\(\int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx\) [217]

Optimal result

Integrand size = 20, antiderivative size = 186 \[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx=-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {2 d (c+d x) \cos (a+b x)}{b^2}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \sin (a+b x)}{b^3}-\frac {(c+d x)^2 \sin (a+b x)}{b} \] Output:

-2*I*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b-2*d*(d*x+c)*cos(b*x+a)/b^2+2*I*d*( 
d*x+c)*polylog(2,-I*exp(I*(b*x+a)))/b^2-2*I*d*(d*x+c)*polylog(2,I*exp(I*(b 
*x+a)))/b^2-2*d^2*polylog(3,-I*exp(I*(b*x+a)))/b^3+2*d^2*polylog(3,I*exp(I 
*(b*x+a)))/b^3+2*d^2*sin(b*x+a)/b^3-(d*x+c)^2*sin(b*x+a)/b
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.69 \[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx=-\frac {2 i b^2 c^2 \arctan \left (e^{i (a+b x)}\right )+2 b c d \cos (a+b x)+2 b d^2 x \cos (a+b x)-2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )-b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )-2 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )+b^2 c^2 \sin (a+b x)-2 d^2 \sin (a+b x)+2 b^2 c d x \sin (a+b x)+b^2 d^2 x^2 \sin (a+b x)}{b^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4907, 3042, 3777, 25, 3042, 3777, 3042, 3117, 4669, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4907

\(\displaystyle \int (c+d x)^2 \sec (a+b x)dx-\int (c+d x)^2 \cos (a+b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx-\int (c+d x)^2 \sin \left (a+b x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {2 d \int -((c+d x) \sin (a+b x))dx}{b}+\int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 d \int (c+d x) \sin (a+b x)dx}{b}+\int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \int (c+d x) \sin (a+b x)dx}{b}+\int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {2 d \left (\frac {d \int \cos (a+b x)dx}{b}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}+\int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx+\frac {2 d \left (\frac {d \int \sin \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3117

\(\displaystyle \int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 4669

\(\displaystyle -\frac {2 d \int (c+d x) \log \left (1-i e^{i (a+b x)}\right )dx}{b}+\frac {2 d \int (c+d x) \log \left (1+i e^{i (a+b x)}\right )dx}{b}-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {i d \int \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )dx}{b}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {i d \int \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )dx}{b}\right )}{b}-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {d \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^2}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {d \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^2}\right )}{b}+\frac {2 d \left (\frac {d \sin (a+b x)}{b^2}-\frac {(c+d x) \cos (a+b x)}{b}\right )}{b}-\frac {(c+d x)^2 \sin (a+b x)}{b}\)

Input:

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

Output:

((-2*I)*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/b + (2*d*((I*(c + d*x)*PolyLo 
g[2, (-I)*E^(I*(a + b*x))])/b - (d*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^2)) 
/b - (2*d*((I*(c + d*x)*PolyLog[2, I*E^(I*(a + b*x))])/b - (d*PolyLog[3, I 
*E^(I*(a + b*x))])/b^2))/b - ((c + d*x)^2*Sin[a + b*x])/b + (2*d*(-(((c + 
d*x)*Cos[a + b*x])/b) + (d*Sin[a + b*x])/b^2))/b
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 4669

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x 
))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 4907

Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> -Int[(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^ 
(p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x] /; Fr 
eeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 511 vs. \(2 (169 ) = 338\).

Time = 0.67 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.75

method	result	size
risch	\(\frac {i \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +2 i b \,d^{2} x +b^{2} c^{2}+2 i b c d -2 d^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{2 b^{3}}-\frac {2 i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 c d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 i c d \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i c d \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {a^{2} d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {2 d^{2} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {i \left (x^{2} d^{2} b^{2}+2 b^{2} c d x -2 i b \,d^{2} x +b^{2} c^{2}-2 i b c d -2 d^{2}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{2 b^{3}}+\frac {4 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 d^{2} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}-\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 c d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}\)	\(512\)

Input:

int((d*x+c)^2*sin(b*x+a)*tan(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (160) = 320\).

Time = 0.12 (sec) , antiderivative size = 656, normalized size of antiderivative = 3.53 \[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-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)) + 4*(b*d^2*x + b 
*c*d)*cos(b*x + a) + 2*(I*b*d^2*x + I*b*c*d)*dilog(I*cos(b*x + a) + sin(b* 
x + a)) + 2*(I*b*d^2*x + I*b*c*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) + 2 
*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + 2*(-I*b*d^ 
2*x - 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) + 2*(b^2*d^2*x^2 + 2*b^2*c* 
d*x + b^2*c^2 - 2*d^2)*sin(b*x + a))/b^3
 

Sympy [F]

\[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right )^{2} \sin {\left (a + b x \right )} \tan {\left (a + b x \right )}\, dx \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (160) = 320\).

Time = 0.20 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.77 \[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(c^2*(log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1) - 2*sin(b*x + a)) 
- 2*a*c*d*(log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1) - 2*sin(b*x + a)) 
/b + a^2*d^2*(log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1) - 2*sin(b*x + 
a))/b^2 + (4*d^2*polylog(3, I*e^(I*b*x + I*a)) - 4*d^2*polylog(3, -I*e^(I* 
b*x + I*a)) - 2*(I*(b*x + a)^2*d^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a))*arct 
an2(cos(b*x + a), sin(b*x + a) + 1) - 2*(I*(b*x + a)^2*d^2 + 2*(I*b*c*d - 
I*a*d^2)*(b*x + a))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 4*(b*c*d + 
(b*x + a)*d^2 - a*d^2)*cos(b*x + a) - 4*(I*b*c*d + I*(b*x + a)*d^2 - I*a*d 
^2)*dilog(I*e^(I*b*x + I*a)) - 4*(-I*b*c*d - I*(b*x + a)*d^2 + I*a*d^2)*di 
log(-I*e^(I*b*x + I*a)) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))* 
log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - ((b*x + a)^2*d 
^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2* 
sin(b*x + a) + 1) - 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - 2*d 
^2)*sin(b*x + a))/b^2)/b
 

Giac [F]

\[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{2} \sin \left (b x + a\right ) \tan \left (b x + a\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx=\int \sin \left (a+b\,x\right )\,\mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^2 \sin (a+b x) \tan (a+b x) \, dx=\frac {\left (\int \sin \left (b x +a \right ) \tan \left (b x +a \right ) x^{2}d x \right ) b \,d^{2}+2 \left (\int \sin \left (b x +a \right ) \tan \left (b x +a \right ) x d x \right ) b c d -\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) c^{2}+\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) c^{2}-\sin \left (b x +a \right ) c^{2}}{b} \] Input:

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

Output:

(int(sin(a + b*x)*tan(a + b*x)*x**2,x)*b*d**2 + 2*int(sin(a + b*x)*tan(a + 
 b*x)*x,x)*b*c*d - log(tan((a + b*x)/2) - 1)*c**2 + log(tan((a + b*x)/2) + 
 1)*c**2 - sin(a + b*x)*c**2)/b