3.6 $$\int x^3 \tan ^2(a+b x) \, dx$$

Optimal. Leaf size=98 $-\frac{3 i x \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac{3 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{x^3 \tan (a+b x)}{b}-\frac{i x^3}{b}-\frac{x^4}{4}$

[Out]

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

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Rubi [A]  time = 0.16711, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.583, Rules used = {3720, 3719, 2190, 2531, 2282, 6589, 30} $-\frac{3 i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^3}+\frac{3 \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{x^3 \tan (a+b x)}{b}-\frac{i x^3}{b}-\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Tan[a + b*x]^2,x]

[Out]

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

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 0.619273, size = 115, normalized size = 1.17 $\frac{6 i b x \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+3 \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )+2 b^2 x^2 \left (\frac{2 i b x}{1+e^{2 i a}}+3 \log \left (1+e^{-2 i (a+b x)}\right )\right )}{2 b^4}+\frac{x^3 \sec (a) \sin (b x) \sec (a+b x)}{b}-\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Tan[a + b*x]^2,x]

[Out]

-x^4/4 + (2*b^2*x^2*(((2*I)*b*x)/(1 + E^((2*I)*a)) + 3*Log[1 + E^((-2*I)*(a + b*x))]) + (6*I)*b*x*PolyLog[2, -
E^((-2*I)*(a + b*x))] + 3*PolyLog[3, -E^((-2*I)*(a + b*x))])/(2*b^4) + (x^3*Sec[a]*Sec[a + b*x]*Sin[b*x])/b

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Maple [A]  time = 0.054, size = 133, normalized size = 1.4 \begin{align*} -{\frac{{x}^{4}}{4}}+{\frac{2\,i{x}^{3}}{b \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }}-{\frac{2\,i{x}^{3}}{b}}+{\frac{6\,i{a}^{2}x}{{b}^{3}}}+{\frac{4\,i{a}^{3}}{{b}^{4}}}+3\,{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{3\,ix{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{3\,{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{4}}}-6\,{\frac{{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^4+2*I*x^3/b/(1+exp(2*I*(b*x+a)))-2*I/b*x^3+6*I/b^3*a^2*x+4*I/b^4*a^3+3*x^2*ln(1+exp(2*I*(b*x+a)))/b^2-3
*I*x*polylog(2,-exp(2*I*(b*x+a)))/b^3+3/2*polylog(3,-exp(2*I*(b*x+a)))/b^4-6/b^4*a^2*ln(exp(I*(b*x+a)))

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Maxima [B]  time = 1.78277, size = 864, normalized size = 8.82 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b*x + a - tan(b*x + a))*a^3 - 3*((b*x + a)^2*cos(2*b*x + 2*a)^2 + (b*x + a)^2*sin(2*b*x + 2*a)^2 + 2*(
b*x + a)^2*cos(2*b*x + 2*a) + (b*x + a)^2 - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)
*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) - 4*(b*x + a)*sin(2*b*x + 2*a))*a^2/(co
s(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 2*(I*(b*x + a)^4 - 4*I*(b*x + a)^3*a + (12*(
b*x + a)^2 - 24*(b*x + a)*a + 12*((b*x + a)^2 - 2*(b*x + a)*a)*cos(2*b*x + 2*a) + (12*I*(b*x + a)^2 - 24*I*(b*
x + a)*a)*sin(2*b*x + 2*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (I*(b*x + a)^4 - 4*(b*x + a)^3*(
I*a + 2) + 24*(b*x + a)^2*a)*cos(2*b*x + 2*a) - 12*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) + b*x)*dilog
(-e^(2*I*b*x + 2*I*a)) + (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a + (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a)*cos(2*b*
x + 2*a) + 6*((b*x + a)^2 - 2*(b*x + a)*a)*sin(2*b*x + 2*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*c
os(2*b*x + 2*a) + 1) + (-6*I*cos(2*b*x + 2*a) + 6*sin(2*b*x + 2*a) - 6*I)*polylog(3, -e^(2*I*b*x + 2*I*a)) - (
(b*x + a)^4 - (b*x + a)^3*(4*a - 8*I) - 24*I*(b*x + a)^2*a)*sin(2*b*x + 2*a))/(-4*I*cos(2*b*x + 2*a) + 4*sin(2
*b*x + 2*a) - 4*I))/b^4

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Fricas [C]  time = 1.56632, size = 614, normalized size = 6.27 \begin{align*} -\frac{b^{4} x^{4} - 4 \, b^{3} x^{3} \tan \left (b x + a\right ) - 6 \, b^{2} x^{2} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, b^{2} x^{2} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 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 ) + 6 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 ) - 3 \,{\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 ) - 3 \,{\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^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^4*x^4 - 4*b^3*x^3*tan(b*x + a) - 6*b^2*x^2*log(-2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 6*b^2*x
^2*log(-2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 6*I*b*x*dilog(2*(I*tan(b*x + a) - 1)/(tan(b*x + a)^2 +
1) + 1) + 6*I*b*x*dilog(2*(-I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1) + 1) - 3*polylog(3, (tan(b*x + a)^2 + 2*
I*tan(b*x + a) - 1)/(tan(b*x + a)^2 + 1)) - 3*polylog(3, (tan(b*x + a)^2 - 2*I*tan(b*x + a) - 1)/(tan(b*x + a)
^2 + 1)))/b^4

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*tan(a + b*x)**2, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*tan(b*x + a)^2, x)