\(\int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx\) [261]

Optimal result

Integrand size = 22, antiderivative size = 145 \[ \int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx=\frac {4 i d (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \cos (a+b x)}{b^3}+\frac {(c+d x)^2 \cos (a+b x)}{b}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {(c+d x)^2 \sec (a+b x)}{b}-\frac {2 d (c+d x) \sin (a+b x)}{b^2} \] Output:

4*I*d*(d*x+c)*arctan(exp(I*(b*x+a)))/b^2-2*d^2*cos(b*x+a)/b^3+(d*x+c)^2*co 
s(b*x+a)/b-2*I*d^2*polylog(2,-I*exp(I*(b*x+a)))/b^3+2*I*d^2*polylog(2,I*ex 
p(I*(b*x+a)))/b^3+(d*x+c)^2*sec(b*x+a)/b-2*d*(d*x+c)*sin(b*x+a)/b^2
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(362\) vs. \(2(145)=290\).

Time = 2.12 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.50 \[ \int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx=\frac {-4 b c d \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )-4 d^2 \arctan (\cot (a)) \text {arctanh}\left (\sin (a)+\cos (a) \tan \left (\frac {b x}{2}\right )\right )+\frac {2 d^2 \csc (a) \left ((b x-\arctan (\cot (a))) \left (\log \left (1-e^{i (b x-\arctan (\cot (a)))}\right )-\log \left (1+e^{i (b x-\arctan (\cot (a)))}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i (b x-\arctan (\cot (a)))}\right )-i \operatorname {PolyLog}\left (2,e^{i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {\csc ^2(a)}}+b^2 (c+d x)^2 \sec (a)+\cos (b x) \left (\left (-2 d^2+b^2 (c+d x)^2\right ) \cos (a)-2 b d (c+d x) \sin (a)\right )-\left (2 b d (c+d x) \cos (a)+\left (-2 d^2+b^2 (c+d x)^2\right ) \sin (a)\right ) \sin (b x)+\frac {b^2 (c+d x)^2 \sin \left (\frac {b x}{2}\right )}{\left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {b^2 (c+d x)^2 \sin \left (\frac {b x}{2}\right )}{\left (\cos \left (\frac {a}{2}\right )+\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}}{b^3} \] Input:

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

Output:

(-4*b*c*d*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] - 4*d^2*ArcTan[Cot[a]]*Arc 
Tanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] + (2*d^2*Csc[a]*((b*x - ArcTan[Cot[a]]) 
*(Log[1 - E^(I*(b*x - ArcTan[Cot[a]]))] - Log[1 + E^(I*(b*x - ArcTan[Cot[a 
]]))]) + I*PolyLog[2, -E^(I*(b*x - ArcTan[Cot[a]]))] - I*PolyLog[2, E^(I*( 
b*x - ArcTan[Cot[a]]))]))/Sqrt[Csc[a]^2] + b^2*(c + d*x)^2*Sec[a] + Cos[b* 
x]*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] - 2*b*d*(c + d*x)*Sin[a]) - (2*b*d*( 
c + d*x)*Cos[a] + (-2*d^2 + b^2*(c + d*x)^2)*Sin[a])*Sin[b*x] + (b^2*(c + 
d*x)^2*Sin[(b*x)/2])/((Cos[a/2] - Sin[a/2])*(Cos[(a + b*x)/2] - Sin[(a + b 
*x)/2])) - (b^2*(c + d*x)^2*Sin[(b*x)/2])/((Cos[a/2] + Sin[a/2])*(Cos[(a + 
 b*x)/2] + Sin[(a + b*x)/2])))/b^3
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {4907, 3042, 3777, 3042, 3777, 25, 3042, 3118, 4909, 3042, 4669, 2715, 2838}

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.

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

\(\Big \downarrow \) 4907

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

\(\Big \downarrow \) 4909

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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 4909

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Simp[(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - 
 Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; FreeQ[{ 
a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (134 ) = 268\).

Time = 0.90 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.38

method	result	size
risch	\(\frac {\left (x^{2} d^{2} b^{2}+2 b^{2} c d x +2 i b \,d^{2} x +b^{2} c^{2}+2 i b c d -2 d^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{2 b^{3}}+\frac {\left (x^{2} d^{2} b^{2}+2 b^{2} c d x -2 i b \,d^{2} x +b^{2} c^{2}-2 i b c d -2 d^{2}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{2 b^{3}}+\frac {2 \,{\mathrm e}^{i \left (b x +a \right )} \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}+\frac {4 i d c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {2 i d^{2} \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {4 i d^{2} a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\)	\(345\)

Input:

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

Output:

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

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. 511 vs. \(2 (127) = 254\).

