3.68 \(\int x^4 (a+b \tan ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=161 \[ \frac{1}{5} x^5 \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{b \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2} c^{5/2}}+\frac{b \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right )}{5 \sqrt{2} c^{5/2}}-\frac{2 b x^3}{15 c} \]

[Out]

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

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Rubi [A]  time = 0.112558, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5033, 321, 297, 1162, 617, 204, 1165, 628} \[ \frac{1}{5} x^5 \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{b \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2} c^{5/2}}+\frac{b \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right )}{5 \sqrt{2} c^{5/2}}-\frac{2 b x^3}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTan[c*x^2]),x]

[Out]

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

Rule 5033

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 321

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 297

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

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]

Rubi steps

\begin{align*} \int x^4 \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \tan ^{-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 \tan ^{-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 \tan ^{-1}\left (c x^2\right )\right )-\frac{b \int \frac{1-c x^2}{1+c^2 x^4} \, dx}{5 c^2}+\frac{b \int \frac{1+c x^2}{1+c^2 x^4} \, dx}{5 c^2}\\ &=-\frac{2 b x^3}{15 c}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{b \int \frac{1}{\frac{1}{c}-\frac{\sqrt{2} x}{\sqrt{c}}+x^2} \, dx}{10 c^3}+\frac{b \int \frac{1}{\frac{1}{c}+\frac{\sqrt{2} x}{\sqrt{c}}+x^2} \, dx}{10 c^3}+\frac{b \int \frac{\frac{\sqrt{2}}{\sqrt{c}}+2 x}{-\frac{1}{c}-\frac{\sqrt{2} x}{\sqrt{c}}-x^2} \, dx}{10 \sqrt{2} c^{5/2}}+\frac{b \int \frac{\frac{\sqrt{2}}{\sqrt{c}}-2 x}{-\frac{1}{c}+\frac{\sqrt{2} x}{\sqrt{c}}-x^2} \, dx}{10 \sqrt{2} c^{5/2}}\\ &=-\frac{2 b x^3}{15 c}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac{b \log \left (1-\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \log \left (1+\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2} c^{5/2}}+\frac{b \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2} c^{5/2}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2} c^{5/2}}\\ &=-\frac{2 b x^3}{15 c}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac{b \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2} c^{5/2}}+\frac{b \tan ^{-1}\left (1+\sqrt{2} \sqrt{c} x\right )}{5 \sqrt{2} c^{5/2}}+\frac{b \log \left (1-\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \log \left (1+\sqrt{2} \sqrt{c} x+c x^2\right )}{10 \sqrt{2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0482874, size = 179, normalized size = 1.11 \[ \frac{a x^5}{5}+\frac{b \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2} c^{5/2}}-\frac{b \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{10 \sqrt{2} c^{5/2}}+\frac{b \tan ^{-1}\left (\frac{2 \sqrt{c} x-\sqrt{2}}{\sqrt{2}}\right )}{5 \sqrt{2} c^{5/2}}+\frac{b \tan ^{-1}\left (\frac{2 \sqrt{c} x+\sqrt{2}}{\sqrt{2}}\right )}{5 \sqrt{2} c^{5/2}}-\frac{2 b x^3}{15 c}+\frac{1}{5} b x^5 \tan ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 140, normalized size = 0.9 \begin{align*}{\frac{a{x}^{5}}{5}}+{\frac{b{x}^{5}\arctan \left ( c{x}^{2} \right ) }{5}}-{\frac{2\,b{x}^{3}}{15\,c}}+{\frac{b\sqrt{2}}{20\,{c}^{3}}\ln \left ({ \left ({x}^{2}-\sqrt [4]{{c}^{-2}}x\sqrt{2}+\sqrt{{c}^{-2}} \right ) \left ({x}^{2}+\sqrt [4]{{c}^{-2}}x\sqrt{2}+\sqrt{{c}^{-2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{c}^{-2}}}}}+{\frac{b\sqrt{2}}{10\,{c}^{3}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{c}^{-2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{c}^{-2}}}}}+{\frac{b\sqrt{2}}{10\,{c}^{3}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{c}^{-2}}}}}-1 \right ){\frac{1}{\sqrt [4]{{c}^{-2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctan(c*x^2)),x)

[Out]

1/5*a*x^5+1/5*b*x^5*arctan(c*x^2)-2/15*b*x^3/c+1/20*b/c^3/(1/c^2)^(1/4)*2^(1/2)*ln((x^2-(1/c^2)^(1/4)*x*2^(1/2
)+(1/c^2)^(1/2))/(x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2)))+1/10*b/c^3/(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/
(1/c^2)^(1/4)*x+1)+1/10*b/c^3/(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*x-1)

