\(\int (d+e x) (a+b \arctan (c x^2)) \, dx\) [22]

Optimal result

Integrand size = 16, antiderivative size = 192 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {b d^2 \arctan \left (c x^2\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}+\frac {b d \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b d \arctan \left (1+\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b d \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b d \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2} \sqrt {c}}-\frac {b e \log \left (1+c^2 x^4\right )}{4 c} \]

[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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {4980, 1845, 303, 1176, 631, 210, 1179, 642, 1262, 649, 209, 266} \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}-\frac {b d^2 \arctan \left (c x^2\right )}{2 e}+\frac {b d \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b d \arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {b e \log \left (c^2 x^4+1\right )}{4 c}-\frac {b d \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b d \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}} \]

[In]

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

[Out]

-1/2*(b*d^2*ArcTan[c*x^2])/e + ((d + e*x)^2*(a + b*ArcTan[c*x^2]))/(2*e) + (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 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 266

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 303

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 631

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 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 649

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

Rule 1176

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 1179

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 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1845

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(c*x)^(m + ii)*((Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2))/(c^ii*(a + b*x^n))), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 4980

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

Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}-\frac {(b c) \int \frac {x (d+e x)^2}{1+c^2 x^4} \, dx}{e} \\ & = \frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}-\frac {(b c) \int \left (\frac {2 d e x^2}{1+c^2 x^4}+\frac {x \left (d^2+e^2 x^2\right )}{1+c^2 x^4}\right ) \, dx}{e} \\ & = \frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}-(2 b c d) \int \frac {x^2}{1+c^2 x^4} \, dx-\frac {(b c) \int \frac {x \left (d^2+e^2 x^2\right )}{1+c^2 x^4} \, dx}{e} \\ & = \frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}+(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 {(b c) \text {Subst}\left (\int \frac {d^2+e^2 x}{1+c^2 x^2} \, dx,x,x^2\right )}{2 e} \\ & = \frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}-\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 {\left (b c d^2\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,x^2\right )}{2 e}-\frac {1}{2} (b c e) \text {Subst}\left (\int \frac {x}{1+c^2 x^2} \, dx,x,x^2\right ) \\ & = -\frac {b d^2 \arctan \left (c x^2\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}-\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) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}+\frac {(b d) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}} \\ & = -\frac {b d^2 \arctan \left (c x^2\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \arctan \left (c x^2\right )\right )}{2 e}+\frac {b d \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b d \arctan \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*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.80 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=a d x+\frac {1}{2} a e x^2+b d x \arctan \left (c x^2\right )+\frac {1}{2} b e x^2 \arctan \left (c x^2\right )-\frac {b d \left (-2 \arctan \left (1-\sqrt {2} \sqrt {c} x\right )+2 \arctan \left (1+\sqrt {2} \sqrt {c} x\right )+\log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )-\log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )\right )}{2 \sqrt {2} \sqrt {c}}-\frac {b e \log \left (1+c^2 x^4\right )}{4 c} \]

[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)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.76

method	result	size
default	\(a \left (\frac {1}{2} e \,x^{2}+d x \right )+b \left (\frac {\arctan \left (c \,x^{2}\right ) x^{2} e}{2}+\arctan \left (c \,x^{2}\right ) d x -c \left (\frac {d \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+\frac {e \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}\right )\right )\)	\(146\)
parts	\(a \left (\frac {1}{2} e \,x^{2}+d x \right )+b \left (\frac {\arctan \left (c \,x^{2}\right ) x^{2} e}{2}+\arctan \left (c \,x^{2}\right ) d x -c \left (\frac {d \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+\frac {e \ln \left (c^{2} x^{4}+1\right )}{4 c^{2}}\right )\right )\)	\(146\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.43 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {2 \, a c e x^{2} + 4 \, a c d x + 2 \, {\left (b c e x^{2} + 2 \, b c d x\right )} \arctan \left (c x^{2}\right ) - {\left (b e - 2 \, c \sqrt {-\sqrt {-\frac {b^{4} d^{4}}{c^{2}}}}\right )} \log \left (b^{3} d^{3} x + \sqrt {-\frac {b^{4} d^{4}}{c^{2}}} c \sqrt {-\sqrt {-\frac {b^{4} d^{4}}{c^{2}}}}\right ) - {\left (b e + 2 \, c \sqrt {-\sqrt {-\frac {b^{4} d^{4}}{c^{2}}}}\right )} \log \left (b^{3} d^{3} x - \sqrt {-\frac {b^{4} d^{4}}{c^{2}}} c \sqrt {-\sqrt {-\frac {b^{4} d^{4}}{c^{2}}}}\right ) - {\left (b e + 2 \, \left (-\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {1}{4}} c\right )} \log \left (b^{3} d^{3} x + \left (-\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {3}{4}} c\right ) - {\left (b e - 2 \, \left (-\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {1}{4}} c\right )} \log \left (b^{3} d^{3} x - \left (-\frac {b^{4} d^{4}}{c^{2}}\right )^{\frac {3}{4}} c\right )}{4 \, c} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.62 (sec) , antiderivative size = 1266, normalized size of antiderivative = 6.59 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.88 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{2} \, a e x^{2} - \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}}\right )} - 4 \, x \arctan \left (c x^{2}\right )\right )} b d + a d x + \frac {{\left (2 \, c x^{2} \arctan \left (c x^{2}\right ) - \log \left (c^{2} x^{4} + 1\right )\right )} b e}{4 \, c} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.96 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{2} \, b e x^{2} \arctan \left (c x^{2}\right ) + \frac {1}{2} \, a e x^{2} + b d x \arctan \left (c x^{2}\right ) + a d x - \frac {\sqrt {2} b c d \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{2 \, {\left | c \right |}^{\frac {3}{2}}} - \frac {\sqrt {2} b c d \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{2 \, {\left | c \right |}^{\frac {3}{2}}} + \frac {{\left (\sqrt {2} b c d \sqrt {{\left | c \right |}} - b c e\right )} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{4 \, c^{2}} - \frac {{\left (\sqrt {2} b c d \sqrt {{\left | c \right |}} + b c e\right )} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{4 \, c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.69 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.06 \[ \int (d+e x) \left (a+b \arctan \left (c x^2\right )\right ) \, dx=a\,d\,x+\frac {a\,e\,x^2}{2}+b\,d\,x\,\mathrm {atan}\left (c\,x^2\right )-\frac {b\,e\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}-1\right )}{4\,c}-\frac {b\,e\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}+1\right )}{4\,c}-\frac {b\,e\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}-1\right )}{4\,c}-\frac {b\,e\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}+1\right )}{4\,c}+\frac {b\,e\,x^2\,\mathrm {atan}\left (c\,x^2\right )}{2}-\frac {b\,d\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}-1\right )\,\sqrt {-c\,1{}\mathrm {i}}}{2\,c}+\frac {b\,d\,\ln \left (x\,\sqrt {-c\,1{}\mathrm {i}}+1\right )\,\sqrt {-c\,1{}\mathrm {i}}}{2\,c}-\frac {b\,d\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}-1\right )\,\sqrt {c\,1{}\mathrm {i}}}{2\,c}+\frac {b\,d\,\ln \left (x\,\sqrt {c\,1{}\mathrm {i}}+1\right )\,\sqrt {c\,1{}\mathrm {i}}}{2\,c} \]

[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)