3.3.31 \(\int (e+f x)^3 \coth ^{-1}(\tan (a+b x)) \, dx\) [231]

Optimal. Leaf size=302 \[ \frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \text {ArcTan}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 f^3 \text {PolyLog}\left (5,-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \text {PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4} \]

[Out]

1/4*(f*x+e)^4*arccoth(tan(b*x+a))/f+1/4*I*(f*x+e)^4*arctan(exp(2*I*(b*x+a)))/f-1/4*I*(f*x+e)^3*polylog(2,-I*ex
p(2*I*(b*x+a)))/b+1/4*I*(f*x+e)^3*polylog(2,I*exp(2*I*(b*x+a)))/b+3/8*f*(f*x+e)^2*polylog(3,-I*exp(2*I*(b*x+a)
))/b^2-3/8*f*(f*x+e)^2*polylog(3,I*exp(2*I*(b*x+a)))/b^2+3/8*I*f^2*(f*x+e)*polylog(4,-I*exp(2*I*(b*x+a)))/b^3-
3/8*I*f^2*(f*x+e)*polylog(4,I*exp(2*I*(b*x+a)))/b^3-3/16*f^3*polylog(5,-I*exp(2*I*(b*x+a)))/b^4+3/16*f^3*polyl
og(5,I*exp(2*I*(b*x+a)))/b^4

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Rubi [A]
time = 0.17, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6387, 4266, 2611, 6744, 2320, 6724} \begin {gather*} \frac {i (e+f x)^4 \text {ArcTan}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {3 f^3 \text {Li}_5\left (-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \text {Li}_5\left (i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^3*ArcCoth[Tan[a + b*x]],x]

[Out]

((e + f*x)^4*ArcCoth[Tan[a + b*x]])/(4*f) + ((I/4)*(e + f*x)^4*ArcTan[E^((2*I)*(a + b*x))])/f - ((I/4)*(e + f*
x)^3*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)^3*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (3*f*
(e + f*x)^2*PolyLog[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (3*f*(e + f*x)^2*PolyLog[3, I*E^((2*I)*(a + b*x))]
)/(8*b^2) + (((3*I)/8)*f^2*(e + f*x)*PolyLog[4, (-I)*E^((2*I)*(a + b*x))])/b^3 - (((3*I)/8)*f^2*(e + f*x)*Poly
Log[4, I*E^((2*I)*(a + b*x))])/b^3 - (3*f^3*PolyLog[5, (-I)*E^((2*I)*(a + b*x))])/(16*b^4) + (3*f^3*PolyLog[5,
I*E^((2*I)*(a + b*x))])/(16*b^4)

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 6387

Int[ArcCoth[Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCoth[
Tan[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)^3 \coth ^{-1}(\tan (a+b x)) \, dx &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}-\frac {b \int (e+f x)^4 \sec (2 a+2 b x) \, dx}{4 f}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac {1}{2} \int (e+f x)^3 \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac {1}{2} \int (e+f x)^3 \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(3 i f) \int (e+f x)^2 \text {Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}-\frac {(3 i f) \int (e+f x)^2 \text {Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {\left (3 f^2\right ) \int (e+f x) \text {Li}_3\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}+\frac {\left (3 f^2\right ) \int (e+f x) \text {Li}_3\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {\left (3 i f^3\right ) \int \text {Li}_4\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{8 b^3}+\frac {\left (3 i f^3\right ) \int \text {Li}_4\left (i e^{i (2 a+2 b x)}\right ) \, dx}{8 b^3}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {\left (3 f^3\right ) \text {Subst}\left (\int \frac {\text {Li}_4(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{16 b^4}+\frac {\left (3 f^3\right ) \text {Subst}\left (\int \frac {\text {Li}_4(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{16 b^4}\\ &=\frac {(e+f x)^4 \coth ^{-1}(\tan (a+b x))}{4 f}+\frac {i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac {i (e+f x)^3 \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x)^3 \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {3 f (e+f x)^2 \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {3 f (e+f x)^2 \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \text {Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \text {Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac {3 f^3 \text {Li}_5\left (-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac {3 f^3 \text {Li}_5\left (i e^{2 i (a+b x)}\right )}{16 b^4}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(654\) vs. \(2(302)=604\).
time = 0.18, size = 654, normalized size = 2.17 \begin {gather*} \frac {1}{4} x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right ) \coth ^{-1}(\tan (a+b x))+\frac {-8 b^4 e^3 x \log \left (1-i e^{2 i (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1-i e^{2 i (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1-i e^{2 i (a+b x)}\right )+8 b^4 e^3 x \log \left (1+i e^{2 i (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1+i e^{2 i (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1+i e^{2 i (a+b x)}\right )-4 i b^3 (e+f x)^3 \text {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+4 i b^3 (e+f x)^3 \text {PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b^2 e^2 f \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+12 b^2 e f^2 x \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )+6 b^2 f^3 x^2 \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b^2 e^2 f \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-12 b^2 e f^2 x \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )-6 b^2 f^3 x^2 \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )+6 i b e f^2 \text {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )+6 i b f^3 x \text {PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-6 i b e f^2 \text {PolyLog}\left (4,i e^{2 i (a+b x)}\right )-6 i b f^3 x \text {PolyLog}\left (4,i e^{2 i (a+b x)}\right )-3 f^3 \text {PolyLog}\left (5,-i e^{2 i (a+b x)}\right )+3 f^3 \text {PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^3*ArcCoth[Tan[a + b*x]],x]

