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

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

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 328, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.80, size = 321, normalized size = 0.98 \[ -\frac {4 a \left (c^2 d^4+e^4\right )+2 b \sqrt {c} \left (2 c^{3/2} d^3-\sqrt {2} c d^2 e+\sqrt {2} e^3\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right ) (d+e x)+2 b \sqrt {c} \left (2 c^{3/2} d^3+\sqrt {2} c d^2 e-\sqrt {2} e^3\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right ) (d+e x)+4 b \left (c^2 d^4+e^4\right ) \tan ^{-1}\left (c x^2\right )-2 b c d e^2 \log \left (c^2 x^4+1\right ) (d+e x)+\sqrt {2} b \sqrt {c} e \left (c d^2+e^2\right ) \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right ) (d+e x)-\sqrt {2} b \sqrt {c} e \left (c d^2+e^2\right ) \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right ) (d+e x)+8 b c d e^2 (d+e x) \log (d+e x)}{4 e \left (c^2 d^4+e^4\right ) (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [B]  time = 3.69, size = 1443, normalized size = 4.40 \[ \text {result too large to display} \]

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*(c*d*e*log(c^2 - 4*c^2*d/(x*e + d) + 6*c^2*d^2/(x*e + d)^2 - 4*c^2*d^3/(x*e + d)^3 + c^2*d^4/(x*e + d)^4 +
e^4/(x*e + d)^4)/(c^2*d^4 + e^4) - 2*arctan(c*x^2)*e^(-1)/(x*e + d) - ((c^6*d^8*e^16 + 2*c^4*d^4*e^20 + c^2*e
^24)^(1/4)*cos(1/2*arctan(1/2*(c^3*d^4*e^8 - c*e^12)*e^(-10)/(c^2*d^2)))*log(64*(2*sqrt(2)*c^(7/2)*d^3 - 2*sqr
t(2)*c^(7/2)*d^4/(x*e + d) + 8*sqrt(c^2)*c^2*d^2*e - 12*sqrt(c^2)*c^2*d^3*e/(x*e + d) + 3*sqrt(2)*sqrt(c^2)*c^
(3/2)*d*e^2 - 7*sqrt(2)*sqrt(c^2)*c^(3/2)*d^2*e^2/(x*e + d) + sqrt(2)*c^(3/2)*e^4/(x*e + d))^2 + 64*(2*sqrt(2)
*sqrt(c^2)*c^(5/2)*d^3 - 2*sqrt(2)*sqrt(c^2)*c^(5/2)*d^4/(x*e + d) - 3*sqrt(2)*c^(5/2)*d*e^2 + 7*sqrt(2)*c^(5/
2)*d^2*e^2/(x*e + d) - 2*sqrt(c^2)*c*e^3 + 8*sqrt(c^2)*c*d*e^3/(x*e + d) + sqrt(2)*sqrt(c^2)*sqrt(c)*e^4/(x*e
