Integrand size = 20, antiderivative size = 159 \[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=\frac {6 i d (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {6 d^3 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}-\frac {6 d^3 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}+\frac {(c+d x)^3 \sec (a+b x)}{b} \]
6*I*d*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b^2-6*I*d^2*(d*x+c)*polylog(2,-I*ex p(I*(b*x+a)))/b^3+6*I*d^2*(d*x+c)*polylog(2,I*exp(I*(b*x+a)))/b^3+6*d^3*po lylog(3,-I*exp(I*(b*x+a)))/b^4-6*d^3*polylog(3,I*exp(I*(b*x+a)))/b^4+(d*x+ c)^3*sec(b*x+a)/b
Time = 0.80 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.61 \[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=-\frac {3 d \left (-2 i b^2 c^2 \arctan \left (e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )-b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )-2 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )\right )}{b^4}+\frac {(c+d x)^3 \sec (a+b x)}{b} \]
(-3*d*((-2*I)*b^2*c^2*ArcTan[E^(I*(a + b*x))] + 2*b^2*c*d*x*Log[1 - I*E^(I *(a + b*x))] + b^2*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] - 2*b^2*c*d*x*Log[1 + I*E^(I*(a + b*x))] - b^2*d^2*x^2*Log[1 + I*E^(I*(a + b*x))] + (2*I)*b*d* (c + d*x)*PolyLog[2, (-I)*E^(I*(a + b*x))] - (2*I)*b*d*(c + d*x)*PolyLog[2 , I*E^(I*(a + b*x))] - 2*d^2*PolyLog[3, (-I)*E^(I*(a + b*x))] + 2*d^2*Poly Log[3, I*E^(I*(a + b*x))]))/b^4 + ((c + d*x)^3*Sec[a + b*x])/b
Time = 0.56 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4909, 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.
| \(\displaystyle \int (c+d x)^3 \tan (a+b x) \sec (a+b x) \, dx\) |
| \(\Big \downarrow \) 4909 |
| \(\displaystyle \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \int (c+d x)^2 \sec (a+b x)dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \int (c+d x)^2 \csc \left (a+b x+\frac {\pi }{2}\right )dx}{b}\) |
| \(\Big \downarrow \) 4669 |
| \(\displaystyle \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \left (-\frac {2 d \int (c+d x) \log \left (1-i e^{i (a+b x)}\right )dx}{b}+\frac {2 d \int (c+d x) \log \left (1+i e^{i (a+b x)}\right )dx}{b}-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}\right )}{b}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \left (\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {i d \int \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )dx}{b}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {i d \int \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )dx}{b}\right )}{b}-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}\right )}{b}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \left (\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {d \int e^{-i (a+b x)} \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )de^{i (a+b x)}}{b^2}\right )}{b}-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}\right )}{b}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {3 d \left (-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}-\frac {d \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^2}\right )}{b}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}-\frac {d \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^2}\right )}{b}\right )}{b}\) |
(-3*d*(((-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*PolyL og[3, I*E^(I*(a + b*x))])/b^2))/b))/b + ((c + d*x)^3*Sec[a + b*x])/b
3.3.48.3.1 Defintions of rubi rules used
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]]]
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]
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]
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]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (142 ) = 284\).
Time = 2.24 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.91
| method | result | size |
| risch | \(\frac {2 \,{\mathrm e}^{i \left (x b +a \right )} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}+\frac {6 i d \,c^{2} \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{2}}+\frac {3 d^{3} \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) x^{2}}{b^{2}}+\frac {6 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}+\frac {6 i c \,d^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {12 i d^{2} c a \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {6 i d^{3} a^{2} \arctan \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}+\frac {6 i d^{3} x \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {3 d^{3} a^{2} \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}+\frac {6 d^{3} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}-\frac {6 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{3}}-\frac {3 d^{3} \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) x^{2}}{b^{2}}-\frac {6 d^{3} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}-\frac {3 d^{3} a^{2} \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}-\frac {6 i c \,d^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}-\frac {6 i d^{3} x \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {6 d^{2} c \ln \left (1+i {\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{3}}-\frac {6 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{2}}\) | \(463\) |
2*exp(I*(b*x+a))*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/b/(exp(2*I*(b*x+a))+1 )+6*I/b^2*d*c^2*arctan(exp(I*(b*x+a)))+3/b^2*d^3*ln(1+I*exp(I*(b*x+a)))*x^ 2+6/b^2*d^2*c*ln(1+I*exp(I*(b*x+a)))*x+6*I*c*d^2*polylog(2,I*exp(I*(b*x+a) ))/b^3-12*I/b^3*d^2*c*a*arctan(exp(I*(b*x+a)))+6*I/b^4*d^3*a^2*arctan(exp( I*(b*x+a)))+6*I*d^3*x*polylog(2,I*exp(I*(b*x+a)))/b^3+3/b^4*d^3*a^2*ln(1-I *exp(I*(b*x+a)))+6*d^3*polylog(3,-I*exp(I*(b*x+a)))/b^4-6/b^3*d^2*c*ln(1-I *exp(I*(b*x+a)))*a-3/b^2*d^3*ln(1-I*exp(I*(b*x+a)))*x^2-6*d^3*polylog(3,I* exp(I*(b*x+a)))/b^4-3/b^4*d^3*a^2*ln(1+I*exp(I*(b*x+a)))-6*I*c*d^2*polylog (2,-I*exp(I*(b*x+a)))/b^3-6*I*d^3*x*polylog(2,-I*exp(I*(b*x+a)))/b^3+6/b^3 *d^2*c*ln(1+I*exp(I*(b*x+a)))*a-6/b^2*d^2*c*ln(1-I*exp(I*(b*x+a)))*x
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (133) = 266\).
