Integrand size = 37, antiderivative size = 237 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {5 C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \]
-5*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A+C )*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)+1/16*(A-15*C)*sec(d *x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/32*(3*A+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))/a ^(5/2)/d*2^(1/2)+1/16*(3*A+35*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*se c(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(603\) vs. \(2(237)=474\).
Time = 5.73 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.54 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {16 (3 A+35 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+16 (3 A+115 C) \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+6 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+70 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+14 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+110 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+32 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)-3 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-115 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-6 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-230 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-3 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)-115 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)}{32 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
(16*(3*A + 35*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^5*Sec[c + d*x]^3*Sin[(c + d*x)/2] + 16*(3*A + 115*C)*ArcSin[Sqrt[Sec[c + d*x]]]*Cos [(c + d*x)/2]^5*Sec[c + d*x]^3*Sin[(c + d*x)/2] + 6*A*Sqrt[1 - Sec[c + d*x ]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 70*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d *x]^(3/2)*Sin[c + d*x] + 14*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Si n[c + d*x] + 110*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] + 32*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x] - 3*Sqrt[2]* A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 115*Sqrt[2]*C*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x] ]]*Tan[c + d*x] - 6*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] - 230*Sqrt[2]*C*ArcTan[(Sqrt[2]* Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] - 3* Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[ c + d*x]^2*Tan[c + d*x] - 115*Sqrt[2]*C*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]] )/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^2*Tan[c + d*x])/(32*d*Sqrt[1 - Sec[ c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.55 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 4573, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {\sec ^{\frac {5}{2}}(c+d x) (a (3 A-5 C)+2 a (A+5 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (3 A-5 C)+2 a (A+5 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (3 A-5 C)+2 a (A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (A-15 C) a^2+2 (3 A+35 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (A-15 C) a^2+2 (3 A+35 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 (A-15 C) a^2+2 (3 A+35 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^3 (3 A+35 C)-80 a^3 C \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (3 A+35 C)-80 a^3 C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4511 |
| \(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-80 a^2 C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-80 a^2 C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {160 a^2 C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {160 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \frac {\frac {\frac {-\frac {2 a^3 (3 A+115 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {160 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\frac {\sqrt {2} a^{5/2} (3 A+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {160 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
-1/4*((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 2)) + ((a*(A - 15*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (((-160*a^(5/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a *Sec[c + d*x]]])/d + (Sqrt[2]*a^(5/2)*(3*A + 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 + 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)
3.3.89.3.1 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_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(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*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(931\) vs. \(2(200)=400\).
Time = 1.25 (sec) , antiderivative size = 932, normalized size of antiderivative = 3.93
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(932\) |
| parts | \(\text {Expression too large to display}\) | \(975\) |
-1/32/a^3/d*(-((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-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*A*(-(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+2*C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1- cos(d*x+c))^5*csc(d*x+c)^5+3*A*(-(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+19*C*(-(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+40*C*arctan(1/2*(csc(d*x+c)-cot(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*c sc(d*x+c)^2+40*C*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)^2-3*A*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)^2-115*C*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)^2-5*A*( -(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-53*C*(-(1 -cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-40*C*arctan( 1/2*(csc(d*x+c)-cot(d*x+c)+1)*2^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^( 1/2))*2^(1/2)-40*C*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)+3*A*arctan(1/(-(1-cos(d*x+c))^2*c sc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))+115*C*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)...
Time = 0.33 (sec) , antiderivative size = 755, normalized size of antiderivative = 3.19 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 115 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 80 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left ({\left (3 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A + 55 \, C\right )} \cos \left (d x + c\right ) + 16 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{64 \, {\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 )}}, -\frac {\sqrt {2} {\left ({\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 115 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 80 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) - \frac {2 \, {\left ({\left (3 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A + 55 \, C\right )} \cos \left (d x + c\right ) + 16 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\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 )}}\right ] \]
[1/64*(sqrt(2)*((3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c )^2 + 3*(3*A + 115*C)*cos(d*x + c) + 3*A + 115*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(co s(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)) + 80*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s in(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((3*A + 35*C)*cos(d*x + c)^2 + (7*A + 55*C)*cos(d*x + c) + 16*C)*sqrt(( a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(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), - 1/32*(sqrt(2)*((3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c) ^2 + 3*(3*A + 115*C)*cos(d*x + c) + 3*A + 115*C)*sqrt(-a)*arctan(sqrt(2)*s qrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin( d*x + c))) + 80*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq rt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) - 2*((3*A + 35*C)*cos(d*x + c)^2 + (7*A + 55*C)*cos(d*x + c) + 16*C)*sqrt(( a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(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)...
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 93381 vs. \(2 (200) = 400\).
Time = 17.67 (sec) , antiderivative size = 93381, normalized size of antiderivative = 394.01 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/32*((512*((2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos (5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 2*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3* c) + (2*cos(2*d*x + 2*c) + cos(d*x + c))*sin(5/2*d*x + 5/2*c) + cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) + 2*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos (5*d*x + 5*c)^2 + 2560*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)* sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2*d* x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)*sin (5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(8/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/ 2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2* d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2 *c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos( 2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c) *sin(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3 *d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(6/5*arctan2(sin(5/2*d*x + 5/2*c), co s(5/2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos( 5/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10* cos(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x...
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]