Integrand size = 33, antiderivative size = 74 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {A x}{a^2}-\frac {(4 A-B-2 C) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
A*x/a^2-1/3*(4*A-B-2*C)*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+C)*tan(d* x+c)/d/(a+a*sec(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(175\) vs. \(2(74)=148\).
Time = 0.77 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.36 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (9 A d x \cos \left (\frac {d x}{2}\right )+9 A d x \cos \left (c+\frac {d x}{2}\right )+3 A d x \cos \left (c+\frac {3 d x}{2}\right )+3 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 A \sin \left (\frac {d x}{2}\right )+6 B \sin \left (\frac {d x}{2}\right )+6 C \sin \left (\frac {d x}{2}\right )+12 A \sin \left (c+\frac {d x}{2}\right )-6 B \sin \left (c+\frac {d x}{2}\right )-10 A \sin \left (c+\frac {3 d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )+2 C \sin \left (c+\frac {3 d x}{2}\right )\right )}{24 a^2 d} \]
(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*A*d*x*Cos[(d*x)/2] + 9*A*d*x*Cos[c + (d*x) /2] + 3*A*d*x*Cos[c + (3*d*x)/2] + 3*A*d*x*Cos[2*c + (3*d*x)/2] - 18*A*Sin [(d*x)/2] + 6*B*Sin[(d*x)/2] + 6*C*Sin[(d*x)/2] + 12*A*Sin[c + (d*x)/2] - 6*B*Sin[c + (d*x)/2] - 10*A*Sin[c + (3*d*x)/2] + 4*B*Sin[c + (3*d*x)/2] + 2*C*Sin[c + (3*d*x)/2]))/(24*a^2*d)
Time = 0.48 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4540, 25, 3042, 4407, 3042, 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.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 4540 |
\(\displaystyle -\frac {\int -\frac {3 a A-a (A-B-2 C) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {3 a A-a (A-B-2 C) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a A-a (A-B-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle \frac {3 A x-a (4 A-B-2 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 A x-a (4 A-B-2 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {3 A x-\frac {a (4 A-B-2 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (3*A*x - (a*( 4*A - B - 2*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2)
3.5.62.3.1 Defintions of rubi rules used
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_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Sim p[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + (b*B*(m + 1) - a*(A*(m + 1) - C*m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-9 A +3 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 d x A}{6 a^{2} d}\) | \(53\) |
risch | \(\frac {A x}{a^{2}}+\frac {2 i \left (-6 A \,{\mathrm e}^{2 i \left (d x +c \right )}+3 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,{\mathrm e}^{i \left (d x +c \right )}+3 B \,{\mathrm e}^{i \left (d x +c \right )}+3 C \,{\mathrm e}^{i \left (d x +c \right )}-5 A +2 B +C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\) | \(98\) |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(99\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +4 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\) | \(99\) |
norman | \(\frac {\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {A x}{a}+\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}+\frac {\left (3 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -2 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(126\) |
Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, A d x \cos \left (d x + c\right )^{2} + 6 \, A d x \cos \left (d x + c\right ) + 3 \, A d x - {\left ({\left (5 \, A - 2 \, B - C\right )} \cos \left (d x + c\right ) + 4 \, A - B - 2 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/3*(3*A*d*x*cos(d*x + c)^2 + 6*A*d*x*cos(d*x + c) + 3*A*d*x - ((5*A - 2*B - C)*cos(d*x + c) + 4*A - B - 2*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
(Integral(A/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x )**2/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (70) = 140\).
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.22 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]
-1/6*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - C*(3*sin( d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - B *(3*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3) /a^2)/d
Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (d x + c\right )} A}{a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
1/6*(6*(d*x + c)*A/a^2 + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 16.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.53 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {B\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )-\frac {5\,A\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}-\frac {3\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {3\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {C\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}+\frac {9\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (c+d\,x\right )}{2}+\frac {3\,A\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (c+d\,x\right )}{2}}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]