\(\int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx\) [223]

Optimal result

Integrand size = 22, antiderivative size = 175 \[ \int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx=\frac {(c+d x)^2}{4 b}+\frac {i (c+d x)^3}{3 d}-\frac {(c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {i d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {d (c+d x) \cos (a+b x) \sin (a+b x)}{2 b^2}+\frac {d^2 \sin ^2(a+b x)}{4 b^3}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(518\) vs. \(2(175)=350\).

Time = 6.34 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.96 \[ \int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx=-\frac {i d^2 e^{-i a} \left (2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 b \left (1+e^{2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )\right ) \sec (a)}{12 b^3}-\frac {c^2 \sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a))}{b \left (\cos ^2(a)+\sin ^2(a)\right )}-\frac {c d \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{b^2 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}+\frac {\cos (2 b x) \left (2 b^2 c^2 \cos (2 a)-d^2 \cos (2 a)+4 b^2 c d x \cos (2 a)+2 b^2 d^2 x^2 \cos (2 a)-2 b c d \sin (2 a)-2 b d^2 x \sin (2 a)\right )}{8 b^3}-\frac {\left (2 b c d \cos (2 a)+2 b d^2 x \cos (2 a)+2 b^2 c^2 \sin (2 a)-d^2 \sin (2 a)+4 b^2 c d x \sin (2 a)+2 b^2 d^2 x^2 \sin (2 a)\right ) \sin (2 b x)}{8 b^3}+\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right ) \tan (a) \] Input:

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

Output:

((-1/12*I)*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^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)) - (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTa 
n[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^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) + (Cos[2*b*x]*(2*b^2*c^2*Cos 
[2*a] - d^2*Cos[2*a] + 4*b^2*c*d*x*Cos[2*a] + 2*b^2*d^2*x^2*Cos[2*a] - 2*b 
*c*d*Sin[2*a] - 2*b*d^2*x*Sin[2*a]))/(8*b^3) - ((2*b*c*d*Cos[2*a] + 2*b*d^ 
2*x*Cos[2*a] + 2*b^2*c^2*Sin[2*a] - d^2*Sin[2*a] + 4*b^2*c*d*x*Sin[2*a] + 
2*b^2*d^2*x^2*Sin[2*a])*Sin[2*b*x])/(8*b^3) + (x*(3*c^2 + 3*c*d*x + d^2*x^ 
2)*Tan[a])/3
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4907, 3042, 4202, 2620, 3011, 2720, 4904, 3042, 3791, 17, 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.

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

\(\Big \downarrow \) 4907

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

\(\displaystyle -2 i \int \frac {e^{2 i (a+b x)} (c+d x)^2}{1+e^{2 i (a+b x)}}dx-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2620

\(\displaystyle -2 i \left (\frac {i d \int (c+d x) \log \left (1+e^{2 i (a+b x)}\right )dx}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3011

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \int \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{2 b}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2720

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )-\int (c+d x)^2 \cos (a+b x) \sin (a+b x)dx+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 4904

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {d \int (c+d x) \sin ^2(a+b x)dx}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {d \int (c+d x) \sin (a+b x)^2dx}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3791

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {d \left (\frac {1}{2} \int (c+d x)dx+\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 17

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {d \left (\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^2}{4 d}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^3}{3 d}\)

\(\Big \downarrow \) 7143

\(\displaystyle -2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {d \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{2 b}\right )+\frac {d \left (\frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^2}{4 d}\right )}{b}-\frac {(c+d x)^2 \sin ^2(a+b x)}{2 b}+\frac {i (c+d x)^3}{3 d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 17

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

rule 2620

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 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 3791

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> 
 Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x 
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n)   Int[(c + d* 
x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 
 1]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[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] && IGt 
Q[m, 0]
 

rule 4904

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x 
_)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) 
, x] - Simp[d*(m/(b*(n + 1)))   Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), 
 x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
 

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. 387 vs. \(2 (155 ) = 310\).

