Integrand size = 32, antiderivative size = 184 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {7 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {79 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {11 A \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {23 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}} \] Output:
7*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(5/2)/d-79/16*A*ar ctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))*2^(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*s ec(d*x+c))^(3/2)+23/8*A*sin(d*x+c)/a^2/d/(a-a*sec(d*x+c))^(1/2)
Time = 1.58 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {A \left (448 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin ^5\left (\frac {1}{2} (c+d x)\right )-316 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin ^5\left (\frac {1}{2} (c+d x)\right )+\sqrt {1+\sec (c+d x)} (8 \sin (c+d x)+(-35+23 \sec (c+d x)) \tan (c+d x))\right )}{8 a^2 d (-1+\sec (c+d x))^2 \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \] Input:
Integrate[(Cos[c + d*x]*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(5/2),x ]
Output:
(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 |
| \(\displaystyle \frac {\frac {\frac {\frac {112 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {79 \sqrt {2} a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{d}}{a}+\frac {46 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{4 a^2}-\frac {11 a A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{3/2}}}{4 a^2}-\frac {A \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}\) |
Input:
Int[(Cos[c + d*x]*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(5/2),x]
Output:
-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)
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. \(453\) vs. \(2(155)=310\).
Time = 2.12 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.47
| method | result | size |
| default | \(\frac {A \left (\left (32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-102 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+66\right ) \sqrt {2}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \left (79 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-79 \csc \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {2}\, \ln \left (\frac {2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right ) \left (-79 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+79 \csc \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\operatorname {arctanh}\left (\frac {\sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right ) \left (-224 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+224 \csc \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{32 d \sqrt {-\frac {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, a^{2}}\) | \(454\) |
Input:
int(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x,method=_RETURNVER BOSE)
Output:
1/32*A/d/(-a/(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/((2*co s(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)/a^2*((32*cos(1/2*d*x +1/2*c)^4-102*cos(1/2*d*x+1/2*c)^2+66)*2^(1/2)*((2*cos(1/2*d*x+1/2*c)^2-1) /(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*cot(1/2*d*x+1/2*c)*csc(1/2*d*x+1/2*c)^2+2 ^(1/2)*arctanh((2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2 *d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*(79*cot(1/2*d*x+1/2*c)-7 9*csc(1/2*d*x+1/2*c))+2^(1/2)*ln(2*(((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d *x+1/2*c)+1)^2)^(1/2)*cos(1/2*d*x+1/2*c)+((2*cos(1/2*d*x+1/2*c)^2-1)/(cos( 1/2*d*x+1/2*c)+1)^2)^(1/2)-2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1)) *(-79*cot(1/2*d*x+1/2*c)+79*csc(1/2*d*x+1/2*c))+arctanh(2^(1/2)*cos(1/2*d* x+1/2*c)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1 /2*c)+1)^2)^(1/2))*(-224*cot(1/2*d*x+1/2*c)+224*csc(1/2*d*x+1/2*c)))
Time = 0.11 (sec) , antiderivative size = 612, normalized size of antiderivative = 3.33 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm= "fricas")
Output:
[-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))]
Timed out. \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm= "maxima")
Output:
integrate((A*sec(d*x + c) + A)*cos(d*x + c)/(-a*sec(d*x + c) + a)^(5/2), x )
Time = 0.63 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {79 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {112 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {16 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )} a^{2}} - \frac {\sqrt {2} {\left (17 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 15 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{16 \, d} \] Input:
integrate(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm= "giac")
Output:
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. \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((cos(c + d*x)*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(5/2),x)
Output:
int((cos(c + d*x)*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(5/2), x)
\[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{3}-3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )-1}d x +\int \frac {\sqrt {-\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )}{\sec \left (d x +c \right )^{3}-3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )-1}d x \right )}{a^{2}} \] Input:
int(cos(d*x+c)*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x)
Output:
( - sqrt(a)*(int((sqrt( - sec(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x))/(se c(c + d*x)**3 - 3*sec(c + d*x)**2 + 3*sec(c + d*x) - 1),x) + int((sqrt( - sec(c + d*x) + 1)*cos(c + d*x))/(sec(c + d*x)**3 - 3*sec(c + d*x)**2 + 3*s ec(c + d*x) - 1),x)))/a**2