### 3.12 $$\int x^2 \tan ^3(a+b x) \, dx$$

Optimal. Leaf size=128 $-\frac{i x \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x \tan (a+b x)}{b^2}-\frac{\log (\cos (a+b x))}{b^3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{x^2}{2 b}-\frac{i x^3}{3}$

[Out]

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

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Rubi [A]  time = 0.180095, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.667, Rules used = {3720, 3475, 30, 3719, 2190, 2531, 2282, 6589} $-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}+\frac{\text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x \tan (a+b x)}{b^2}-\frac{\log (\cos (a+b x))}{b^3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{x^2}{2 b}-\frac{i x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Tan[a + b*x]^3,x]

[Out]

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

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3719

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

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 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 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 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^2 \tan ^3(a+b x) \, dx &=\frac{x^2 \tan ^2(a+b x)}{2 b}-\frac{\int x \tan ^2(a+b x) \, dx}{b}-\int x^2 \tan (a+b x) \, dx\\ &=-\frac{i x^3}{3}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}} \, dx+\frac{\int \tan (a+b x) \, dx}{b^2}+\frac{\int x \, dx}{b}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}-\frac{2 \int x \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{i \int \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^3}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}+\frac{\text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 2.4844, size = 172, normalized size = 1.34 $\frac{e^{-i a} \sec (a) \left (6 i \left (1+e^{2 i a}\right ) b x \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+3 \left (1+e^{2 i a}\right ) \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )+2 b^2 x^2 \left (3 \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )+2 i b x\right )\right )-4 b^3 x^3 \tan (a)+6 b^2 x^2 \sec ^2(a+b x)-12 b x \sec (a) \sin (b x) \sec (a+b x)-12 (b x \tan (a)+\log (\cos (a+b x)))}{12 b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Tan[a + b*x]^3,x]

[Out]

(((2*b^2*x^2*((2*I)*b*x + 3*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) + (6*I)*b*(1 + E^((2*I)*a))*x*Pol
yLog[2, -E^((-2*I)*(a + b*x))] + 3*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/E^(I*a) + 6*b^
2*x^2*Sec[a + b*x]^2 - 12*b*x*Sec[a]*Sec[a + b*x]*Sin[b*x] - 4*b^3*x^3*Tan[a] - 12*(Log[Cos[a + b*x]] + b*x*Ta
n[a]))/(12*b^3)

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Maple [A]  time = 0.058, size = 180, normalized size = 1.4 \begin{align*} -{\frac{i}{3}}{x}^{3}+2\,{\frac{x \left ( bx{{\rm e}^{2\,i \left ( bx+a \right ) }}-i{{\rm e}^{2\,i \left ( bx+a \right ) }}-i \right ) }{{b}^{2} \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) ^{2}}}-2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{2\,i{a}^{2}x}{{b}^{2}}}+{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{ix{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-{\frac{\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{{\frac{4\,i}{3}}{a}^{3}}{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*tan(b*x+a)^3,x)

[Out]

-1/3*I*x^3+2*x*(b*x*exp(2*I*(b*x+a))-I*exp(2*I*(b*x+a))-I)/b^2/(1+exp(2*I*(b*x+a)))^2-2/b^3*a^2*ln(exp(I*(b*x+
a)))+2*I/b^2*a^2*x+x^2*ln(1+exp(2*I*(b*x+a)))/b-I*x*polylog(2,-exp(2*I*(b*x+a)))/b^2+1/2*polylog(3,-exp(2*I*(b
*x+a)))/b^3-1/b^3*ln(1+exp(2*I*(b*x+a)))+2/b^3*ln(exp(I*(b*x+a)))+4/3*I/b^3*a^3

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Maxima [B]  time = 1.98648, size = 999, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*tan(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/2*(a^2*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1)) + 2*(2*(b*x + a)^3 - 6*(b*x + a)^2*a - (6*(b*x +
a)^2 - 12*(b*x + a)*a + 6*((b*x + a)^2 - 2*(b*x + a)*a - 1)*cos(4*b*x + 4*a) + 12*((b*x + a)^2 - 2*(b*x + a)*a
- 1)*cos(2*b*x + 2*a) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a - 6*I)*sin(4*b*x + 4*a) + (12*I*(b*x + a)^2 - 24*
I*(b*x + a)*a - 12*I)*sin(2*b*x + 2*a) - 6)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 2*((b*x + a)^3 -
3*(b*x + a)^2*a - 6*b*x - 6*a)*cos(4*b*x + 4*a) + (4*(b*x + a)^3 - (b*x + a)^2*(12*a - 12*I) + 12*(b*x + a)*(
-2*I*a - 1) - 12*a)*cos(2*b*x + 2*a) + (6*b*x*cos(4*b*x + 4*a) + 12*b*x*cos(2*b*x + 2*a) + 6*I*b*x*sin(4*b*x +
4*a) + 12*I*b*x*sin(2*b*x + 2*a) + 6*b*x)*dilog(-e^(2*I*b*x + 2*I*a)) - (-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a +
(-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a + 3*I)*cos(4*b*x + 4*a) + (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a + 6*I)*cos
(2*b*x + 2*a) + 3*((b*x + a)^2 - 2*(b*x + a)*a - 1)*sin(4*b*x + 4*a) + 6*((b*x + a)^2 - 2*(b*x + a)*a - 1)*sin
(2*b*x + 2*a) + 3*I)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - (-3*I*cos(4*b*x +
4*a) - 6*I*cos(2*b*x + 2*a) + 3*sin(4*b*x + 4*a) + 6*sin(2*b*x + 2*a) - 3*I)*polylog(3, -e^(2*I*b*x + 2*I*a))
- (-2*I*(b*x + a)^3 + 6*I*(b*x + a)^2*a + 12*I*b*x + 12*I*a)*sin(4*b*x + 4*a) - (-4*I*(b*x + a)^3 - 12*(b*x +
a)^2*(-I*a - 1) - (b*x + a)*(24*a - 12*I) + 12*I*a)*sin(2*b*x + 2*a) - 12*a)/(-6*I*cos(4*b*x + 4*a) - 12*I*co
s(2*b*x + 2*a) + 6*sin(4*b*x + 4*a) + 12*sin(2*b*x + 2*a) - 6*I))/b^3

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Fricas [C]  time = 1.75364, size = 657, normalized size = 5.13 \begin{align*} \frac{2 \, b^{2} x^{2} \tan \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} + 2 i \, b x{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 2 i \, b x{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 4 \, b x \tan \left (b x + a\right ) + 2 \,{\left (b^{2} x^{2} - 1\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \,{\left (b^{2} x^{2} - 1\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) +{\rm polylog}\left (3, \frac{\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) +{\rm polylog}\left (3, \frac{\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{4 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*tan(b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \tan ^{3}{\left (a + b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*tan(b*x+a)**3,x)

[Out]

Integral(x**2*tan(a + b*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \tan \left (b x + a\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*tan(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*tan(b*x + a)^3, x)