Time = 1.17 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.22

method	result	size
risch	\(\frac {2 i c d \,a^{2}}{b^{2}}-\frac {2 i d^{2} a^{2} x}{b^{2}}-\frac {i c^{3}}{3 d}+\frac {i d c \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}+\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 i b \,d^{2} x +2 b^{2} c^{2}+2 i b c d -d^{2}\right ) {\mathrm e}^{2 i \left (b x +a \right )}}{16 b^{3}}+\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x -2 i b \,d^{2} x +2 b^{2} c^{2}-2 i b c d -d^{2}\right ) {\mathrm e}^{-2 i \left (b x +a \right )}}{16 b^{3}}-\frac {c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {2 c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {i d^{2} x^{3}}{3}+\frac {2 d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b}+i d c \,x^{2}-i c^{2} x +\frac {i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {4 i d^{2} a^{3}}{3 b^{3}}+\frac {4 i c d a x}{b}-\frac {4 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b}-\frac {d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}\)	\(388\)

Input:

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

Output:

2*I/b^2*c*d*a^2-2*I/b^2*d^2*a^2*x-1/3*I/d*c^3+I/b^2*d*c*polylog(2,-exp(2*I 
*(b*x+a)))+1/16*(2*x^2*d^2*b^2+2*I*b*d^2*x+4*b^2*c*d*x+2*I*b*c*d+2*b^2*c^2 
-d^2)/b^3*exp(2*I*(b*x+a))+1/16*(2*x^2*d^2*b^2-2*I*b*d^2*x+4*b^2*c*d*x-2*I 
*b*c*d+2*b^2*c^2-d^2)/b^3*exp(-2*I*(b*x+a))-1/b*c^2*ln(exp(2*I*(b*x+a))+1) 
+2/b*c^2*ln(exp(I*(b*x+a)))+1/3*I*d^2*x^3+2/b^3*d^2*a^2*ln(exp(I*(b*x+a))) 
-2/b*d*c*ln(exp(2*I*(b*x+a))+1)*x+I*d*c*x^2-I*c^2*x+I/b^2*d^2*polylog(2,-e 
xp(2*I*(b*x+a)))*x-4/3*I/b^3*d^2*a^3+4*I/b*c*d*a*x-4/b^2*c*d*a*ln(exp(I*(b 
*x+a)))-1/b*d^2*ln(exp(2*I*(b*x+a))+1)*x^2-1/2*d^2*polylog(3,-exp(2*I*(b*x 
+a)))/b^3
 

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. 688 vs. \(2 (152) = 304\).

Time = 0.13 (sec) , antiderivative size = 688, normalized size of antiderivative = 3.93 \[ \int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(b^2*d^2*x^2 + 2*b^2*c*d*x - (2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 
 - d^2)*cos(b*x + a)^2 + 4*d^2*polylog(3, I*cos(b*x + a) + sin(b*x + a)) + 
 4*d^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 4*d^2*polylog(3, -I*cos 
(b*x + a) + sin(b*x + a)) + 4*d^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a 
)) + 2*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a) + 4*(I*b*d^2*x + I*b*c* 
d)*dilog(I*cos(b*x + a) + sin(b*x + a)) + 4*(-I*b*d^2*x - I*b*c*d)*dilog(I 
*cos(b*x + a) - sin(b*x + a)) + 4*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x 
+ a) + sin(b*x + a)) + 4*(I*b*d^2*x + I*b*c*d)*dilog(-I*cos(b*x + a) - sin 
(b*x + a)) + 2*(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*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 + 2*a*b*c*d - a^2*d^2)*log(I*c 
os(b*x + a) + sin(b*x + a) + 1) + 2*(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) + 2*(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) + 2 
*(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) - s 
in(b*x + a) + 1) + 2*(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*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) - 
 I*sin(b*x + a) + I))/b^3
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*sin(a + b*x)**2*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. 383 vs. \(2 (152) = 304\).

