3.22 \(\int (d+e x) (a+b \tan ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=192 \[ \frac {(d+e x)^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{2 e}-\frac {b e \log \left (c^2 x^4+1\right )}{4 c}-\frac {b d^2 \tan ^{-1}\left (c x^2\right )}{2 e}-\frac {b d \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b d \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b d \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b d \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}} \]

[Out]

-1/2*b*d^2*arctan(c*x^2)/e+1/2*(e*x+d)^2*(a+b*arctan(c*x^2))/e-1/4*b*e*ln(c^2*x^4+1)/c-1/2*b*d*arctan(-1+x*2^(
1/2)*c^(1/2))*2^(1/2)/c^(1/2)-1/2*b*d*arctan(1+x*2^(1/2)*c^(1/2))*2^(1/2)/c^(1/2)-1/4*b*d*ln(1+c*x^2-x*2^(1/2)
*c^(1/2))*2^(1/2)/c^(1/2)+1/4*b*d*ln(1+c*x^2+x*2^(1/2)*c^(1/2))*2^(1/2)/c^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 191, normalized size of antiderivative = 0.99, number of steps used = 16, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6742, 5027, 297, 1162, 617, 204, 1165, 628, 5033, 260} \[ \frac {a (d+e x)^2}{2 e}-\frac {b e \log \left (c^2 x^4+1\right )}{4 c}-\frac {b d \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b d \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+b d x \tan ^{-1}\left (c x^2\right )+\frac {b d \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b d \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}+\frac {1}{2} b e x^2 \tan ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcTan[c*x^2]),x]

[Out]

(a*(d + e*x)^2)/(2*e) + b*d*x*ArcTan[c*x^2] + (b*e*x^2*ArcTan[c*x^2])/2 + (b*d*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/
(Sqrt[2]*Sqrt[c]) - (b*d*ArcTan[1 + Sqrt[2]*Sqrt[c]*x])/(Sqrt[2]*Sqrt[c]) - (b*d*Log[1 - Sqrt[2]*Sqrt[c]*x + c
*x^2])/(2*Sqrt[2]*Sqrt[c]) + (b*d*Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2])/(2*Sqrt[2]*Sqrt[c]) - (b*e*Log[1 + c^2*x
^4])/(4*c)

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 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 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]

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 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 5027

Int[ArcTan[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTan[c*x^n], x] - Dist[c*n, Int[x^n/(1 + c^2*x^(2*n)), x],
x] /; FreeQ[{c, n}, x]

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 153, normalized size = 0.80 \[ a d x+\frac {1}{2} a e x^2-\frac {b e \log \left (c^2 x^4+1\right )}{4 c}+b d x \tan ^{-1}\left (c x^2\right )-\frac {b d \left (\log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )-\log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )-2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )+2 \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )\right )}{2 \sqrt {2} \sqrt {c}}+\frac {1}{2} b e x^2 \tan ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (a*e*x^2)/2 + b*d*x*ArcTan[c*x^2] + (b*e*x^2*ArcTan[c*x^2])/2 - (b*d*(-2*ArcTan[1 - Sqrt[2]*Sqrt[c]*x]
+ 2*ArcTan[1 + Sqrt[2]*Sqrt[c]*x] + Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2] - Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2]))
/(2*Sqrt[2]*Sqrt[c]) - (b*e*Log[1 + c^2*x^4])/(4*c)

