3.1.30 \(\int (c+d x)^2 \sec (a+b x) \, dx\) [30]

3.1.30.1 Optimal result

Integrand size = 14, antiderivative size = 137 \[ \int (c+d x)^2 \sec (a+b x) \, dx=-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3} \]

output

-2*I*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b+2*I*d*(d*x+c)*polylog(2,-I*exp(I*( 
b*x+a)))/b^2-2*I*d*(d*x+c)*polylog(2,I*exp(I*(b*x+a)))/b^2-2*d^2*polylog(3 
,-I*exp(I*(b*x+a)))/b^3+2*d^2*polylog(3,I*exp(I*(b*x+a)))/b^3
 

3.1.30.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.95 \[ \int (c+d x)^2 \sec (a+b x) \, dx=-\frac {2 i \left (b^2 (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )-d \left (b (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+i d \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )\right )+d \left (b (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+i d \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )\right )\right )}{b^3} \]

input

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

output

((-2*I)*(b^2*(c + d*x)^2*ArcTan[E^(I*(a + b*x))] - d*(b*(c + d*x)*PolyLog[ 
2, (-I)*E^(I*(a + b*x))] + I*d*PolyLog[3, (-I)*E^(I*(a + b*x))]) + d*(b*(c 
 + d*x)*PolyLog[2, I*E^(I*(a + b*x))] + I*d*PolyLog[3, I*E^(I*(a + b*x))]) 
))/b^3
 

3.1.30.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4669, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

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

output

((-2*I)*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/b + (2*d*((I*(c + d*x)*PolyLo 
g[2, (-I)*E^(I*(a + b*x))])/b - (d*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^2)) 
/b - (2*d*((I*(c + d*x)*PolyLog[2, I*E^(I*(a + b*x))])/b - (d*PolyLog[3, I 
*E^(I*(a + b*x))])/b^2))/b
 

3.1.30.3.1 Defintions of rubi rules used

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 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 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]
 

3.1.30.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (120 ) = 240\).

Time = 1.47 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.86

method	result	size
risch	\(\frac {4 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {a^{2} d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}+\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {2 d^{2} \operatorname {Li}_{3}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \operatorname {Li}_{3}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {2 c d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {2 i c d \,\operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {2 i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} \operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 i d^{2} \operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 i c d \,\operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(392\)

input

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

output

4*I/b^2*c*d*a*arctan(exp(I*(b*x+a)))-2*I/b*c^2*arctan(exp(I*(b*x+a)))+1/b^ 
3*a^2*d^2*ln(1+I*exp(I*(b*x+a)))-1/b*d^2*ln(1+I*exp(I*(b*x+a)))*x^2+1/b*d^ 
2*ln(1-I*exp(I*(b*x+a)))*x^2-2*d^2*polylog(3,-I*exp(I*(b*x+a)))/b^3+2*d^2* 
polylog(3,I*exp(I*(b*x+a)))/b^3-2/b*c*d*ln(1+I*exp(I*(b*x+a)))*x+2/b*c*d*l 
n(1-I*exp(I*(b*x+a)))*x-2/b^2*c*d*ln(1+I*exp(I*(b*x+a)))*a-2*I/b^2*c*d*pol 
ylog(2,I*exp(I*(b*x+a)))-1/b^3*a^2*d^2*ln(1-I*exp(I*(b*x+a)))+2/b^2*c*d*ln 
(1-I*exp(I*(b*x+a)))*a-2*I/b^3*d^2*a^2*arctan(exp(I*(b*x+a)))-2*I/b^2*d^2* 
polylog(2,I*exp(I*(b*x+a)))*x+2*I/b^2*d^2*polylog(2,-I*exp(I*(b*x+a)))*x+2 
*I/b^2*c*d*polylog(2,-I*exp(I*(b*x+a)))
 

3.1.30.5 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. 598 vs. \(2 (111) = 222\).

Time = 0.32 (sec) , antiderivative size = 598, normalized size of antiderivative = 4.36 \[ \int (c+d x)^2 \sec (a+b x) \, dx=-\frac {2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 2 \, d^{2} {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 \, {\left (i \, b d^{2} x + i \, b c d\right )} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + 2 \, {\left (-i \, b d^{2} x - i \, b c d\right )} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{3}} \]

input

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

output

-1/2*(2*d^2*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 2*d^2*polylog(3, I 
*cos(b*x + a) - sin(b*x + a)) + 2*d^2*polylog(3, -I*cos(b*x + a) + sin(b*x 
 + a)) - 2*d^2*polylog(3, -I*cos(b*x + a) - sin(b*x + a)) + 2*(I*b*d^2*x + 
 I*b*c*d)*dilog(I*cos(b*x + a) + sin(b*x + a)) + 2*(I*b*d^2*x + I*b*c*d)*d 
ilog(I*cos(b*x + a) - sin(b*x + a)) + 2*(-I*b*d^2*x - I*b*c*d)*dilog(-I*co 
s(b*x + a) + sin(b*x + a)) + 2*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x + a 
) - sin(b*x + a)) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x + a) + I*s 
in(b*x + a) + I) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(cos(b*x + a) - I*si 
n(b*x + a) + I) - (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I* 
cos(b*x + a) + sin(b*x + a) + 1) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d 
- a^2*d^2)*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b^2*d^2*x^2 + 2*b^2*c 
*d*x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b^2 
*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) - sin(b* 
x + a) + 1) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) + I*sin(b* 
x + a) + I) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(-cos(b*x + a) - I*sin(b* 
x + a) + I))/b^3
 

3.1.30.6 Sympy [F]

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

input

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

output

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

3.1.30.7 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. 402 vs. \(2 (111) = 222\).

Time = 0.42 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.93 \[ \int (c+d x)^2 \sec (a+b x) \, dx=\frac {2 \, c^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right ) - \frac {4 \, a c d \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b} + \frac {2 \, a^{2} d^{2} \log \left (\sec \left (b x + a\right ) + \tan \left (b x + a\right )\right )}{b^{2}} + \frac {4 \, d^{2} {\rm Li}_{3}(i \, e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2} {\rm Li}_{3}(-i \, e^{\left (i \, b x + i \, a\right )}) - 2 \, {\left (i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (i \, b c d - i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), \sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (i \, {\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (i \, b c d - i \, a d^{2}\right )} {\left (b x + a\right )}\right )} \arctan \left (\cos \left (b x + a\right ), -\sin \left (b x + a\right ) + 1\right ) - 4 \, {\left (i \, b c d + i \, {\left (b x + a\right )} d^{2} - i \, a d^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, b x + i \, a\right )}\right ) - 4 \, {\left (-i \, b c d - i \, {\left (b x + a\right )} d^{2} + i \, a d^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, b x + i \, a\right )}\right ) + {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) + 1\right ) - {\left ({\left (b x + a\right )}^{2} d^{2} + 2 \, {\left (b c d - a d^{2}\right )} {\left (b x + a\right )}\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \sin \left (b x + a\right ) + 1\right )}{b^{2}}}{2 \, b} \]

input

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

output

1/2*(2*c^2*log(sec(b*x + a) + tan(b*x + a)) - 4*a*c*d*log(sec(b*x + a) + t 
an(b*x + a))/b + 2*a^2*d^2*log(sec(b*x + a) + tan(b*x + a))/b^2 + (4*d^2*p 
olylog(3, I*e^(I*b*x + I*a)) - 4*d^2*polylog(3, -I*e^(I*b*x + I*a)) - 2*(I 
*(b*x + a)^2*d^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a))*arctan2(cos(b*x + a), 
sin(b*x + a) + 1) - 2*(I*(b*x + a)^2*d^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a) 
)*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 4*(I*b*c*d + I*(b*x + a)*d^2 
- I*a*d^2)*dilog(I*e^(I*b*x + I*a)) - 4*(-I*b*c*d - I*(b*x + a)*d^2 + I*a* 
d^2)*dilog(-I*e^(I*b*x + I*a)) + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x 
 + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - ((b*x + 
 a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a) 
^2 - 2*sin(b*x + a) + 1))/b^2)/b
 

3.1.30.8 Giac [F]

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

input

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

output

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

3.1.30.9 Mupad [F(-1)]

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

input

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

output

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