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

Optimal. Leaf size=328 \[ -\frac{a+b \tan ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{b c d e \log \left (c^2 x^4+1\right )}{2 \left (c^2 d^4+e^4\right )}-\frac{b \sqrt{c} \left (c d^2+e^2\right ) \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{2 \sqrt{2} \left (c^2 d^4+e^4\right )}+\frac{b \sqrt{c} \left (c d^2+e^2\right ) \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{2 \sqrt{2} \left (c^2 d^4+e^4\right )}+\frac{b c^2 d^3 \tan ^{-1}\left (c x^2\right )}{e \left (c^2 d^4+e^4\right )}-\frac{2 b c d e \log (d+e x)}{c^2 d^4+e^4}+\frac{b \sqrt{c} \left (c d^2-e^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{\sqrt{2} \left (c^2 d^4+e^4\right )}-\frac{b \sqrt{c} \left (c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right )}{\sqrt{2} \left (c^2 d^4+e^4\right )} \]

[Out]

(b*c^2*d^3*ArcTan[c*x^2])/(e*(c^2*d^4 + e^4)) - (a + b*ArcTan[c*x^2])/(e*(d + e*x)) + (b*Sqrt[c]*(c*d^2 - e^2)
*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/(Sqrt[2]*(c^2*d^4 + e^4)) - (b*Sqrt[c]*(c*d^2 - e^2)*ArcTan[1 + Sqrt[2]*Sqrt[c
]*x])/(Sqrt[2]*(c^2*d^4 + e^4)) - (2*b*c*d*e*Log[d + e*x])/(c^2*d^4 + e^4) - (b*Sqrt[c]*(c*d^2 + e^2)*Log[1 -
Sqrt[2]*Sqrt[c]*x + c*x^2])/(2*Sqrt[2]*(c^2*d^4 + e^4)) + (b*Sqrt[c]*(c*d^2 + e^2)*Log[1 + Sqrt[2]*Sqrt[c]*x +
c*x^2])/(2*Sqrt[2]*(c^2*d^4 + e^4)) + (b*c*d*e*Log[1 + c^2*x^4])/(2*(c^2*d^4 + e^4))

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Rubi [A]  time = 0.522159, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {5205, 12, 6725, 1876, 1168, 1162, 617, 204, 1165, 628, 1248, 635, 203, 260} \[ -\frac{a+b \tan ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{b c d e \log \left (c^2 x^4+1\right )}{2 \left (c^2 d^4+e^4\right )}-\frac{b \sqrt{c} \left (c d^2+e^2\right ) \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right )}{2 \sqrt{2} \left (c^2 d^4+e^4\right )}+\frac{b \sqrt{c} \left (c d^2+e^2\right ) \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right )}{2 \sqrt{2} \left (c^2 d^4+e^4\right )}+\frac{b c^2 d^3 \tan ^{-1}\left (c x^2\right )}{e \left (c^2 d^4+e^4\right )}-\frac{2 b c d e \log (d+e x)}{c^2 d^4+e^4}+\frac{b \sqrt{c} \left (c d^2-e^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right )}{\sqrt{2} \left (c^2 d^4+e^4\right )}-\frac{b \sqrt{c} \left (c d^2-e^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right )}{\sqrt{2} \left (c^2 d^4+e^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c^2*d^3*ArcTan[c*x^2])/(e*(c^2*d^4 + e^4)) - (a + b*ArcTan[c*x^2])/(e*(d + e*x)) + (b*Sqrt[c]*(c*d^2 - e^2)
*ArcTan[1 - Sqrt[2]*Sqrt[c]*x])/(Sqrt[2]*(c^2*d^4 + e^4)) - (b*Sqrt[c]*(c*d^2 - e^2)*ArcTan[1 + Sqrt[2]*Sqrt[c
]*x])/(Sqrt[2]*(c^2*d^4 + e^4)) - (2*b*c*d*e*Log[d + e*x])/(c^2*d^4 + e^4) - (b*Sqrt[c]*(c*d^2 + e^2)*Log[1 -
Sqrt[2]*Sqrt[c]*x + c*x^2])/(2*Sqrt[2]*(c^2*d^4 + e^4)) + (b*Sqrt[c]*(c*d^2 + e^2)*Log[1 + Sqrt[2]*Sqrt[c]*x +
c*x^2])/(2*Sqrt[2]*(c^2*d^4 + e^4)) + (b*c*d*e*Log[1 + c^2*x^4])/(2*(c^2*d^4 + e^4))