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fricas [B]  time = 0.49, size = 534, normalized size = 2.78 \[ \frac {2 \, a b^{4} c d^{4} e x^{2} + 4 \, a b^{4} c d^{5} x + 4 \, \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {1}{4}} b^{4} c d^{4} \arctan \left (-\frac {b^{8} d^{8} + \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {5}{4}} b^{3} c^{3} d^{3} x - \sqrt {2} \sqrt {b^{6} d^{6} x^{2} + \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c d^{3} x + \sqrt {\frac {b^{4} d^{4}}{c^{2}}} b^{4} d^{4}} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {5}{4}} c^{3}}{b^{8} d^{8}}\right ) + 4 \, \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {1}{4}} b^{4} c d^{4} \arctan \left (\frac {b^{8} d^{8} - \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {5}{4}} b^{3} c^{3} d^{3} x + \sqrt {2} \sqrt {b^{6} d^{6} x^{2} - \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c d^{3} x + \sqrt {\frac {b^{4} d^{4}}{c^{2}}} b^{4} d^{4}} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {5}{4}} c^{3}}{b^{8} d^{8}}\right ) + 2 \, {\left (b^{5} c d^{4} e x^{2} + 2 \, b^{5} c d^{5} x\right )} \arctan \left (c x^{2}\right ) - {\left (b^{5} d^{4} e - \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {1}{4}} b^{4} c d^{4}\right )} \log \left (b^{6} d^{6} x^{2} + \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c d^{3} x + \sqrt {\frac {b^{4} d^{4}}{c^{2}}} b^{4} d^{4}\right ) - {\left (b^{5} d^{4} e + \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {1}{4}} b^{4} c d^{4}\right )} \log \left (b^{6} d^{6} x^{2} - \sqrt {2} \left (\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c d^{3} x + \sqrt {\frac {b^{4} d^{4}}{c^{2}}} b^{4} d^{4}\right )}{4 \, b^{4} c d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*b^4*c*d^4*e*x^2 + 4*a*b^4*c*d^5*x + 4*sqrt(2)*(b^4*d^4/c^2)^(1/4)*b^4*c*d^4*arctan(-(b^8*d^8 + sqrt(2
)*(b^4*d^4/c^2)^(5/4)*b^3*c^3*d^3*x - sqrt(2)*sqrt(b^6*d^6*x^2 + sqrt(2)*(b^4*d^4/c^2)^(3/4)*b^3*c*d^3*x + sqr
t(b^4*d^4/c^2)*b^4*d^4)*(b^4*d^4/c^2)^(5/4)*c^3)/(b^8*d^8)) + 4*sqrt(2)*(b^4*d^4/c^2)^(1/4)*b^4*c*d^4*arctan((
b^8*d^8 - sqrt(2)*(b^4*d^4/c^2)^(5/4)*b^3*c^3*d^3*x + sqrt(2)*sqrt(b^6*d^6*x^2 - sqrt(2)*(b^4*d^4/c^2)^(3/4)*b
^3*c*d^3*x + sqrt(b^4*d^4/c^2)*b^4*d^4)*(b^4*d^4/c^2)^(5/4)*c^3)/(b^8*d^8)) + 2*(b^5*c*d^4*e*x^2 + 2*b^5*c*d^5
*x)*arctan(c*x^2) - (b^5*d^4*e - sqrt(2)*(b^4*d^4/c^2)^(1/4)*b^4*c*d^4)*log(b^6*d^6*x^2 + sqrt(2)*(b^4*d^4/c^2
)^(3/4)*b^3*c*d^3*x + sqrt(b^4*d^4/c^2)*b^4*d^4) - (b^5*d^4*e + sqrt(2)*(b^4*d^4/c^2)^(1/4)*b^4*c*d^4)*log(b^6
*d^6*x^2 - sqrt(2)*(b^4*d^4/c^2)^(3/4)*b^3*c*d^3*x + sqrt(b^4*d^4/c^2)*b^4*d^4))/(b^4*c*d^4)

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giac [A]  time = 5.10, size = 201, normalized size = 1.05 \[ -\frac {1}{4} \, b c^{3} d {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | c \right |}} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{4}} + \frac {2 \, \sqrt {2} \sqrt {{\left | c \right |}} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{4}} - \frac {\sqrt {2} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{2} {\left | c \right |}^{\frac {3}{2}}} + \frac {\sqrt {2} \sqrt {{\left | c \right |}} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{4}}\right )} + \frac {2 \, b c x^{2} \arctan \left (c x^{2}\right ) e + 4 \, b c d x \arctan \left (c x^{2}\right ) + 2 \, a c x^{2} e + 4 \, a c d x - b e \log \left (c^{2} x^{4} + 1\right )}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*c^3*d*(2*sqrt(2)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^4 + 2*sqr
t(2)*sqrt(abs(c))*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt(abs(c)))*sqrt(abs(c)))/c^4 - sqrt(2)*log(x^2 + sqrt(2
)*x/sqrt(abs(c)) + 1/abs(c))/(c^2*abs(c)^(3/2)) + sqrt(2)*sqrt(abs(c))*log(x^2 - sqrt(2)*x/sqrt(abs(c)) + 1/ab
s(c))/c^4) + 1/4*(2*b*c*x^2*arctan(c*x^2)*e + 4*b*c*d*x*arctan(c*x^2) + 2*a*c*x^2*e + 4*a*c*d*x - b*e*log(c^2*
x^4 + 1))/c