[Out]

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*ArcCoth[Tan[a + b*x]])/4 + (-8*b^4*e^3*x*Log[1 - I*E^((2*I)*(a
+ b*x))] - 12*b^4*e^2*f*x^2*Log[1 - I*E^((2*I)*(a + b*x))] - 8*b^4*e*f^2*x^3*Log[1 - I*E^((2*I)*(a + b*x))] -
2*b^4*f^3*x^4*Log[1 - I*E^((2*I)*(a + b*x))] + 8*b^4*e^3*x*Log[1 + I*E^((2*I)*(a + b*x))] + 12*b^4*e^2*f*x^2*L
og[1 + I*E^((2*I)*(a + b*x))] + 8*b^4*e*f^2*x^3*Log[1 + I*E^((2*I)*(a + b*x))] + 2*b^4*f^3*x^4*Log[1 + I*E^((2
*I)*(a + b*x))] - (4*I)*b^3*(e + f*x)^3*PolyLog[2, (-I)*E^((2*I)*(a + b*x))] + (4*I)*b^3*(e + f*x)^3*PolyLog[2
, I*E^((2*I)*(a + b*x))] + 6*b^2*e^2*f*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] + 12*b^2*e*f^2*x*PolyLog[3, (-I)*E
^((2*I)*(a + b*x))] + 6*b^2*f^3*x^2*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] - 6*b^2*e^2*f*PolyLog[3, I*E^((2*I)*(
a + b*x))] - 12*b^2*e*f^2*x*PolyLog[3, I*E^((2*I)*(a + b*x))] - 6*b^2*f^3*x^2*PolyLog[3, I*E^((2*I)*(a + b*x))
] + (6*I)*b*e*f^2*PolyLog[4, (-I)*E^((2*I)*(a + b*x))] + (6*I)*b*f^3*x*PolyLog[4, (-I)*E^((2*I)*(a + b*x))] -
(6*I)*b*e*f^2*PolyLog[4, I*E^((2*I)*(a + b*x))] - (6*I)*b*f^3*x*PolyLog[4, I*E^((2*I)*(a + b*x))] - 3*f^3*Poly
Log[5, (-I)*E^((2*I)*(a + b*x))] + 3*f^3*PolyLog[5, I*E^((2*I)*(a + b*x))])/(16*b^4)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 19.10, size = 7429, normalized size = 24.60

method	result	size
risch	\(\text {Expression too large to display}\)	\(7429\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*arccoth(tan(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)^3*arccoth(tan(b*x+a)),x, algorithm="maxima")

[Out]

1/16*(f^3*x^4 + 4*f^2*x^3*e + 6*f*x^2*e^2 + 4*x*e^3)*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) - 1/16*(f^3*x^4 + 4*f^2*x^3*e + 6*f*x^2*e^2 + 4*x*e^3)*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(1/2*((b*f^3*x^4 + 4*b*f^2*x^3*e + 6*b*f*x^2*e^2 + 4*b*x*e^3)*c
os(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f^3*x^4 + 4*b*f^2*x^3*e + 6*b*f*x^2*e^2 + 4*b*x*e^3)*sin(4*b*x + 4*a)*si
n(2*b*x + 2*a) + (b*f^3*x^4 + 4*b*f^2*x^3*e + 6*b*f*x^2*e^2 + 4*b*x*e^3)*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. 3040 vs. \(2 (244) = 488\).
time = 0.47, size = 3040, normalized size = 10.07 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arccoth(tan(b*x+a)),x, algorithm="fricas")

[Out]

