\(\int (c+d x)^3 \tan ^2(a+b x) \, dx\) [254]

Optimal result

Integrand size = 16, antiderivative size = 128 \[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^4}{4 d}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d^3 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac {(c+d x)^3 \tan (a+b x)}{b} \] Output:

-I*(d*x+c)^3/b-1/4*(d*x+c)^4/d+3*d*(d*x+c)^2*ln(1+exp(2*I*(b*x+a)))/b^2-3* 
I*d^2*(d*x+c)*polylog(2,-exp(2*I*(b*x+a)))/b^3+3/2*d^3*polylog(3,-exp(2*I* 
(b*x+a)))/b^4+(d*x+c)^3*tan(b*x+a)/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. \(424\) vs. \(2(128)=256\).

Time = 6.32 (sec) , antiderivative size = 424, normalized size of antiderivative = 3.31 \[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=-\frac {1}{4} x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )+\frac {i d^3 e^{-i a} \left (2 b^2 x^2 \left (2 b x-3 i \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 b \left (1+e^{2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-3 i \left (1+e^{2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )\right ) \sec (a)}{4 b^4}+\frac {3 c^2 d \sec (a) (\cos (a) \log (\cos (a) \cos (b x)-\sin (a) \sin (b x))+b x \sin (a))}{b^2 \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac {3 c d^2 \csc (a) \left (b^2 e^{-i \arctan (\cot (a))} x^2-\frac {\cot (a) \left (i b x (-\pi -2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )\right )}{\sqrt {1+\cot ^2(a)}}\right ) \sec (a)}{b^3 \sqrt {\csc ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}}+\frac {\sec (a) \sec (a+b x) \left (c^3 \sin (b x)+3 c^2 d x \sin (b x)+3 c d^2 x^2 \sin (b x)+d^3 x^3 \sin (b x)\right )}{b} \] Input:

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

Output:

-1/4*(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)) + ((I/4)*d^3*(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^4*E^(I*a)) + (3*c^2*d*S 
ec[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[C 
os[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
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 4203, 17, 3042, 4202, 2620, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4203

\(\displaystyle -\frac {3 d \int (c+d x)^2 \tan (a+b x)dx}{b}-\int (c+d x)^3dx+\frac {(c+d x)^3 \tan (a+b x)}{b}\)

\(\Big \downarrow \) 17

\(\displaystyle -\frac {3 d \int (c+d x)^2 \tan (a+b x)dx}{b}+\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {(c+d x)^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {3 d \int (c+d x)^2 \tan (a+b x)dx}{b}+\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {(c+d x)^4}{4 d}\)

\(\Big \downarrow \) 4202

\(\displaystyle -\frac {3 d \left (\frac {i (c+d x)^3}{3 d}-2 i \int \frac {e^{2 i (a+b x)} (c+d x)^2}{1+e^{2 i (a+b x)}}dx\right )}{b}+\frac {(c+d x)^3 \tan (a+b x)}{b}-\frac {(c+d x)^4}{4 d}\)

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

-1/4*(c + d*x)^4/d - (3*d*(((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)))/b + ((c + d*x)^3*Tan[a + b*x])/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 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 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 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. 355 vs. \(2 (116 ) = 232\).

Time = 0.50 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.78

method	result	size
risch	\(-\frac {d^{3} x^{4}}{4}-d^{2} c \,x^{3}-\frac {3 d \,c^{2} x^{2}}{2}-c^{3} x -\frac {c^{4}}{4 d}-\frac {12 i d^{2} c a x}{b^{2}}-\frac {2 i d^{3} x^{3}}{b}-\frac {6 i d^{2} c \,a^{2}}{b^{3}}+\frac {4 i d^{3} a^{3}}{b^{4}}-\frac {6 i d^{2} c \,x^{2}}{b}-\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {6 i d^{3} a^{2} x}{b^{3}}+\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}+\frac {12 d^{2} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 d^{2} c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 i d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}+\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{4}}\)	\(356\)

Input:

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

Output:

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

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. 373 vs. \(2 (113) = 226\).

Time = 0.08 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.91 \[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=-\frac {b^{4} d^{3} x^{4} + 4 \, b^{4} c d^{2} x^{3} + 6 \, b^{4} c^{2} d x^{2} + 4 \, b^{4} c^{3} x - 3 \, d^{3} {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 \, d^{3} {\rm polylog}\left (3, \frac {\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, {\left (-i \, b d^{3} x - i \, b c d^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 6 \, {\left (i \, b d^{3} x + i \, b c d^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - 6 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 6 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) - 4 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \tan \left (b x + a\right )}{4 \, b^{4}} \] Input:

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

Output:

-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*tan(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 - 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 + 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
 

Sympy [F]

\[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=\int \left (c + d x\right )^{3} \tan ^{2}{\left (a + b x \right )}\, dx \] Input:

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

Output:

Integral((c + d*x)**3*tan(a + b*x)**2, 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. 1363 vs. \(2 (113) = 226\).

Time = 0.41 (sec) , antiderivative size = 1363, normalized size of antiderivative = 10.65 \[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-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 + si 
n(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 + I*a*d^3)*(b*x + a)^3 + 12*((b*x 
 + a)^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a) + ((b*x + a)^2*d^3 + 2*(b*c*d^ 
2 - a*d^3)*(b*x + a))*cos(2*b*x + 2*a) - (-I*(b*x + a)^2*d^3 + 2*(-I*b*...
 

Giac [F]

\[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \tan \left (b x + a\right )^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=\int {\mathrm {tan}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 \tan ^2(a+b x) \, dx=\frac {-12 \left (\int \tan \left (b x +a \right ) x^{2}d x \right ) b \,d^{3}-24 \left (\int \tan \left (b x +a \right ) x d x \right ) b c \,d^{2}-6 \,\mathrm {log}\left (\tan \left (b x +a \right )^{2}+1\right ) c^{2} d +4 \tan \left (b x +a \right ) b \,c^{3}+12 \tan \left (b x +a \right ) b \,c^{2} d x +12 \tan \left (b x +a \right ) b c \,d^{2} x^{2}+4 \tan \left (b x +a \right ) b \,d^{3} x^{3}-4 b^{2} c^{3} x -6 b^{2} c^{2} d \,x^{2}-4 b^{2} c \,d^{2} x^{3}-b^{2} d^{3} x^{4}}{4 b^{2}} \] Input:

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

Output:

( - 12*int(tan(a + b*x)*x**2,x)*b*d**3 - 24*int(tan(a + b*x)*x,x)*b*c*d**2 
 - 6*log(tan(a + b*x)**2 + 1)*c**2*d + 4*tan(a + b*x)*b*c**3 + 12*tan(a + 
b*x)*b*c**2*d*x + 12*tan(a + b*x)*b*c*d**2*x**2 + 4*tan(a + b*x)*b*d**3*x* 
*3 - 4*b**2*c**3*x - 6*b**2*c**2*d*x**2 - 4*b**2*c*d**2*x**3 - b**2*d**3*x 
**4)/(4*b**2)