\(\int (c+d x)^2 \sec (a+b x) \tan ^2(a+b x) \, dx\) [299]

Optimal result

Integrand size = 22, antiderivative size = 193 \[ \int (c+d x)^2 \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {d^2 \text {arctanh}(\sin (a+b x))}{b^3}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}+\frac {i d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}+\frac {d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {d (c+d x) \sec (a+b x)}{b^2}+\frac {(c+d x)^2 \sec (a+b x) \tan (a+b x)}{2 b} \] Output:

I*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b+d^2*arctanh(sin(b*x+a))/b^3-I*d*(d*x+ 
c)*polylog(2,-I*exp(I*(b*x+a)))/b^2+I*d*(d*x+c)*polylog(2,I*exp(I*(b*x+a)) 
)/b^2+d^2*polylog(3,-I*exp(I*(b*x+a)))/b^3-d^2*polylog(3,I*exp(I*(b*x+a))) 
/b^3-d*(d*x+c)*sec(b*x+a)/b^2+1/2*(d*x+c)^2*sec(b*x+a)*tan(b*x+a)/b
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(443\) vs. \(2(193)=386\).

Time = 6.75 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.30 \[ \int (c+d x)^2 \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {4 i b c^2 \arctan \left (e^{i (a+b x)}\right )-\frac {8 i d^2 \arctan \left (e^{i (a+b x)}\right )}{b}-4 b c d x \log \left (1-i e^{i (a+b x)}\right )-2 b d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )+4 b c d x \log \left (1+i e^{i (a+b x)}\right )+2 b d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )-4 i d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+4 i d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+\frac {4 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b}-\frac {4 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b}-4 d (c+d x) \sec (a)+\frac {b (c+d x)^2}{\left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}-\frac {4 d (c+d x) \sin \left (\frac {b x}{2}\right )}{\left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {b (c+d x)^2}{\left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}+\frac {4 d (c+d x) \sin \left (\frac {b x}{2}\right )}{\left (\cos \left (\frac {a}{2}\right )+\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (a+b x)\right )+\sin \left (\frac {1}{2} (a+b x)\right )\right )}}{4 b^2} \] Input:

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

Output:

((4*I)*b*c^2*ArcTan[E^(I*(a + b*x))] - ((8*I)*d^2*ArcTan[E^(I*(a + b*x))]) 
/b - 4*b*c*d*x*Log[1 - I*E^(I*(a + b*x))] - 2*b*d^2*x^2*Log[1 - I*E^(I*(a 
+ b*x))] + 4*b*c*d*x*Log[1 + I*E^(I*(a + b*x))] + 2*b*d^2*x^2*Log[1 + I*E^ 
(I*(a + b*x))] - (4*I)*d*(c + d*x)*PolyLog[2, (-I)*E^(I*(a + b*x))] + (4*I 
)*d*(c + d*x)*PolyLog[2, I*E^(I*(a + b*x))] + (4*d^2*PolyLog[3, (-I)*E^(I* 
(a + b*x))])/b - (4*d^2*PolyLog[3, I*E^(I*(a + b*x))])/b - 4*d*(c + d*x)*S 
ec[a] + (b*(c + d*x)^2)/(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])^2 - (4*d*(c 
+ d*x)*Sin[(b*x)/2])/((Cos[a/2] - Sin[a/2])*(Cos[(a + b*x)/2] - Sin[(a + b 
*x)/2])) - (b*(c + d*x)^2)/(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])^2 + (4*d* 
(c + d*x)*Sin[(b*x)/2])/((Cos[a/2] + Sin[a/2])*(Cos[(a + b*x)/2] + Sin[(a 
+ b*x)/2])))/(4*b^2)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.82, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {4913, 3042, 4669, 3011, 2720, 4674, 3042, 4257, 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.

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

\(\Big \downarrow \) 4913

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}+\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}+\frac {1}{2} \int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx+\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {d^2 \text {arctanh}(\sin (a+b x))}{b^3}-\frac {d (c+d x) \sec (a+b x)}{b^2}+\frac {(c+d x)^2 \tan (a+b x) \sec (a+b x)}{2 b}\)

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, 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 4674

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

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]
 

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. 583 vs. \(2 (174 ) = 348\).

Time = 1.02 (sec) , antiderivative size = 584, normalized size of antiderivative = 3.03

method	result	size
risch	\(-\frac {i d^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {i \left (d^{2} x^{2} b \,{\mathrm e}^{3 i \left (b x +a \right )}+2 c d x b \,{\mathrm e}^{3 i \left (b x +a \right )}+c^{2} b \,{\mathrm e}^{3 i \left (b x +a \right )}-d^{2} x^{2} b \,{\mathrm e}^{i \left (b x +a \right )}-2 c d x b \,{\mathrm e}^{i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{3 i \left (b x +a \right )}-c^{2} b \,{\mathrm e}^{i \left (b x +a \right )}-2 i c d \,{\mathrm e}^{3 i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{i \left (b x +a \right )}-2 i c d \,{\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 d \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {i c d \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {a^{2} d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b^{3}}-\frac {c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {d^{2} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {d^{2} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{2 b}+\frac {i d^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {c d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}-\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{2 b}-\frac {c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {c d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {2 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}\)	\(584\)

Input:

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

Output:

-I/b^2*d^2*polylog(2,-I*exp(I*(b*x+a)))*x-I/b^2/(exp(2*I*(b*x+a))+1)^2*(d^ 
2*x^2*b*exp(3*I*(b*x+a))+2*c*d*x*b*exp(3*I*(b*x+a))+c^2*b*exp(3*I*(b*x+a)) 
-d^2*x^2*b*exp(I*(b*x+a))-2*c*d*x*b*exp(I*(b*x+a))-2*I*d^2*x*exp(3*I*(b*x+ 
a))-c^2*b*exp(I*(b*x+a))-2*I*c*d*exp(3*I*(b*x+a))-2*I*d^2*x*exp(I*(b*x+a)) 
-2*I*c*d*exp(I*(b*x+a)))+I/b^2*c*d*polylog(2,I*exp(I*(b*x+a)))-I/b^2*c*d*p 
olylog(2,-I*exp(I*(b*x+a)))-1/2/b^3*a^2*d^2*ln(I*exp(I*(b*x+a))+1)-1/b^2*c 
*d*ln(1-I*exp(I*(b*x+a)))*a-d^2*polylog(3,I*exp(I*(b*x+a)))/b^3+I/b^3*d^2* 
a^2*arctan(exp(I*(b*x+a)))+1/2/b^3*a^2*d^2*ln(1-I*exp(I*(b*x+a)))+d^2*poly 
log(3,-I*exp(I*(b*x+a)))/b^3+1/2/b*d^2*ln(I*exp(I*(b*x+a))+1)*x^2+I/b^2*d^ 
2*polylog(2,I*exp(I*(b*x+a)))*x+I/b*c^2*arctan(exp(I*(b*x+a)))+1/b^2*c*d*l 
n(I*exp(I*(b*x+a))+1)*a-1/2/b*d^2*ln(1-I*exp(I*(b*x+a)))*x^2-1/b*c*d*ln(1- 
I*exp(I*(b*x+a)))*x+1/b*c*d*ln(I*exp(I*(b*x+a))+1)*x-2*I/b^2*c*d*a*arctan( 
exp(I*(b*x+a)))-2*I/b^3*d^2*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. 795 vs. \(2 (165) = 330\).

Time = 0.13 (sec) , antiderivative size = 795, normalized size of antiderivative = 4.12 \[ \int (c+d x)^2 \sec (a+b x) \tan ^2(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*tan(a + b*x)**2*sec(a + b*x), 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. 1891 vs. \(2 (165) = 330\).

Time = 0.70 (sec) , antiderivative size = 1891, normalized size of antiderivative = 9.80 \[ \int (c+d x)^2 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(c^2*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - l 
og(sin(b*x + a) - 1)) - 2*a*c*d*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log 
(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b + a^2*d^2*(2*sin(b*x + a)/(s 
in(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b^2 - 
4*(2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - 2*d^2 + ((b*x + a)^2 
*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - 2*d^2)*cos(4*b*x + 4*a) + 2*((b*x + a 
)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - 2*d^2)*cos(2*b*x + 2*a) + (I*(b*x 
+ a)^2*d^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a) - 2*I*d^2)*sin(4*b*x + 4*a) + 
 2*(I*(b*x + a)^2*d^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a) - 2*I*d^2)*sin(2*b 
*x + 2*a))*arctan2(cos(b*x + a), sin(b*x + a) + 1) + 2*((b*x + a)^2*d^2 + 
2*(b*c*d - a*d^2)*(b*x + a) - 2*d^2 + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2) 
*(b*x + a) - 2*d^2)*cos(4*b*x + 4*a) + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d 
^2)*(b*x + a) - 2*d^2)*cos(2*b*x + 2*a) + (I*(b*x + a)^2*d^2 + 2*(I*b*c*d 
- I*a*d^2)*(b*x + a) - 2*I*d^2)*sin(4*b*x + 4*a) + 2*(I*(b*x + a)^2*d^2 + 
2*(I*b*c*d - I*a*d^2)*(b*x + a) - 2*I*d^2)*sin(2*b*x + 2*a))*arctan2(cos(b 
*x + a), -sin(b*x + a) + 1) - 4*((b*x + a)^2*d^2 - 2*I*b*c*d + 2*I*a*d^2 + 
 2*(b*c*d - (a + I)*d^2)*(b*x + a))*cos(3*b*x + 3*a) + 4*((b*x + a)^2*d^2 
+ 2*I*b*c*d - 2*I*a*d^2 + 2*(b*c*d - (a - I)*d^2)*(b*x + a))*cos(b*x + a) 
+ 4*(b*c*d + (b*x + a)*d^2 - a*d^2 + (b*c*d + (b*x + a)*d^2 - a*d^2)*cos(4 
*b*x + 4*a) + 2*(b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (I*b...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

(2*int(sec(a + b*x)*tan(a + b*x)**2*x**2,x)*sin(a + b*x)**2*b*d**2 - 2*int 
(sec(a + b*x)*tan(a + b*x)**2*x**2,x)*b*d**2 + 4*int(sec(a + b*x)*tan(a + 
b*x)**2*x,x)*sin(a + b*x)**2*b*c*d - 4*int(sec(a + b*x)*tan(a + b*x)**2*x, 
x)*b*c*d + log(tan((a + b*x)/2) - 1)*sin(a + b*x)**2*c**2 - log(tan((a + b 
*x)/2) - 1)*c**2 - log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2*c**2 + log(ta 
n((a + b*x)/2) + 1)*c**2 - sin(a + b*x)*c**2)/(2*b*(sin(a + b*x)**2 - 1))