-1/32*(3*f^3*polylog(5, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) - 3*f^3*polylog(5, (I*ta
n(b*x + a)^2 - 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + 3*f^3*polylog(5, (-I*tan(b*x + a)^2 + 2*tan(b*x + a
) + I)/(tan(b*x + a)^2 + 1)) - 3*f^3*polylog(5, (-I*tan(b*x + a)^2 - 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1))
+ 4*(-I*b^3*f^3*x^3 - 3*I*b^3*f^2*x^2*cosh(1) - 3*I*b^3*f*x*cosh(1)^2 - I*b^3*cosh(1)^3 - I*b^3*sinh(1)^3 - 3
*I*(b^3*f*x + b^3*cosh(1))*sinh(1)^2 - 3*I*(b^3*f^2*x^2 + 2*b^3*f*x*cosh(1) + b^3*cosh(1)^2)*sinh(1))*dilog(-(
(I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(-I*b^3*f^3*x^3 - 3*I*b^3*f^2*x
^2*cosh(1) - 3*I*b^3*f*x*cosh(1)^2 - I*b^3*cosh(1)^3 - I*b^3*sinh(1)^3 - 3*I*(b^3*f*x + b^3*cosh(1))*sinh(1)^2
- 3*I*(b^3*f^2*x^2 + 2*b^3*f*x*cosh(1) + b^3*cosh(1)^2)*sinh(1))*dilog(-((I + 1)*tan(b*x + a)^2 - 2*tan(b*x +
a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 4*(I*b^3*f^3*x^3 + 3*I*b^3*f^2*x^2*cosh(1) + 3*I*b^3*f*x*cosh(1)^2 +
I*b^3*cosh(1)^3 + I*b^3*sinh(1)^3 + 3*I*(b^3*f*x + b^3*cosh(1))*sinh(1)^2 + 3*I*(b^3*f^2*x^2 + 2*b^3*f*x*cosh(
1) + b^3*cosh(1)^2)*sinh(1))*dilog(-(-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) +
1) + 4*(I*b^3*f^3*x^3 + 3*I*b^3*f^2*x^2*cosh(1) + 3*I*b^3*f*x*cosh(1)^2 + I*b^3*cosh(1)^3 + I*b^3*sinh(1)^3 +
3*I*(b^3*f*x + b^3*cosh(1))*sinh(1)^2 + 3*I*(b^3*f^2*x^2 + 2*b^3*f*x*cosh(1) + b^3*cosh(1)^2)*sinh(1))*dilog(-
(-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(b^4*f^3*x^4 - a^4*f^3 + 4*(b
^4*x + a*b^3)*cosh(1)^3 + 4*(b^4*x + a*b^3)*sinh(1)^3 + 6*(b^4*f*x^2 - a^2*b^2*f)*cosh(1)^2 + 6*(b^4*f*x^2 - a
^2*b^2*f + 2*(b^4*x + a*b^3)*cosh(1))*sinh(1)^2 + 4*(b^4*f^2*x^3 + a^3*b*f^2)*cosh(1) + 4*(b^4*f^2*x^3 + a^3*b
*f^2 + 3*(b^4*x + a*b^3)*cosh(1)^2 + 3*(b^4*f*x^2 - a^2*b^2*f)*cosh(1))*sinh(1))*log(((I + 1)*tan(b*x + a)^2 +
2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 2*(a^4*f^3 - 4*a^3*b*f^2*cosh(1) + 6*a^2*b^2*f*cosh(1)^2 - 4*
a*b^3*cosh(1)^3 - 4*a*b^3*sinh(1)^3 + 6*(a^2*b^2*f - 2*a*b^3*cosh(1))*sinh(1)^2 - 4*(a^3*b*f^2 - 3*a^2*b^2*f*c
osh(1) + 3*a*b^3*cosh(1)^2)*sinh(1))*log(((I + 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 +
1)) - 2*(a^4*f^3 - 4*a^3*b*f^2*cosh(1) + 6*a^2*b^2*f*cosh(1)^2 - 4*a*b^3*cosh(1)^3 - 4*a*b^3*sinh(1)^3 + 6*(a
^2*b^2*f - 2*a*b^3*cosh(1))*sinh(1)^2 - 4*(a^3*b*f^2 - 3*a^2*b^2*f*cosh(1) + 3*a*b^3*cosh(1)^2)*sinh(1))*log((
(I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) - 2*(b^4*f^3*x^4 - a^4*f^3 + 4*(b^4*x
+ a*b^3)*cosh(1)^3 + 4*(b^4*x + a*b^3)*sinh(1)^3 + 6*(b^4*f*x^2 - a^2*b^2*f)*cosh(1)^2 + 6*(b^4*f*x^2 - a^2*b
^2*f + 2*(b^4*x + a*b^3)*cosh(1))*sinh(1)^2 + 4*(b^4*f^2*x^3 + a^3*b*f^2)*cosh(1) + 4*(b^4*f^2*x^3 + a^3*b*f^2
+ 3*(b^4*x + a*b^3)*cosh(1)^2 + 3*(b^4*f*x^2 - a^2*b^2*f)*cosh(1))*sinh(1))*log(((I + 1)*tan(b*x + a)^2 - 2*t
