3.2.84 \(\int (e+f x)^2 \cot ^{-1}(\tanh (a+b x)) \, dx\) [184]

Optimal. Leaf size=229 \[ \frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {(e+f x)^3 \text {ArcTan}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {i (e+f x)^2 \text {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x)^2 \text {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}+\frac {i f (e+f x) \text {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f (e+f x) \text {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f^2 \text {PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}+\frac {i f^2 \text {PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3} \]

[Out]

1/3*(f*x+e)^3*arccot(tanh(b*x+a))/f+1/3*(f*x+e)^3*arctan(exp(2*b*x+2*a))/f-1/4*I*(f*x+e)^2*polylog(2,-I*exp(2*
b*x+2*a))/b+1/4*I*(f*x+e)^2*polylog(2,I*exp(2*b*x+2*a))/b+1/4*I*f*(f*x+e)*polylog(3,-I*exp(2*b*x+2*a))/b^2-1/4
*I*f*(f*x+e)*polylog(3,I*exp(2*b*x+2*a))/b^2-1/8*I*f^2*polylog(4,-I*exp(2*b*x+2*a))/b^3+1/8*I*f^2*polylog(4,I*
exp(2*b*x+2*a))/b^3

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Rubi [A]
time = 0.12, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5292, 4265, 2611, 6744, 2320, 6724} \begin {gather*} \frac {(e+f x)^3 \text {ArcTan}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {i f^2 \text {Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}+\frac {i f^2 \text {Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}+\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*ArcCot[Tanh[a + b*x]],x]

[Out]

((e + f*x)^3*ArcCot[Tanh[a + b*x]])/(3*f) + ((e + f*x)^3*ArcTan[E^(2*a + 2*b*x)])/(3*f) - ((I/4)*(e + f*x)^2*P
olyLog[2, (-I)*E^(2*a + 2*b*x)])/b + ((I/4)*(e + f*x)^2*PolyLog[2, I*E^(2*a + 2*b*x)])/b + ((I/4)*f*(e + f*x)*
PolyLog[3, (-I)*E^(2*a + 2*b*x)])/b^2 - ((I/4)*f*(e + f*x)*PolyLog[3, I*E^(2*a + 2*b*x)])/b^2 - ((I/8)*f^2*Pol
yLog[4, (-I)*E^(2*a + 2*b*x)])/b^3 + ((I/8)*f^2*PolyLog[4, I*E^(2*a + 2*b*x)])/b^3

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5292

Int[ArcCot[Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCot[T
anh[a + b*x]]/(f*(m + 1))), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Sech[2*a + 2*b*x], x], x] /; FreeQ[
{a, b, e, f}, x] && IGtQ[m, 0]

Rule 6724

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

Rule 6744

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

Rubi steps

\begin {align*} \int (e+f x)^2 \cot ^{-1}(\tanh (a+b x)) \, dx &=\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {b \int (e+f x)^3 \text {sech}(2 a+2 b x) \, dx}{3 f}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {1}{2} i \int (e+f x)^2 \log \left (1-i e^{2 a+2 b x}\right ) \, dx+\frac {1}{2} i \int (e+f x)^2 \log \left (1+i e^{2 a+2 b x}\right ) \, dx\\ &=\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac {(i f) \int (e+f x) \text {Li}_2\left (-i e^{2 a+2 b x}\right ) \, dx}{2 b}-\frac {(i f) \int (e+f x) \text {Li}_2\left (i e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}-\frac {\left (i f^2\right ) \int \text {Li}_3\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b^2}+\frac {\left (i f^2\right ) \int \text {Li}_3\left (i e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}-\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}+\frac {\left (i f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac {(e+f x)^3 \cot ^{-1}(\tanh (a+b x))}{3 f}+\frac {(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac {i f (e+f x) \text {Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f (e+f x) \text {Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}-\frac {i f^2 \text {Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}+\frac {i f^2 \text {Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 375, normalized size = 1.64 \begin {gather*} \frac {1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \cot ^{-1}(\tanh (a+b x))+\frac {i \left (12 b^3 e^2 x \log \left (1-i e^{2 (a+b x)}\right )+12 b^3 e f x^2 \log \left (1-i e^{2 (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )-12 b^3 e^2 x \log \left (1+i e^{2 (a+b x)}\right )-12 b^3 e f x^2 \log \left (1+i e^{2 (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )-6 b^2 (e+f x)^2 \text {PolyLog}\left (2,-i e^{2 (a+b x)}\right )+6 b^2 (e+f x)^2 \text {PolyLog}\left (2,i e^{2 (a+b x)}\right )+6 b e f \text {PolyLog}\left (3,-i e^{2 (a+b x)}\right )+6 b f^2 x \text {PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b e f \text {PolyLog}\left (3,i e^{2 (a+b x)}\right )-6 b f^2 x \text {PolyLog}\left (3,i e^{2 (a+b x)}\right )-3 f^2 \text {PolyLog}\left (4,-i e^{2 (a+b x)}\right )+3 f^2 \text {PolyLog}\left (4,i e^{2 (a+b x)}\right )\right )}{24 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^2*ArcCot[Tanh[a + b*x]],x]

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcCot[Tanh[a + b*x]])/3 + ((I/24)*(12*b^3*e^2*x*Log[1 - I*E^(2*(a + b*x))] + 1
2*b^3*e*f*x^2*Log[1 - I*E^(2*(a + b*x))] + 4*b^3*f^2*x^3*Log[1 - I*E^(2*(a + b*x))] - 12*b^3*e^2*x*Log[1 + I*E
^(2*(a + b*x))] - 12*b^3*e*f*x^2*Log[1 + I*E^(2*(a + b*x))] - 4*b^3*f^2*x^3*Log[1 + I*E^(2*(a + b*x))] - 6*b^2
*(e + f*x)^2*PolyLog[2, (-I)*E^(2*(a + b*x))] + 6*b^2*(e + f*x)^2*PolyLog[2, I*E^(2*(a + b*x))] + 6*b*e*f*Poly
Log[3, (-I)*E^(2*(a + b*x))] + 6*b*f^2*x*PolyLog[3, (-I)*E^(2*(a + b*x))] - 6*b*e*f*PolyLog[3, I*E^(2*(a + b*x
))] - 6*b*f^2*x*PolyLog[3, I*E^(2*(a + b*x))] - 3*f^2*PolyLog[4, (-I)*E^(2*(a + b*x))] + 3*f^2*PolyLog[4, I*E^
(2*(a + b*x))]))/b^3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 13.66, size = 5425, normalized size = 23.69

method	result	size
risch	\(\text {Expression too large to display}\)	\(5425\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*arccot(tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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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((f*x+e)^2*arccot(tanh(b*x+a)),x, algorithm="maxima")

[Out]

1/3*(f^2*x^3 + 3*f*x^2*e + 3*x*e^2)*arctan2(e^(2*b*x + 2*a) + 1, e^(2*b*x + 2*a) - 1) + integrate(2/3*(b*f^2*x
^3*e^(2*a) + 3*b*f*x^2*e^(2*a + 1) + 3*b*x*e^(2*a + 2))*e^(2*b*x)/(e^(4*b*x + 4*a) + 1), x)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1511 vs. \(2 (186) = 372\).
time = 2.90, size = 1511, normalized size = 6.60 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*arccot(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/6*(6*I*f^2*polylog(4, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 6*I*f^2*polylog(4, -1/2*sqrt(4*I)*(co
sh(b*x + a) + sinh(b*x + a))) - 6*I*f^2*polylog(4, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 6*I*f^2*p
olylog(4, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*(b^3*f^2*x^3 + 3*b^3*f*x^2*cosh(1) + 3*b^3*x*co
sh(1)^2 + 3*b^3*x*sinh(1)^2 + 3*(b^3*f*x^2 + 2*b^3*x*cosh(1))*sinh(1))*arctan(cosh(b*x + a)/sinh(b*x + a)) - 3
*(-I*b^2*f^2*x^2 - 2*I*b^2*f*x*cosh(1) - I*b^2*cosh(1)^2 - I*b^2*sinh(1)^2 - 2*I*(b^2*f*x + b^2*cosh(1))*sinh(
1))*dilog(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 3*(-I*b^2*f^2*x^2 - 2*I*b^2*f*x*cosh(1) - I*b^2*cos
h(1)^2 - I*b^2*sinh(1)^2 - 2*I*(b^2*f*x + b^2*cosh(1))*sinh(1))*dilog(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x
+ a))) - 3*(I*b^2*f^2*x^2 + 2*I*b^2*f*x*cosh(1) + I*b^2*cosh(1)^2 + I*b^2*sinh(1)^2 + 2*I*(b^2*f*x + b^2*cosh
(1))*sinh(1))*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 3*(I*b^2*f^2*x^2 + 2*I*b^2*f*x*cosh(1) +
I*b^2*cosh(1)^2 + I*b^2*sinh(1)^2 + 2*I*(b^2*f*x + b^2*cosh(1))*sinh(1))*dilog(-1/2*sqrt(-4*I)*(cosh(b*x + a)
+ sinh(b*x + a))) + (I*b^3*f^2*x^3 + I*a^3*f^2 + 3*I*(b^3*x + a*b^2)*cosh(1)^2 + 3*I*(b^3*x + a*b^2)*sinh(1)^
2 + 3*I*(b^3*f*x^2 - a^2*b*f)*cosh(1) + 3*I*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log(1/2
*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^3*f^2*x^3 + I*a^3*f^2 + 3*I*(b^3*x + a*b^2)*cosh(1)^2 +
3*I*(b^3*x + a*b^2)*sinh(1)^2 + 3*I*(b^3*f*x^2 - a^2*b*f)*cosh(1) + 3*I*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b
^2)*cosh(1))*sinh(1))*log(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b^3*f^2*x^3 - I*a^3*f^2 -
3*I*(b^3*x + a*b^2)*cosh(1)^2 - 3*I*(b^3*x + a*b^2)*sinh(1)^2 - 3*I*(b^3*f*x^2 - a^2*b*f)*cosh(1) - 3*I*(b^3*f
*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) +
(-I*b^3*f^2*x^3 - I*a^3*f^2 - 3*I*(b^3*x + a*b^2)*cosh(1)^2 - 3*I*(b^3*x + a*b^2)*sinh(1)^2 - 3*I*(b^3*f*x^2
- a^2*b*f)*cosh(1) - 3*I*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log(-1/2*sqrt(-4*I)*(cosh(
b*x + a) + sinh(b*x + a)) + 1) + (-I*a^3*f^2 + 3*I*a^2*b*f*cosh(1) - 3*I*a*b^2*cosh(1)^2 - 3*I*a*b^2*sinh(1)^2
+ 3*I*(a^2*b*f - 2*a*b^2*cosh(1))*sinh(1))*log(I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-I*a^3*f^2
+ 3*I*a^2*b*f*cosh(1) - 3*I*a*b^2*cosh(1)^2 - 3*I*a*b^2*sinh(1)^2 + 3*I*(a^2*b*f - 2*a*b^2*cosh(1))*sinh(1))*
log(-I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (I*a^3*f^2 - 3*I*a^2*b*f*cosh(1) + 3*I*a*b^2*cosh(1)^2
+ 3*I*a*b^2*sinh(1)^2 - 3*I*(a^2*b*f - 2*a*b^2*cosh(1))*sinh(1))*log(I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh(
b*x + a)) + (I*a^3*f^2 - 3*I*a^2*b*f*cosh(1) + 3*I*a*b^2*cosh(1)^2 + 3*I*a*b^2*sinh(1)^2 - 3*I*(a^2*b*f - 2*a*
b^2*cosh(1))*sinh(1))*log(-I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) - 6*(I*b*f^2*x + I*b*f*cosh(1) +
I*b*f*sinh(1))*polylog(3, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 6*(I*b*f^2*x + I*b*f*cosh(1) + I*b*
f*sinh(1))*polylog(3, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 6*(-I*b*f^2*x - I*b*f*cosh(1) - I*b*f*
sinh(1))*polylog(3, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 6*(-I*b*f^2*x - I*b*f*cosh(1) - I*b*f*si
nh(1))*polylog(3, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*acot(tanh(b*x+a)),x)

[Out]

Integral((e + f*x)**2*acot(tanh(a + b*x)), 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((f*x+e)^2*arccot(tanh(b*x+a)),x, algorithm="giac")

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(tanh(a + b*x))*(e + f*x)^2,x)

[Out]

int(acot(tanh(a + b*x))*(e + f*x)^2, x)

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