Integrand size = 32, antiderivative size = 184 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {7 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {79 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {23 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}} \]
7*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/2*A*sin( d*x+c)/d/(a-a*sec(d*x+c))^(5/2)-11/8*A*sin(d*x+c)/a/d/(a-a*sec(d*x+c))^(3/ 2)-79/16*A*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))/a ^(5/2)/d*2^(1/2)+23/8*A*sin(d*x+c)/a^2/d/(a-a*sec(d*x+c))^(1/2)
Time = 1.75 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {A \left (448 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin ^5\left (\frac {1}{2} (c+d x)\right )-316 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin ^5\left (\frac {1}{2} (c+d x)\right )+\sqrt {1+\sec (c+d x)} (8 \sin (c+d x)+(-35+23 \sec (c+d x)) \tan (c+d x))\right )}{8 a^2 d (-1+\sec (c+d x))^2 \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \]
(A*(448*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*Cos[(c + d*x)/2]*Sec[c + d*x]^3*Si n[(c + d*x)/2]^5 - 316*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]]*Cos [(c + d*x)/2]*Sec[c + d*x]^3*Sin[(c + d*x)/2]^5 + Sqrt[1 + Sec[c + d*x]]*( 8*Sin[c + d*x] + (-35 + 23*Sec[c + d*x])*Tan[c + d*x])))/(8*a^2*d*(-1 + Se c[c + d*x])^2*Sqrt[1 + Sec[c + d*x]]*Sqrt[a - a*Sec[c + d*x]])
Time = 1.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.469, Rules used = {3042, 4508, 3042, 4508, 27, 3042, 4510, 25, 3042, 4408, 3042, 4261, 216, 4282, 216}
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.
| \(\displaystyle \int \frac {\cos (c+d x) (A \sec (c+d x)+A)}{(a-a \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (6 a A+5 a \sec (c+d x) A)}{(a-a \sec (c+d x))^{3/2}}dx}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {6 a A+5 a \csc \left (c+d x+\frac {\pi }{2}\right ) A}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (46 A a^2+33 A \sec (c+d x) a^2\right )}{2 \sqrt {a-a \sec (c+d x)}}dx}{2 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (46 A a^2+33 A \sec (c+d x) a^2\right )}{\sqrt {a-a \sec (c+d x)}}dx}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {46 A a^2+33 A \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}-\frac {\int -\frac {56 A a^3+23 A \sec (c+d x) a^3}{\sqrt {a-a \sec (c+d x)}}dx}{a}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {56 A a^3+23 A \sec (c+d x) a^3}{\sqrt {a-a \sec (c+d x)}}dx}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {56 A a^3+23 A \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {\frac {\frac {79 a^3 A \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}}dx+56 a^2 A \int \sqrt {a-a \sec (c+d x)}dx}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {79 a^3 A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+56 a^2 A \int \sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {\frac {79 a^3 A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {112 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{a-a \sec (c+d x)}+a}d\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}}{d}}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {79 a^3 A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {112 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {\frac {\frac {112 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {158 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{a-a \sec (c+d x)}+2 a}d\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}}{d}}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {112 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {79 \sqrt {2} a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{d}}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
-1/2*(A*Sin[c + d*x])/(d*(a - a*Sec[c + d*x])^(5/2)) + ((-11*a*A*Sin[c + d *x])/(2*d*(a - a*Sec[c + d*x])^(3/2)) + (((112*a^(5/2)*A*ArcTan[(Sqrt[a]*T an[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/d - (79*Sqrt[2]*a^(5/2)*A*ArcTan[( Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/d)/a + (46*a^2* A*Sin[c + d*x])/(d*Sqrt[a - a*Sec[c + d*x]]))/(4*a^2))/(4*a^2)
3.2.76.3.1 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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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*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[(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]
Leaf count of result is larger than twice the leaf count of optimal. \(374\) vs. \(2(155)=310\).
Time = 31.65 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.04
| method | result | size |
| default | \(-\frac {A \sqrt {2}\, \left (8 \cos \left (d x +c \right )^{3} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+56 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {2}\, \cos \left (d x +c \right )^{2}-27 \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-112 \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-12 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )+79 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+56 \sqrt {2}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+23 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-158 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+79 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right ) \csc \left (d x +c \right )}{16 a^{2} d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )-1\right ) \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}}\) | \(375\) |
-1/16*A/a^2/d*2^(1/2)*(8*cos(d*x+c)^3*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2)+56*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)^2- 27*cos(d*x+c)^2*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-112*2^(1/2)*cos (d*x+c)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-12*(-cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*2^(1/2)*cos(d*x+c)+79*arctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos( d*x+c)+1))^(1/2))*cos(d*x+c)^2+56*2^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2))+23*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-158*cos(d*x+c)*ar ctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+79*arctan(1/2*2^(1/2) /(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/( cos(d*x+c)-1)/(-a*(sec(d*x+c)-1))^(1/2)*csc(d*x+c)
Time = 0.32 (sec) , antiderivative size = 612, normalized size of antiderivative = 3.33 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {79 \, \sqrt {2} {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{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 )}} + {\left (3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 112 \, {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{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 )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, {\left (8 \, A \cos \left (d x + c\right )^{4} - 27 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} + 23 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )} \sin \left (d x + c\right )}, \frac {79 \, \sqrt {2} {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 112 \, {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (8 \, A \cos \left (d x + c\right )^{4} - 27 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} + 23 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/32*(79*sqrt(2)*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(-a)*log( (2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d*x + c) + 112*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos( d*x + c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) + 4*(8*A*cos(d*x + c)^4 - 27*A*cos(d*x + c)^3 - 12*A*c os(d*x + c)^2 + 23*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) )/((a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c)), 1/ 16*(79*sqrt(2)*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(a)*arctan(sq rt(2)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d* x + c)))*sin(d*x + c) - 112*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt (a)*arctan(sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*s in(d*x + c)))*sin(d*x + c) - 2*(8*A*cos(d*x + c)^4 - 27*A*cos(d*x + c)^3 - 12*A*cos(d*x + c)^2 + 23*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d* x + c)))/((a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c))]
\[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=A \left (\int \frac {\cos {\left (c + d x \right )}}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]
A*(Integral(cos(c + d*x)/(a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x)**2 - 2*a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a**2*sqrt(-a*sec(c + d*x) + a)), x) + Integral(cos(c + d*x)*sec(c + d*x)/(a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x)**2 - 2*a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a* *2*sqrt(-a*sec(c + d*x) + a)), x))
\[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 1.24 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {79 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {112 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {16 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )} a^{2}} - \frac {\sqrt {2} {\left (17 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 15 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{16 \, d} \]
1/16*(79*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(5 /2) - 112*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a)) /a^(5/2) - 16*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A/((a*tan(1/2*d*x + 1/2*c)^2 + a)*a^2) - sqrt(2)*(17*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 15*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A*a)/(a^4*tan(1/2*d*x + 1/2*c)^4) )/d
Timed out. \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]