Integrand size = 37, antiderivative size = 268 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(15 A+7 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(49 A+25 C) \sin (c+d x)}{10 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(9 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}} \]
-1/2*(A+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/4*(15*A+ 7*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^(3/2)/d*2^(1/2)+1/10*(9*A+5 *C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/10*(49*A+25*C )*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-1/10*(13*A+5*C)*s in(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(1/2)
Time = 4.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (5 \sqrt {2} (15 A+7 C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)+\sqrt {1-\sec (c+d x)} (38 A+20 C-2 A \cos (2 (c+d x)) (-1+\sec (c+d x))+(47 A+25 C) \sec (c+d x))\right ) \sin (c+d x)}{10 d \sqrt {-1+\cos (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]
((5*Sqrt[2]*(15*A + 7*C)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[ c + d*x]]]*Cos[(c + d*x)/2]^2*Sec[c + d*x]^(3/2) + Sqrt[1 - Sec[c + d*x]]* (38*A + 20*C - 2*A*Cos[2*(c + d*x)]*(-1 + Sec[c + d*x]) + (47*A + 25*C)*Se c[c + d*x]))*Sin[c + d*x])/(10*d*Sqrt[-1 + Cos[c + d*x]]*(a*(1 + Sec[c + d *x]))^(3/2))
Time = 1.64 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 4753, 3042, 4573, 27, 3042, 4510, 27, 3042, 4510, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^{5/2} \left (A+C \sec (c+d x)^2\right )}{(a \sec (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sec ^2(c+d x)+A}{\sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int -\frac {a (9 A+5 C)-2 a (3 A+C) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+5 C)-2 a (3 A+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A+5 C)-2 a (3 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {3 a^2 (13 A+5 C)-4 a^2 (9 A+5 C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {3 a^2 (13 A+5 C)-4 a^2 (9 A+5 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {3 a^2 (13 A+5 C)-4 a^2 (9 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 \left (a^3 (49 A+25 C)-2 a^3 (13 A+5 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (49 A+25 C)-2 a^3 (13 A+5 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (49 A+25 C)-2 a^3 (13 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-5 a^3 (15 A+7 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-5 a^3 (15 A+7 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {10 a^3 (15 A+7 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (9 A+5 C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (13 A+5 C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {5 \sqrt {2} a^{5/2} (15 A+7 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{5 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A + C)*Sin[c + d*x])/(d*Sec[ c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((2*a*(9*A + 5*C)*Sin[c + d*x ])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((2*a^2*(13*A + 5*C )*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-5*Sqr t[2]*a^(5/2)*(15*A + 7*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^3*(49*A + 25*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(5*a))/(4*a^2))
3.12.61.3.1 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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d *n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[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. \(485\) vs. \(2(227)=454\).
Time = 0.77 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.81
| method | result | size |
| default | \(\frac {\left (-75 A \sqrt {2}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{2}-35 C \sqrt {2}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{2}+8 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )-150 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \cos \left (d x +c \right )-70 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \cos \left (d x +c \right )-8 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-75 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-35 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+72 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+40 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+98 A \sin \left (d x +c \right )+50 C \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{20 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2}}\) | \(486\) |
1/20/a^2/d*(-75*A*2^(1/2)*(-1/(1+cos(d*x+c)))^(1/2)*arctan(1/2*sin(d*x+c)* 2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2-35*C*2^(1/2 )*(-1/(1+cos(d*x+c)))^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/( -1/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2+8*A*cos(d*x+c)^3*sin(d*x+c)-150*A*a rctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1 /(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)-70*C*arctan(1/2*sin(d*x+c)*2^(1/ 2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1/(1+cos(d*x+c)))^(1/2)*2^( 1/2)*cos(d*x+c)-8*A*cos(d*x+c)^2*sin(d*x+c)-75*A*arctan(1/2*sin(d*x+c)*2^( 1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-1/(1+cos(d*x+c)))^(1/2)*2 ^(1/2)-35*C*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c) ))^(1/2))*(-1/(1+cos(d*x+c)))^(1/2)*2^(1/2)+72*A*cos(d*x+c)*sin(d*x+c)+40* C*cos(d*x+c)*sin(d*x+c)+98*A*sin(d*x+c)+50*C*sin(d*x+c))*cos(d*x+c)^(1/2)* (a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))^2
Time = 0.31 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 7 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 4 \, A \cos \left (d x + c\right )^{2} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 49 \, A + 25 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 7 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 4 \, A \cos \left (d x + c\right )^{2} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 49 \, A + 25 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]
[1/40*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^2 + 2*(15*A + 7*C)*cos(d*x + c ) + 15*A + 7*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*(4*A*cos(d*x + c)^3 - 4*A*cos(d*x + c)^2 + 4*(9*A + 5*C)*cos(d*x + c) + 49*A + 25*C)*sqr t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2 *d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/20*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^2 + 2*(15*A + 7*C)*cos(d*x + c) + 15*A + 7*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*(4*A*cos(d*x + c)^3 - 4*A*cos(d*x + c)^2 + 4*(9*A + 5*C)*cos(d*x + c) + 49*A + 25*C)*sqrt((a*cos(d*x + c) + a)/cos(d *x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d* cos(d*x + c) + a^2*d)]
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]