}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \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*ArcTan[c*x^3]),x]

[Out]

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

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Maple [A]
time = 0.10, size = 167, normalized size = 0.95

method	result	size
default	\(\frac {x^{8} a}{8}+\frac {x^{8} b \arctan \left (c \,x^{3}\right )}{8}-\frac {3 b \,x^{5}}{40 c}+\frac {b \sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} \ln \left (x^{2}-\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{32 c}+\frac {b \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{16 c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\frac {b \sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} \ln \left (x^{2}+\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{32 c}+\frac {b \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{16 c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\frac {b \arctan \left (\frac {x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}\right )}{8 c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 152, normalized size = 0.86 \begin {gather*} \frac {1}{8} \, a x^{8} + \frac {1}{160} \, {\left (20 \, x^{8} \arctan \left (c x^{3}\right ) - {\left (\frac {12 \, x^{5}}{c^{2}} + \frac {5 \, {\left (\frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} + \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} - \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {4 \, \arctan \left (c^{\frac {1}{3}} x\right )}{c^{\frac {5}{3}}} - \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x + \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} - \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x - \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )}}{c^{2}}\right )} c\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (128) = 256\).
time = 1.74, size = 439, normalized size = 2.49 \begin {gather*} \frac {20 \, b c x^{8} \arctan \left (c x^{3}\right ) + 20 \, a c x^{8} - 12 \, b x^{5} - 5 \, \sqrt {3} c \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} \log \left (\sqrt {3} b^{5} c^{13} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {5}{6}} + b^{6} c^{10} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {2}{3}} + b^{10} x^{2}\right ) + 5 \, \sqrt {3} c \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} \log \left (-\sqrt {3} b^{5} c^{13} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {5}{6}} + b^{6} c^{10} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {2}{3}} + b^{10} x^{2}\right ) - 20 \, c \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} \arctan \left (-\frac {2 \, b^{5} c^{3} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} + \sqrt {3} b^{6} - 2 \, \sqrt {\sqrt {3} b^{5} c^{13} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {5}{6}} + b^{6} c^{10} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {2}{3}} + b^{10} x^{2}} c^{3} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}}}{b^{6}}\right ) - 20 \, c \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} \arctan \left (-\frac {2 \, b^{5} c^{3} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} - \sqrt {3} b^{6} - 2 \, \sqrt {-\sqrt {3} b^{5} c^{13} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {5}{6}} + b^{6} c^{10} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {2}{3}} + b^{10} x^{2}} c^{3} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}}}{b^{6}}\right ) - 40 \, c \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} \arctan \left (-\frac {b^{5} c^{3} x \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}} - \sqrt {b^{6} c^{10} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {2}{3}} + b^{10} x^{2}} c^{3} \left (\frac {b^{6}}{c^{16}}\right )^{\frac {1}{6}}}{b^{6}}\right )}{160 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*(20*b*c*x^8*arctan(c*x^3) + 20*a*c*x^8 - 12*b*x^5 - 5*sqrt(3)*c*(b^6/c^16)^(1/6)*log(sqrt(3)*b^5*c^13*x*
(b^6/c^16)^(5/6) + b^6*c^10*(b^6/c^16)^(2/3) + b^10*x^2) + 5*sqrt(3)*c*(b^6/c^16)^(1/6)*log(-sqrt(3)*b^5*c^13*
x*(b^6/c^16)^(5/6) + b^6*c^10*(b^6/c^16)^(2/3) + b^10*x^2) - 20*c*(b^6/c^16)^(1/6)*arctan(-(2*b^5*c^3*x*(b^6/c
^16)^(1/6) + sqrt(3)*b^6 - 2*sqrt(sqrt(3)*b^5*c^13*x*(b^6/c^16)^(5/6) + b^6*c^10*(b^6/c^16)^(2/3) + b^10*x^2)*
c^3*(b^6/c^16)^(1/6))/b^6) - 20*c*(b^6/c^16)^(1/6)*arctan(-(2*b^5*c^3*x*(b^6/c^16)^(1/6) - sqrt(3)*b^6 - 2*sqr
t(-sqrt(3)*b^5*c^13*x*(b^6/c^16)^(5/6) + b^6*c^10*(b^6/c^16)^(2/3) + b^10*x^2)*c^3*(b^6/c^16)^(1/6))/b^6) - 40
*c*(b^6/c^16)^(1/6)*arctan(-(b^5*c^3*x*(b^6/c^16)^(1/6) - sqrt(b^6*c^10*(b^6/c^16)^(2/3) + b^10*x^2)*c^3*(b^6/
c^16)^(1/6))/b^6))/c

