3.13.50 \(\int (d+e x^2) (a+b \text {ArcTan}(c x))^2 \, dx\) [1250]

Optimal. Leaf size=231 \[ \frac {b^2 e x}{3 c^2}-\frac {b^2 e \text {ArcTan}(c x)}{3 c^3}-\frac {b e x^2 (a+b \text {ArcTan}(c x))}{3 c}+\frac {i d (a+b \text {ArcTan}(c x))^2}{c}-\frac {i e (a+b \text {ArcTan}(c x))^2}{3 c^3}+d x (a+b \text {ArcTan}(c x))^2+\frac {1}{3} e x^3 (a+b \text {ArcTan}(c x))^2+\frac {2 b d (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {2 b e (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^3}+\frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c}-\frac {i b^2 e \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^3} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {5034, 4930, 5040, 4964, 2449, 2352, 4946, 5036, 327, 209} \begin {gather*} -\frac {i e (a+b \text {ArcTan}(c x))^2}{3 c^3}-\frac {2 b e \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{3 c^3}+d x (a+b \text {ArcTan}(c x))^2+\frac {i d (a+b \text {ArcTan}(c x))^2}{c}+\frac {2 b d \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{c}+\frac {1}{3} e x^3 (a+b \text {ArcTan}(c x))^2-\frac {b e x^2 (a+b \text {ArcTan}(c x))}{3 c}-\frac {b^2 e \text {ArcTan}(c x)}{3 c^3}-\frac {i b^2 e \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{3 c^3}+\frac {b^2 e x}{3 c^2}+\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*e*x)/(3*c^2) - (b^2*e*ArcTan[c*x])/(3*c^3) - (b*e*x^2*(a + b*ArcTan[c*x]))/(3*c) + (I*d*(a + b*ArcTan[c*x
])^2)/c - ((I/3)*e*(a + b*ArcTan[c*x])^2)/c^3 + d*x*(a + b*ArcTan[c*x])^2 + (e*x^3*(a + b*ArcTan[c*x])^2)/3 +
(2*b*d*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c - (2*b*e*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) + (I
*b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - ((I/3)*b^2*e*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 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 5034

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcTan[c*x])^p, (d + e*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[q] && IGtQ[p, 0]

Rule 5036

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

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 208, normalized size = 0.90 \begin {gather*} \frac {3 a^2 c^3 d x+b^2 c e x-a b c^2 e x^2+a^2 c^3 e x^3+b^2 \left (-3 i c^2 d+i e+c^3 \left (3 d x+e x^3\right )\right ) \text {ArcTan}(c x)^2-b \text {ArcTan}(c x) \left (-2 a c^3 x \left (3 d+e x^2\right )+b \left (e+c^2 e x^2\right )+2 b \left (-3 c^2 d+e\right ) \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-3 a b c^2 d \log \left (1+c^2 x^2\right )+a b e \log \left (1+c^2 x^2\right )-i b^2 \left (3 c^2 d-e\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*x + b^2*c*e*x - a*b*c^2*e*x^2 + a^2*c^3*e*x^3 + b^2*((-3*I)*c^2*d + I*e + c^3*(3*d*x + e*x^3))*Ar
cTan[c*x]^2 - b*ArcTan[c*x]*(-2*a*c^3*x*(3*d + e*x^2) + b*(e + c^2*e*x^2) + 2*b*(-3*c^2*d + e)*Log[1 + E^((2*I
)*ArcTan[c*x])]) - 3*a*b*c^2*d*Log[1 + c^2*x^2] + a*b*e*Log[1 + c^2*x^2] - I*b^2*(3*c^2*d - e)*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. 549 vs. \(2 (209 ) = 418\).
time = 0.31, size = 550, normalized size = 2.38 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

1/3*a^2*x^3*e + 36*b^2*c^2*e*integrate(1/48*x^4*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*c^2*e*integrate(1/48*x
^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 4*b^2*c^2*e*integrate(1/48*x^4*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) +
36*b^2*c^2*d*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*c^2*d*integrate(1/48*x^2*log(c^2*x^2 +
1)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*d*integrate(1/48*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 1/4*b^2*d*arcta
n(c*x)^3/c - 8*b^2*c*e*integrate(1/48*x^3*arctan(c*x)/(c^2*x^2 + 1), x) - 24*b^2*c*d*integrate(1/48*x*arctan(c
*x)/(c^2*x^2 + 1), x) + a^2*d*x + 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*e + 36*b^2*
e*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*e*integrate(1/48*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2
+ 1), x) + 3*b^2*d*integrate(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/c + 1/12*(b^2*x^3*e + 3*b^2*d*x)*arctan(c*x)^2 - 1/48*(b^2*x^3*e + 3*b^2*d*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^2+d)*(a+b*arctan(c*x))^2,x, algorithm="fricas")

[Out]

integral(a^2*x^2*e + a^2*d + (b^2*x^2*e + b^2*d)*arctan(c*x)^2 + 2*(a*b*x^2*e + a*b*d)*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^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(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^2+d)*(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 (e\,x^2+d\right ) \,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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