Integrand size = 45, antiderivative size = 294 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {(2 B-5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {(3 A-43 B+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \]
-1/4*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2)+1/16*(A+ 7*B-15*C)*sin(d*x+c)/a/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2)+(2*B-5*C) *arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d *x+c)^(1/2)/a^(5/2)/d+1/32*(3*A-43*B+115*C)*arctanh(1/2*sin(d*x+c)*a^(1/2) *sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x +c)^(1/2)/a^(5/2)/d*2^(1/2)+1/16*(3*A-11*B+35*C)*sin(d*x+c)/a^2/d/cos(d*x+ c)^(3/2)/(a+a*sec(d*x+c))^(1/2)
Time = 9.09 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((6 A-86 B+230 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 \sqrt {2} (2 B-5 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} (3 A-11 B+67 C+2 (7 A-15 B+55 C) \cos (c+d x)+(3 A-11 B+35 C) \cos (2 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d \sqrt {\cos (c+d x)} (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{5/2}} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(5/2)),x]
(Cos[(c + d*x)/2]^5*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((6*A - 86*B + 230*C)*ArcTanh[Sin[(c + d*x)/2]] + 32*Sqrt[2]*(2*B - 5*C)*ArcTanh[Sqrt[2] *Sin[(c + d*x)/2]] + ((3*A - 11*B + 67*C + 2*(7*A - 15*B + 55*C)*Cos[c + d *x] + (3*A - 11*B + 35*C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*Sec[c + d*x ]*Tan[(c + d*x)/2])/2))/(4*d*Sqrt[Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x ] + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.96 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.99, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 4753, 3042, 4572, 27, 3042, 4507, 27, 3042, 4509, 3042, 4511, 3042, 4288, 222, 4295, 219}
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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{5/2} (a \sec (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{(\sec (c+d x) a+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (3 A+5 B-5 C)+2 a (A-B+5 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (3 A+5 B-5 C)+2 a (A-B+5 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (3 A+5 B-5 C)+2 a (A-B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (A+7 B-15 C) a^2+2 (3 A-11 B+35 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (A+7 B-15 C) a^2+2 (3 A-11 B+35 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 (A+7 B-15 C) a^2+2 (3 A-11 B+35 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((3 A-11 B+35 C) a^3+16 (2 B-5 C) \sec (c+d x) a^3\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((3 A-11 B+35 C) a^3+16 (2 B-5 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4511 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx+16 a^2 (2 B-5 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+16 a^2 (2 B-5 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {32 a^2 (2 B-5 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {a^3 (3 A-43 B+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {32 a^{5/2} (2 B-5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\frac {32 a^{5/2} (2 B-5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^3 (3 A-43 B+115 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\frac {\sqrt {2} a^{5/2} (3 A-43 B+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {32 a^{5/2} (2 B-5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec [c + d*x])^(5/2)),x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B + C)*Sec[c + d*x]^(7/2 )*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + ((a*(A + 7*B - 15*C)*Sec[ c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (((32*a^(5 /2)*(2*B - 5*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/ d + (Sqrt[2]*a^(5/2)*(3*A - 43*B + 115*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d* x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (2*a^2*(3*A - 11*B + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x] ]))/(4*a^2))/(8*a^2))
3.13.91.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[d/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b , d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(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*((d*Csc[e + f*x])^n/(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*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(1355\) vs. \(2(249)=498\).
Time = 1.03 (sec) , antiderivative size = 1356, normalized size of antiderivative = 4.61
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2 ),x,method=_RETURNVERBOSE)
1/a^2/d*(-1/32*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2*(-2*a/((1-cos(d*x+c)) ^2*csc(d*x+c)^2-1))^(1/2)*(2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-c os(d*x+c))^5*csc(d*x+c)^5+40*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2^(1/2) /(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*2^(1/2)*(1-cos(d*x+c))^2*csc(d* x+c)^2+40*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)*2^(1/2)/(-(1-cos(d*x+c))^2 *csc(d*x+c)^2-1)^(1/2))*2^(1/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2+19*(-(1-cos( d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c)^3-115*arctan(1 /(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(1-cos (d*x+c))^2*csc(d*x+c)^2-40*2^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2 ^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-40*2^(1/2)*arctan(1/2*(-c ot(d*x+c)+csc(d*x+c)+1)*2^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))- 53*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))+115*a rctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))) )/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)/a/((1-cos(d*x+c))^2*csc(d*x+c)^ 2+1)^2/(-((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/((1-cos(d*x+c))^2*csc(d*x+c)^2+ 1))^(5/2)-1/32*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2*(-2*a/((1-cos(d*x+c)) ^2*csc(d*x+c)^2-1))^(1/2)*(2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-c os(d*x+c))^3*csc(d*x+c)^3+16*2^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1) *2^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+16*2^(1/2)*arctan(1/2*( -cot(d*x+c)+csc(d*x+c)+1)*2^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1...
Time = 0.38 (sec) , antiderivative size = 884, normalized size of antiderivative = 3.01 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) )^(5/2),x, algorithm="fricas")
[1/64*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^4 + 3*(3*A - 43*B + 115* C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + (3*A - 43*B + 115*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sq rt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2* a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((3*A - 1 1*B + 35*C)*cos(d*x + c)^2 + (7*A - 15*B + 55*C)*cos(d*x + c) + 16*C)*sqrt ((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 16*( (2*B - 5*C)*cos(d*x + c)^4 + 3*(2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)* cos(d*x + c)^2 + (2*B - 5*C)*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 + 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt (cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d *cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/32*(sqrt(2)*((3*A - 43*B + 115*C )*cos(d*x + c)^4 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + (3*A - 43*B + 115*C)*cos(d*x + c))*sqrt(-a)*arcta n(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) - 2*((3*A - 11*B + 35*C)*cos(d*x + c)^2 + (7*A - 15* B + 55*C)*cos(d*x + c) + 16*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqr t(cos(d*x + c))*sin(d*x + c) - 16*((2*B - 5*C)*cos(d*x + c)^4 + 3*(2*B - 5 *C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*x + c)^2 + (2*B - 5*C)*cos(d*x...
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) )^(5/2),x, algorithm="maxima")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) )^(5/2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(5 /2)*cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos (c + d*x))^(5/2)),x)