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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4498, 4266, 2317, 2438, 4270} \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]

[In]

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

[Out]

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

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 4266

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] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[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 4270

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] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 4498

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]

Rubi steps \begin{align*} \text {integral}& = -\int (c+d x) \sec (a+b x) \, dx+\int (c+d x) \sec ^3(a+b x) \, dx \\ & = \frac {2 i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}+\frac {1}{2} \int (c+d x) \sec (a+b x) \, dx+\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 {i (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}-\frac {(i d) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}+\frac {(i d) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}-\frac {d \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {d \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{2 b} \\ & = \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 )}{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) \sec (a+b x) \tan (a+b x)}{2 b}+\frac {(i d) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac {(i d) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2} \\ & = \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} \\ \end{align*}

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 = 7.03 (sec) , antiderivative size = 555, normalized size of antiderivative = 4.74 \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx=-\frac {c \text {arctanh}(\sin (a+b x))}{2 b}+\frac {d x \left (a \log \left (1-\tan \left (\frac {1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (-1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )-i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac {1}{2}+\frac {i}{2}\right ) \left (-1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )-i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )-a \log \left (1+\tan \left (\frac {1}{2} (a+b x)\right )\right )-i \operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1+i)-(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+i \operatorname {PolyLog}\left (2,\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i+\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )-i \operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1+i)+(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+i \operatorname {PolyLog}\left (2,\frac {1}{2} \left ((1-i)+(1+i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{2 b \left (a-i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )}+\frac {d x}{4 b \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}-\frac {d \sin \left (\frac {1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {d x}{4 b \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}+\frac {d \sin \left (\frac {1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}+\frac {c \sec (a+b x) \tan (a+b x)}{2 b} \]

[In]

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

[Out]

-1/2*(c*ArcTanh[Sin[a + b*x]])/b + (d*x*(a*Log[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])] - I*Log[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])] + 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*Tan[(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*(C
os[(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)

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 = 0.70 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.28

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

[In]

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

[Out]

-I/b^2/(exp(2*I*(b*x+a))+1)^2*(d*x*b*exp(3*I*(b*x+a))-I*d*exp(3*I*(b*x+a))+c*b*exp(3*I*(b*x+a))-d*x*b*exp(I*(b
*x+a))-I*d*exp(I*(b*x+a))-c*b*exp(I*(b*x+a)))+I/b*c*arctan(exp(I*(b*x+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/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(1+I*exp(I*(b*x+a)))+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 = 0.30 (sec) , antiderivative size = 435, normalized size of antiderivative = 3.72 \[ \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 2 \, d \cos \left (b x + a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )}{4 \, b^{2} \cos \left (b x + a\right )^{2}} \]

[In]

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

[Out]

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)*cos(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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-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))*co
s(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))*integ
rate((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*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) + 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 + 2*a))*sin(3*b*x + 3*a) - 8*((b*d*x + b*c)*cos(b*x + a) - d*s
in(b*x + a))*sin(2*b*x + 2*a) + 4*(b*d*x + b*c)*sin(b*x + a))/(b^2*cos(4*b*x + 4*a)^2 + 4*b^2*cos(2*b*x + 2*a)
^2 + b^2*sin(4*b*x + 4*a)^2 + 4*b^2*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*b^2*sin(2*b*x + 2*a)^2 + 4*b^2*cos(2
*b*x + 2*a) + b^2 + 2*(2*b^2*cos(2*b*x + 2*a) + b^2)*cos(4*b*x + 4*a))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

\text{Hanged}