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maple [A]  time = 0.03, size = 167, normalized size = 0.87 \[ \frac {a \,x^{2} e}{2}+a d x +\frac {b \arctan \left (c \,x^{2}\right ) x^{2} e}{2}+b \arctan \left (c \,x^{2}\right ) d x -\frac {b d \sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b d \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b d \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b e \ln \left (c^{2} x^{4}+1\right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*e+a*d*x+1/2*b*arctan(c*x^2)*x^2*e+b*arctan(c*x^2)*d*x-1/4*b*d/c/(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)))-1/2*b*d/c/(1/c^2)^(1/4)*2^(1/2)*
arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)-1/2*b*d/c/(1/c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/c^2)^(1/4)*x-1)-1/4*b*e*ln
(c^2*x^4+1)/c

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maxima [A]  time = 0.41, size = 168, normalized size = 0.88 \[ \frac {1}{2} \, a e x^{2} - \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}}\right )} - 4 \, x \arctan \left (c x^{2}\right )\right )} b d + a d x + \frac {{\left (2 \, c x^{2} \arctan \left (c x^{2}\right ) - \log \left (c^{2} x^{4} + 1\right )\right )} b e}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*e*x^2 - 1/4*(c*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqrt(c))/c^(3/2) + 2*sqrt(2)*arct
an(1/2*sqrt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqrt(c))/c^(3/2) - sqrt(2)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1)/c^(3/2)
+ sqrt(2)*log(c*x^2 - sqrt(2)*sqrt(c)*x + 1)/c^(3/2)) - 4*x*arctan(c*x^2))*b*d + a*d*x + 1/4*(2*c*x^2*arctan(
c*x^2) - log(c^2*x^4 + 1))*b*e/c

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mupad [B]  time = 2.53, size = 203, normalized size = 1.06 \[ a\,d\,x+\frac {a\,e\,x^2}{2}+b\,d\,x\,\mathrm {atan}\left (c\,x^2\right )-\frac {b\,e\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}-1\right )}{4\,c}-\frac {b\,e\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}+1\right )}{4\,c}-\frac {b\,e\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}-1\right )}{4\,c}-\frac {b\,e\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}+1\right )}{4\,c}+\frac {b\,e\,x^2\,\mathrm {atan}\left (c\,x^2\right )}{2}-\frac {b\,d\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}-1\right )\,\sqrt {-c\,1{}\mathrm {i}}}{2\,c}+\frac {b\,d\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}+1\right )\,\sqrt {-c\,1{}\mathrm {i}}}{2\,c}-\frac {b\,d\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}-1\right )\,\sqrt {c\,1{}\mathrm {i}}}{2\,c}+\frac {b\,d\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}+1\right )\,\sqrt {c\,1{}\mathrm {i}}}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))*(d + e*x),x)

[Out]

a*d*x + (a*e*x^2)/2 + b*d*x*atan(c*x^2) - (b*e*log(x*(-c*1i)^(1/2) - 1))/(4*c) - (b*e*log(x*(-c*1i)^(1/2) + 1)
)/(4*c) - (b*e*log(x*(c*1i)^(1/2) - 1))/(4*c) - (b*e*log(x*(c*1i)^(1/2) + 1))/(4*c) + (b*e*x^2*atan(c*x^2))/2
- (b*d*log(x*(-c*1i)^(1/2) - 1)*(-c*1i)^(1/2))/(2*c) + (b*d*log(x*(-c*1i)^(1/2) + 1)*(-c*1i)^(1/2))/(2*c) - (b
*d*log(x*(c*1i)^(1/2) - 1)*(c*1i)^(1/2))/(2*c) + (b*d*log(x*(c*1i)^(1/2) + 1)*(c*1i)^(1/2))/(2*c)

