Integrand size = 43, antiderivative size = 268 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(9 A-49 B+119 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B+33 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(3 A-13 B+33 C) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(9 A-49 B+119 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac {(B-2 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {(9 A-49 B+119 C) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \]
1/10*(9*A-49*B+119*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli pticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(3*A-13*B+33*C)*(cos(1/2*d*x+1 /2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^ 3/d+1/6*(3*A-13*B+33*C)*sin(d*x+c)/a^3/d/cos(d*x+c)^(3/2)-1/5*(A-B+C)*sin( d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3+1/3*(B-2*C)*sin(d*x+c)/a/d/co s(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-1/30*(9*A-49*B+119*C)*sin(d*x+c)/d/cos(d *x+c)^(3/2)/(a^3+a^3*cos(d*x+c))-1/10*(9*A-49*B+119*C)*sin(d*x+c)/a^3/d/co s(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 16.74 (sec) , antiderivative size = 1894, normalized size of antiderivative = 7.07 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]
(-4*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C* Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c] ]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2 *d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) + (52*B*Cos[c/2 + (d*x)/ 2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c] ]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d* x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Co t[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c ]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[ c]^2]*(a + a*Sec[c + d*x])^3) - (44*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hyperg eometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]* Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d *x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d *x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(( -4*(-20*B + 60*C + 9*A*Cos[c] - 29*B*Cos[c] + 59*C*Cos[c])*Csc[c/2]*Sec[c/ 2]*Sec[c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B...
Time = 1.50 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4600, 3042, 3520, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}
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.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{7/2} (a \sec (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 4600 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {a (3 A-3 B+13 C)+a (3 A+7 B-7 C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (3 A-3 B+13 C)+a (3 A+7 B-7 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (3 A-3 B+13 C)+a (3 A+7 B-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {3 (3 A-8 B+23 C) a^2+25 (B-2 C) \cos (c+d x) a^2}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {3 (3 A-8 B+23 C) a^2+25 (B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (5 a^3 (3 A-13 B+33 C)-a^3 (9 A-49 B+119 C) \cos (c+d x)\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (3 A-13 B+33 C)-a^3 (9 A-49 B+119 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (3 A-13 B+33 C)-a^3 (9 A-49 B+119 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-a^3 (9 A-49 B+119 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-a^3 (9 A-49 B+119 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (3 A-13 B+33 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^3 (9 A-49 B+119 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}-\frac {a^2 (9 A-49 B+119 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)}}{3 a^2}+\frac {10 a (B-2 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3}\) |
-1/5*((A - B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x]) ^3) + ((10*a*(B - 2*C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) + (-((a^2*(9*A - 49*B + 119*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3 /2)*(a + a*Cos[c + d*x]))) + (3*(5*a^3*(3*A - 13*B + 33*C)*((2*EllipticF[( c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) - a^3*( 9*A - 49*B + 119*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/( d*Sqrt[Cos[c + d*x]]))))/(2*a^2))/(3*a^2))/(10*a^2)
3.13.38.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(1012\) vs. \(2(296)=592\).
Time = 2.93 (sec) , antiderivative size = 1013, normalized size of antiderivative = 3.78
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, method=_RETURNVERBOSE)
-1/4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^3*(1/3*(- 2*B+4*C)*(2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)* (2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^ (1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(2*sin(1/2*d*x+1/2*c)^2- 1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1 /2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/ 2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x +1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(sin(1/2*d*x+ 1/2*c)^2-1)+(-4*B+12*C)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*( 2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-Ell ipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 *c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 1/2)+(A-B+C)*(1/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos (1/2*d*x+1/2*c)^5+4/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) /cos(1/2*d*x+1/2*c)^3+18/5*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^ (1/2)/cos(1/2*d*x+1/2*c)-8/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+ 1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+18/5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c os(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2 )^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.25 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (9 \, A - 49 \, B + 119 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (33 \, A - 188 \, B + 453 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (9 \, A - 59 \, B + 139 \, C\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right ) - 20 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (3 i \, A - 13 i \, B + 33 i \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (3 i \, A - 13 i \, B + 33 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (3 i \, A - 13 i \, B + 33 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (3 i \, A - 13 i \, B + 33 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-3 i \, A + 13 i \, B - 33 i \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (-3 i \, A + 13 i \, B - 33 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-3 i \, A + 13 i \, B - 33 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-3 i \, A + 13 i \, B - 33 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-9 i \, A + 49 i \, B - 119 i \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (-9 i \, A + 49 i \, B - 119 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-9 i \, A + 49 i \, B - 119 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-9 i \, A + 49 i \, B - 119 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (9 i \, A - 49 i \, B + 119 i \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (9 i \, A - 49 i \, B + 119 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (9 i \, A - 49 i \, B + 119 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (9 i \, A - 49 i \, B + 119 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) )^3,x, algorithm="fricas")
-1/60*(2*(3*(9*A - 49*B + 119*C)*cos(d*x + c)^4 + 2*(33*A - 188*B + 453*C) *cos(d*x + c)^3 + 5*(9*A - 59*B + 139*C)*cos(d*x + c)^2 - 60*(B - 2*C)*cos (d*x + c) - 20*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(3*I*A - 13 *I*B + 33*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(3*I*A - 13*I*B + 33*I*C)*cos(d* x + c)^4 + 3*sqrt(2)*(3*I*A - 13*I*B + 33*I*C)*cos(d*x + c)^3 + sqrt(2)*(3 *I*A - 13*I*B + 33*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-3*I*A + 13*I*B - 33*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(-3*I*A + 13*I*B - 33*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-3 *I*A + 13*I*B - 33*I*C)*cos(d*x + c)^3 + sqrt(2)*(-3*I*A + 13*I*B - 33*I*C )*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) ) + 3*(sqrt(2)*(-9*I*A + 49*I*B - 119*I*C)*cos(d*x + c)^5 + 3*sqrt(2)*(-9* I*A + 49*I*B - 119*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-9*I*A + 49*I*B - 119* I*C)*cos(d*x + c)^3 + sqrt(2)*(-9*I*A + 49*I*B - 119*I*C)*cos(d*x + c)^2)* weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(9*I*A - 49*I*B + 119*I*C)*cos(d*x + c)^5 + 3*sqrt(2) *(9*I*A - 49*I*B + 119*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(9*I*A - 49*I*B + 1 19*I*C)*cos(d*x + c)^3 + sqrt(2)*(9*I*A - 49*I*B + 119*I*C)*cos(d*x + c)^2 )*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c))))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d* x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) )^3,x, algorithm="maxima")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c) )^3,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^3* cos(d*x + c)^(7/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]