Time = 0.19 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.19 \[ \int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx=-\frac {12 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} c^{2} - \frac {24 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a c d}{b} + \frac {12 \, {\left (\sin \left (b x + a\right )^{2} + \log \left (\sin \left (b x + a\right )^{2} - 1\right )\right )} a^{2} d^{2}}{b^{2}} + \frac {-8 i \, {\left (b x + a\right )}^{3} d^{2} - 24 \, {\left (i \, b c d - i \, a d^{2}\right )} {\left (b x + a\right )}^{2} + 12 \, d^{2} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 24 \, {\left (-i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (-i \, b c d + i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 3 \, {\left (2 \, {\left (b x + a\right )}^{2} d^{2} + 4 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )} - d^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) - 24 \, {\left (i \, b c d + i \, {\left (b x + a\right )} d^{2} - i \, a d^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 12 \, {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{b^{2}}}{24 \, b} \] Input:

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

Output:

-1/24*(12*(sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1))*c^2 - 24*(sin(b*x + a 
)^2 + log(sin(b*x + a)^2 - 1))*a*c*d/b + 12*(sin(b*x + a)^2 + log(sin(b*x 
+ a)^2 - 1))*a^2*d^2/b^2 + (-8*I*(b*x + a)^3*d^2 - 24*(I*b*c*d - I*a*d^2)* 
(b*x + a)^2 + 12*d^2*polylog(3, -e^(2*I*b*x + 2*I*a)) - 24*(-I*(b*x + a)^2 
*d^2 + 2*(-I*b*c*d + I*a*d^2)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b 
*x + 2*a) + 1) - 3*(2*(b*x + a)^2*d^2 + 4*(b*c*d - a*d^2)*(b*x + a) - d^2) 
*cos(2*b*x + 2*a) - 24*(I*b*c*d + I*(b*x + a)*d^2 - I*a*d^2)*dilog(-e^(2*I 
*b*x + 2*I*a)) + 12*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(co 
s(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 6*(b*c*d 
 + (b*x + a)*d^2 - a*d^2)*sin(2*b*x + 2*a))/b^2)/b
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^2 \sin ^2(a+b x) \tan (a+b x) \, dx=\frac {-2 \cos \left (b x +a \right ) \sin \left (b x +a \right ) c d +8 \cos \left (b x +a \right ) b c d x +4 \cos \left (b x +a \right ) b \,d^{2} x^{2}-16 \left (\int \frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}d x \right ) b \,d^{2}-16 \left (\int \frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x^{2}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}d x \right ) b^{2} d^{2}-32 \left (\int \frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}d x \right ) b^{2} c d +16 \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}+\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}d x \right ) b \,d^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) b \,c^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) b \,c^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) b \,c^{2}-\sin \left (b x +a \right )^{2} b \,c^{2}-4 \sin \left (b x +a \right )^{2} b c d x -2 \sin \left (b x +a \right )^{2} b \,d^{2} x^{2}-8 \sin \left (b x +a \right ) c d +10 a c d +2 b c d x +4 b \,d^{2} x^{2}}{2 b^{2}} \] Input:

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

Output:

( - 2*cos(a + b*x)*sin(a + b*x)*c*d + 8*cos(a + b*x)*b*c*d*x + 4*cos(a + b 
*x)*b*d**2*x**2 - 16*int((tan((a + b*x)/2)**2*x)/(tan((a + b*x)/2)**6 + ta 
n((a + b*x)/2)**4 - tan((a + b*x)/2)**2 - 1),x)*b*d**2 - 16*int((tan((a + 
b*x)/2)*x**2)/(tan((a + b*x)/2)**6 + tan((a + b*x)/2)**4 - tan((a + b*x)/2 
)**2 - 1),x)*b**2*d**2 - 32*int((tan((a + b*x)/2)*x)/(tan((a + b*x)/2)**6 
+ tan((a + b*x)/2)**4 - tan((a + b*x)/2)**2 - 1),x)*b**2*c*d + 16*int(x/(t 
an((a + b*x)/2)**6 + tan((a + b*x)/2)**4 - tan((a + b*x)/2)**2 - 1),x)*b*d 
**2 + 2*log(tan((a + b*x)/2)**2 + 1)*b*c**2 - 2*log(tan((a + b*x)/2) - 1)* 
b*c**2 - 2*log(tan((a + b*x)/2) + 1)*b*c**2 - sin(a + b*x)**2*b*c**2 - 4*s 
in(a + b*x)**2*b*c*d*x - 2*sin(a + b*x)**2*b*d**2*x**2 - 8*sin(a + b*x)*c* 
d + 10*a*c*d + 2*b*c*d*x + 4*b*d**2*x**2)/(2*b**2)