Time = 0.29 (sec) , antiderivative size = 783, normalized size of antiderivative = 4.92 \[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=\frac {2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} + 6 \, b^{3} c^{2} d x + 2 \, b^{3} c^{3} + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 \, d^{3} \cos \left (b x + a\right ) {\rm polylog}\left (3, -i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 6 \, {\left (-i \, b d^{3} x - i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 \, {\left (-i \, b d^{3} x - i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 6 \, {\left (i \, b d^{3} x + i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - 6 \, {\left (i \, b d^{3} x + i \, b c d^{2}\right )} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + 2 \, a b c d^{2} - a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + 3 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{4} \cos \left (b x + a\right )} \]
1/2*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 6*b^3*c^2*d*x + 2*b^3*c^3 + 6*d^3*c os(b*x + a)*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 6*d^3*cos(b*x + a) *polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 6*d^3*cos(b*x + a)*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) - 6*d^3*cos(b*x + a)*polylog(3, -I*cos(b* x + a) - sin(b*x + a)) - 6*(-I*b*d^3*x - I*b*c*d^2)*cos(b*x + a)*dilog(I*c os(b*x + a) + sin(b*x + a)) - 6*(-I*b*d^3*x - I*b*c*d^2)*cos(b*x + a)*dilo g(I*cos(b*x + a) - sin(b*x + a)) - 6*(I*b*d^3*x + I*b*c*d^2)*cos(b*x + a)* dilog(-I*cos(b*x + a) + sin(b*x + a)) - 6*(I*b*d^3*x + I*b*c*d^2)*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + I) + 3*(b^2*c^2* d - 2*a*b*c*d^2 + a^2*d^3)*cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + I) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cos(b*x + a )*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cos(b*x + a)* log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cos(b*x + a)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cos(b*x + a) *log(-cos(b*x + a) - I*sin(b*x + a) + I))/(b^4*cos(b*x + a))
\[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=\int \left (c + d x\right )^{3} \tan {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1770 vs. \(2 (133) = 266\).
Time = 0.44 (sec) , antiderivative size = 1770, normalized size of antiderivative = 11.13 \[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=\text {Too large to display} \]
1/2*(3*(4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin(2*b*x + 2*a)*sin(b*x + a) + 4*(b*x + a)*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)*log(cos(b*x + a)^2 + sin(b*x + a )^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*c os(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*c^2*d/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*b) - 6*(4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin( 2*b*x + 2*a)*sin(b*x + a) + 4*(b*x + a)*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)*log(cos(b*x + a)^2 + sin(b *x + a)^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*a*c*d^2/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b* x + 2*a) + 1)*b^2) + 3*(4*(b*x + a)*cos(2*b*x + 2*a)*cos(b*x + a) + 4*(b*x + a)*sin(2*b*x + 2*a)*sin(b*x + a) + 4*(b*x + a)*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)*log(cos(b*x + a) ^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) + (cos(2*b*x + 2*a)^2 + sin(2*b* x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))*a^2*d^3/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1)*b^3) + 2*c^3/cos(b*x + a) - 6*a*c^2*d/(b*cos(b*x + a)) + 6*a^2*c*d^2/(b^2*cos(b*x + a)) - 2*a^3*d^3/(b^3*cos(b*x + a)) + ...
\[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right ) \tan \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x)^3 \sec (a+b x) \tan (a+b x) \, dx=\int \frac {\mathrm {tan}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \]