\(\int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx\) [291]

Optimal result

Integrand size = 22, antiderivative size = 139 \[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {2 i d (c+d x)^3}{b^2}-\frac {6 d^2 (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac {6 i d^3 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^4}-\frac {3 d^4 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^5}+\frac {(c+d x)^4 \sec ^2(a+b x)}{2 b}-\frac {2 d (c+d x)^3 \tan (a+b x)}{b^2} \] Output:

2*I*d*(d*x+c)^3/b^2-6*d^2*(d*x+c)^2*ln(1+exp(2*I*(b*x+a)))/b^3+6*I*d^3*(d* 
x+c)*polylog(2,-exp(2*I*(b*x+a)))/b^4-3*d^4*polylog(3,-exp(2*I*(b*x+a)))/b 
^5+1/2*(d*x+c)^4*sec(b*x+a)^2/b-2*d*(d*x+c)^3*tan(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. \(324\) vs. \(2(139)=278\).

Time = 6.28 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.33 \[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\frac {-d^4 e^{-i a} \left (2 b^2 x^2 \left (2 i b x+3 \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )\right )+6 i b \left (1+e^{2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+3 \left (1+e^{2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i (a+b x)}\right )\right ) \sec (a)+b^4 (c+d x)^4 \sec ^2(a+b x)-4 b^3 d (c+d x)^3 \sec (a) \sec (a+b x) \sin (b x)-12 b^2 c^2 d^2 (\log (\cos (a+b x))+b x \tan (a))+12 b c d^3 \left (-i b x (\pi +2 \arctan (\cot (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x-\arctan (\cot (a))) \log \left (1-e^{2 i (b x-\arctan (\cot (a)))}\right )+\pi \log (\cos (b x))-2 \arctan (\cot (a)) \log (\sin (b x-\arctan (\cot (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x-\arctan (\cot (a)))}\right )-b^2 e^{-i \arctan (\cot (a))} x^2 \sqrt {\csc ^2(a)} \tan (a)\right )}{2 b^5} \] Input:

Integrate[(c + d*x)^4*Sec[a + b*x]^2*Tan[a + b*x],x]
 

Output:

(-((d^4*(2*b^2*x^2*((2*I)*b*x + 3*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + 
 b*x))]) + (6*I)*b*(1 + E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] + 
 3*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/E^(I*a)) + 
 b^4*(c + d*x)^4*Sec[a + b*x]^2 - 4*b^3*d*(c + d*x)^3*Sec[a]*Sec[a + b*x]* 
Sin[b*x] - 12*b^2*c^2*d^2*(Log[Cos[a + b*x]] + b*x*Tan[a]) + 12*b*c*d^3*(( 
-I)*b*x*(Pi + 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - Ar 
cTan[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]]))] - (b^2*x^2*Sqrt[Csc[a]^2]*Tan[a])/E^(I*ArcTan[Co 
t[a]])))/(2*b^5)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {4909, 3042, 4672, 25, 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.

\(\displaystyle \int (c+d x)^4 \tan (a+b x) \sec ^2(a+b x) \, dx\)

\(\Big \downarrow \) 4909

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c+d x)^4 \sec ^2(a+b x)}{2 b}-\frac {2 d \int (c+d x)^3 \csc \left (a+b x+\frac {\pi }{2}\right )^2dx}{b}\)

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[(c + d*x)^4*Sec[a + b*x]^2*Tan[a + b*x],x]
 

Output:

((c + d*x)^4*Sec[a + b*x]^2)/(2*b) - (2*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))/b
 

Defintions of rubi rules used

rule 25

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

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 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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]
 

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. 488 vs. \(2 (129 ) = 258\).

