3.318 \(\int x \tanh ^{-1}(c+d \tan (a+b x)) \, dx\)

Optimal. Leaf size=295 \[ \frac {\text {Li}_3\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{8 b^2}-\frac {\text {Li}_3\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{8 b^2}-\frac {i x \text {Li}_2\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b}+\frac {i x \text {Li}_2\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b}+\frac {1}{4} x^2 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )-\frac {1}{4} x^2 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )+\frac {1}{2} x^2 \tanh ^{-1}(d \tan (a+b x)+c) \]

[Out]

1/2*x^2*arctanh(c+d*tan(b*x+a))+1/4*x^2*ln(1+(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))-1/4*x^2*ln(1+(1+c-I*d)*ex
p(2*I*a+2*I*b*x)/(1+c+I*d))-1/4*I*x*polylog(2,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b+1/4*I*x*polylog(2,-(1
+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b+1/8*polylog(3,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b^2-1/8*polylog
(3,-(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b^2

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Rubi [A]  time = 0.40, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6267, 2190, 2531, 2282, 6589} \[ \frac {\text {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{8 b^2}-\frac {\text {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{8 b^2}-\frac {i x \text {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b}+\frac {i x \text {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b}+\frac {1}{4} x^2 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )-\frac {1}{4} x^2 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )+\frac {1}{2} x^2 \tanh ^{-1}(d \tan (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[c + d*Tan[a + b*x]],x]

[Out]

(x^2*ArcTanh[c + d*Tan[a + b*x]])/2 + (x^2*Log[1 + ((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d)])/4 -
(x^2*Log[1 + ((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d)])/4 - ((I/4)*x*PolyLog[2, -(((1 - c + I*d)
*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/b + ((I/4)*x*PolyLog[2, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(
1 + c + I*d))])/b + PolyLog[3, -(((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))]/(8*b^2) - PolyLog[3,
-(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d))]/(8*b^2)

Rule 2190

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

Rule 2282

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 2531

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 6267

Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcTanh[c + d*Tan[a + b*x]])/(f*(m + 1)), x] + (-Dist[(I*b*(1 + c - I*d))/(f*(m + 1)), Int[((e + f*x)^(m
+ 1)*E^(2*I*a + 2*I*b*x))/(1 + c + I*d + (1 + c - I*d)*E^(2*I*a + 2*I*b*x)), x], x] + Dist[(I*b*(1 - c + I*d)
)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*E^(2*I*a + 2*I*b*x))/(1 - c - I*d + (1 - c + I*d)*E^(2*I*a + 2*I*b*x)),
x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(c + I*d)^2, 1]

Rule 6589

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]

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 257, normalized size = 0.87 \[ \frac {1}{2} x^2 \tanh ^{-1}(d \tan (a+b x)+c)+\frac {2 b^2 x^2 \log \left (1+\frac {(c-i d-1) e^{2 i (a+b x)}}{c+i d-1}\right )-2 b^2 x^2 \log \left (1+\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )-2 i b x \text {Li}_2\left (\frac {(-c+i d+1) e^{2 i (a+b x)}}{c+i d-1}\right )+2 i b x \text {Li}_2\left (-\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )+\text {Li}_3\left (\frac {(-c+i d+1) e^{2 i (a+b x)}}{c+i d-1}\right )-\text {Li}_3\left (-\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[c + d*Tan[a + b*x]],x]

[Out]

(x^2*ArcTanh[c + d*Tan[a + b*x]])/2 + (2*b^2*x^2*Log[1 + ((-1 + c - I*d)*E^((2*I)*(a + b*x)))/(-1 + c + I*d)]
- 2*b^2*x^2*Log[1 + ((1 + c - I*d)*E^((2*I)*(a + b*x)))/(1 + c + I*d)] - (2*I)*b*x*PolyLog[2, ((1 - c + I*d)*E
^((2*I)*(a + b*x)))/(-1 + c + I*d)] + (2*I)*b*x*PolyLog[2, -(((1 + c - I*d)*E^((2*I)*(a + b*x)))/(1 + c + I*d)
)] + PolyLog[3, ((1 - c + I*d)*E^((2*I)*(a + b*x)))/(-1 + c + I*d)] - PolyLog[3, -(((1 + c - I*d)*E^((2*I)*(a
+ b*x)))/(1 + c + I*d))])/(8*b^2)

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fricas [C]  time = 1.06, size = 1689, normalized size = 5.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(c+d*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/16*(4*b^2*x^2*log(-(d*tan(b*x + a) + c + 1)/(d*tan(b*x + a) + c - 1)) + 2*I*b*x*dilog(-((2*I*(c + 1)*d + 2*d
^2)*tan(b*x + a)^2 + 2*c^2 - 2*I*(c + 1)*d + (2*I*c^2 + 4*(c + 1)*d - 2*I*d^2 + 4*I*c + 2*I)*tan(b*x + a) + 4*
c + 2)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1) - 2*I*b*x*dilog(-((-2*I*(c + 1)*d + 2
*d^2)*tan(b*x + a)^2 + 2*c^2 + 2*I*(c + 1)*d + (-2*I*c^2 + 4*(c + 1)*d + 2*I*d^2 - 4*I*c - 2*I)*tan(b*x + a) +
4*c + 2)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1) + 2*I*b*x*dilog((2*(I*(c - 1)*d -
d^2)*tan(b*x + a)^2 - 2*c^2 - 2*I*(c - 1)*d - (-2*I*c^2 + 4*(c - 1)*d + 2*I*d^2 + 4*I*c - 2*I)*tan(b*x + a) +
4*c - 2)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1) + 1) - 2*I*b*x*dilog((2*(-I*(c - 1)*d -
d^2)*tan(b*x + a)^2 - 2*c^2 + 2*I*(c - 1)*d - (2*I*c^2 + 4*(c - 1)*d - 2*I*d^2 - 4*I*c + 2*I)*tan(b*x + a) + 4
*c - 2)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1) + 1) - 2*a^2*log(((I*(c + 1)*d + d^2)*tan
(b*x + a)^2 - c^2 + I*(c + 1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/(tan(b*x + a)^2 + 1)) -
2*a^2*log(((I*(c + 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c + 1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) +
2*c + 1)/(tan(b*x + a)^2 + 1)) + 2*a^2*log(((I*(c - 1)*d + d^2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (I*c^2 +
I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/(tan(b*x + a)^2 + 1)) + 2*a^2*log(((I*(c - 1)*d - d^2)*tan(b*x + a
)^2 + c^2 + I*(c - 1)*d + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x + a) - 2*c + 1)/(tan(b*x + a)^2 + 1)) - 2*(b^2*x
^2 - a^2)*log(((2*I*(c + 1)*d + 2*d^2)*tan(b*x + a)^2 + 2*c^2 - 2*I*(c + 1)*d + (2*I*c^2 + 4*(c + 1)*d - 2*I*d
^2 + 4*I*c + 2*I)*tan(b*x + a) + 4*c + 2)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1)) - 2*(b
^2*x^2 - a^2)*log(((-2*I*(c + 1)*d + 2*d^2)*tan(b*x + a)^2 + 2*c^2 + 2*I*(c + 1)*d + (-2*I*c^2 + 4*(c + 1)*d +
2*I*d^2 - 4*I*c - 2*I)*tan(b*x + a) + 4*c + 2)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1))
+ 2*(b^2*x^2 - a^2)*log(-(2*(I*(c - 1)*d - d^2)*tan(b*x + a)^2 - 2*c^2 - 2*I*(c - 1)*d - (-2*I*c^2 + 4*(c - 1)
*d + 2*I*d^2 + 4*I*c - 2*I)*tan(b*x + a) + 4*c - 2)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c +
1)) + 2*(b^2*x^2 - a^2)*log(-(2*(-I*(c - 1)*d - d^2)*tan(b*x + a)^2 - 2*c^2 + 2*I*(c - 1)*d - (2*I*c^2 + 4*(c
- 1)*d - 2*I*d^2 - 4*I*c + 2*I)*tan(b*x + a) + 4*c - 2)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*
c + 1)) - polylog(3, ((c^2 + 2*I*(c + 1)*d - d^2 + 2*c + 1)*tan(b*x + a)^2 - c^2 - 2*I*(c + 1)*d + d^2 + (2*I*
c^2 - 4*(c + 1)*d - 2*I*d^2 + 4*I*c + 2*I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1)*tan(b*x + a)^2 + c^2
+ d^2 + 2*c + 1)) - polylog(3, ((c^2 - 2*I*(c + 1)*d - d^2 + 2*c + 1)*tan(b*x + a)^2 - c^2 + 2*I*(c + 1)*d +
d^2 + (-2*I*c^2 - 4*(c + 1)*d + 2*I*d^2 - 4*I*c - 2*I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1)*tan(b*x
+ a)^2 + c^2 + d^2 + 2*c + 1)) + polylog(3, ((c^2 + 2*I*(c - 1)*d - d^2 - 2*c + 1)*tan(b*x + a)^2 - c^2 - 2*I*
(c - 1)*d + d^2 + (2*I*c^2 - 4*(c - 1)*d - 2*I*d^2 - 4*I*c + 2*I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c +
1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1)) + polylog(3, ((c^2 - 2*I*(c - 1)*d - d^2 - 2*c + 1)*tan(b*x + a)^2 -
c^2 + 2*I*(c - 1)*d + d^2 + (-2*I*c^2 - 4*(c - 1)*d + 2*I*d^2 + 4*I*c - 2*I)*tan(b*x + a) + 2*c - 1)/((c^2 +
d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*c + 1)))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {artanh}\left (d \tan \left (b x + a\right ) + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(c+d*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arctanh(d*tan(b*x + a) + c), x)

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maple [C]  time = 4.95, size = 6593, normalized size = 22.35 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(c+d*tan(b*x+a)),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, b d \int -\frac {2 \, {\left (c^{2} + d^{2} - 1\right )} x^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c d x^{2} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (c^{2} + d^{2} - 1\right )} x^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + {\left (c^{2} - d^{2} - 1\right )} x^{2} \cos \left (2 \, b x + 2 \, a\right ) - {\left (2 \, c d x^{2} \sin \left (2 \, b x + 2 \, a\right ) - {\left (c^{2} - d^{2} - 1\right )} x^{2} \cos \left (2 \, b x + 2 \, a\right )\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (2 \, c d x^{2} \cos \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} - d^{2} - 1\right )} x^{2} \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (4 \, b x + 4 \, a\right )}{c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} + {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, c^{2} + 2 \, {\left (c^{4} + d^{4} - 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} - 2 \, c^{2} + 2 \, {\left (c^{4} - d^{4} - 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (c d^{3} + {\left (c^{3} - c\right )} d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + 4 \, {\left (c^{4} - d^{4} - 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (2 \, c d^{3} - 2 \, {\left (c^{3} - c\right )} d - 2 \, {\left (c d^{3} + {\left (c^{3} - c\right )} d\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (c^{4} - d^{4} - 2 \, c^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (4 \, b x + 4 \, a\right ) + 8 \, {\left (c d^{3} + {\left (c^{3} - c\right )} d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 1}\,{d x} + \frac {1}{8} \, x^{2} \log \left ({\left (c^{2} + d^{2} + 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + 4 \, {\left (c + 1\right )} d \sin \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} + d^{2} + 2 \, c + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + c^{2} + d^{2} + 2 \, {\left (c^{2} - d^{2} + 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, c + 1\right ) - \frac {1}{8} \, x^{2} \log \left ({\left (c^{2} + d^{2} - 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + 4 \, {\left (c - 1\right )} d \sin \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} + d^{2} - 2 \, c + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + c^{2} + d^{2} + 2 \, {\left (c^{2} - d^{2} - 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, c + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(c+d*tan(b*x+a)),x, algorithm="maxima")

[Out]

-2*b*d*integrate(-(2*(c^2 + d^2 - 1)*x^2*cos(2*b*x + 2*a)^2 + 2*c*d*x^2*sin(2*b*x + 2*a) + 2*(c^2 + d^2 - 1)*x
^2*sin(2*b*x + 2*a)^2 + (c^2 - d^2 - 1)*x^2*cos(2*b*x + 2*a) - (2*c*d*x^2*sin(2*b*x + 2*a) - (c^2 - d^2 - 1)*x
^2*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*d*x^2*cos(2*b*x + 2*a) + (c^2 - d^2 - 1)*x^2*sin(2*b*x + 2*a))*si
n(4*b*x + 4*a))/(c^4 + d^4 + 2*(c^2 + 1)*d^2 + (c^4 + d^4 + 2*(c^2 + 1)*d^2 - 2*c^2 + 1)*cos(4*b*x + 4*a)^2 +
4*(c^4 + d^4 + 2*(c^2 - 1)*d^2 - 2*c^2 + 1)*cos(2*b*x + 2*a)^2 + (c^4 + d^4 + 2*(c^2 + 1)*d^2 - 2*c^2 + 1)*sin
(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 - 1)*d^2 - 2*c^2 + 1)*sin(2*b*x + 2*a)^2 - 2*c^2 + 2*(c^4 + d^4 - 2*(3
*c^2 - 1)*d^2 - 2*c^2 + 2*(c^4 - d^4 - 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(c*d^3 + (c^3 - c)*d)*sin(2*b*x + 2*a)
+ 1)*cos(4*b*x + 4*a) + 4*(c^4 - d^4 - 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(2*c*d^3 - 2*(c^3 - c)*d - 2*(c*d^3 + (
c^3 - c)*d)*cos(2*b*x + 2*a) - (c^4 - d^4 - 2*c^2 + 1)*sin(2*b*x + 2*a))*sin(4*b*x + 4*a) + 8*(c*d^3 + (c^3 -
c)*d)*sin(2*b*x + 2*a) + 1), x) + 1/8*x^2*log((c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c + 1)*d*sin(2*b*x
+ 2*a) + (c^2 + d^2 + 2*c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 + 2*(c^2 - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + 2*
c + 1) - 1/8*x^2*log((c^2 + d^2 - 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c - 1)*d*sin(2*b*x + 2*a) + (c^2 + d^2 - 2*
c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 + 2*(c^2 - d^2 - 2*c + 1)*cos(2*b*x + 2*a) - 2*c + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,\mathrm {atanh}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(c + d*tan(a + b*x)),x)

[Out]

int(x*atanh(c + d*tan(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {atanh}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(c+d*tan(b*x+a)),x)

[Out]

Integral(x*atanh(c + d*tan(a + b*x)), x)

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