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Sympy [A]
time = 61.13, size = 264, normalized size = 1.50 \begin {gather*} \begin {cases} \frac {a x^{8}}{8} + \frac {b x^{8} \operatorname {atan}{\left (c x^{3} \right )}}{8} - \frac {3 b x^{5}}{40 c} + \frac {3 b \log {\left (4 x^{2} - 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{32 c^{3} \sqrt [6]{- \frac {1}{c^{2}}}} - \frac {3 b \log {\left (4 x^{2} + 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{32 c^{3} \sqrt [6]{- \frac {1}{c^{2}}}} + \frac {\sqrt {3} b \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [6]{- \frac {1}{c^{2}}}} - \frac {\sqrt {3}}{3} \right )}}{16 c^{3} \sqrt [6]{- \frac {1}{c^{2}}}} + \frac {\sqrt {3} b \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [6]{- \frac {1}{c^{2}}}} + \frac {\sqrt {3}}{3} \right )}}{16 c^{3} \sqrt [6]{- \frac {1}{c^{2}}}} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{8 c^{4} \left (- \frac {1}{c^{2}}\right )^{\frac {2}{3}}} & \text {for}\: c \neq 0 \\\frac {a x^{8}}{8} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**8/8 + b*x**8*atan(c*x**3)/8 - 3*b*x**5/(40*c) + 3*b*log(4*x**2 - 4*x*(-1/c**2)**(1/6) + 4*(-1/
c**2)**(1/3))/(32*c**3*(-1/c**2)**(1/6)) - 3*b*log(4*x**2 + 4*x*(-1/c**2)**(1/6) + 4*(-1/c**2)**(1/3))/(32*c**
3*(-1/c**2)**(1/6)) + sqrt(3)*b*atan(2*sqrt(3)*x/(3*(-1/c**2)**(1/6)) - sqrt(3)/3)/(16*c**3*(-1/c**2)**(1/6))
+ sqrt(3)*b*atan(2*sqrt(3)*x/(3*(-1/c**2)**(1/6)) + sqrt(3)/3)/(16*c**3*(-1/c**2)**(1/6)) - b*atan(c*x**3)/(8*
c**4*(-1/c**2)**(2/3)), Ne(c, 0)), (a*x**8/8, True))

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Giac [A]
time = 0.49, size = 171, normalized size = 0.97 \begin {gather*} -\frac {1}{32} \, b c^{15} {\left (\frac {\sqrt {3} \log \left (x^{2} + \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{16} {\left | c \right |}^{\frac {5}{3}}} - \frac {\sqrt {3} \log \left (x^{2} - \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{c^{16} {\left | c \right |}^{\frac {5}{3}}} - \frac {2 \, {\left | c \right |}^{\frac {1}{3}} \arctan \left ({\left (2 \, x + \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{18}} - \frac {2 \, {\left | c \right |}^{\frac {1}{3}} \arctan \left ({\left (2 \, x - \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{c^{18}} - \frac {4 \, {\left | c \right |}^{\frac {1}{3}} \arctan \left (x {\left | c \right |}^{\frac {1}{3}}\right )}{c^{18}}\right )} + \frac {5 \, b c x^{8} \arctan \left (c x^{3}\right ) + 5 \, a c x^{8} - 3 \, b x^{5}}{40 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*b*c^15*(sqrt(3)*log(x^2 + sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/(c^16*abs(c)^(5/3)) - sqrt(3)*log(x^2
- sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/(c^16*abs(c)^(5/3)) - 2*abs(c)^(1/3)*arctan((2*x + sqrt(3)/abs(c)^
(1/3))*abs(c)^(1/3))/c^18 - 2*abs(c)^(1/3)*arctan((2*x - sqrt(3)/abs(c)^(1/3))*abs(c)^(1/3))/c^18 - 4*abs(c)^(
1/3)*arctan(x*abs(c)^(1/3))/c^18) + 1/40*(5*b*c*x^8*arctan(c*x^3) + 5*a*c*x^8 - 3*b*x^5)/c

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Mupad [B]
time = 1.00, size = 122, normalized size = 0.69 \begin {gather*} \frac {a\,x^8}{8}-\frac {3\,b\,x^5}{40\,c}-\frac {b\,\left (\mathrm {atan}\left ({\left (-1\right )}^{2/3}\,c^{1/3}\,x\right )+\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )+2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\right )}{16\,c^{8/3}}+\frac {b\,x^8\,\mathrm {atan}\left (c\,x^3\right )}{8}+\frac {\sqrt {3}\,b\,\left (\mathrm {atan}\left ({\left (-1\right )}^{2/3}\,c^{1/3}\,x\right )-\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,c^{1/3}\,x\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\right )\,1{}\mathrm {i}}{16\,c^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*atan(c*x^3)),x)

[Out]

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

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