Time = 4.75 (sec) , antiderivative size = 489, normalized size of antiderivative = 3.52

method	result	size
risch	\(\frac {2 b \,d^{4} x^{4} {\mathrm e}^{2 i \left (b x +a \right )}+8 b c \,d^{3} x^{3} {\mathrm e}^{2 i \left (b x +a \right )}+12 b \,c^{2} d^{2} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}+8 b \,c^{3} d x \,{\mathrm e}^{2 i \left (b x +a \right )}-4 i d^{4} x^{3} {\mathrm e}^{2 i \left (b x +a \right )}+2 b \,c^{4} {\mathrm e}^{2 i \left (b x +a \right )}-12 i c \,d^{3} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}-12 i c^{2} d^{2} x \,{\mathrm e}^{2 i \left (b x +a \right )}-4 i d^{4} x^{3}-4 i c^{3} d \,{\mathrm e}^{2 i \left (b x +a \right )}-12 i c \,d^{3} x^{2}-12 i c^{2} d^{2} x -4 i c^{3} d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {6 d^{2} c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b^{3}}+\frac {12 d^{2} c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {8 i d^{4} a^{3}}{b^{5}}+\frac {24 i d^{3} c a x}{b^{3}}-\frac {24 d^{3} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 d^{4} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b^{3}}-\frac {3 d^{4} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{5}}+\frac {4 i d^{4} x^{3}}{b^{2}}+\frac {6 i d^{3} c \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{4}}+\frac {12 d^{4} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{5}}-\frac {12 d^{3} c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b^{3}}+\frac {12 i d^{3} c \,a^{2}}{b^{4}}-\frac {12 i d^{4} a^{2} x}{b^{4}}+\frac {6 i d^{4} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{4}}+\frac {12 i d^{3} c \,x^{2}}{b^{2}}\)	\(489\)

Input:

int((d*x+c)^4*sec(b*x+a)^2*tan(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

2*(b*d^4*x^4*exp(2*I*(b*x+a))+4*b*c*d^3*x^3*exp(2*I*(b*x+a))+6*b*c^2*d^2*x 
^2*exp(2*I*(b*x+a))+4*b*c^3*d*x*exp(2*I*(b*x+a))-2*I*d^4*x^3*exp(2*I*(b*x+ 
a))+b*c^4*exp(2*I*(b*x+a))-6*I*c*d^3*x^2*exp(2*I*(b*x+a))-6*I*c^2*d^2*x*ex 
p(2*I*(b*x+a))-2*I*d^4*x^3-2*I*c^3*d*exp(2*I*(b*x+a))-6*I*c*d^3*x^2-6*I*c^ 
2*d^2*x-2*I*c^3*d)/b^2/(exp(2*I*(b*x+a))+1)^2-6/b^3*d^2*c^2*ln(exp(2*I*(b* 
x+a))+1)+12/b^3*d^2*c^2*ln(exp(I*(b*x+a)))-8*I/b^5*d^4*a^3+24*I/b^3*d^3*c* 
a*x-24/b^4*d^3*c*a*ln(exp(I*(b*x+a)))-6/b^3*d^4*ln(exp(2*I*(b*x+a))+1)*x^2 
-3*d^4*polylog(3,-exp(2*I*(b*x+a)))/b^5+4*I/b^2*d^4*x^3+6*I/b^4*d^3*c*poly 
log(2,-exp(2*I*(b*x+a)))+12/b^5*d^4*a^2*ln(exp(I*(b*x+a)))-12/b^3*d^3*c*ln 
(exp(2*I*(b*x+a))+1)*x+12*I/b^4*d^3*c*a^2-12*I/b^4*d^4*a^2*x+6*I/b^4*d^4*p 
olylog(2,-exp(2*I*(b*x+a)))*x+12*I/b^2*d^3*c*x^2
 

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. 892 vs. \(2 (126) = 252\).