+ d))^2) - (c^6*d^8*e^16 + 2*c^4*d^4*e^20 + c^2*e^24)^(1/4)*cos(1/2*arctan(1/2*(c^3*d^4*e^8 - c*e^12)*e^(-10)/
(c^2*d^2)))*log(64*(2*sqrt(2)*c^(7/2)*d^3 - 2*sqrt(2)*c^(7/2)*d^4/(x*e + d) - 8*sqrt(c^2)*c^2*d^2*e + 12*sqrt(
c^2)*c^2*d^3*e/(x*e + d) + 3*sqrt(2)*sqrt(c^2)*c^(3/2)*d*e^2 - 7*sqrt(2)*sqrt(c^2)*c^(3/2)*d^2*e^2/(x*e + d) +
sqrt(2)*c^(3/2)*e^4/(x*e + d))^2 + 64*(2*sqrt(2)*sqrt(c^2)*c^(5/2)*d^3 - 2*sqrt(2)*sqrt(c^2)*c^(5/2)*d^4/(x*e
+ d) - 3*sqrt(2)*c^(5/2)*d*e^2 + 7*sqrt(2)*c^(5/2)*d^2*e^2/(x*e + d) + 2*sqrt(c^2)*c*e^3 - 8*sqrt(c^2)*c*d*e^
3/(x*e + d) + sqrt(2)*sqrt(c^2)*sqrt(c)*e^4/(x*e + d))^2) - 2*(c^2*d^3*e^3 + (c^6*d^8*e^16 + 2*c^4*d^4*e^20 +
c^2*e^24)^(1/4)*sin(1/2*arctan(1/2*(c^3*d^4*e^8 - c*e^12)*e^(-10)/(c^2*d^2))))*arctan(-(2*sqrt(2)*sqrt(c^2)*c^
(5/2)*d^3 - 2*sqrt(2)*sqrt(c^2)*c^(5/2)*d^4/(x*e + d) - 3*sqrt(2)*c^(5/2)*d*e^2 + 7*sqrt(2)*c^(5/2)*d^2*e^2/(x
*e + d) + 2*sqrt(c^2)*c*e^3 - 8*sqrt(c^2)*c*d*e^3/(x*e + d) + sqrt(2)*sqrt(c^2)*sqrt(c)*e^4/(x*e + d))/(2*sqrt
(2)*c^(7/2)*d^3 - 2*sqrt(2)*c^(7/2)*d^4/(x*e + d) - 8*sqrt(c^2)*c^2*d^2*e + 12*sqrt(c^2)*c^2*d^3*e/(x*e + d) +
3*sqrt(2)*sqrt(c^2)*c^(3/2)*d*e^2 - 7*sqrt(2)*sqrt(c^2)*c^(3/2)*d^2*e^2/(x*e + d) + sqrt(2)*c^(3/2)*e^4/(x*e
+ d))) - 2*(c^2*d^3*e^3 - (c^6*d^8*e^16 + 2*c^4*d^4*e^20 + c^2*e^24)^(1/4)*sin(1/2*arctan(1/2*(c^3*d^4*e^8 - c
*e^12)*e^(-10)/(c^2*d^2))))*arctan(-(2*sqrt(2)*sqrt(c^2)*c^(5/2)*d^3 - 2*sqrt(2)*sqrt(c^2)*c^(5/2)*d^4/(x*e +
d) - 3*sqrt(2)*c^(5/2)*d*e^2 + 7*sqrt(2)*c^(5/2)*d^2*e^2/(x*e + d) - 2*sqrt(c^2)*c*e^3 + 8*sqrt(c^2)*c*d*e^3/(
x*e + d) + sqrt(2)*sqrt(c^2)*sqrt(c)*e^4/(x*e + d))/(2*sqrt(2)*c^(7/2)*d^3 - 2*sqrt(2)*c^(7/2)*d^4/(x*e + d) +
8*sqrt(c^2)*c^2*d^2*e - 12*sqrt(c^2)*c^2*d^3*e/(x*e + d) + 3*sqrt(2)*sqrt(c^2)*c^(3/2)*d*e^2 - 7*sqrt(2)*sqrt
(c^2)*c^(3/2)*d^2*e^2/(x*e + d) + sqrt(2)*c^(3/2)*e^4/(x*e + d))))/(c^2*d^4*e^4 + e^8))*b - a*e^(-1)/(x*e + d)

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maple [A]  time = 0.04, size = 433, normalized size = 1.32 \[ -\frac {a}{\left (e x +d \right ) e}-\frac {b \arctan \left (c \,x^{2}\right )}{\left (e x +d \right ) e}-\frac {2 b c d e \ln \left (e x +d \right )}{c^{2} d^{4}+e^{4}}+\frac {b \,e^{2} c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{4 c^{2} d^{4}+4 e^{4}}+\frac {b \,e^{2} c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 c^{2} d^{4}+2 e^{4}}+\frac {b \,e^{2} c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 c^{2} d^{4}+2 e^{4}}+\frac {b \,c^{3} d^{3} \arctan \left (x^{2} \sqrt {c^{2}}\right )}{e \left (c^{2} d^{4}+e^{4}\right ) \sqrt {c^{2}}}-\frac {b c \,d^{2} \sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{4 \left (c^{2} d^{4}+e^{4}\right ) \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b c \,d^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 \left (c^{2} d^{4}+e^{4}\right ) \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b c \,d^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 \left (c^{2} d^{4}+e^{4}\right ) \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+\frac {b c d e \ln \left (c^{2} x^{4}+1\right )}{2 c^{2} d^{4}+2 e^{4}} \]

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)+1/4*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)))+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)+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 = 0.54, size = 287, normalized size = 0.88 \[ -\frac {1}{4} \, {\left ({\left (\frac {8 \, d e \log \left (e x + d\right )}{c^{2} d^{4} + e^{4}} - \frac {\frac {\sqrt {2} {\left (c d^{2} e + \sqrt {2} \sqrt {c} d e^{2} + e^{3}\right )} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} {\left (c d^{2} e - \sqrt {2} \sqrt {c} d e^{2} + e^{3}\right )} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {2 \, {\left (2 \, c^{2} d^{3} + \sqrt {2} c^{\frac {3}{2}} d^{2} e - \sqrt {2} \sqrt {c} e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c} + \frac {2 \, {\left (2 \, c^{2} d^{3} - \sqrt {2} c^{\frac {3}{2}} d^{2} e + \sqrt {2} \sqrt {c} e^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c}}{c^{2} d^{4} e + e^{5}}\right )} c + \frac {4 \, \arctan \left (c x^{2}\right )}{e^{2} x + d e}\right )} b - \frac {a}{e^{2} x + d e} \]

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)*(c*d^2*e + sqrt(2)*sqrt(c)*d*e^2 + e^3)*log(c*x^2 + sqrt(
2)*sqrt(c)*x + 1)/sqrt(c) - sqrt(2)*(c*d^2*e - sqrt(2)*sqrt(c)*d*e^2 + e^3)*log(c*x^2 - sqrt(2)*sqrt(c)*x + 1)
/sqrt(c) - 2*(2*c^2*d^3 + sqrt(2)*c^(3/2)*d^2*e - sqrt(2)*sqrt(c)*e^3)*arctan(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqr
t(c))/sqrt(c))/c + 2*(2*c^2*d^3 - sqrt(2)*c^(3/2)*d^2*e + sqrt(2)*sqrt(c)*e^3)*arctan(1/2*sqrt(2)*(2*c*x - sqr
t(2)*sqrt(c))/sqrt(c))/c)/(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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mupad [B]  time = 0.66, size = 883, normalized size = 2.69 \[ \left (\sum _{k=1}^4\ln \left (\frac {{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^8\,e^9\,x\,320-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^{10}\,d^5\,e^4\,128+16\,b^4\,c^{10}\,e\,x-\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )\,b^3\,c^9\,e^3\,8+{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^8\,d\,e^8\,384+\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )\,b^3\,c^{11}\,d^3\,x\,8-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^3\,b\,c^9\,d^2\,e^5\,320-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^4\,c^{10}\,d^4\,e^5\,x\,192+{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^3\,b\,c^{11}\,d^5\,e^2\,x\,32+{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^2\,b^2\,c^{10}\,d^2\,e^3\,x\,64-{\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )}^3\,b\,c^9\,d\,e^6\,x\,416}{e^2}\right )\,\mathrm {root}\left (16\,c^2\,d^4\,e^4\,z^4+16\,e^8\,z^4-32\,b\,c\,d\,e^5\,z^3+8\,b^2\,c^2\,d^2\,e^2\,z^2+b^4\,c^2,z,k\right )\right )-\frac {a}{x\,e^2+d\,e}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{x\,e^2+d\,e}-\frac {2\,b\,c\,d\,e\,\ln \left (d+e\,x\right )}{c^2\,d^4+e^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log((320*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z,
k)^4*c^8*e^9*x - 128*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2
, z, k)^4*c^10*d^5*e^4 + 16*b^4*c^10*e*x - 8*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c
^2*d^2*e^2*z^2 + b^4*c^2, z, k)*b^3*c^9*e^3 + 384*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*
b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^4*c^8*d*e^8 + 8*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 +
8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)*b^3*c^11*d^3*x - 320*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e
^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^3*b*c^9*d^2*e^5 - 192*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 3
2*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^4*c^10*d^4*e^5*x + 32*root(16*c^2*d^4*e^4*z^4 + 16*e^
8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^3*b*c^11*d^5*e^2*x + 64*root(16*c^2*d^4*e^4*
z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^2*b^2*c^10*d^2*e^3*x - 416*root(1
6*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k)^3*b*c^9*d*e^6*x)/e^
2)*root(16*c^2*d^4*e^4*z^4 + 16*e^8*z^4 - 32*b*c*d*e^5*z^3 + 8*b^2*c^2*d^2*e^2*z^2 + b^4*c^2, z, k), k, 1, 4)
- a/(d*e + e^2*x) - (b*atan(c*x^2))/(d*e + e^2*x) - (2*b*c*d*e*log(d + e*x))/(e^4 + c^2*d^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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