### 3.40 $$\int (c+d x)^2 (a+b \tan (e+f x)) \, dx$$

Optimal. Leaf size=115 $\frac{i b d (c+d x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b (c+d x)^3}{3 d}$

[Out]

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

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Rubi [A]  time = 0.208222, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {3722, 3719, 2190, 2531, 2282, 6589} $\frac{a (c+d x)^3}{3 d}+\frac{i b d (c+d x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac{b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{i b (c+d x)^3}{3 d}-\frac{b d^2 \text{Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c + d*x)^2*(a + b*Tan[e + f*x]),x]

[Out]

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

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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]

Rubi steps

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

Mathematica [A]  time = 0.115045, size = 191, normalized size = 1.66 $\frac{i b c d \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}+\frac{i b d^2 x \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac{b d^2 \text{PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3}+a c^2 x+a c d x^2+\frac{1}{3} a d^2 x^3-\frac{b c^2 \log (\cos (e+f x))}{f}-\frac{2 b c d x \log \left (1+e^{2 i (e+f x)}\right )}{f}+i b c d x^2-\frac{b d^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac{1}{3} i b d^2 x^3$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c + d*x)^2*(a + b*Tan[e + f*x]),x]

[Out]

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

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Maple [B]  time = 0.074, size = 295, normalized size = 2.6 \begin{align*}{\frac{4\,ibcdex}{f}}-{\frac{{\frac{4\,i}{3}}b{d}^{2}{e}^{3}}{{f}^{3}}}+{\frac{a{d}^{2}{x}^{3}}{3}}+ibcd{x}^{2}+acd{x}^{2}+a{c}^{2}x-{\frac{b{c}^{2}\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ) }{f}}+2\,{\frac{b{c}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}+2\,{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}-ib{c}^{2}x+{\frac{ibcd{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-2\,{\frac{bcd\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ) x}{f}}+{\frac{2\,ibcd{e}^{2}}{{f}^{2}}}-{\frac{b{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( fx+e \right ) }}+1 \right ){x}^{2}}{f}}+{\frac{ib{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}-{\frac{b{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }{2\,{f}^{3}}}-4\,{\frac{bcde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ib{d}^{2}{e}^{2}x}{{f}^{2}}}+{\frac{i}{3}}b{d}^{2}{x}^{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*(a+b*tan(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 1.6821, size = 498, normalized size = 4.33 \begin{align*} \frac{6 \,{\left (f x + e\right )} a c^{2} + \frac{2 \,{\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac{6 \,{\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac{6 \,{\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac{6 \,{\left (f x + e\right )}^{2} a c d}{f} - \frac{12 \,{\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right )\right ) + \frac{6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac{12 \, b c d e \log \left (\sec \left (f x + e\right )\right )}{f} - \frac{-2 i \,{\left (f x + e\right )}^{3} b d^{2} + 3 \, b d^{2}{\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )}) +{\left (6 i \, b d^{2} e - 6 i \, b c d f\right )}{\left (f x + e\right )}^{2} +{\left (6 i \,{\left (f x + e\right )}^{2} b d^{2} +{\left (-12 i \, b d^{2} e + 12 i \, b c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) +{\left (-6 i \,{\left (f x + e\right )} b d^{2} + 6 i \, b d^{2} e - 6 i \, b c d f\right )}{\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 3 \,{\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \,{\left (b d^{2} e - b c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}{f^{2}}}{6 \, f} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.68288, size = 814, normalized size = 7.08 \begin{align*} \frac{4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x - 3 \, b d^{2}{\rm polylog}\left (3, \frac{\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b d^{2}{\rm polylog}\left (3, \frac{\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) +{\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )}{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) +{\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )}{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, f^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*a*d^2*f^3*x^3 + 12*a*c*d*f^3*x^2 + 12*a*c^2*f^3*x - 3*b*d^2*polylog(3, (tan(f*x + e)^2 + 2*I*tan(f*x +
e) - 1)/(tan(f*x + e)^2 + 1)) - 3*b*d^2*polylog(3, (tan(f*x + e)^2 - 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 +
1)) + (-6*I*b*d^2*f*x - 6*I*b*c*d*f)*dilog(2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) + (6*I*b*d^2*f*x +
6*I*b*c*d*f)*dilog(2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) - 6*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x + b*c
^2*f^2)*log(-2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x + b*c^2*f^2)*log(
-2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)))/f^3

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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