Time = 0.12 (sec) , antiderivative size = 892, normalized size of antiderivative = 6.42 \[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*sec(b*x+a)^2*tan(b*x+a),x, algorithm="fricas")
 

Output:

1/2*(b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + b 
^4*c^4 - 12*d^4*cos(b*x + a)^2*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 
 12*d^4*cos(b*x + a)^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) - 12*d^4* 
cos(b*x + a)^2*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) - 12*d^4*cos(b*x 
 + a)^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a)) - 12*(I*b*d^4*x + I*b*c 
*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a)) - 12*(-I*b*d^4*x 
 - I*b*c*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a)) - 12*(-I 
*b*d^4*x - I*b*c*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) + sin(b*x + a)) 
 - 12*(I*b*d^4*x + I*b*c*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b 
*x + a)) - 6*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*cos(b*x + a)^2*log(cos( 
b*x + a) + I*sin(b*x + a) + I) - 6*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*c 
os(b*x + a)^2*log(cos(b*x + a) - I*sin(b*x + a) + I) - 6*(b^2*d^4*x^2 + 2* 
b^2*c*d^3*x + 2*a*b*c*d^3 - a^2*d^4)*cos(b*x + a)^2*log(I*cos(b*x + a) + s 
in(b*x + a) + 1) - 6*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + 2*a*b*c*d^3 - a^2*d^4) 
*cos(b*x + a)^2*log(I*cos(b*x + a) - sin(b*x + a) + 1) - 6*(b^2*d^4*x^2 + 
2*b^2*c*d^3*x + 2*a*b*c*d^3 - a^2*d^4)*cos(b*x + a)^2*log(-I*cos(b*x + a) 
+ sin(b*x + a) + 1) - 6*(b^2*d^4*x^2 + 2*b^2*c*d^3*x + 2*a*b*c*d^3 - a^2*d 
^4)*cos(b*x + a)^2*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - 6*(b^2*c^2*d^ 
2 - 2*a*b*c*d^3 + a^2*d^4)*cos(b*x + a)^2*log(-cos(b*x + a) + I*sin(b*x + 
a) + I) - 6*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*cos(b*x + a)^2*log(-c...
 

Sympy [F]

\[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right )^{4} \tan {\left (a + b x \right )} \sec ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)**4*sec(b*x+a)**2*tan(b*x+a),x)
 

Output:

Integral((c + d*x)**4*tan(a + b*x)*sec(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. 3446 vs. \(2 (126) = 252\).

Time = 0.29 (sec) , antiderivative size = 3446, normalized size of antiderivative = 24.79 \[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*sec(b*x+a)^2*tan(b*x+a),x, algorithm="maxima")
 

Output:

1/2*(c^4*tan(b*x + a)^2 - 4*a*c^3*d*tan(b*x + a)^2/b + 6*a^2*c^2*d^2*tan(b 
*x + a)^2/b^2 - 4*a^3*c*d^3*tan(b*x + a)^2/b^3 + a^4*d^4*tan(b*x + a)^2/b^ 
4 + 8*(4*(b*x + a)*cos(2*b*x + 2*a)^2 + 4*(b*x + a)*sin(2*b*x + 2*a)^2 + ( 
2*(b*x + a)*cos(2*b*x + 2*a) + sin(2*b*x + 2*a))*cos(4*b*x + 4*a) + 2*(b*x 
 + a)*cos(2*b*x + 2*a) + (2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a) 
- 1)*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*c^3*d/((2*(2*cos(2*b*x + 2*a) + 
1)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4*b* 
x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*sin(2*b*x + 2*a)^2 + 
4*cos(2*b*x + 2*a) + 1)*b) - 24*(4*(b*x + a)*cos(2*b*x + 2*a)^2 + 4*(b*x + 
 a)*sin(2*b*x + 2*a)^2 + (2*(b*x + a)*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) 
*cos(4*b*x + 4*a) + 2*(b*x + a)*cos(2*b*x + 2*a) + (2*(b*x + a)*sin(2*b*x 
+ 2*a) - cos(2*b*x + 2*a) - 1)*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*a*c^2* 
d^2/((2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 4 
*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 
2*a) + 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) + 1)*b^2) + 24*(4*(b*x + 
a)*cos(2*b*x + 2*a)^2 + 4*(b*x + a)*sin(2*b*x + 2*a)^2 + (2*(b*x + a)*cos( 
2*b*x + 2*a) + sin(2*b*x + 2*a))*cos(4*b*x + 4*a) + 2*(b*x + a)*cos(2*b*x 
+ 2*a) + (2*(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a) - 1)*sin(4*b*x + 
 4*a) - sin(2*b*x + 2*a))*a^2*c*d^3/((2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x 
 + 4*a) + cos(4*b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^...
 

