Integrand size = 34, antiderivative size = 194 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {31 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}} \] Output:
31/4*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(3/2)/d-11/2*A* arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))*2^(1/2)/a^(3 /2)/d-A*cos(d*x+c)*sin(d*x+c)/d/(a-a*sec(d*x+c))^(3/2)+13/4*A*sin(d*x+c)/a /d/(a-a*sec(d*x+c))^(1/2)+3/2*A*cos(d*x+c)*sin(d*x+c)/a/d/(a-a*sec(d*x+c)) ^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.92 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {A \sec (c+d x) \left (-8 \cos ^3(c+d x)-40 (-1+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1+\sec (c+d x)\right )+\frac {11 (-1+\cos (c+d x)) \left (7 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (1+2 \cos (c+d x)) \sqrt {1+\sec (c+d x)}\right )}{\sqrt {1+\sec (c+d x)}}\right ) \tan (c+d x)}{8 d (a-a \sec (c+d x))^{3/2}} \] Input:
Integrate[(Cos[c + d*x]^2*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2) ,x]
Output:
(A*Sec[c + d*x]*(-8*Cos[c + d*x]^3 - 40*(-1 + Cos[c + d*x])*Hypergeometric 2F1[1/2, 3, 3/2, 1 + Sec[c + d*x]] + (11*(-1 + Cos[c + d*x])*(7*ArcTanh[Sq rt[1 + Sec[c + d*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(1 + 2*Cos[c + d*x])*Sqrt[1 + Sec[c + d*x]]))/Sqrt[1 + Sec[ c + d*x]])*Tan[c + d*x])/(8*d*(a - a*Sec[c + d*x])^(3/2))
Time = 1.27 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.441, Rules used = {3042, 4508, 3042, 4510, 25, 3042, 4510, 27, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) (A \sec (c+d x)+A)}{(a-a \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (6 a A+5 a \sec (c+d x) A)}{\sqrt {a-a \sec (c+d x)}}dx}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {6 a A+5 a \csc \left (c+d x+\frac {\pi }{2}\right ) A}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}-\frac {\int -\frac {\cos (c+d x) \left (13 A a^2+9 A \sec (c+d x) a^2\right )}{\sqrt {a-a \sec (c+d x)}}dx}{2 a}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (13 A a^2+9 A \sec (c+d x) a^2\right )}{\sqrt {a-a \sec (c+d x)}}dx}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {13 A a^2+9 A \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}-\frac {\int -\frac {31 A a^3+13 A \sec (c+d x) a^3}{2 \sqrt {a-a \sec (c+d x)}}dx}{a}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {31 A a^3+13 A \sec (c+d x) a^3}{\sqrt {a-a \sec (c+d x)}}dx}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {31 A a^3+13 A \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {\frac {\frac {44 a^3 A \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}}dx+31 a^2 A \int \sqrt {a-a \sec (c+d x)}dx}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {44 a^3 A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+31 a^2 A \int \sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {\frac {44 a^3 A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {62 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{a-a \sec (c+d x)}+a}d\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}}{d}}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {44 a^3 A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {62 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {\frac {\frac {62 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {88 a^3 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{a-a \sec (c+d x)}+2 a}d\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}}{d}}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {62 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {44 \sqrt {2} a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{d}}{2 a}+\frac {13 a^2 A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a}+\frac {3 a A \sin (c+d x) \cos (c+d x)}{d \sqrt {a-a \sec (c+d x)}}}{2 a^2}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}}\) |
Input:
Int[(Cos[c + d*x]^2*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]
Output:
-((A*Cos[c + d*x]*Sin[c + d*x])/(d*(a - a*Sec[c + d*x])^(3/2))) + ((3*a*A* Cos[c + d*x]*Sin[c + d*x])/(d*Sqrt[a - a*Sec[c + d*x]]) + (((62*a^(5/2)*A* ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/d - (44*Sqrt[2]*a ^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])] )/d)/(2*a) + (13*a^2*A*Sin[c + d*x])/(d*Sqrt[a - a*Sec[c + d*x]]))/(2*a))/ (2*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. \(830\) vs. \(2(167)=334\).
