Integrand size = 40, antiderivative size = 125 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(2 B-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(4 B-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
C*arctanh(sin(d*x+c))/a^3/d+1/5*(B-C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d *x+c))^3-1/15*(2*B-7*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/15*(4*B-29*C)* tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
Time = 1.82 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 C \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 (2 B-7 C) \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+24 (B-C) \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )+2 (B-11 C) \tan \left (\frac {1}{2} (c+d x)\right )}{15 a^3 d} \] Input:
Integrate[(Sec[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[ c + d*x])^3,x]
Output:
(15*C*(-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*(2*B - 7*C)*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 24* (B - C)*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6 + 2*(B - 11*C)*Tan[(c + d*x)/2]) /(15*a^3*d)
Time = 1.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4560, 3042, 4507, 3042, 4496, 25, 3042, 4486, 3042, 4257, 4281}
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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (2 a (B-C)+5 a C \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a (B-C)+5 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (2 (2 B-7 C) a^2+15 C \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (2 (2 B-7 C) a^2+15 C \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 (2 B-7 C) a^2+15 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {a^2 (4 B-29 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx+15 a C \int \sec (c+d x)dx}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (4 B-29 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+15 a C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a^2 (4 B-29 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {15 a C \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {\frac {\frac {a^2 (4 B-29 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}+\frac {15 a C \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {a (2 B-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(Sec[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d* x])^3,x]
Output:
((B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + (-1/3 *(a*(2*B - 7*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + ((15*a*C*ArcTan h[Sin[c + d*x]])/d + (a^2*(4*B - 29*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x ])))/(3*a^2))/(5*a^2)
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\frac {-20 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+20 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {10 \left (B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+5 B -35 C \right )}{20 a^{3} d}\) | \(92\) |
| derivativedivides | \(\frac {4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d \,a^{3}}\) | \(119\) |
| default | \(\frac {4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d \,a^{3}}\) | \(119\) |
| risch | \(-\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}+75 C \,{\mathrm e}^{3 i \left (d x +c \right )}-20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+145 C \,{\mathrm e}^{2 i \left (d x +c \right )}-10 B \,{\mathrm e}^{i \left (d x +c \right )}+95 C \,{\mathrm e}^{i \left (d x +c \right )}-2 B +22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\) | \(146\) |
| norman | \(\frac {\frac {\left (B -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}-\frac {\left (B -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {\left (B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}-\frac {\left (3 B -43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}+\frac {\left (-59 C +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(209\) |
Input:
int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method =_RETURNVERBOSE)
Output:
1/20*(-20*C*ln(tan(1/2*d*x+1/2*c)-1)+20*C*ln(tan(1/2*d*x+1/2*c)+1)+tan(1/2 *d*x+1/2*c)*((B-C)*tan(1/2*d*x+1/2*c)^4+10/3*(B-2*C)*tan(1/2*d*x+1/2*c)^2+ 5*B-35*C))/a^3/d
Time = 0.08 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (B - 11 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B - 17 \, C\right )} \cos \left (d x + c\right ) + 7 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
1/30*(15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*lo g(sin(d*x + c) + 1) - 15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos( d*x + c) + C)*log(-sin(d*x + c) + 1) + 2*(2*(B - 11*C)*cos(d*x + c)^2 + 3* (2*B - 17*C)*cos(d*x + c) + 7*B - 32*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^ 3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**2*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3 ,x)
Output:
(Integral(B*sec(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**4/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Time = 0.05 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d* x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d* x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1 ) - 1)/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(c os(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d
Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(60*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*C*log(abs(tan(1/2*d *x + 1/2*c) - 1))/a^3 + (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/ 2*d*x + 1/2*c)^5 + 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 20*C*a^12*tan(1/2*d* x + 1/2*c)^3 + 15*B*a^12*tan(1/2*d*x + 1/2*c) - 105*C*a^12*tan(1/2*d*x + 1 /2*c))/a^15)/d
Time = 12.47 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{12\,a^3}+\frac {B-3\,C}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{4\,a^3}+\frac {B-3\,C}{4\,a^3}-\frac {B+3\,C}{4\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}+\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d} \] Input:
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + a/cos(c + d*x ))^3),x)
Output:
(tan(c/2 + (d*x)/2)^3*((B - C)/(12*a^3) + (B - 3*C)/(12*a^3)))/d + (tan(c/ 2 + (d*x)/2)*((B - C)/(4*a^3) + (B - 3*C)/(4*a^3) - (B + 3*C)/(4*a^3)))/d + (tan(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d) + (2*C*atanh(tan(c/2 + (d*x)/2 )))/(a^3*d)
Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{60 a^{3} d} \] Input:
int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)
Output:
( - 60*log(tan((c + d*x)/2) - 1)*c + 60*log(tan((c + d*x)/2) + 1)*c + 3*ta n((c + d*x)/2)**5*b - 3*tan((c + d*x)/2)**5*c + 10*tan((c + d*x)/2)**3*b - 20*tan((c + d*x)/2)**3*c + 15*tan((c + d*x)/2)*b - 105*tan((c + d*x)/2)*c )/(60*a**3*d)