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Maxima [B]  time = 1.52734, size = 375, normalized size = 2.33 \begin{align*} \frac{1}{5} \, a x^{5} + \frac{1}{60} \,{\left (12 \, x^{5} \arctan \left (c x^{2}\right ) - c{\left (\frac{8 \, x^{3}}{c^{2}} + \frac{3 \,{\left (\frac{\sqrt{2} \log \left (\sqrt{c^{2}} x^{2} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{3}{4}}} - \frac{\sqrt{2} \log \left (\sqrt{c^{2}} x^{2} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{3}{4}}} - \frac{\sqrt{2} \log \left (\frac{2 \, \sqrt{c^{2}} x - \sqrt{2} \sqrt{-\sqrt{c^{2}}} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}{2 \, \sqrt{c^{2}} x + \sqrt{2} \sqrt{-\sqrt{c^{2}}} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{c^{2}} \sqrt{-\sqrt{c^{2}}}} - \frac{\sqrt{2} \log \left (\frac{2 \, \sqrt{c^{2}} x - \sqrt{2} \sqrt{-\sqrt{c^{2}}} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}{2 \, \sqrt{c^{2}} x + \sqrt{2} \sqrt{-\sqrt{c^{2}}} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}}}\right )}{\sqrt{c^{2}} \sqrt{-\sqrt{c^{2}}}}\right )}}{c^{2}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5 + 1/60*(12*x^5*arctan(c*x^2) - c*(8*x^3/c^2 + 3*(sqrt(2)*log(sqrt(c^2)*x^2 + sqrt(2)*(c^2)^(1/4)*x +
1)/(c^2)^(3/4) - sqrt(2)*log(sqrt(c^2)*x^2 - sqrt(2)*(c^2)^(1/4)*x + 1)/(c^2)^(3/4) - sqrt(2)*log((2*sqrt(c^2
)*x - sqrt(2)*sqrt(-sqrt(c^2)) + sqrt(2)*(c^2)^(1/4))/(2*sqrt(c^2)*x + sqrt(2)*sqrt(-sqrt(c^2)) + sqrt(2)*(c^2
)^(1/4)))/(sqrt(c^2)*sqrt(-sqrt(c^2))) - sqrt(2)*log((2*sqrt(c^2)*x - sqrt(2)*sqrt(-sqrt(c^2)) - sqrt(2)*(c^2)
^(1/4))/(2*sqrt(c^2)*x + sqrt(2)*sqrt(-sqrt(c^2)) - sqrt(2)*(c^2)^(1/4)))/(sqrt(c^2)*sqrt(-sqrt(c^2))))/c^2))*
b

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Fricas [B]  time = 2.85234, size = 894, normalized size = 5.55 \begin{align*} \frac{12 \, b c x^{5} \arctan \left (c x^{2}\right ) + 12 \, a c x^{5} - 8 \, b x^{3} - 12 \, \sqrt{2} c \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2} b^{3} c^{3} x \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} - \sqrt{2} \sqrt{\sqrt{2} b^{3} c^{7} x \left (\frac{b^{4}}{c^{10}}\right )^{\frac{3}{4}} + b^{4} c^{4} \sqrt{\frac{b^{4}}{c^{10}}} + b^{6} x^{2}} c^{3} \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} + b^{4}}{b^{4}}\right ) - 12 \, \sqrt{2} c \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2} b^{3} c^{3} x \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} - \sqrt{2} \sqrt{-\sqrt{2} b^{3} c^{7} x \left (\frac{b^{4}}{c^{10}}\right )^{\frac{3}{4}} + b^{4} c^{4} \sqrt{\frac{b^{4}}{c^{10}}} + b^{6} x^{2}} c^{3} \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} - b^{4}}{b^{4}}\right ) - 3 \, \sqrt{2} c \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} \log \left (\sqrt{2} b^{3} c^{7} x \left (\frac{b^{4}}{c^{10}}\right )^{\frac{3}{4}} + b^{4} c^{4} \sqrt{\frac{b^{4}}{c^{10}}} + b^{6} x^{2}\right ) + 3 \, \sqrt{2} c \left (\frac{b^{4}}{c^{10}}\right )^{\frac{1}{4}} \log \left (-\sqrt{2} b^{3} c^{7} x \left (\frac{b^{4}}{c^{10}}\right )^{\frac{3}{4}} + b^{4} c^{4} \sqrt{\frac{b^{4}}{c^{10}}} + b^{6} x^{2}\right )}{60 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*b*c*x^5*arctan(c*x^2) + 12*a*c*x^5 - 8*b*x^3 - 12*sqrt(2)*c*(b^4/c^10)^(1/4)*arctan(-(sqrt(2)*b^3*c^3
*x*(b^4/c^10)^(1/4) - sqrt(2)*sqrt(sqrt(2)*b^3*c^7*x*(b^4/c^10)^(3/4) + b^4*c^4*sqrt(b^4/c^10) + b^6*x^2)*c^3*
(b^4/c^10)^(1/4) + b^4)/b^4) - 12*sqrt(2)*c*(b^4/c^10)^(1/4)*arctan(-(sqrt(2)*b^3*c^3*x*(b^4/c^10)^(1/4) - sqr
t(2)*sqrt(-sqrt(2)*b^3*c^7*x*(b^4/c^10)^(3/4) + b^4*c^4*sqrt(b^4/c^10) + b^6*x^2)*c^3*(b^4/c^10)^(1/4) - b^4)/
b^4) - 3*sqrt(2)*c*(b^4/c^10)^(1/4)*log(sqrt(2)*b^3*c^7*x*(b^4/c^10)^(3/4) + b^4*c^4*sqrt(b^4/c^10) + b^6*x^2)
+ 3*sqrt(2)*c*(b^4/c^10)^(1/4)*log(-sqrt(2)*b^3*c^7*x*(b^4/c^10)^(3/4) + b^4*c^4*sqrt(b^4/c^10) + b^6*x^2))/c

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Sympy [A]  time = 55.2367, size = 184, normalized size = 1.14 \begin{align*} \begin{cases} \frac{a x^{5}}{5} + \frac{b x^{5} \operatorname{atan}{\left (c x^{2} \right )}}{5} - \frac{2 b x^{3}}{15 c} - \frac{\sqrt [4]{-1} b \operatorname{atan}{\left (c x^{2} \right )}}{5 c^{8} \left (\frac{1}{c^{2}}\right )^{\frac{11}{4}}} - \frac{\left (-1\right )^{\frac{3}{4}} b \log{\left (x - \sqrt [4]{-1} \sqrt [4]{\frac{1}{c^{2}}} \right )}}{5 c^{13} \left (\frac{1}{c^{2}}\right )^{\frac{21}{4}}} + \frac{\left (-1\right )^{\frac{3}{4}} b \log{\left (x^{2} + i \sqrt{\frac{1}{c^{2}}} \right )}}{10 c^{13} \left (\frac{1}{c^{2}}\right )^{\frac{21}{4}}} + \frac{\left (-1\right )^{\frac{3}{4}} b \operatorname{atan}{\left (\frac{\left (-1\right )^{\frac{3}{4}} x}{\sqrt [4]{\frac{1}{c^{2}}}} \right )}}{5 c^{13} \left (\frac{1}{c^{2}}\right )^{\frac{21}{4}}} & \text{for}\: c \neq 0 \\\frac{a x^{5}}{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**5/5 + b*x**5*atan(c*x**2)/5 - 2*b*x**3/(15*c) - (-1)**(1/4)*b*atan(c*x**2)/(5*c**8*(c**(-2))**
(11/4)) - (-1)**(3/4)*b*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(5*c**13*(c**(-2))**(21/4)) + (-1)**(3/4)*b*log(
x**2 + I*sqrt(c**(-2)))/(10*c**13*(c**(-2))**(21/4)) + (-1)**(3/4)*b*atan((-1)**(3/4)*x/(c**(-2))**(1/4))/(5*c
**13*(c**(-2))**(21/4)), Ne(c, 0)), (a*x**5/5, True))

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Giac [A]  time = 1.30486, size = 228, normalized size = 1.42 \begin{align*} \frac{1}{20} \, b c^{9}{\left (\frac{2 \, \sqrt{2} \sqrt{{\left | c \right |}} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \frac{\sqrt{2}}{\sqrt{{\left | c \right |}}}\right )} \sqrt{{\left | c \right |}}\right )}{c^{12}} + \frac{2 \, \sqrt{2} \sqrt{{\left | c \right |}} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \frac{\sqrt{2}}{\sqrt{{\left | c \right |}}}\right )} \sqrt{{\left | c \right |}}\right )}{c^{12}} - \frac{\sqrt{2} \sqrt{{\left | c \right |}} \log \left (x^{2} + \frac{\sqrt{2} x}{\sqrt{{\left | c \right |}}} + \frac{1}{{\left | c \right |}}\right )}{c^{12}} + \frac{\sqrt{2} \sqrt{{\left | c \right |}} \log \left (x^{2} - \frac{\sqrt{2} x}{\sqrt{{\left | c \right |}}} + \frac{1}{{\left | c \right |}}\right )}{c^{12}}\right )} + \frac{3 \, b c x^{5} \arctan \left (c x^{2}\right ) + 3 \, a c x^{5} - 2 \, b x^{3}}{15 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*b*c^9*(2*sqrt(2)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^12 + 2*sqrt
(2)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^12 - sqrt(2)*sqrt(abs(c))*log
(x^2 + sqrt(2)*x/sqrt(abs(c)) + 1/abs(c))/c^12 + sqrt(2)*sqrt(abs(c))*log(x^2 - sqrt(2)*x/sqrt(abs(c)) + 1/abs
(c))/c^12) + 1/15*(3*b*c*x^5*arctan(c*x^2) + 3*a*c*x^5 - 2*b*x^3)/c