Rule 5205

Int[((a_.) + ArcTan[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 + u^2), x], x]
, x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m +
1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 1876

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

Rule 1168

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

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]

Rule 1248

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 635

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.744945, size = 321, normalized size = 0.98 \[ -\frac{4 a \left (c^2 d^4+e^4\right )+4 b \left (c^2 d^4+e^4\right ) \tan ^{-1}\left (c x^2\right )+2 b \sqrt{c} \left (2 c^{3/2} d^3-\sqrt{2} c d^2 e+\sqrt{2} e^3\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{c} x\right ) (d+e x)+2 b \sqrt{c} \left (2 c^{3/2} d^3+\sqrt{2} c d^2 e-\sqrt{2} e^3\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{c} x+1\right ) (d+e x)-2 b c d e^2 \log \left (c^2 x^4+1\right ) (d+e x)+\sqrt{2} b \sqrt{c} e \left (c d^2+e^2\right ) \log \left (c x^2-\sqrt{2} \sqrt{c} x+1\right ) (d+e x)-\sqrt{2} b \sqrt{c} e \left (c d^2+e^2\right ) \log \left (c x^2+\sqrt{2} \sqrt{c} x+1\right ) (d+e x)+8 b c d e^2 (d+e x) \log (d+e x)}{4 e \left (c^2 d^4+e^4\right ) (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*a*(c^2*d^4 + e^4) + 4*b*(c^2*d^4 + e^4)*ArcTan[c*x^2] + 2*b*Sqrt[c]*(2*c^(3/2)*d^3 - Sqrt[2]*c*d^2*e + Sqr
t[2]*e^3)*(d + e*x)*ArcTan[1 - Sqrt[2]*Sqrt[c]*x] + 2*b*Sqrt[c]*(2*c^(3/2)*d^3 + Sqrt[2]*c*d^2*e - Sqrt[2]*e^3
)*(d + e*x)*ArcTan[1 + Sqrt[2]*Sqrt[c]*x] + 8*b*c*d*e^2*(d + e*x)*Log[d + e*x] + Sqrt[2]*b*Sqrt[c]*e*(c*d^2 +
e^2)*(d + e*x)*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2] - Sqrt[2]*b*Sqrt[c]*e*(c*d^2 + e^2)*(d + e*x)*Log[1 + Sqrt[2
]*Sqrt[c]*x + c*x^2] - 2*b*c*d*e^2*(d + e*x)*Log[1 + c^2*x^4])/(4*e*(c^2*d^4 + e^4)*(d + e*x))

