3.1.10 \(\int (d+e x)^2 (a+b \text {ArcTan}(c x))^2 \, dx\) [10]

Optimal. Leaf size=270 \[ -\frac {2 a b d e x}{c}+\frac {b^2 e^2 x}{3 c^2}-\frac {b^2 e^2 \text {ArcTan}(c x)}{3 c^3}-\frac {2 b^2 d e x \text {ArcTan}(c x)}{c}-\frac {b e^2 x^2 (a+b \text {ArcTan}(c x))}{3 c}+\frac {i \left (3 c^2 d^2-e^2\right ) (a+b \text {ArcTan}(c x))^2}{3 c^3}-\frac {d \left (d^2-\frac {3 e^2}{c^2}\right ) (a+b \text {ArcTan}(c x))^2}{3 e}+\frac {(d+e x)^3 (a+b \text {ArcTan}(c x))^2}{3 e}+\frac {2 b \left (3 c^2 d^2-e^2\right ) (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^3}+\frac {b^2 d e \log \left (1+c^2 x^2\right )}{c^2}+\frac {i b^2 \left (3 c^2 d^2-e^2\right ) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^3} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4974, 4930, 266, 4946, 327, 209, 5104, 5004, 5040, 4964, 2449, 2352} \begin {gather*} -\frac {d \left (d^2-\frac {3 e^2}{c^2}\right ) (a+b \text {ArcTan}(c x))^2}{3 e}+\frac {i \left (3 c^2 d^2-e^2\right ) (a+b \text {ArcTan}(c x))^2}{3 c^3}+\frac {2 b \left (3 c^2 d^2-e^2\right ) \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{3 c^3}+\frac {(d+e x)^3 (a+b \text {ArcTan}(c x))^2}{3 e}-\frac {b e^2 x^2 (a+b \text {ArcTan}(c x))}{3 c}-\frac {2 a b d e x}{c}-\frac {b^2 e^2 \text {ArcTan}(c x)}{3 c^3}-\frac {2 b^2 d e x \text {ArcTan}(c x)}{c}+\frac {b^2 d e \log \left (c^2 x^2+1\right )}{c^2}+\frac {b^2 e^2 x}{3 c^2}+\frac {i b^2 \left (3 c^2 d^2-e^2\right ) \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{3 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 327

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4930

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

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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 4974

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

Rule 5004

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

Rule 5040

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*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rule 5104

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]
time = 0.35, size = 312, normalized size = 1.16 \begin {gather*} \frac {3 a^2 c^3 d^2 x-6 a b c^2 d e x+b^2 c e^2 x+3 a^2 c^3 d e x^2-a b c^2 e^2 x^2+a^2 c^3 e^2 x^3+b^2 \left (-3 i c^2 d^2+3 c d e+i e^2+c^3 x \left (3 d^2+3 d e x+e^2 x^2\right )\right ) \text {ArcTan}(c x)^2+b \text {ArcTan}(c x) \left (6 a c d e-b e \left (e+6 c^2 d x+c^2 e x^2\right )+2 a c^3 x \left (3 d^2+3 d e x+e^2 x^2\right )+2 b \left (3 c^2 d^2-e^2\right ) \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-3 a b c^2 d^2 \log \left (1+c^2 x^2\right )+3 b^2 c d e \log \left (1+c^2 x^2\right )+a b e^2 \log \left (1+c^2 x^2\right )-i b^2 \left (3 c^2 d^2-e^2\right ) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )}{3 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*c^3*d^2*x - 6*a*b*c^2*d*e*x + b^2*c*e^2*x + 3*a^2*c^3*d*e*x^2 - a*b*c^2*e^2*x^2 + a^2*c^3*e^2*x^3 + b^2
*((-3*I)*c^2*d^2 + 3*c*d*e + I*e^2 + c^3*x*(3*d^2 + 3*d*e*x + e^2*x^2))*ArcTan[c*x]^2 + b*ArcTan[c*x]*(6*a*c*d
*e - b*e*(e + 6*c^2*d*x + c^2*e*x^2) + 2*a*c^3*x*(3*d^2 + 3*d*e*x + e^2*x^2) + 2*b*(3*c^2*d^2 - e^2)*Log[1 + E
^((2*I)*ArcTan[c*x])]) - 3*a*b*c^2*d^2*Log[1 + c^2*x^2] + 3*b^2*c*d*e*Log[1 + c^2*x^2] + a*b*e^2*Log[1 + c^2*x
^2] - I*b^2*(3*c^2*d^2 - e^2)*PolyLog[2, -E^((2*I)*ArcTan[c*x])])/(3*c^3)

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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. 699 vs. \(2 (250 ) = 500\).
time = 0.23, size = 700, normalized size = 2.59 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

72*b^2*c^2*d*e*integrate(1/48*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^2*c^2*d*e*integrate(1/48*x^3*log(c^2*x
^2 + 1)^2/(c^2*x^2 + 1), x) + 36*b^2*c^2*d^2*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*d
*e*integrate(1/48*x^3*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3*b^2*c^2*d^2*integrate(1/48*x^2*log(c^2*x^2 + 1)^2
/(c^2*x^2 + 1), x) + 12*b^2*c^2*d^2*integrate(1/48*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 1/4*b^2*d^2*arctan
(c*x)^3/c + 1/3*a^2*x^3*e^2 + a^2*d*x^2*e + 36*b^2*c^2*e^2*integrate(1/48*x^4*arctan(c*x)^2/(c^2*x^2 + 1), x)
+ 3*b^2*c^2*e^2*integrate(1/48*x^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 4*b^2*c^2*e^2*integrate(1/48*x^4*log
(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - 24*b^2*c*d*e*integrate(1/48*x^2*arctan(c*x)/(c^2*x^2 + 1), x) - 24*b^2*c*d^2
*integrate(1/48*x*arctan(c*x)/(c^2*x^2 + 1), x) + a^2*d^2*x + 2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3)
)*a*b*d*e - 8*b^2*c*e^2*integrate(1/48*x^3*arctan(c*x)/(c^2*x^2 + 1), x) + 72*b^2*d*e*integrate(1/48*x*arctan(
c*x)^2/(c^2*x^2 + 1), x) + 6*b^2*d*e*integrate(1/48*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 3*b^2*d^2*integra
te(1/48*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d^2/c + 1/3*(2*x^3*a
rctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*e^2 + 36*b^2*e^2*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x^
2 + 1), x) + 3*b^2*e^2*integrate(1/48*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 1/12*(b^2*x^3*e^2 + 3*b^2*d*x
^2*e + 3*b^2*d^2*x)*arctan(c*x)^2 - 1/48*(b^2*x^3*e^2 + 3*b^2*d*x^2*e + 3*b^2*d^2*x)*log(c^2*x^2 + 1)^2

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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