3.211 \(\int (c+d x)^2 \tan (a+b x) \, dx\)

Optimal. Leaf size=96 \[ \frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.152983, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3719, 2190, 2531, 2282, 6589} \[ \frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tan[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0409372, size = 100, normalized size = 1.04 \[ \frac{6 i b d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )-3 d^3 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )+2 i b^2 (c+d x)^2 \left (b (c+d x)+3 i d \log \left (1+e^{2 i (a+b x)}\right )\right )}{6 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Tan[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.277, size = 257, normalized size = 2.7 \begin{align*} icd{x}^{2}+{\frac{2\,i{a}^{2}cd}{{b}^{2}}}-i{c}^{2}x-{\frac{{c}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}+2\,{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+2\,{\frac{{a}^{2}{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{2\,i{a}^{2}{d}^{2}x}{{b}^{2}}}+{\frac{i}{3}}{d}^{2}{x}^{3}-2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}+{\frac{idc{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-4\,{\frac{cda\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{{\frac{4\,i}{3}}{a}^{3}{d}^{2}}{{b}^{3}}}+{\frac{4\,iacdx}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*sec(b*x+a)*sin(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.85091, size = 378, normalized size = 3.94 \begin{align*} -\frac{3 \, c^{2} \log \left (-\sin \left (b x + a\right )^{2} + 1\right ) - \frac{6 \, a c d \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b} + \frac{3 \, a^{2} d^{2} \log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{b^{2}} + \frac{-2 i \,{\left (b x + a\right )}^{3} d^{2} +{\left (-6 i \, b c d + 6 i \, a d^{2}\right )}{\left (b x + a\right )}^{2} + 3 \, d^{2}{\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) +{\left (6 i \,{\left (b x + a\right )}^{2} d^{2} +{\left (12 i \, b c d - 12 i \, a d^{2}\right )}{\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-6 i \, b c d - 6 i \,{\left (b x + a\right )} d^{2} + 6 i \, a d^{2}\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 3 \,{\left ({\left (b x + a\right )}^{2} d^{2} + 2 \,{\left (b c d - a d^{2}\right )}{\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )}{b^{2}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sec(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 0.597114, size = 1534, normalized size = 15.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*sec(b*x+a)*sin(b*x+a),x)

[Out]

Integral((c + d*x)**2*sin(a + b*x)*sec(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sec(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^2*sec(b*x + a)*sin(b*x + a), x)