Integrand size = 31, antiderivative size = 138 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \]
-1/7*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(4*A-11*B)*tan(d*x+c)/a/d/ (a+a*sec(d*x+c))^3+1/105*(8*A+13*B)*tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))^2+1/ 105*(8*A+13*B)*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))
Time = 0.55 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.59 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\left (13 A+8 B+4 (13 A+8 B) \sec (c+d x)+4 (8 A+13 B) \sec ^2(c+d x)+(8 A+13 B) \sec ^3(c+d x)\right ) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^4} \]
((13*A + 8*B + 4*(13*A + 8*B)*Sec[c + d*x] + 4*(8*A + 13*B)*Sec[c + d*x]^2 + (8*A + 13*B)*Sec[c + d*x]^3)*Tan[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x] )^4)
Time = 0.72 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4496, 25, 3042, 4488, 3042, 4283, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
\(\Big \downarrow \) 4496 |
\(\displaystyle -\frac {\int -\frac {\sec (c+d x) (4 a (A-B)+7 a B \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sec (c+d x) (4 a (A-B)+7 a B \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (4 a (A-B)+7 a B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4488 |
\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {1}{5} (8 A+13 B) \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
-1/7*((A - B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + ((a*(4*A - 11*B)* Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((8*A + 13*B)*(Tan[c + d*x]/( 3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec[c + d*x]))))/ 5)/(7*a^2)
3.2.12.3.1 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_), x_ Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(m + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) ] && IntegerQ[2*m]
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 = 0.78 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.58
method | result | size |
parallelrisch | \(\frac {\left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {7 \left (-A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+\frac {7 \left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+7 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{56 a^{4} d}\) | \(80\) |
derivativedivides | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (-A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{8 d \,a^{4}}\) | \(88\) |
default | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (-A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{8 d \,a^{4}}\) | \(88\) |
risch | \(\frac {2 i \left (105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+175 A \,{\mathrm e}^{4 i \left (d x +c \right )}+140 B \,{\mathrm e}^{4 i \left (d x +c \right )}+280 A \,{\mathrm e}^{3 i \left (d x +c \right )}+140 B \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+91 \,{\mathrm e}^{i \left (d x +c \right )} A +56 B \,{\mathrm e}^{i \left (d x +c \right )}+13 A +8 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\) | \(138\) |
norman | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{56 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (7 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}+\frac {\left (11 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (11 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{420 a d}-\frac {\left (17 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{3}}\) | \(167\) |
1/56*((A-B)*tan(1/2*d*x+1/2*c)^6+7/5*(-A-B)*tan(1/2*d*x+1/2*c)^4+7/3*(-A+B )*tan(1/2*d*x+1/2*c)^2+7*A+7*B)*tan(1/2*d*x+1/2*c)/a^4/d
Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left ({\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (8 \, A + 13 \, B\right )} \cos \left (d x + c\right ) + 8 \, A + 13 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
1/105*((13*A + 8*B)*cos(d*x + c)^3 + 4*(13*A + 8*B)*cos(d*x + c)^2 + 4*(8* A + 13*B)*cos(d*x + c) + 8*A + 13*B)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a ^4*d)
\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) )/a**4
Time = 0.22 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]
1/840*(B*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/ (cos(d*x + c) + 1)^7)/a^4 + A*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*si n(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]
1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 - 21*A*ta n(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/ 2*c)^3 + 35*B*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105*B* tan(1/2*d*x + 1/2*c))/(a^4*d)
Time = 13.82 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+B\right )}{40\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{24\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{8\,a^4}}{d} \]