3.4.30 \(\int (e+f x)^2 \tanh ^{-1}(\cot (a+b x)) \, dx\) [330]

Optimal. Leaf size=234 \[ \frac {i (e+f x)^3 \text {ArcTan}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}-\frac {i (e+f x)^2 \text {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \text {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f (e+f x) \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac {i f^2 \text {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {i f^2 \text {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3} \]

[Out]

1/3*I*(f*x+e)^3*arctan(exp(2*I*(b*x+a)))/f+1/3*(f*x+e)^3*arctanh(cot(b*x+a))/f-1/4*I*(f*x+e)^2*polylog(2,-I*ex
p(2*I*(b*x+a)))/b+1/4*I*(f*x+e)^2*polylog(2,I*exp(2*I*(b*x+a)))/b+1/4*f*(f*x+e)*polylog(3,-I*exp(2*I*(b*x+a)))
/b^2-1/4*f*(f*x+e)*polylog(3,I*exp(2*I*(b*x+a)))/b^2+1/8*I*f^2*polylog(4,-I*exp(2*I*(b*x+a)))/b^3-1/8*I*f^2*po
lylog(4,I*exp(2*I*(b*x+a)))/b^3

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Rubi [A]
time = 0.12, antiderivative size = 234, 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 = {6388, 4266, 2611, 6744, 2320, 6724} \begin {gather*} \frac {i (e+f x)^3 \text {ArcTan}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {i f^2 \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {i f^2 \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}+\frac {f (e+f x) \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*ArcTanh[Cot[a + b*x]],x]

[Out]

((I/3)*(e + f*x)^3*ArcTan[E^((2*I)*(a + b*x))])/f + ((e + f*x)^3*ArcTanh[Cot[a + b*x]])/(3*f) - ((I/4)*(e + f*
x)^2*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)^2*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (f*(e
+ f*x)*PolyLog[3, (-I)*E^((2*I)*(a + b*x))])/(4*b^2) - (f*(e + f*x)*PolyLog[3, I*E^((2*I)*(a + b*x))])/(4*b^2
) + ((I/8)*f^2*PolyLog[4, (-I)*E^((2*I)*(a + b*x))])/b^3 - ((I/8)*f^2*PolyLog[4, I*E^((2*I)*(a + 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 4266

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

Rule 6388

Int[ArcTanh[Cot[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[
Cot[a + b*x]]/(f*(m + 1))), x] - Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Sec[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 \tanh ^{-1}(\cot (a+b x)) \, dx &=\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}-\frac {b \int (e+f x)^3 \sec (2 a+2 b x) \, dx}{3 f}\\ &=\frac {i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}+\frac {1}{2} \int (e+f x)^2 \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac {1}{2} \int (e+f x)^2 \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac {i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(i f) \int (e+f x) \text {Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{2 b}-\frac {(i f) \int (e+f x) \text {Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{2 b}\\ &=\frac {i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {f (e+f x) \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f^2 \int \text {Li}_3\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}+\frac {f^2 \int \text {Li}_3\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}\\ &=\frac {i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {f (e+f x) \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \text {Li}_3\left (i e^{2 i (a+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^{i (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^{i (2 a+2 b x)}\right )}{8 b^3}\\ &=\frac {i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac {(e+f x)^3 \tanh ^{-1}(\cot (a+b x))}{3 f}-\frac {i (e+f x)^2 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^2 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {f (e+f x) \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac {f (e+f x) \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}+\frac {i f^2 \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {i f^2 \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 409, normalized size = 1.75 \begin {gather*} \frac {1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \tanh ^{-1}(\cot (a+b x))+\frac {-12 b^3 e^2 x \log \left (1-i e^{2 i (a+b x)}\right )-12 b^3 e f x^2 \log \left (1-i e^{2 i (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e^2 x \log \left (1+i e^{2 i (a+b x)}\right )+12 b^3 e f x^2 \log \left (1+i e^{2 i (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )-6 i b^2 (e+f x)^2 \text {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+6 i b^2 (e+f x)^2 \text {PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b e f \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+6 b f^2 x \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b e f \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-6 b f^2 x \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )+3 i f^2 \text {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-3 i f^2 \text {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{24 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^2*ArcTanh[Cot[a + b*x]],x]

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcTanh[Cot[a + b*x]])/3 + (-12*b^3*e^2*x*Log[1 - I*E^((2*I)*(a + b*x))] - 12*b
^3*e*f*x^2*Log[1 - I*E^((2*I)*(a + b*x))] - 4*b^3*f^2*x^3*Log[1 - I*E^((2*I)*(a + b*x))] + 12*b^3*e^2*x*Log[1
+ I*E^((2*I)*(a + b*x))] + 12*b^3*e*f*x^2*Log[1 + I*E^((2*I)*(a + b*x))] + 4*b^3*f^2*x^3*Log[1 + I*E^((2*I)*(a
+ b*x))] - (6*I)*b^2*(e + f*x)^2*PolyLog[2, (-I)*E^((2*I)*(a + b*x))] + (6*I)*b^2*(e + f*x)^2*PolyLog[2, I*E^
((2*I)*(a + b*x))] + 6*b*e*f*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] + 6*b*f^2*x*PolyLog[3, (-I)*E^((2*I)*(a + b*
x))] - 6*b*e*f*PolyLog[3, I*E^((2*I)*(a + b*x))] - 6*b*f^2*x*PolyLog[3, I*E^((2*I)*(a + b*x))] + (3*I)*f^2*Pol
yLog[4, (-I)*E^((2*I)*(a + b*x))] - (3*I)*f^2*PolyLog[4, I*E^((2*I)*(a + b*x))])/(24*b^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 11.23, size = 5543, normalized size = 23.69

method	result	size
risch	\(\text {Expression too large to display}\)	\(5543\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*arctanh(cot(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*arctanh(cot(b*x+a)),x, algorithm="maxima")