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Maple [A]  time = 0.036, size = 433, normalized size = 1.3 \begin{align*} -{\frac{a}{ \left ( ex+d \right ) e}}-{\frac{b\arctan \left ( c{x}^{2} \right ) }{ \left ( ex+d \right ) e}}-2\,{\frac{bcde\ln \left ( ex+d \right ) }{{c}^{2}{d}^{4}+{e}^{4}}}+{\frac{b{e}^{2}c\sqrt{2}}{2\,{c}^{2}{d}^{4}+2\,{e}^{4}}\sqrt [4]{{c}^{-2}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{c}^{-2}}}}}+1 \right ) }+{\frac{b{e}^{2}c\sqrt{2}}{2\,{c}^{2}{d}^{4}+2\,{e}^{4}}\sqrt [4]{{c}^{-2}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{c}^{-2}}}}}-1 \right ) }+{\frac{b{e}^{2}c\sqrt{2}}{4\,{c}^{2}{d}^{4}+4\,{e}^{4}}\sqrt [4]{{c}^{-2}}\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{b{c}^{3}{d}^{3}}{e \left ({c}^{2}{d}^{4}+{e}^{4} \right ) }\arctan \left ({x}^{2}\sqrt{{c}^{2}} \right ){\frac{1}{\sqrt{{c}^{2}}}}}-{\frac{bc{d}^{2}\sqrt{2}}{4\,{c}^{2}{d}^{4}+4\,{e}^{4}}\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{bc{d}^{2}\sqrt{2}}{2\,{c}^{2}{d}^{4}+2\,{e}^{4}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{c}^{-2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{c}^{-2}}}}}-{\frac{bc{d}^{2}\sqrt{2}}{2\,{c}^{2}{d}^{4}+2\,{e}^{4}}\arctan \left ({\sqrt{2}x{\frac{1}{\sqrt [4]{{c}^{-2}}}}}-1 \right ){\frac{1}{\sqrt [4]{{c}^{-2}}}}}+{\frac{bcde\ln \left ({c}^{2}{x}^{4}+1 \right ) }{2\,{c}^{2}{d}^{4}+2\,{e}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.48466, size = 649, normalized size = 1.98 \begin{align*} -\frac{1}{4} \,{\left ({\left (\frac{8 \, d e \log \left (e x + d\right )}{c^{2} d^{4} + e^{4}} - \frac{\frac{\sqrt{2}{\left (\sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} c^{2} d e^{2} + \sqrt{c^{2}} c^{2} d^{2} e + c^{2} e^{3}\right )} \log \left (\sqrt{c^{2}} x^{2} + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{4}} c^{2}} + \frac{\sqrt{2}{\left (\sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} c^{2} d e^{2} - \sqrt{c^{2}} c^{2} d^{2} e - c^{2} e^{3}\right )} \log \left (\sqrt{c^{2}} x^{2} - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} x + 1\right )}{{\left (c^{2}\right )}^{\frac{1}{4}} c^{2}} - \frac{{\left (2 \, c^{4} d^{3} + \sqrt{2}{\left (c^{2}\right )}^{\frac{3}{4}} c^{2} d^{2} e - \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} c^{2} e^{3}\right )} \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 )}{{\left (c^{2}\right )}^{\frac{1}{4}} c^{2} \sqrt{-\sqrt{c^{2}}}} + \frac{{\left (2 \, c^{4} d^{3} - \sqrt{2}{\left (c^{2}\right )}^{\frac{3}{4}} c^{2} d^{2} e + \sqrt{2}{\left (c^{2}\right )}^{\frac{1}{4}} c^{2} e^{3}\right )} \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 )}{{\left (c^{2}\right )}^{\frac{1}{4}} c^{2} \sqrt{-\sqrt{c^{2}}}}}{c^{2} d^{4} e + e^{5}}\right )} c + \frac{4 \, \arctan \left (c x^{2}\right )}{e^{2} x + d e}\right )} b - \frac{a}{e^{2} x + d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 14.2967, size = 512, normalized size = 1.56 \begin{align*} \frac{1}{2} \,{\left ({\left (\frac{d e \log \left (c^{2} x^{4} + 1\right )}{c^{2} d^{4} + e^{4}} - \frac{4 \, d e^{2} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e + e^{5}} - \frac{2 \,{\left (\sqrt{2} c^{2} d^{3}{\left | c \right |} + c^{2} d^{2} \sqrt{{\left | c \right |}} e -{\left | c \right |}^{\frac{3}{2}} e^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \frac{\sqrt{2}}{\sqrt{{\left | c \right |}}}\right )} \sqrt{{\left | c \right |}}\right )}{\sqrt{2} c^{4} d^{4} e + \sqrt{2} c^{2} e^{5}} + \frac{2 \,{\left (\sqrt{2} c^{2} d^{3}{\left | c \right |} - c^{2} d^{2} \sqrt{{\left | c \right |}} e +{\left | c \right |}^{\frac{3}{2}} e^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \frac{\sqrt{2}}{\sqrt{{\left | c \right |}}}\right )} \sqrt{{\left | c \right |}}\right )}{\sqrt{2} c^{4} d^{4} e + \sqrt{2} c^{2} e^{5}} + \frac{{\left (c^{2} d^{2} \sqrt{{\left | c \right |}} +{\left | c \right |}^{\frac{3}{2}} e^{2}\right )} \log \left (x^{2} + \frac{\sqrt{2} x}{\sqrt{{\left | c \right |}}} + \frac{1}{{\left | c \right |}}\right )}{\sqrt{2} c^{4} d^{4} + \sqrt{2} c^{2} e^{4}} - \frac{{\left (c^{2} d^{2} \sqrt{{\left | c \right |}} +{\left | c \right |}^{\frac{3}{2}} e^{2}\right )} \log \left (x^{2} - \frac{\sqrt{2} x}{\sqrt{{\left | c \right |}}} + \frac{1}{{\left | c \right |}}\right )}{\sqrt{2} c^{4} d^{4} + \sqrt{2} c^{2} e^{4}}\right )} c - \frac{2 \, \arctan \left (c x^{2}\right ) e^{\left (-1\right )}}{x e + d}\right )} b - \frac{a e^{\left (-1\right )}}{x e + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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