3.1.13 \(\int \frac {(a+b \text {ArcTan}(c+d x))^2}{(c e+d e x)^4} \, dx\) [13]

Optimal. Leaf size=194 \[ -\frac {b^2}{3 d e^4 (c+d x)}-\frac {b^2 \text {ArcTan}(c+d x)}{3 d e^4}-\frac {b (a+b \text {ArcTan}(c+d x))}{3 d e^4 (c+d x)^2}+\frac {i (a+b \text {ArcTan}(c+d x))^2}{3 d e^4}-\frac {(a+b \text {ArcTan}(c+d x))^2}{3 d e^4 (c+d x)^3}-\frac {2 b (a+b \text {ArcTan}(c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{3 d e^4}+\frac {i b^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{3 d e^4} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5151, 12, 4946, 5038, 331, 209, 5044, 4988, 2497} \begin {gather*} -\frac {b (a+b \text {ArcTan}(c+d x))}{3 d e^4 (c+d x)^2}-\frac {(a+b \text {ArcTan}(c+d x))^2}{3 d e^4 (c+d x)^3}+\frac {i (a+b \text {ArcTan}(c+d x))^2}{3 d e^4}-\frac {2 b \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))}{3 d e^4}-\frac {b^2 \text {ArcTan}(c+d x)}{3 d e^4}+\frac {i b^2 \text {Li}_2\left (\frac {2}{1-i (c+d x)}-1\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

-1/3*b^2/(d*e^4*(c + d*x)) - (b^2*ArcTan[c + d*x])/(3*d*e^4) - (b*(a + b*ArcTan[c + d*x]))/(3*d*e^4*(c + d*x)^
2) + ((I/3)*(a + b*ArcTan[c + d*x])^2)/(d*e^4) - (a + b*ArcTan[c + d*x])^2/(3*d*e^4*(c + d*x)^3) - (2*b*(a + b
*ArcTan[c + d*x])*Log[2 - 2/(1 - I*(c + d*x))])/(3*d*e^4) + ((I/3)*b^2*PolyLog[2, -1 + 2/(1 - I*(c + d*x))])/(
d*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 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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4988

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])
^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))
]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 5038

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x
^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 5044

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*d*(p + 1))), x] + Dist[I/d, Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]

Rule 5151

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[(f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{(c e+d e x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {(2 b) \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^3 \left (1+x^2\right )} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {(2 b) \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^3} \, dx,x,c+d x\right )}{3 d e^4}-\frac {(2 b) \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {(2 i b) \text {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x (i+x)} \, dx,x,c+d x\right )}{3 d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right )} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{3 d e^4}-\frac {b^2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,c+d x\right )}{3 d e^4}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{3 d e^4}\\ &=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b^2 \tan ^{-1}(c+d x)}{3 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}-\frac {2 b \left (a+b \tan ^{-1}(c+d x)\right ) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{3 d e^4}+\frac {i b^2 \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{3 d e^4}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 163, normalized size = 0.84 \begin {gather*} -\frac {a b+\frac {a^2}{(c+d x)^3}+\frac {a b}{(c+d x)^2}+\frac {b^2}{c+d x}+b^2 \left (-i+\frac {1}{(c+d x)^3}\right ) \text {ArcTan}(c+d x)^2+b \text {ArcTan}(c+d x) \left (b+\frac {2 a}{(c+d x)^3}+\frac {b}{(c+d x)^2}+2 b \log \left (1-e^{2 i \text {ArcTan}(c+d x)}\right )\right )+2 a b \log \left (\frac {c+d x}{\sqrt {1+(c+d x)^2}}\right )-i b^2 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c+d x)}\right )}{3 d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*b + a^2/(c + d*x)^3 + (a*b)/(c + d*x)^2 + b^2/(c + d*x) + b^2*(-I + (c + d*x)^(-3))*ArcTan[c + d*x]^2
+ b*ArcTan[c + d*x]*(b + (2*a)/(c + d*x)^3 + b/(c + d*x)^2 + 2*b*Log[1 - E^((2*I)*ArcTan[c + d*x])]) + 2*a*b*L
og[(c + d*x)/Sqrt[1 + (c + d*x)^2]] - I*b^2*PolyLog[2, E^((2*I)*ArcTan[c + d*x])])/(d*e^4)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (176 ) = 352\).
time = 0.63, size = 482, normalized size = 2.48

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arctan \left (d x +c \right )^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{3 e^{4}}-\frac {b^{2} \arctan \left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{2}}-\frac {2 b^{2} \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{3 e^{4}}-\frac {i b^{2} \dilog \left (1+i \left (d x +c \right )\right )}{3 e^{4}}-\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{6 e^{4}}-\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{3 e^{4}}-\frac {i b^{2} \ln \left (d x +c -i\right )^{2}}{12 e^{4}}+\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{6 e^{4}}+\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{6 e^{4}}+\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{6 e^{4}}+\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{3 e^{4}}-\frac {b^{2}}{3 e^{4} \left (d x +c \right )}-\frac {b^{2} \arctan \left (d x +c \right )}{3 e^{4}}+\frac {i b^{2} \ln \left (d x +c +i\right )^{2}}{12 e^{4}}-\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{6 e^{4}}+\frac {i b^{2} \dilog \left (1-i \left (d x +c \right )\right )}{3 e^{4}}-\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{6 e^{4}}-\frac {2 a b \arctan \left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{3}}+\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right )}{3 e^{4}}-\frac {a b}{3 e^{4} \left (d x +c \right )^{2}}-\frac {2 a b \ln \left (d x +c \right )}{3 e^{4}}}{d}\)	\(482\)
default	\(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arctan \left (d x +c \right )^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{3 e^{4}}-\frac {b^{2} \arctan \left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{2}}-\frac {2 b^{2} \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{3 e^{4}}-\frac {i b^{2} \dilog \left (1+i \left (d x +c \right )\right )}{3 e^{4}}-\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{6 e^{4}}-\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{3 e^{4}}-\frac {i b^{2} \ln \left (d x +c -i\right )^{2}}{12 e^{4}}+\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{6 e^{4}}+\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{6 e^{4}}+\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{6 e^{4}}+\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{3 e^{4}}-\frac {b^{2}}{3 e^{4} \left (d x +c \right )}-\frac {b^{2} \arctan \left (d x +c \right )}{3 e^{4}}+\frac {i b^{2} \ln \left (d x +c +i\right )^{2}}{12 e^{4}}-\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{6 e^{4}}+\frac {i b^{2} \dilog \left (1-i \left (d x +c \right )\right )}{3 e^{4}}-\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{6 e^{4}}-\frac {2 a b \arctan \left (d x +c \right )}{3 e^{4} \left (d x +c \right )^{3}}+\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right )}{3 e^{4}}-\frac {a b}{3 e^{4} \left (d x +c \right )^{2}}-\frac {2 a b \ln \left (d x +c \right )}{3 e^{4}}}{d}\)	\(482\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/3*a^2/e^4/(d*x+c)^3-1/3*b^2/e^4/(d*x+c)^3*arctan(d*x+c)^2+1/3*b^2/e^4*arctan(d*x+c)*ln(1+(d*x+c)^2)-1/
3*b^2/e^4*arctan(d*x+c)/(d*x+c)^2-2/3*b^2/e^4*ln(d*x+c)*arctan(d*x+c)+1/6*I*b^2/e^4*dilog(1/2*I*(d*x+c-I))-1/6
*I*b^2/e^4*ln(d*x+c-I)*ln(-1/2*I*(d*x+c+I))-1/3*I*b^2/e^4*ln(d*x+c)*ln(1+I*(d*x+c))-1/12*I*b^2/e^4*ln(d*x+c-I)
^2+1/6*I*b^2/e^4*ln(d*x+c+I)*ln(1/2*I*(d*x+c-I))+1/6*I*b^2/e^4*ln(d*x+c-I)*ln(1+(d*x+c)^2)+1/3*I*b^2/e^4*dilog
(1-I*(d*x+c))+1/3*I*b^2/e^4*ln(d*x+c)*ln(1-I*(d*x+c))-1/3*b^2/e^4/(d*x+c)-1/3*b^2/e^4*arctan(d*x+c)+1/12*I*b^2
/e^4*ln(d*x+c+I)^2-1/6*I*b^2/e^4*ln(d*x+c+I)*ln(1+(d*x+c)^2)-1/3*I*b^2/e^4*dilog(1+I*(d*x+c))-1/6*I*b^2/e^4*di
log(-1/2*I*(d*x+c+I))-2/3*a*b/e^4/(d*x+c)^3*arctan(d*x+c)+1/3*a*b/e^4*ln(1+(d*x+c)^2)-1/3*a*b/e^4/(d*x+c)^2-2/
3*a*b/e^4*ln(d*x+c))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d*(e^(-4)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2 - 2*e^(-4)*log(d*x + c)/d^2 - 1/(d^4*x^2*e^4 + 2*c*d^3*x*e
^4 + c^2*d^2*e^4)) - 2*arctan(d*x + c)/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4))*a*b - 1/
48*(4*arctan(d*x + c)^2 - 48*(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4)*integrate(1/48*(36*
(d^2*x^2 + 2*c*d*x + c^2 + 1)*arctan(d*x + c)^2 + 3*(d^2*x^2 + 2*c*d*x + c^2 + 1)*log(d^2*x^2 + 2*c*d*x + c^2
+ 1)^2 + 8*(d*x + c)*arctan(d*x + c) - 4*(d^2*x^2 + 2*c*d*x + c^2)*log(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^6*x^6*
e^4 + 6*c*d^5*x^5*e^4 + (15*c^2*e^4 + e^4)*d^4*x^4 + 4*(5*c^3*e^4 + c*e^4)*d^3*x^3 + c^6*e^4 + 3*(5*c^4*e^4 +
2*c^2*e^4)*d^2*x^2 + c^4*e^4 + 2*(3*c^5*e^4 + 2*c^3*e^4)*d*x), x) - log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2)*b^2/(d
^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4) - 1/3*a^2/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d
^2*x*e^4 + c^3*d*e^4)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctan(d*x + c)^2 + 2*a*b*arctan(d*x + c) + a^2)*e^(-4)/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 +
4*c^3*d*x + c^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**2*atan(c +
d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(2*a*b*atan(c + d*x)
/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^2/(c*e + d*e*x)^4,x)

[Out]

int((a + b*atan(c + d*x))^2/(c*e + d*e*x)^4, x)

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