Time = 0.11 (sec) , antiderivative size = 511, normalized size of antiderivative = 3.52 \[ \int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx=\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{b^{3} \cos \left (b x + a\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*(2*((3*b^2*d^2*x^2 + 6*b^2*c*d*x + 3*b^2*c^2 - 2*d^2)*cos(2*b*x + 3*a) 
*cos(b*x + 2*a) + (3*b^2*d^2*x^2 + 6*b^2*c*d*x + 3*b^2*c^2 - 2*d^2)*sin(2* 
b*x + 3*a)*sin(b*x + 2*a) + (3*b^2*d^2*x^2*cos(a) + 6*b^2*c*d*x*cos(a) + 3 
*b^2*c^2*cos(a) - 2*d^2*cos(a))*cos(b*x + 2*a) + (3*b^2*d^2*x^2*sin(a) + 6 
*b^2*c*d*x*sin(a) + 3*b^2*c^2*sin(a) - 2*d^2*sin(a))*sin(b*x + 2*a))*cos(3 
*b*x + 3*a)^2 + ((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a 
) - 2*(b*d^2*x + b*c*d)*sin(b*x + a))*cos(2*b*x + 3*a)^2 + 2*((3*b^2*d^2*x 
^2 + 6*b^2*c*d*x + 3*b^2*c^2 - 2*d^2)*cos(2*b*x + 3*a)*cos(b*x + 2*a) + (3 
*b^2*d^2*x^2 + 6*b^2*c*d*x + 3*b^2*c^2 - 2*d^2)*sin(2*b*x + 3*a)*sin(b*x + 
 2*a) + (3*b^2*d^2*x^2*cos(a) + 6*b^2*c*d*x*cos(a) + 3*b^2*c^2*cos(a) - 2* 
d^2*cos(a))*cos(b*x + 2*a) + (3*b^2*d^2*x^2*sin(a) + 6*b^2*c*d*x*sin(a) + 
3*b^2*c^2*sin(a) - 2*d^2*sin(a))*sin(b*x + 2*a))*sin(3*b*x + 3*a)^2 + ((b^ 
2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a) - 2*(b*d^2*x + b*c 
*d)*sin(b*x + a))*sin(2*b*x + 3*a)^2 + ((b^2*d^2*x^2*cos(a) + b^2*c^2*cos( 
a) + 2*b*c*d*sin(a) - 2*d^2*cos(a) + 2*(b^2*c*d*cos(a) + b*d^2*sin(a))*x + 
 (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(2*b*x + 3*a) + 2*(b*d^2 
*x + b*c*d)*sin(2*b*x + 3*a))*cos(3*b*x + 3*a)^2 + (b^2*d^2*x^2*cos(a) + b 
^2*c^2*cos(a) + 2*b*c*d*sin(a) - 2*d^2*cos(a) + 2*(b^2*c*d*cos(a) + b*d^2* 
sin(a))*x)*cos(b*x + a)^2 + (b^2*d^2*x^2*cos(a) + b^2*c^2*cos(a) + 2*b*c*d 
*sin(a) - 2*d^2*cos(a) + 2*(b^2*c*d*cos(a) + b*d^2*sin(a))*x + (b^2*d^2...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx=\frac {-16 \cos \left (b x +a \right ) \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}d x \right ) b^{2} d^{2}-2 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) d^{2}+6 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) b c d +2 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) d^{2}-6 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) b c d +2 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) d^{2}-6 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b c d -4 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b \,d^{2} x -12 \cos \left (b x +a \right ) a b c d -6 \cos \left (b x +a \right ) b^{2} c^{2}+3 \cos \left (b x +a \right ) b^{2} d^{2} x^{2}+4 \cos \left (b x +a \right ) d^{2}-3 \sin \left (b x +a \right )^{2} b^{2} c^{2}-6 \sin \left (b x +a \right )^{2} b^{2} c d x -3 \sin \left (b x +a \right )^{2} b^{2} d^{2} x^{2}+4 \sin \left (b x +a \right )^{2} d^{2}+2 \sin \left (b x +a \right ) b \,d^{2} x +6 b^{2} c^{2}+12 b^{2} c d x +6 b^{2} d^{2} x^{2}-4 d^{2}}{3 \cos \left (b x +a \right ) b^{3}} \] Input:

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

Output:

( - 16*cos(a + b*x)*int(x/(tan((a + b*x)/2)**6 - tan((a + b*x)/2)**4 - tan 
((a + b*x)/2)**2 + 1),x)*b**2*d**2 - 2*cos(a + b*x)*log(tan((a + b*x)/2)** 
2 + 1)*d**2 + 6*cos(a + b*x)*log(tan((a + b*x)/2) - 1)*b*c*d + 2*cos(a + b 
*x)*log(tan((a + b*x)/2) - 1)*d**2 - 6*cos(a + b*x)*log(tan((a + b*x)/2) + 
 1)*b*c*d + 2*cos(a + b*x)*log(tan((a + b*x)/2) + 1)*d**2 - 6*cos(a + b*x) 
*sin(a + b*x)*b*c*d - 4*cos(a + b*x)*sin(a + b*x)*b*d**2*x - 12*cos(a + b* 
x)*a*b*c*d - 6*cos(a + b*x)*b**2*c**2 + 3*cos(a + b*x)*b**2*d**2*x**2 + 4* 
cos(a + b*x)*d**2 - 3*sin(a + b*x)**2*b**2*c**2 - 6*sin(a + b*x)**2*b**2*c 
*d*x - 3*sin(a + b*x)**2*b**2*d**2*x**2 + 4*sin(a + b*x)**2*d**2 + 2*sin(a 
 + b*x)*b*d**2*x + 6*b**2*c**2 + 12*b**2*c*d*x + 6*b**2*d**2*x**2 - 4*d**2 
)/(3*cos(a + b*x)*b**3)