Time = 1.92 (sec) , antiderivative size = 831, normalized size of antiderivative = 4.28
Input:
int(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x,method=_RETURNV ERBOSE)
Output:
A*(1/4/d*((7*cos(1/2*d*x+1/2*c)-7)*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))+(20*cos(1/2*d*x+1/2*c)-20)*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))+(-7*cos(1/2*d*x+1/2*c)+7)*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))+cos(1/2*d*x+1/2*c)*(4*cos(1/2*d*x+1/2*c)^2-6)*2^(1/2)*((2*c os(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))/a/(-a/(2*cos(1/2*d *x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/((2*cos(1/2*d*x+1/2*c)^2-1)/(co s(1/2*d*x+1/2*c)+1)^2)^(1/2)*csc(1/2*d*x+1/2*c)-1/4/d*(2^(1/2)*cos(1/2*d*x +1/2*c)*(4*cos(1/2*d*x+1/2*c)^4+7*cos(1/2*d*x+1/2*c)^2-15)*((2*cos(1/2*d*x +1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)+(-18*cos(1/2*d*x+1/2*c)+18)*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))+(18*cos(1/2*d*x+1/2*c)-1 8)*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))+(51*cos(1/2*d*x+1/ 2*c)-51)*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)))/a/(-a/(2*cos(1/2...
Time = 0.11 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {22 \, \sqrt {2} {\left (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 ) + 31 \, {\left (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 ) + 2 \, {\left (2 \, A \cos \left (d x + c\right )^{4} + 9 \, A \cos \left (d x + c\right )^{3} - 6 \, A \cos \left (d x + c\right )^{2} - 13 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{8 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac {22 \, \sqrt {2} {\left (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 ) - 31 \, {\left (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 ) - {\left (2 \, A \cos \left (d x + c\right )^{4} + 9 \, A \cos \left (d x + c\right )^{3} - 6 \, A \cos \left (d x + c\right )^{2} - 13 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \] Input:
integrate(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorith m="fricas")
Output:
[-1/8*(22*sqrt(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) + 31*(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) + 2*(2*A*cos(d*x + c)^4 + 9*A*cos(d*x + c)^3 - 6*A*cos(d*x + c)^2 - 13*A*cos(d*x + c))*sqrt((a*cos (d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c)), 1/4*(22*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(a)*arctan(sqrt(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) - 31*(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)*sin(d*x + c)))*sin(d*x + c) - (2*A*cos(d*x + c)^4 + 9*A*cos(d*x + c)^3 - 6*A*cos(d*x + c)^2 - 13*A*cos(d*x + c))*sqr t((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d* x + c))]
\[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=A \left (\int \frac {\cos ^{2}{\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \] Input:
integrate(cos(d*x+c)**2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(3/2),x)
Output:
A*(Integral(cos(c + d*x)**2/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a *sqrt(-a*sec(c + d*x) + a)), x) + Integral(cos(c + d*x)**2*sec(c + d*x)/(- a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x ))
\[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorith m="maxima")
Output:
integrate((A*sec(d*x + c) + A)*cos(d*x + c)^2/(-a*sec(d*x + c) + a)^(3/2), x)
Time = 0.58 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {22 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {31 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {\sqrt {2} {\left (7 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 18 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{2} a} - \frac {2 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorith m="giac")
Output:
1/4*(22*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(3/ 2) - 31*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a ^(3/2) - sqrt(2)*(7*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 18*sqrt(a*tan (1/2*d*x + 1/2*c)^2 - a)*A*a)/((a*tan(1/2*d*x + 1/2*c)^2 + a)^2*a) - 2*sqr t(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A/(a^2*tan(1/2*d*x + 1/2*c)^2))/d
Timed out. \[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((cos(c + d*x)^2*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(3/2),x)
Output:
int((cos(c + d*x)^2*(A + A/cos(c + d*x)))/(a - a/cos(c + d*x))^(3/2), x)
\[ \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2}-2 \sec \left (d x +c \right )+1}d x +\int \frac {\sqrt {-\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}-2 \sec \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x)
Output:
(sqrt(a)*(int((sqrt( - sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x))/(se c(c + d*x)**2 - 2*sec(c + d*x) + 1),x) + int((sqrt( - sec(c + d*x) + 1)*co s(c + d*x)**2)/(sec(c + d*x)**2 - 2*sec(c + d*x) + 1),x)))/a