an(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 2*(b^4*f^3*x^4 - a^4*f^3 + 4*(b^4*x + a*b^3)*cosh(1)^3 + 4*(b^4*x
+ a*b^3)*sinh(1)^3 + 6*(b^4*f*x^2 - a^2*b^2*f)*cosh(1)^2 + 6*(b^4*f*x^2 - a^2*b^2*f + 2*(b^4*x + a*b^3)*cosh(
1))*sinh(1)^2 + 4*(b^4*f^2*x^3 + a^3*b*f^2)*cosh(1) + 4*(b^4*f^2*x^3 + a^3*b*f^2 + 3*(b^4*x + a*b^3)*cosh(1)^2
+ 3*(b^4*f*x^2 - a^2*b^2*f)*cosh(1))*sinh(1))*log((-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x
+ a)^2 + 1)) - 2*(b^4*f^3*x^4 - a^4*f^3 + 4*(b^4*x + a*b^3)*cosh(1)^3 + 4*(b^4*x + a*b^3)*sinh(1)^3 + 6*(b^4*
f*x^2 - a^2*b^2*f)*cosh(1)^2 + 6*(b^4*f*x^2 - a^2*b^2*f + 2*(b^4*x + a*b^3)*cosh(1))*sinh(1)^2 + 4*(b^4*f^2*x^
3 + a^3*b*f^2)*cosh(1) + 4*(b^4*f^2*x^3 + a^3*b*f^2 + 3*(b^4*x + a*b^3)*cosh(1)^2 + 3*(b^4*f*x^2 - a^2*b^2*f)*
cosh(1))*sinh(1))*log((-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) + 2*(a^4*f^3 -
4*a^3*b*f^2*cosh(1) + 6*a^2*b^2*f*cosh(1)^2 - 4*a*b^3*cosh(1)^3 - 4*a*b^3*sinh(1)^3 + 6*(a^2*b^2*f - 2*a*b^3*c
osh(1))*sinh(1)^2 - 4*(a^3*b*f^2 - 3*a^2*b^2*f*cosh(1) + 3*a*b^3*cosh(1)^2)*sinh(1))*log(((I - 1)*tan(b*x + a)
^2 + 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) - 2*(a^4*f^3 - 4*a^3*b*f^2*cosh(1) + 6*a^2*b^2*f*cosh(1)^
2 - 4*a*b^3*cosh(1)^3 - 4*a*b^3*sinh(1)^3 + 6*(a^2*b^2*f - 2*a*b^3*cosh(1))*sinh(1)^2 - 4*(a^3*b*f^2 - 3*a^2*b
^2*f*cosh(1) + 3*a*b^3*cosh(1)^2)*sinh(1))*log(((I - 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I + 1)/(tan(b*x +
a)^2 + 1)) - 4*(b^4*f^3*x^4 + 4*b^4*f^2*x^3*cosh(1) + 6*b^4*f*x^2*cosh(1)^2 + 4*b^4*x*cosh(1)^3 + 4*b^4*x*sinh
(1)^3 + 6*(b^4*f*x^2 + 2*b^4*x*cosh(1))*sinh(1)^2 + 4*(b^4*f^2*x^3 + 3*b^4*f*x^2*cosh(1) + 3*b^4*x*cosh(1)^2)*
sinh(1))*log((tan(b*x + a) + 1)/(tan(b*x + a) - 1)) + 6*(-I*b*f^3*x - I*b*f^2*cosh(1) - I*b*f^2*sinh(1))*polyl
og(4, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + 6*(-I*b*f^3*x - I*b*f^2*cosh(1) - I*b*f^
2*sinh(1))*polylog(4, (I*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + 6*(I*b*f^3*x + I*b*f^2*c
osh(1) + I*b*f^2*sinh(1))*polylog(4, (-I*tan(b*x + a)^2 + 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) + 6*(I*b*f
^3*x + I*b*f^2*cosh(1) + I*b*f^2*sinh(1))*polyl...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e + f x\right )^{3} \operatorname {acoth}{\left (\tan {\left (a + b x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*acoth(tan(b*x+a)),x)

[Out]

Integral((e + f*x)**3*acoth(tan(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)^3*arccoth(tan(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*arccoth(tan(b*x + a)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \mathrm {acoth}\left (\mathrm {tan}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(tan(a + b*x))*(e + f*x)^3,x)

[Out]

int(acoth(tan(a + b*x))*(e + f*x)^3, x)

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