\(\int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx\) [300]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b} \] Output:

I*(d*x+c)*arctan(exp(I*(b*x+a)))/b-1/2*I*d*polylog(2,-I*exp(I*(b*x+a)))/b^ 
2+1/2*I*d*polylog(2,I*exp(I*(b*x+a)))/b^2-1/2*d*sec(b*x+a)/b^2+1/2*(d*x+c) 
*sec(b*x+a)*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. \(555\) vs. \(2(117)=234\).

Time = 6.56 (sec) , antiderivative size = 555, normalized size of antiderivative = 4.74 \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(c*ArcTanh[Sin[a + b*x]])/b + (d*x*(a*Log[1 - Tan[(a + b*x)/2]] + I*L 
og[1 + I*Tan[(a + b*x)/2]]*Log[(-1/2 - I/2)*(-1 + Tan[(a + b*x)/2])] - I*L 
og[1 - I*Tan[(a + b*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[(a + b*x)/2])] - I*L 
og[1 + I*Tan[(a + b*x)/2]]*Log[(1/2 - I/2)*(1 + Tan[(a + b*x)/2])] + I*Log 
[1 - I*Tan[(a + b*x)/2]]*Log[(1/2 + I/2)*(1 + Tan[(a + b*x)/2])] - a*Log[1 
 + Tan[(a + b*x)/2]] - I*PolyLog[2, ((1 + I) - (1 - I)*Tan[(a + b*x)/2])/2 
] + I*PolyLog[2, (-1/2 - I/2)*(I + Tan[(a + b*x)/2])] - I*PolyLog[2, ((1 + 
 I) + (1 - I)*Tan[(a + b*x)/2])/2] + I*PolyLog[2, ((1 - I) + (1 + I)*Tan[( 
a + b*x)/2])/2]))/(2*b*(a - I*Log[1 - I*Tan[(a + b*x)/2]] + I*Log[1 + I*Ta 
n[(a + b*x)/2]])) + (d*x)/(4*b*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])^2) - 
(d*Sin[(a + b*x)/2])/(2*b^2*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])) - (d*x) 
/(4*b*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])^2) + (d*Sin[(a + b*x)/2])/(2*b 
^2*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])) + (c*Sec[a + b*x]*Tan[a + b*x])/ 
(2*b)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.64, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4913, 3042, 4669, 2715, 2838, 4673, 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.

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

\(\Big \downarrow \) 4913

\(\displaystyle \int (c+d x) \sec ^3(a+b x)dx-\int (c+d x) \sec (a+b x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 4673

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 4673

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

rule 4913

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]*Tan[(a_.) + (b_.)*(x 
_)]^(p_), x_Symbol] :> -Int[(c + d*x)^m*Sec[a + b*x]*Tan[a + b*x]^(p - 2), 
x] + Int[(c + d*x)^m*Sec[a + b*x]^3*Tan[a + b*x]^(p - 2), x] /; FreeQ[{a, b 
, c, d, m}, x] && IGtQ[p/2, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (98 ) = 196\).

Time = 0.52 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.28

method	result	size
risch	\(-\frac {i \left (b d x \,{\mathrm e}^{3 i \left (b x +a \right )}-i d \,{\mathrm e}^{3 i \left (b x +a \right )}+c b \,{\mathrm e}^{3 i \left (b x +a \right )}-b d x \,{\mathrm e}^{i \left (b x +a \right )}-i d \,{\mathrm e}^{i \left (b x +a \right )}-b c \,{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {i c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{2 b}+\frac {d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{2 b^{2}}-\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}-\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}-\frac {i d \operatorname {dilog}\left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b^{2}}+\frac {i d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {i d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(267\)

Input:

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

Output:

-I/b^2/(exp(2*I*(b*x+a))+1)^2*(b*d*x*exp(3*I*(b*x+a))-I*d*exp(3*I*(b*x+a)) 
+c*b*exp(3*I*(b*x+a))-b*d*x*exp(I*(b*x+a))-I*d*exp(I*(b*x+a))-b*c*exp(I*(b 
*x+a)))+I/b*c*arctan(exp(I*(b*x+a)))+1/2/b*d*ln(I*exp(I*(b*x+a))+1)*x+1/2/ 
b^2*d*ln(I*exp(I*(b*x+a))+1)*a-1/2/b*d*ln(1-I*exp(I*(b*x+a)))*x-1/2/b^2*d* 
ln(1-I*exp(I*(b*x+a)))*a-1/2*I/b^2*d*dilog(I*exp(I*(b*x+a))+1)+1/2*I/b^2*d 
*dilog(1-I*exp(I*(b*x+a)))-I/b^2*d*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. 435 vs. \(2 (93) = 186\).

