Integrand size = 32, antiderivative size = 125 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {5 a^3 (4 B+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (4 B+3 C) \tan (c+d x)}{d}+\frac {3 a^3 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {a^3 (4 B+3 C) \tan ^3(c+d x)}{12 d} \] Output:
5/8*a^3*(4*B+3*C)*arctanh(sin(d*x+c))/d+a^3*(4*B+3*C)*tan(d*x+c)/d+3/8*a^3 *(4*B+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/4*C*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1 /12*a^3*(4*B+3*C)*tan(d*x+c)^3/d
Time = 0.46 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (24 B \coth ^{-1}(\sin (c+d x))+9 (4 B+5 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (96 (B+C)+9 (4 B+5 C) \sec (c+d x)+6 C \sec ^3(c+d x)+8 (B+3 C) \tan ^2(c+d x)\right )\right )}{24 d} \] Input:
Integrate[(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(a^3*(24*B*ArcCoth[Sin[c + d*x]] + 9*(4*B + 5*C)*ArcTanh[Sin[c + d*x]] + T an[c + d*x]*(96*(B + C) + 9*(4*B + 5*C)*Sec[c + d*x] + 6*C*Sec[c + d*x]^3 + 8*(B + 3*C)*Tan[c + d*x]^2)))/(24*d)
Time = 0.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 4542, 27, 3042, 4278, 2009}
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.
| \(\displaystyle \int (a \sec (c+d x)+a)^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {\int a (4 B+3 C) \sec (c+d x) (\sec (c+d x) a+a)^3dx}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} (4 B+3 C) \int \sec (c+d x) (\sec (c+d x) a+a)^3dx+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (4 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3dx+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \frac {1}{4} (4 B+3 C) \int \left (a^3 \sec ^4(c+d x)+3 a^3 \sec ^3(c+d x)+3 a^3 \sec ^2(c+d x)+a^3 \sec (c+d x)\right )dx+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} (4 B+3 C) \left (\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
Input:
Int[(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + ((4*B + 3*C)*((5*a^3*ArcTa nh[Sin[c + d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[c + d*x]*Tan [c + d*x])/(2*d) + (a^3*Tan[c + d*x]^3)/(3*d)))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(b*(m + 1)) I nt[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 0.75 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40
| method | result | size |
| norman | \(\frac {-\frac {73 a^{3} \left (4 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {55 a^{3} \left (4 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {5 a^{3} \left (4 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {a^{3} \left (49 C +44 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {5 a^{3} \left (4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {5 a^{3} \left (4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(175\) |
| parallelrisch | \(\frac {26 a^{3} \left (-\frac {15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{13}+\frac {15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{13}+\left (B +\frac {15 C}{13}\right ) \sin \left (2 d x +2 c \right )+\frac {9 \left (B +\frac {5 C}{4}\right ) \sin \left (3 d x +3 c \right )}{26}+\frac {\left (11 B +9 C \right ) \sin \left (4 d x +4 c \right )}{26}+\frac {9 \sin \left (d x +c \right ) \left (B +\frac {23 C}{12}\right )}{26}\right )}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(178\) |
| parts | \(-\frac {\left (B \,a^{3}+3 a^{3} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{3}+a^{3} C \right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{3}+3 a^{3} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{3}}{d}+\frac {a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(181\) |
| derivativedivides | \(\frac {B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \tan \left (d x +c \right )+3 B \,a^{3} \tan \left (d x +c \right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(219\) |
| default | \(\frac {B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \tan \left (d x +c \right )+3 B \,a^{3} \tan \left (d x +c \right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(219\) |
| risch | \(-\frac {i a^{3} \left (36 B \,{\mathrm e}^{7 i \left (d x +c \right )}+45 C \,{\mathrm e}^{7 i \left (d x +c \right )}-72 B \,{\mathrm e}^{6 i \left (d x +c \right )}-24 C \,{\mathrm e}^{6 i \left (d x +c \right )}+36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+69 C \,{\mathrm e}^{5 i \left (d x +c \right )}-264 B \,{\mathrm e}^{4 i \left (d x +c \right )}-216 C \,{\mathrm e}^{4 i \left (d x +c \right )}-36 B \,{\mathrm e}^{3 i \left (d x +c \right )}-69 C \,{\mathrm e}^{3 i \left (d x +c \right )}-280 B \,{\mathrm e}^{2 i \left (d x +c \right )}-264 C \,{\mathrm e}^{2 i \left (d x +c \right )}-36 B \,{\mathrm e}^{i \left (d x +c \right )}-45 C \,{\mathrm e}^{i \left (d x +c \right )}-88 B -72 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) | \(287\) |
Input:
int((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBO SE)
Output:
