[next] [prev] [prev-tail] [tail] [up]

3.4.17 \(\int x^2 \tanh ^{-1}(c+d \tan (a+b x)) \, dx\) [317]

Optimal. Leaf size=395 \[ \frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \text {PolyLog}\left (2,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x^2 \text {PolyLog}\left (2,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {x \text {PolyLog}\left (3,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {PolyLog}\left (3,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}+\frac {i \text {PolyLog}\left (4,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{8 b^3}-\frac {i \text {PolyLog}\left (4,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{8 b^3} \]

[Out]

1/3*x^3*arctanh(c+d*tan(b*x+a))+1/6*x^3*ln(1+(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))-1/6*x^3*ln(1+(1+c-I*d)*ex
p(2*I*a+2*I*b*x)/(1+c+I*d))-1/4*I*x^2*polylog(2,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b+1/4*I*x^2*polylog(2
,-(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b+1/4*x*polylog(3,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d))/b^2-1/4*x
*polylog(3,-(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b^2+1/8*I*polylog(4,-(1-c+I*d)*exp(2*I*a+2*I*b*x)/(1-c-I*d
))/b^3-1/8*I*polylog(4,-(1+c-I*d)*exp(2*I*a+2*I*b*x)/(1+c+I*d))/b^3

________________________________________________________________________________________

Rubi [A]
time = 0.36, antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6402, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {i \text {Li}_4\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{8 b^3}-\frac {i \text {Li}_4\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{8 b^3}+\frac {x \text {Li}_3\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b^2}-\frac {i x^2 \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^2 \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}{6} x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )+\frac {1}{3} x^3 \tanh ^{-1}(d \tan (a+b x)+c) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[c + d*Tan[a + b*x]],x]

[Out]

(x^3*ArcTanh[c + d*Tan[a + b*x]])/3 + (x^3*Log[1 + ((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d)])/6 -
 (x^3*Log[1 + ((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d)])/6 - ((I/4)*x^2*PolyLog[2, -(((1 - c + I*
d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/b + ((I/4)*x^2*PolyLog[2, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x
))/(1 + c + I*d))])/b + (x*PolyLog[3, -(((1 - c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/(4*b^2) - (x*
PolyLog[3, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c + I*d))])/(4*b^2) + ((I/8)*PolyLog[4, -(((1 - c +
I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c - I*d))])/b^3 - ((I/8)*PolyLog[4, -(((1 + c - I*d)*E^((2*I)*a + (2*I)*b*x
))/(1 + c + I*d))])/b^3

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 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 6402

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 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 x^2 \tanh ^{-1}(c+d \tan (a+b x)) \, dx &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{3} (b (i (1-c)-d)) \int \frac {e^{2 i a+2 i b x} x^3}{1-c-i d+(1-c+i d) e^{2 i a+2 i b x}} \, dx-\frac {1}{3} (b (i+i c+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+c+i d+(1+c-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \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^2 \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^2 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right ) \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \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^2 \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 x \text {Li}_2\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right ) \, dx}{2 b}-\frac {i \int x \text {Li}_2\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \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^2 \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 {x \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}-\frac {\int \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right ) \, dx}{4 b^2}+\frac {\int \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \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^2 \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 {x \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}+\frac {i \text {Subst}\left (\int \frac {\text {Li}_3\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^3}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_3\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^3}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \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^2 \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 {x \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}+\frac {i \text {Li}_4\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{8 b^3}-\frac {i \text {Li}_4\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{8 b^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.65, size = 346, normalized size = 0.88 \begin {gather*} \frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {4 b^3 x^3 \log \left (1+\frac {(-1+c-i d) e^{2 i (a+b x)}}{-1+c+i d}\right )-4 b^3 x^3 \log \left (1+\frac {(1+c-i d) e^{2 i (a+b x)}}{1+c+i d}\right )-6 i b^2 x^2 \text {PolyLog}\left (2,\frac {(1-c+i d) e^{2 i (a+b x)}}{-1+c+i d}\right )+6 i b^2 x^2 \text {PolyLog}\left (2,-\frac {(1+c-i d) e^{2 i (a+b x)}}{1+c+i d}\right )+6 b x \text {PolyLog}\left (3,\frac {(1-c+i d) e^{2 i (a+b x)}}{-1+c+i d}\right )-6 b x \text {PolyLog}\left (3,-\frac {(1+c-i d) e^{2 i (a+b x)}}{1+c+i d}\right )+3 i \text {PolyLog}\left (4,\frac {(1-c+i d) e^{2 i (a+b x)}}{-1+c+i d}\right )-3 i \text {PolyLog}\left (4,-\frac {(1+c-i d) e^{2 i (a+b x)}}{1+c+i d}\right )}{24 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[c + d*Tan[a + b*x]],x]

[Out]