Time = 0.10 (sec) , antiderivative size = 435, normalized size of antiderivative = 3.72 \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 2 \, d \cos \left (b x + a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )}{4 \, b^{2} \cos \left (b x + a\right )^{2}} \] Input:

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

Output:

1/4*(I*d*cos(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a)) + I*d*cos(b*x 
 + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a)) - I*d*cos(b*x + a)^2*dilog(-I 
*cos(b*x + a) + sin(b*x + a)) - I*d*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - 
 sin(b*x + a)) - (b*c - a*d)*cos(b*x + a)^2*log(cos(b*x + a) + I*sin(b*x + 
 a) + I) + (b*c - a*d)*cos(b*x + a)^2*log(cos(b*x + a) - I*sin(b*x + a) + 
I) - (b*d*x + a*d)*cos(b*x + a)^2*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 
 (b*d*x + a*d)*cos(b*x + a)^2*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b* 
d*x + a*d)*cos(b*x + a)^2*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b*d*x 
 + a*d)*cos(b*x + a)^2*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (b*c - a* 
d)*cos(b*x + a)^2*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b*c - a*d)*co 
s(b*x + a)^2*log(-cos(b*x + a) - I*sin(b*x + a) + I) - 2*d*cos(b*x + a) + 
2*(b*d*x + b*c)*sin(b*x + a))/(b^2*cos(b*x + a)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-1/4*(4*(d*cos(3*b*x + 3*a) + d*cos(b*x + a) - (b*d*x + b*c)*sin(3*b*x + 3 
*a) + (b*d*x + b*c)*sin(b*x + a))*cos(4*b*x + 4*a) + 4*(2*d*cos(2*b*x + 2* 
a) + 2*(b*d*x + b*c)*sin(2*b*x + 2*a) + d)*cos(3*b*x + 3*a) + 8*(d*cos(b*x 
 + a) + (b*d*x + b*c)*sin(b*x + a))*cos(2*b*x + 2*a) + 4*d*cos(b*x + a) + 
4*(b^2*d*cos(4*b*x + 4*a)^2 + 4*b^2*d*cos(2*b*x + 2*a)^2 + b^2*d*sin(4*b*x 
 + 4*a)^2 + 4*b^2*d*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*b^2*d*sin(2*b*x 
+ 2*a)^2 + 4*b^2*d*cos(2*b*x + 2*a) + b^2*d + 2*(2*b^2*d*cos(2*b*x + 2*a) 
+ b^2*d)*cos(4*b*x + 4*a))*integrate((x*cos(2*b*x + 2*a)*cos(b*x + a) + x* 
sin(2*b*x + 2*a)*sin(b*x + a) + x*cos(b*x + a))/(cos(2*b*x + 2*a)^2 + sin( 
2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1), x) + (b*c*cos(4*b*x + 4*a)^2 + 4 
*b*c*cos(2*b*x + 2*a)^2 + b*c*sin(4*b*x + 4*a)^2 + 4*b*c*sin(4*b*x + 4*a)* 
sin(2*b*x + 2*a) + 4*b*c*sin(2*b*x + 2*a)^2 + 4*b*c*cos(2*b*x + 2*a) + b*c 
 + 2*(2*b*c*cos(2*b*x + 2*a) + b*c)*cos(4*b*x + 4*a))*log(cos(b*x + a)^2 + 
 sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - (b*c*cos(4*b*x + 4*a)^2 + 4*b*c*co 
s(2*b*x + 2*a)^2 + b*c*sin(4*b*x + 4*a)^2 + 4*b*c*sin(4*b*x + 4*a)*sin(2*b 
*x + 2*a) + 4*b*c*sin(2*b*x + 2*a)^2 + 4*b*c*cos(2*b*x + 2*a) + b*c + 2*(2 
*b*c*cos(2*b*x + 2*a) + b*c)*cos(4*b*x + 4*a))*log(cos(b*x + a)^2 + sin(b* 
x + a)^2 - 2*sin(b*x + a) + 1) + 4*((b*d*x + b*c)*cos(3*b*x + 3*a) - (b*d* 
x + b*c)*cos(b*x + a) + d*sin(3*b*x + 3*a) + d*sin(b*x + a))*sin(4*b*x + 4 
*a) - 4*(b*d*x + b*c + 2*(b*d*x + b*c)*cos(2*b*x + 2*a) - 2*d*sin(2*b*x...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

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