Integrand size = 31, antiderivative size = 102 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(3 A-8 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(3 A+7 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
-1/5*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(3*A-8*B)*tan(d*x+c)/a/d/( a+a*sec(d*x+c))^2+1/15*(3*A+7*B)*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\left (3 A+2 B+(9 A+6 B) \sec (c+d x)+(3 A+7 B) \sec ^2(c+d x)\right ) \tan (c+d x)}{15 a^3 d (1+\sec (c+d x))^3} \]
((3*A + 2*B + (9*A + 6*B)*Sec[c + d*x] + (3*A + 7*B)*Sec[c + d*x]^2)*Tan[c + d*x])/(15*a^3*d*(1 + Sec[c + d*x])^3)
Time = 0.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3042, 4496, 25, 3042, 4488, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 4496 |
\(\displaystyle -\frac {\int -\frac {\sec (c+d x) (3 a (A-B)+5 a B \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sec (c+d x) (3 a (A-B)+5 a B \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a (A-B)+5 a B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4488 |
\(\displaystyle \frac {\frac {1}{3} (3 A+7 B) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx+\frac {a (3 A-8 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} (3 A+7 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a (3 A-8 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {\frac {(3 A+7 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)}+\frac {a (3 A-8 B) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*((A - B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((a*(3*A - 8*B)*T an[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((3*A + 7*B)*Tan[c + d*x])/(3* d*(a + a*Sec[c + d*x])))/(5*a^2)
3.2.2.3.1 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_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(a*b*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
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)]
Time = 0.74 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
parallelrisch | \(-\frac {\left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {10 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-5 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a^{3} d}\) | \(56\) |
derivativedivides | \(\frac {\frac {\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{4 d \,a^{3}}\) | \(64\) |
default | \(\frac {\frac {\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{4 d \,a^{3}}\) | \(64\) |
risch | \(\frac {2 i \left (15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+15 A \,{\mathrm e}^{2 i \left (d x +c \right )}+20 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )} A +10 B \,{\mathrm e}^{i \left (d x +c \right )}+3 A +2 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(90\) |
norman | \(\frac {-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 a d}+\frac {\left (6 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{2}}\) | \(143\) |
-1/20*((A-B)*tan(1/2*d*x+1/2*c)^4-10/3*B*tan(1/2*d*x+1/2*c)^2-5*A-5*B)*tan (1/2*d*x+1/2*c)/a^3/d
Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {{\left ({\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A + 7 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/15*((3*A + 2*B)*cos(d*x + c)^2 + 3*(3*A + 2*B)*cos(d*x + c) + 3*A + 7*B) *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)
\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + 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))/a**3
Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
1/60*(B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*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 + 3*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]
-1/60*(3*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 - 10*B*tan( 1/2*d*x + 1/2*c)^3 - 15*A*tan(1/2*d*x + 1/2*c) - 15*B*tan(1/2*d*x + 1/2*c) )/(a^3*d)
Time = 13.69 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B-3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+10\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]