3.305 \(\int (c+d x)^2 \tan ^3(a+b x) \, dx\)

Optimal. Leaf size=169 \[ -\frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{d (c+d x) \tan (a+b x)}{b^2}-\frac{d^2 \log (\cos (a+b x))}{b^3}+\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}+\frac{c d x}{b}+\frac{d^2 x^2}{2 b}-\frac{i (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.221749, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3720, 3475, 3719, 2190, 2531, 2282, 6589} \[ -\frac{i d (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{d^2 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{d (c+d x) \tan (a+b x)}{b^2}-\frac{d^2 \log (\cos (a+b x))}{b^3}+\frac{(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}+\frac{c d x}{b}+\frac{d^2 x^2}{2 b}-\frac{i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tan[a + b*x]^3,x]

[Out]

(c*d*x)/b + (d^2*x^2)/(2*b) - ((I/3)*(c + d*x)^3)/d + ((c + d*x)^2*Log[1 + E^((2*I)*(a + b*x))])/b - (d^2*Log[
Cos[a + b*x]])/b^3 - (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) - (d*(c + d*x)*Tan[a + b*x])/b^2 + ((c + d*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 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 ^3(a+b x) \, dx &=\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac{d \int (c+d x) \tan ^2(a+b x) \, dx}{b}-\int (c+d x)^2 \tan (a+b x) \, dx\\ &=-\frac{i (c+d x)^3}{3 d}-\frac{d (c+d x) \tan (a+b x)}{b^2}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} (c+d x)^2}{1+e^{2 i (a+b x)}} \, dx+\frac{d \int (c+d x) \, dx}{b}+\frac{d^2 \int \tan (a+b x) \, dx}{b^2}\\ &=\frac{c d x}{b}+\frac{d^2 x^2}{2 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{d^2 \log (\cos (a+b x))}{b^3}-\frac{d (c+d x) \tan (a+b x)}{b^2}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}-\frac{(2 d) \int (c+d x) \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{c d x}{b}+\frac{d^2 x^2}{2 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{d^2 \log (\cos (a+b x))}{b^3}-\frac{i d (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{d (c+d x) \tan (a+b x)}{b^2}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}+\frac{\left (i d^2\right ) \int \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{c d x}{b}+\frac{d^2 x^2}{2 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{d^2 \log (\cos (a+b x))}{b^3}-\frac{i d (c+d x) \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{d (c+d x) \tan (a+b x)}{b^2}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}+\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{c d x}{b}+\frac{d^2 x^2}{2 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{d^2 \log (\cos (a+b x))}{b^3}-\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}-\frac{d (c+d x) \tan (a+b x)}{b^2}+\frac{(c+d x)^2 \tan ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [B]  time = 6.64045, size = 454, normalized size = 2.69 \[ \frac{c d \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{b^2 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac{i e^{-i a} d^2 \sec (a) \left (6 \left (1+e^{2 i a}\right ) b x \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )+2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )\right )}{12 b^3}+\frac{\sec (a) \sec (a+b x) \left (d^2 (-x) \sin (b x)-c d \sin (b x)\right )}{b^2}-\frac{d^2 \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b^3 \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac{c^2 \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac{(c+d x)^2 \sec ^2(a+b x)}{2 b}-\frac{1}{3} x \tan (a) \left (3 c^2+3 c d x+d^2 x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Tan[a + b*x]^3,x]

[Out]

((I/12)*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) + 6*b*(1 + E^((2*I)*a))
*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^3
*E^(I*a)) + ((c + d*x)^2*Sec[a + b*x]^2)/(2*b) + (c^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] +
b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) - (d^2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[
a]))/(b^3*(Cos[a]^2 + Sin[a]^2)) + (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTa
n[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] +
Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[
a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) + (Sec[a]*Sec[a + b*x]*(-(c*d*
Sin[b*x]) - d^2*x*Sin[b*x]))/b^2 - (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Tan[a])/3