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sympy [A]  time = 18.15, size = 1734, normalized size = 9.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a - b*atan((-sqrt(2)/2 - sqrt(2)*I/2)**(-2)))*(d*x + e*x**2/2), Eq(c, -1/(x**2*(-sqrt(2)/2 - sqrt(
2)*I/2)**2))), ((a - b*atan((-sqrt(2)/2 + sqrt(2)*I/2)**(-2)))*(d*x + e*x**2/2), Eq(c, -1/(x**2*(-sqrt(2)/2 +
sqrt(2)*I/2)**2))), ((a - b*atan((sqrt(2)/2 - sqrt(2)*I/2)**(-2)))*(d*x + e*x**2/2), Eq(c, -1/(x**2*(sqrt(2)/2
- sqrt(2)*I/2)**2))), ((a - b*atan((sqrt(2)/2 + sqrt(2)*I/2)**(-2)))*(d*x + e*x**2/2), Eq(c, -1/(x**2*(sqrt(2
)/2 + sqrt(2)*I/2)**2))), (a*(d*x + e*x**2/2), Eq(c, 0)), (2*(-1)**(3/4)*a*c**5*d*x**5*(c**(-2))**(11/4)/(2*(-
1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + (-1)**(3/4)*a*c**5*e*x**6*(c**
(-2))**(11/4)/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + 2*(-1)**(3/
4)*a*c**3*d*x*(c**(-2))**(11/4)/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11
/4)) + (-1)**(3/4)*a*c**3*e*x**2*(c**(-2))**(11/4)/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*
c**3*(c**(-2))**(11/4)) + 2*(-1)**(3/4)*b*c**5*d*x**5*(c**(-2))**(11/4)*atan(c*x**2)/(2*(-1)**(3/4)*c**5*x**4*
(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + (-1)**(3/4)*b*c**5*e*x**6*(c**(-2))**(11/4)*atan(c
*x**2)/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) - 2*I*b*c**4*d*x**4*
(c**(-2))**(5/2)*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3
/4)*c**3*(c**(-2))**(11/4)) + I*b*c**4*d*x**4*(c**(-2))**(5/2)*log(x**2 + I*sqrt(c**(-2)))/(2*(-1)**(3/4)*c**5
*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + 2*I*b*c**4*d*x**4*(c**(-2))**(5/2)*atan((-1)
**(3/4)*x/(c**(-2))**(1/4))/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4))
- (-1)**(3/4)*b*c**4*e*x**4*(c**(-2))**(11/4)*log(x**2 + I*sqrt(c**(-2)))/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))*
*(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + 2*(-1)**(3/4)*b*c**3*d*x*(c**(-2))**(11/4)*atan(c*x**2)/(2*(
-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + (-1)**(3/4)*b*c**3*e*x**2*(c*
*(-2))**(11/4)*atan(c*x**2)/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4))
- 2*I*b*c**2*d*(c**(-2))**(5/2)*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11
/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + I*b*c**2*d*(c**(-2))**(5/2)*log(x**2 + I*sqrt(c**(-2)))/(2*(-1)*
*(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) + 2*I*b*c**2*d*(c**(-2))**(5/2)*ata
n((-1)**(3/4)*x/(c**(-2))**(1/4))/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(
11/4)) - (-1)**(3/4)*b*c**2*e*(c**(-2))**(11/4)*log(x**2 + I*sqrt(c**(-2)))/(2*(-1)**(3/4)*c**5*x**4*(c**(-2))
**(11/4) + 2*(-1)**(3/4)*c**3*(c**(-2))**(11/4)) - 2*b*d*x**4*atan(c*x**2)/(2*(-1)**(3/4)*c**6*x**4*(c**(-2))*
*(11/4) + 2*(-1)**(3/4)*c**4*(c**(-2))**(11/4)) - 2*b*d*atan(c*x**2)/(2*(-1)**(3/4)*c**8*x**4*(c**(-2))**(11/4
) + 2*(-1)**(3/4)*c**6*(c**(-2))**(11/4)) + (-1)**(1/4)*b*e*x**4*(c**(-2))**(1/4)*atan(c*x**2)/(2*(-1)**(3/4)*
c**6*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**4*(c**(-2))**(11/4)) + (-1)**(1/4)*b*e*(c**(-2))**(1/4)*atan(c*
x**2)/(2*(-1)**(3/4)*c**8*x**4*(c**(-2))**(11/4) + 2*(-1)**(3/4)*c**6*(c**(-2))**(11/4)), True))

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