Giac [F]

\[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{4} \sec \left (b x + a\right )^{2} \tan \left (b x + a\right ) \,d x } \] Input:

integrate((d*x+c)^4*sec(b*x+a)^2*tan(b*x+a),x, algorithm="giac")
 

Output:

integrate((d*x + c)^4*sec(b*x + a)^2*tan(b*x + a), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\int \frac {\mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^4}{{\cos \left (a+b\,x\right )}^2} \,d x \] Input:

int((tan(a + b*x)*(c + d*x)^4)/cos(a + b*x)^2,x)
 

Output:

int((tan(a + b*x)*(c + d*x)^4)/cos(a + b*x)^2, x)
 

Reduce [F]

\[ \int (c+d x)^4 \sec ^2(a+b x) \tan (a+b x) \, dx=\text {too large to display} \] Input:

int((d*x+c)^4*sec(b*x+a)^2*tan(b*x+a),x)
 

Output:

( - 14*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*b**4*c*d**3*x**4 + 12 
*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*b**2*c**3*d - 24*cos(a + b* 
x)*sin(a + b*x)*tan((a + b*x)/2)**4*b**2*c**2*d**2*x - 24*cos(a + b*x)*sin 
(a + b*x)*tan((a + b*x)/2)**4*b**2*c*d**3*x**2 - 8*cos(a + b*x)*sin(a + b* 
x)*tan((a + b*x)/2)**4*b**2*d**4*x**3 + 24*cos(a + b*x)*sin(a + b*x)*tan(( 
a + b*x)/2)**4*c*d**3 + 28*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b 
**4*c*d**3*x**4 - 24*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*c* 
*3*d + 48*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*c**2*d**2*x + 
 48*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*c*d**3*x**2 + 16*co 
s(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b**2*d**4*x**3 - 48*cos(a + b* 
x)*sin(a + b*x)*tan((a + b*x)/2)**2*c*d**3 - 14*cos(a + b*x)*sin(a + b*x)* 
b**4*c*d**3*x**4 + 12*cos(a + b*x)*sin(a + b*x)*b**2*c**3*d - 24*cos(a + b 
*x)*sin(a + b*x)*b**2*c**2*d**2*x - 24*cos(a + b*x)*sin(a + b*x)*b**2*c*d* 
*3*x**2 - 8*cos(a + b*x)*sin(a + b*x)*b**2*d**4*x**3 + 24*cos(a + b*x)*sin 
(a + b*x)*c*d**3 - 56*cos(a + b*x)*tan((a + b*x)/2)**4*b**3*c*d**3*x**3 - 
14*cos(a + b*x)*tan((a + b*x)/2)**4*b**3*d**4*x**4 + 24*cos(a + b*x)*tan(( 
a + b*x)/2)**4*b*c**2*d**2 - 48*cos(a + b*x)*tan((a + b*x)/2)**4*b*c*d**3* 
x - 24*cos(a + b*x)*tan((a + b*x)/2)**4*b*d**4*x**2 + 112*cos(a + b*x)*tan 
((a + b*x)/2)**2*b**3*c*d**3*x**3 + 28*cos(a + b*x)*tan((a + b*x)/2)**2*b* 
*3*d**4*x**4 - 48*cos(a + b*x)*tan((a + b*x)/2)**2*b*c**2*d**2 + 96*cos...