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Maple [B]  time = 0.321, size = 400, normalized size = 2.4 \begin{align*}{\frac{-i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-icd{x}^{2}+i{c}^{2}x+2\,{\frac{b{d}^{2}{x}^{2}{{\rm e}^{2\,i \left ( bx+a \right ) }}-i{d}^{2}x{{\rm e}^{2\,i \left ( bx+a \right ) }}+2\,bcdx{{\rm e}^{2\,i \left ( bx+a \right ) }}-icd{{\rm e}^{2\,i \left ( bx+a \right ) }}+b{c}^{2}{{\rm e}^{2\,i \left ( bx+a \right ) }}-i{d}^{2}x-idc}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) ^{2}}}-{\frac{4\,iacdx}{b}}-{\frac{idc{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{i}{3}}{d}^{2}{x}^{3}+{\frac{2\,i{a}^{2}{d}^{2}x}{{b}^{2}}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}-{\frac{2\,i{a}^{2}cd}{{b}^{2}}}-2\,{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{b}}-2\,{\frac{{d}^{2}{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{{\frac{4\,i}{3}}{a}^{3}{d}^{2}}{{b}^{3}}}+2\,{\frac{cd\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{b}}+{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}+2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{{d}^{2}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{3}}}+4\,{\frac{cda\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.44249, size = 1655, normalized size = 9.79 \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*tan(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/2*(c^2*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1)) - 2*a*c*d*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x +
a)^2 - 1))/b + a^2*d^2*(1/(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 - 1))/b^2 + 2*(2*(b*x + a)^3*d^2 + 6*(b*c
*d - a*d^2)*(b*x + a)^2 + 12*b*c*d - 12*a*d^2 - (6*(b*x + a)^2*d^2 + 12*(b*c*d - a*d^2)*(b*x + a) - 6*d^2 + 6*
((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - d^2)*cos(4*b*x + 4*a) + 12*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^
2)*(b*x + a) - d^2)*cos(2*b*x + 2*a) + (6*I*(b*x + a)^2*d^2 + (12*I*b*c*d - 12*I*a*d^2)*(b*x + a) - 6*I*d^2)*s
in(4*b*x + 4*a) + (12*I*(b*x + a)^2*d^2 + (24*I*b*c*d - 24*I*a*d^2)*(b*x + a) - 12*I*d^2)*sin(2*b*x + 2*a))*ar
ctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + 2*((b*x + a)^3*d^2 + 3*(b*c*d - a*d^2)*(b*x + a)^2 - 6*(b*x +
a)*d^2)*cos(4*b*x + 4*a) + (4*(b*x + a)^3*d^2 + (12*b*c*d - (12*a - 12*I)*d^2)*(b*x + a)^2 + 12*b*c*d - 12*a*d
^2 - (-24*I*b*c*d - 12*(-2*I*a - 1)*d^2)*(b*x + a))*cos(2*b*x + 2*a) + (6*b*c*d + 6*(b*x + a)*d^2 - 6*a*d^2 +
6*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(4*b*x + 4*a) + 12*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) - (-6
*I*b*c*d - 6*I*(b*x + a)*d^2 + 6*I*a*d^2)*sin(4*b*x + 4*a) - (-12*I*b*c*d - 12*I*(b*x + a)*d^2 + 12*I*a*d^2)*s
in(2*b*x + 2*a))*dilog(-e^(2*I*b*x + 2*I*a)) - (-3*I*(b*x + a)^2*d^2 + (-6*I*b*c*d + 6*I*a*d^2)*(b*x + a) + 3*
I*d^2 + (-3*I*(b*x + a)^2*d^2 + (-6*I*b*c*d + 6*I*a*d^2)*(b*x + a) + 3*I*d^2)*cos(4*b*x + 4*a) + (-6*I*(b*x +
a)^2*d^2 + (-12*I*b*c*d + 12*I*a*d^2)*(b*x + a) + 6*I*d^2)*cos(2*b*x + 2*a) + 3*((b*x + a)^2*d^2 + 2*(b*c*d -
a*d^2)*(b*x + a) - d^2)*sin(4*b*x + 4*a) + 6*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - d^2)*sin(2*b*x +
2*a))*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*d^2*cos(4*b*x + 4*a) - 6*
I*d^2*cos(2*b*x + 2*a) + 3*d^2*sin(4*b*x + 4*a) + 6*d^2*sin(2*b*x + 2*a) - 3*I*d^2)*polylog(3, -e^(2*I*b*x + 2
*I*a)) - (-2*I*(b*x + a)^3*d^2 + (-6*I*b*c*d + 6*I*a*d^2)*(b*x + a)^2 + 12*I*(b*x + a)*d^2)*sin(4*b*x + 4*a) -
(-4*I*(b*x + a)^3*d^2 + (-12*I*b*c*d - 12*(-I*a - 1)*d^2)*(b*x + a)^2 - 12*I*b*c*d + 12*I*a*d^2 + (24*b*c*d -
(24*a - 12*I)*d^2)*(b*x + a))*sin(2*b*x + 2*a))/(-6*I*b^2*cos(4*b*x + 4*a) - 12*I*b^2*cos(2*b*x + 2*a) + 6*b^
2*sin(4*b*x + 4*a) + 12*b^2*sin(2*b*x + 2*a) - 6*I*b^2))/b

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Fricas [C]  time = 0.518016, size = 883, normalized size = 5.22 \begin{align*} \frac{2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + d^{2}{\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 ) + d^{2}{\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 ) + 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \tan \left (b x + a\right )^{2} +{\left (2 i \, b d^{2} x + 2 i \, b c d\right )}{\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 ) +{\left (-2 i \, b d^{2} x - 2 i \, b c d\right )}{\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 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - d^{2}\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} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - d^{2}\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 4 \,{\left (b d^{2} x + b c d\right )} \tan \left (b x + a\right )}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)**2*tan(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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