(-73/12*a^3*(4*B+3*C)/d*tan(1/2*d*x+1/2*c)^3+55/12*a^3*(4*B+3*C)/d*tan(1/2 *d*x+1/2*c)^5-5/4*a^3*(4*B+3*C)/d*tan(1/2*d*x+1/2*c)^7+1/4*a^3*(49*C+44*B) /d*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2-1)^4-5/8*a^3*(4*B+3*C)/d*ln(t an(1/2*d*x+1/2*c)-1)+5/8*a^3*(4*B+3*C)/d*ln(tan(1/2*d*x+1/2*c)+1)
Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.16 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (4 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="f ricas")
Output:
1/48*(15*(4*B + 3*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(4*B + 3*C)*a^3*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(11*B + 9*C)*a^3*cos (d*x + c)^3 + 9*(4*B + 5*C)*a^3*cos(d*x + c)^2 + 8*(B + 3*C)*a^3*cos(d*x + c) + 6*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{3} \left (\int B \sec {\left (c + d x \right )}\, dx + \int 3 B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
Output:
a**3*(Integral(B*sec(c + d*x), x) + Integral(3*B*sec(c + d*x)**2, x) + Int egral(3*B*sec(c + d*x)**3, x) + Integral(B*sec(c + d*x)**4, x) + Integral( C*sec(c + d*x)**2, x) + Integral(3*C*sec(c + d*x)**3, x) + Integral(3*C*se c(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x))
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (117) = 234\).
Time = 0.04 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.10 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \] Input:
integrate((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="m axima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48*(tan(d*x + c)^3 + 3* tan(d*x + c))*C*a^3 - 3*C*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin( d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d *x + c) - 1)) - 36*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* x + c) + 1) + log(sin(d*x + c) - 1)) - 36*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*B*a^3*log (sec(d*x + c) + tan(d*x + c)) + 144*B*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.70 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="g iac")
Output:
1/24*(15*(4*B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B* a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(60*B*a^3*tan(1/2*d* x + 1/2*c)^7 + 45*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 220*B*a^3*tan(1/2*d*x + 1 /2*c)^5 - 165*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 292*B*a^3*tan(1/2*d*x + 1/2*c )^3 + 219*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 132*B*a^3*tan(1/2*d*x + 1/2*c) - 147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 14.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.48 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (-5\,B\,a^3-\frac {15\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,B\,a^3}{3}+\frac {55\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,B\,a^3}{3}-\frac {73\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,B\,a^3+\frac {49\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,B+3\,C\right )}{4\,d} \] Input:
int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)*(11*B*a^3 + (49*C*a^3)/4) - tan(c/2 + (d*x)/2)^7*(5*B* a^3 + (15*C*a^3)/4) + tan(c/2 + (d*x)/2)^5*((55*B*a^3)/3 + (55*C*a^3)/4) - tan(c/2 + (d*x)/2)^3*((73*B*a^3)/3 + (73*C*a^3)/4))/(d*(6*tan(c/2 + (d*x) /2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/ 2)^8 + 1)) + (5*a^3*atanh(tan(c/2 + (d*x)/2))*(4*B + 3*C))/(4*d)
Time = 0.16 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.03 \[ \int (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3} \left (-88 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b -72 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +96 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +96 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b -45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +90 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b +45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -90 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -36 \sin \left (d x +c \right )^{3} b -45 \sin \left (d x +c \right )^{3} c +36 \sin \left (d x +c \right ) b +51 \sin \left (d x +c \right ) c \right )}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**3*( - 88*cos(c + d*x)*sin(c + d*x)**3*b - 72*cos(c + d*x)*sin(c + d*x) **3*c + 96*cos(c + d*x)*sin(c + d*x)*b + 96*cos(c + d*x)*sin(c + d*x)*c - 60*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b - 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 120*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 90*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 60*log(tan((c + d*x)/2) - 1)*b - 45*log(tan((c + d*x)/2) - 1)*c + 60*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**4*b + 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 120*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 90*log(tan((c + d*x)/2) + 1)*sin( c + d*x)**2*c + 60*log(tan((c + d*x)/2) + 1)*b + 45*log(tan((c + d*x)/2) + 1)*c - 36*sin(c + d*x)**3*b - 45*sin(c + d*x)**3*c + 36*sin(c + d*x)*b + 51*sin(c + d*x)*c))/(24*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))