[Out]

1/12*(f^2*x^3 + 3*f*x^2*e + 3*x*e^2)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 + 4*sin(2*b*x + 2*a) + 2)
- 1/12*(f^2*x^3 + 3*f*x^2*e + 3*x*e^2)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 - 4*sin(2*b*x + 2*a) +
2) - integrate(2/3*((b*f^2*x^3 + 3*b*f*x^2*e + 3*b*x*e^2)*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f^2*x^3 + 3*
b*f*x^2*e + 3*b*x*e^2)*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + (b*f^2*x^3 + 3*b*f*x^2*e + 3*b*x*e^2)*cos(2*b*x + 2
*a))/(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(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. 1590 vs. \(2 (186) = 372\).
time = 0.46, size = 1590, normalized size = 6.79 \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*arctanh(cot(b*x+a)),x, algorithm="fricas")

[Out]

1/48*(-3*I*f^2*polylog(4, I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 3*I*f^2*polylog(4, I*cos(2*b*x + 2*a) - sin
(2*b*x + 2*a)) + 3*I*f^2*polylog(4, -I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) + 3*I*f^2*polylog(4, -I*cos(2*b*x
+ 2*a) - sin(2*b*x + 2*a)) - 6*(-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(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 6*(-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(I*cos(2*b*x + 2*a)
- sin(2*b*x + 2*a)) - 6*(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(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 6*(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(-I*cos(2*b*x + 2*a) - sin
(2*b*x + 2*a)) + 8*(b^3*f^2*x^3 + 3*b^3*f*x^2*cosh(1) + 3*b^3*x*cosh(1)^2 + 3*b^3*x*sinh(1)^2 + 3*(b^3*f*x^2 +
2*b^3*x*cosh(1))*sinh(1))*log(-(cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1)/(cos(2*b*x + 2*a) - sin(2*b*x + 2*a)
+ 1)) + 4*(a^3*f^2 - 3*a^2*b*f*cosh(1) + 3*a*b^2*cosh(1)^2 + 3*a*b^2*sinh(1)^2 - 3*(a^2*b*f - 2*a*b^2*cosh(1)
)*sinh(1))*log(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + I) - 4*(a^3*f^2 - 3*a^2*b*f*cosh(1) + 3*a*b^2*cosh(1)^2
+ 3*a*b^2*sinh(1)^2 - 3*(a^2*b*f - 2*a*b^2*cosh(1))*sinh(1))*log(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) -
4*(b^3*f^2*x^3 + a^3*f^2 + 3*(b^3*x + a*b^2)*cosh(1)^2 + 3*(b^3*x + a*b^2)*sinh(1)^2 + 3*(b^3*f*x^2 - a^2*b*f
)*cosh(1) + 3*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log(I*cos(2*b*x + 2*a) + sin(2*b*x +
2*a) + 1) + 4*(b^3*f^2*x^3 + a^3*f^2 + 3*(b^3*x + a*b^2)*cosh(1)^2 + 3*(b^3*x + a*b^2)*sinh(1)^2 + 3*(b^3*f*x^
2 - a^2*b*f)*cosh(1) + 3*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log(I*cos(2*b*x + 2*a) - s
in(2*b*x + 2*a) + 1) - 4*(b^3*f^2*x^3 + a^3*f^2 + 3*(b^3*x + a*b^2)*cosh(1)^2 + 3*(b^3*x + a*b^2)*sinh(1)^2 +
3*(b^3*f*x^2 - a^2*b*f)*cosh(1) + 3*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log(-I*cos(2*b*
x + 2*a) + sin(2*b*x + 2*a) + 1) + 4*(b^3*f^2*x^3 + a^3*f^2 + 3*(b^3*x + a*b^2)*cosh(1)^2 + 3*(b^3*x + a*b^2)*
sinh(1)^2 + 3*(b^3*f*x^2 - a^2*b*f)*cosh(1) + 3*(b^3*f*x^2 - a^2*b*f + 2*(b^3*x + a*b^2)*cosh(1))*sinh(1))*log
(-I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a) + 1) + 4*(a^3*f^2 - 3*a^2*b*f*cosh(1) + 3*a*b^2*cosh(1)^2 + 3*a*b^2*si
nh(1)^2 - 3*(a^2*b*f - 2*a*b^2*cosh(1))*sinh(1))*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + I) - 4*(a^3*f^2
- 3*a^2*b*f*cosh(1) + 3*a*b^2*cosh(1)^2 + 3*a*b^2*sinh(1)^2 - 3*(a^2*b*f - 2*a*b^2*cosh(1))*sinh(1))*log(-cos(
2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) + 6*(b*f^2*x + b*f*cosh(1) + b*f*sinh(1))*polylog(3, I*cos(2*b*x + 2*a)
+ sin(2*b*x + 2*a)) - 6*(b*f^2*x + b*f*cosh(1) + b*f*sinh(1))*polylog(3, I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a
)) + 6*(b*f^2*x + b*f*cosh(1) + b*f*sinh(1))*polylog(3, -I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 6*(b*f^2*x +
b*f*cosh(1) + b*f*sinh(1))*polylog(3, -I*cos(2*b*x + 2*a) - sin(2*b*x + 2*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 {atanh}{\left (\cot {\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*atanh(cot(b*x+a)),x)

[Out]

Integral((e + f*x)**2*atanh(cot(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*arctanh(cot(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*arctanh(cot(b*x + a)), x)

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

[Out]

int(atanh(cot(a + b*x))*(e + f*x)^2, x)

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