Integrand size = 37, antiderivative size = 219 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {5 (15 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \]
-1/4*(A+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2)-1/16*(13*A -3*C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2)-5/32*(15*A-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*(49*A+C)*s in(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)
Time = 5.59 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {32 A \sqrt {1-\sec (c+d x)} \sin (c+d x)+20 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+20 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+10 \sqrt {2} (15 A-C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+85 A \sqrt {1-\sec (c+d x)} \tan (c+d x)+5 C \sqrt {1-\sec (c+d x)} \tan (c+d x)+49 A \sqrt {1-\sec (c+d x)} \sec (c+d x) \tan (c+d x)+C \sqrt {1-\sec (c+d x)} \sec (c+d x) \tan (c+d x)}{16 d \sqrt {-1+\cos (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
(32*A*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 20*C*ArcSin[Sqrt[1 - Sec[c + d *x]]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*Sin[c + d*x] + 20*C*ArcSin[Sqr t[Sec[c + d*x]]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*Sin[c + d*x] + 10*S qrt[2]*(15*A - C)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x ]]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*Sin[c + d*x] + 85*A*Sqrt[1 - Sec [c + d*x]]*Tan[c + d*x] + 5*C*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x] + 49*A*S qrt[1 - Sec[c + d*x]]*Sec[c + d*x]*Tan[c + d*x] + C*Sqrt[1 - Sec[c + d*x]] *Sec[c + d*x]*Tan[c + d*x])/(16*d*Sqrt[-1 + Cos[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4753, 3042, 4573, 27, 3042, 4508, 27, 3042, 4501, 3042, 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.
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec (c+d x)^2\right )}{(a \sec (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sec ^2(c+d x)+A}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int -\frac {a (9 A+C)-4 a (A-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+C)-4 a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+C)-4 a (A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A+C)-2 a^2 (13 A-3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A+C)-2 a^2 (13 A-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A+C)-2 a^2 (13 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-5 a^2 (15 A-C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-5 a^2 (15 A-C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (15 A-C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {5 \sqrt {2} a^{3/2} (15 A-C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (13 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A + C)*Sqrt[Sec[c + d*x]]*Si n[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(13*A - 3*C)*Sqrt[Se c[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + ((-5*Sqrt[2]*a^ (3/2)*(15*A - C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2 ]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^2*(49*A + C)*Sqrt[Sec[c + d*x]]*Sin [c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))/(8*a^2))
3.12.69.3.1 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[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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -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))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 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)]
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. \(552\) vs. \(2(184)=368\).
Time = 0.83 (sec) , antiderivative size = 553, normalized size of antiderivative = 2.53
| method | result | size |
| default | \(\frac {-\frac {C \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-3 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+5 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 a \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}-\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{2} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+75 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-83 \csc \left (d x +c \right )+83 \cot \left (d x +c \right )\right )}{32 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2} a}}{a^{2} d}\) | \(553\) |
1/a^2/d*(-1/32*C*(-((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/((1-cos(d*x+c))^2*csc (d*x+c)^2+1))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)*(-2*a/((1-cos(d*x+c) )^2*csc(d*x+c)^2-1))^(1/2)/a*(2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*( 1-cos(d*x+c))^3*csc(d*x+c)^3-3*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(- cot(d*x+c)+csc(d*x+c))+5*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)^2-1)^(1/2)-1/32* A*(-((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*(1-cos(d*x+c))^5*csc(d*x+c)^5-17*(1-cos(d*x+c))^3*csc(d*x +c)^3+75*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+cs c(d*x+c)))*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)-83*csc(d*x+c)+83*cot(d *x+c))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2/a)
Time = 0.30 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.28 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - 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 ) - 4 \, {\left (32 \, A \cos \left (d x + c\right )^{2} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 49 \, A + C\right )} \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 )}{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 {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - 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 ) + 2 \, {\left (32 \, A \cos \left (d x + c\right )^{2} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 49 \, A + C\right )} \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 )}{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*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^2 + 3*(15*A - C)*cos(d*x + c) + 15*A - 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(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*(32*A*cos(d*x + c)^2 + 5*(17*A + C)*cos(d*x + c) + 49*A + C)*sq rt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*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 ), 1/32*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^ 2 + 3*(15*A - C)*cos(d*x + c) + 15*A - C)*sqrt(-a)*arctan(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*(32*A*cos(d*x + c)^2 + 5*(17*A + C)*cos(d*x + c) + 49*A + C)*sqrt( (a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*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)]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 261332 vs. \(2 (184) = 368\).
Time = 3.23 (sec) , antiderivative size = 261332, normalized size of antiderivative = 1193.30 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/32*((576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si n(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/ 2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*cos(5/2*d* x + 5/2*c)^6 + 14400*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2 *d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*c os(3/2*d*x + 3/2*c)^6 + 187500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin (1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^6 + 576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2* d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/ 2*c)^6 + 5184*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(3/2* d*x + 3/2*c)^6 + 262500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d* x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^4*sin(1/2 *d*x + 1/2*c)^2 + 77700*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) ^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2...
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]