(x^3*ArcTanh[c + d*Tan[a + b*x]])/3 + (4*b^3*x^3*Log[1 + ((-1 + c - I*d)*E^((2*I)*(a + b*x)))/(-1 + c + I*d)]
- 4*b^3*x^3*Log[1 + ((1 + c - I*d)*E^((2*I)*(a + b*x)))/(1 + c + I*d)] - (6*I)*b^2*x^2*PolyLog[2, ((1 - c + I*
d)*E^((2*I)*(a + b*x)))/(-1 + c + I*d)] + (6*I)*b^2*x^2*PolyLog[2, -(((1 + c - I*d)*E^((2*I)*(a + b*x)))/(1 +
c + I*d))] + 6*b*x*PolyLog[3, ((1 - c + I*d)*E^((2*I)*(a + b*x)))/(-1 + c + I*d)] - 6*b*x*PolyLog[3, -(((1 + c
 - I*d)*E^((2*I)*(a + b*x)))/(1 + c + I*d))] + (3*I)*PolyLog[4, ((1 - c + I*d)*E^((2*I)*(a + b*x)))/(-1 + c +
I*d)] - (3*I)*PolyLog[4, -(((1 + c - I*d)*E^((2*I)*(a + b*x)))/(1 + c + I*d))])/(24*b^3)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 25.55, size = 6895, normalized size = 17.46

method result size
risch \(\text {Expression too large to display}\) \(6895\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

[Out]

1/12*x^3*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)*s
in(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) - 1/12*x^3*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) - 4*b*d*integrate(-1/3*(2*(c^2 + d^2 - 1)*x^3*cos(
2*b*x + 2*a)^2 + 2*c*d*x^3*sin(2*b*x + 2*a) + 2*(c^2 + d^2 - 1)*x^3*sin(2*b*x + 2*a)^2 + (c^2 - d^2 - 1)*x^3*c
os(2*b*x + 2*a) - (2*c*d*x^3*sin(2*b*x + 2*a) - (c^2 - d^2 - 1)*x^3*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*
d*x^3*cos(2*b*x + 2*a) + (c^2 - d^2 - 1)*x^3*sin(2*b*x + 2*a))*sin(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)*c
os(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)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2165 vs. \(2 (279) = 558\).
time = 0.45, size = 2165, normalized size = 5.48 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(c+d*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/48*(8*b^3*x^3*log(-(d*tan(b*x + a) + c + 1)/(d*tan(b*x + a) + c - 1)) - 6*I*b^2*x^2*dilog(2*((I*(c + 1)*d -
d^2)*tan(b*x + a)^2 - c^2 - I*(c + 1)*d + (I*c^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + 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) + 1) + 6*I*b^2*x^2*dilog(2*((-I*(c + 1)*d - d^2)*ta
n(b*x + a)^2 - c^2 + I*(c + 1)*d + (-I*c^2 - 2*(c + 1)*d + I*d^2 - 2*I*c - 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) + 1) + 6*I*b^2*x^2*dilog(2*((I*(c - 1)*d - d^2)*tan(b*x +
 a)^2 - c^2 - I*(c - 1)*d + (I*c^2 - 2*(c - 1)*d - I*d^2 - 2*I*c + 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) + 1) - 6*I*b^2*x^2*dilog(2*((-I*(c - 1)*d - d^2)*tan(b*x + a)^2 -
 c^2 + I*(c - 1)*d + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - 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) + 1) + 4*a^3*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)) + 4*a^3*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)
) - 4*a^3*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)) - 4*a^3*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)) - 6*b*x*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 + 2*(c + 1)*d + I*d^2 - 2*I*c - I)*ta
n(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)) - 6*b*x*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 + 2*(c + 1)*d - I*d^2 +
2*I*c + 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)) + 6*b*x*polyl
og(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 + 2*(c - 1
)*d + I*d^2 + 2*I*c - 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))
 + 6*b*x*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 + 2*(c - 1)*d - I*d^2 - 2*I*c + 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)) - 4*(b^3*x^3 + a^3)*log(-2*((I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c + 1)*d + (I*c^2 - 2*
(c + 1)*d - I*d^2 + 2*I*c + 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)) - 4*(b^3*x^3 + a^3)*log(-2*((-I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1)*d + (-I*c^2 - 2*(c +
1)*d + I*d^2 - 2*I*c - 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)
) + 4*(b^3*x^3 + a^3)*log(-2*((I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c - 1)*d + (I*c^2 - 2*(c - 1)*d -
I*d^2 - 2*I*c + 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)) + 4*(
b^3*x^3 + a^3)*log(-2*((-I*(c - 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c - 1)*d + (-I*c^2 - 2*(c - 1)*d + I*d^2
 + 2*I*c - 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)) + 3*I*poly
log(4, ((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 + 2*(c +
1)*d + I*d^2 - 2*I*c - 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)
) - 3*I*polylog(4, ((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 + 2*(c + 1)*d - I*d^2 + 2*I*c + 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)) - 3*I*polylog(4, ((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 + 2*(c - 1)*d + I*d^2 + 2*I*c - 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)) + 3*I*polylog(4, ((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 + 2*(c - 1)*d - I*d^2 - 2*I*c + 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^3

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {atanh}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(c+d*tan(b*x+a)),x)

[Out]

Integral(x**2*atanh(c + d*tan(a + b*x)), x)

________________________________________________________________________________________

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(x^2*arctanh(c+d*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*arctanh(d*tan(b*x + a) + c), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(c + d*tan(a + b*x)),x)

[Out]

int(x^2*atanh(c + d*tan(a + b*x)), x)

________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]