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

Optimal. Leaf size=128 \[ -\frac{3 i d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac{3 d^3 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{(c+d x)^3 \tan (a+b x)}{b}-\frac{i (c+d x)^3}{b}-\frac{(c+d x)^4}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.210269, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3720, 3719, 2190, 2531, 2282, 6589, 32} \[ -\frac{3 i d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac{3 d^3 \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}+\frac{(c+d x)^3 \tan (a+b x)}{b}-\frac{i (c+d x)^3}{b}-\frac{(c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Tan[a + b*x]^2,x]

[Out]

((-I)*(c + d*x)^3)/b - (c + d*x)^4/(4*d) + (3*d*(c + d*x)^2*Log[1 + E^((2*I)*(a + b*x))])/b^2 - ((3*I)*d^2*(c
+ d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b^3 + (3*d^3*PolyLog[3, -E^((2*I)*(a + b*x))])/(2*b^4) + ((c + d*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 6.56442, size = 424, normalized size = 3.31 \[ \frac{3 c d^2 \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{b^3 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac{i e^{-i a} d^3 \sec (a) \left (6 \left (1+e^{2 i a}\right ) b x \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )+2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )\right )}{4 b^4}+\frac{3 c^2 d \sec (a) (b x \sin (a)+\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x)))}{b^2 \left (\sin ^2(a)+\cos ^2(a)\right )}+\frac{\sec (a) \sec (a+b x) \left (3 c^2 d x \sin (b x)+c^3 \sin (b x)+3 c d^2 x^2 \sin (b x)+d^3 x^3 \sin (b x)\right )}{b}-\frac{1}{4} x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3))/4 + ((I/4)*d^3*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Lo
g[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^4*E^(I*a)) + (3*c^2*d*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] -
Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b^2*(Cos[a]^2 + Sin[a]^2)) + (3*c*d^2*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]])
- (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[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^3*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)])
+ (Sec[a]*Sec[a + b*x]*(c^3*Sin[b*x] + 3*c^2*d*x*Sin[b*x] + 3*c*d^2*x^2*Sin[b*x] + d^3*x^3*Sin[b*x]))/b

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Maple [B]  time = 0.197, size = 348, normalized size = 2.7 \begin{align*} -{\frac{{d}^{3}{x}^{4}}{4}}-c{d}^{2}{x}^{3}-{\frac{3\,{c}^{2}d{x}^{2}}{2}}-{c}^{3}x-{\frac{3\,i{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) x}{{b}^{3}}}+3\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-6\,{\frac{{d}^{3}{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-{\frac{3\,i{d}^{2}c{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{6\,i{d}^{3}{a}^{2}x}{{b}^{3}}}-{\frac{6\,i{d}^{2}c{a}^{2}}{{b}^{3}}}+{\frac{2\,i \left ({d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) }}+3\,{\frac{{d}^{3}\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{{b}^{2}}}-{\frac{2\,i{d}^{3}{x}^{3}}{b}}+{\frac{3\,{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{4}}}+12\,{\frac{{d}^{2}ca\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+6\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}+{\frac{4\,i{d}^{3}{a}^{3}}{{b}^{4}}}-{\frac{12\,i{d}^{2}cax}{{b}^{2}}}-{\frac{6\,i{d}^{2}c{x}^{2}}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d^3*x^4-c*d^2*x^3-3/2*c^2*d*x^2-c^3*x-3*I*d^3/b^3*polylog(2,-exp(2*I*(b*x+a)))*x+3*d/b^2*c^2*ln(exp(2*I*(
b*x+a))+1)-6*d/b^2*c^2*ln(exp(I*(b*x+a)))-6*d^3/b^4*a^2*ln(exp(I*(b*x+a)))-3*I*d^2/b^3*c*polylog(2,-exp(2*I*(b
*x+a)))+6*I*d^3/b^3*a^2*x-6*I*d^2/b^3*c*a^2+2*I*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/b/(exp(2*I*(b*x+a))+1)+3*d
^3/b^2*ln(exp(2*I*(b*x+a))+1)*x^2-2*I*d^3/b*x^3+3/2*d^3*polylog(3,-exp(2*I*(b*x+a)))/b^4+12*d^2/b^3*c*a*ln(exp
(I*(b*x+a)))+6*d^2/b^2*c*ln(exp(2*I*(b*x+a))+1)*x+4*I*d^3/b^4*a^3-12*I*d^2/b^2*c*a*x-6*I*d^2/b*c*x^2

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Maxima [B]  time = 1.93025, size = 1840, normalized size = 14.38 \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)^3*tan(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 0.502994, size = 913, normalized size = 7.13 \begin{align*} -\frac{b^{4} d^{3} x^{4} + 4 \, b^{4} c d^{2} x^{3} + 6 \, b^{4} c^{2} d x^{2} + 4 \, b^{4} c^{3} x - 3 \, d^{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 \, d^{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 ) -{\left (6 i \, b d^{3} x + 6 i \, b c d^{2}\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 (-6 i \, b d^{3} x - 6 i \, b c d^{2}\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 ) - 6 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \,{\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\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^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \tan \left (b x + a\right )}{4 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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