Integrand size = 31, antiderivative size = 170 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {(9 A-5 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-B) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \]
-(3*A-2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-1/2 *(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/4*(9*A-5*B)*arctan(1/2*a^(1/2 )*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/2*(3*A-B) *sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)
Time = 1.00 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-8 (3 A-2 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+2 \sqrt {2} (9 A-5 B) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+2 \cos (c+d x) (3 A-B+2 A \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \sec (c+d x) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]
((-8*(3*A - 2*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2 + 2*Sq rt[2]*(9*A - 5*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2] ^2 + 2*Cos[c + d*x]*(3*A - B + 2*A*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*S ec[c + d*x]*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x] ))^(3/2))
Time = 0.98 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {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.
| \(\displaystyle \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (3 A-B)-3 a (A-B) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (3 A-B)-3 a (A-B) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (3 A-B)-3 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\int -\frac {2 a^2 (3 A-2 B)-a^2 (3 A-B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {2 a^2 (3 A-2 B)-a^2 (3 A-B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {2 a^2 (3 A-2 B)-a^2 (3 A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \sqrt {\sec (c+d x) a+a}dx-a^2 (9 A-5 B) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^2 (9 A-5 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^2 (9 A-5 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {4 a^2 (3 A-2 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^2 (9 A-5 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (9 A-5 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {4 a^{3/2} (3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {2 a (3 A-B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (9 A-5 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
-1/2*((A - B)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-(((4*a^(3/2 )*(3*A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a^(3/2)*(9*A - 5*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[ a + a*Sec[c + d*x]])])/d)/a) + (2*a*(3*A - B)*Sin[c + d*x])/(d*Sqrt[a + a* Sec[c + d*x]]))/(4*a^2)
3.2.57.3.1 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. \(827\) vs. \(2(145)=290\).
Time = 31.53 (sec) , antiderivative size = 828, normalized size of antiderivative = 4.87
-1/4/a^2/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(6*A*2^(1/2)*arc tanh(2^(1/2)/((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-cos(d*x+c))^2*csc(d*x+c)^2 +A*(1-cos(d*x+c))^5*csc(d*x+c)^5-4*B*2^(1/2)*arctanh(2^(1/2)/((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-cos(d*x+c))^2*csc(d*x+c)^2-B*(1-cos(d*x+c))^5*csc(d* x+c)^5-9*A*ln(csc(d*x+c)-cot(d*x+c)+((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)^(1/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2+5 *B*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-c os(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2+6*A*2^(1/ 2)*arctanh(2^(1/2)/((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)+4*A*(1-cos(d*x+c))^3*cs c(d*x+c)^3-4*B*2^(1/2)*arctanh(2^(1/2)/((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)-9*A *((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos( d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+5*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/ 2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-5*A*( -cot(d*x+c)+csc(d*x+c))+B*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc( d*x+c)^2+1)
Time = 2.31 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.58 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 9 \, A - 5 \, B\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 2 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\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 {\sqrt {2} {\left ({\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 9 \, A - 5 \, B\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 ) - 4 \, {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 2 \, B\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 ) - 2 \, {\left (2 \, A \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\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 ] \]
[1/8*(sqrt(2)*((9*A - 5*B)*cos(d*x + c)^2 + 2*(9*A - 5*B)*cos(d*x + c) + 9 *A - 5*B)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos( d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c ) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((3*A - 2*B)*cos(d*x + c )^2 + 2*(3*A - 2*B)*cos(d*x + c) + 3*A - 2*B)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin (d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 4*(2*A*cos(d*x + c)^ 2 + (3*A - B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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), -1/4*(sqrt( 2)*((9*A - 5*B)*cos(d*x + c)^2 + 2*(9*A - 5*B)*cos(d*x + c) + 9*A - 5*B)*s qrt(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))) - 4*((3*A - 2*B)*cos(d*x + c)^2 + 2*(3*A - 2*B)*c os(d*x + c) + 3*A - 2*B)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*(2*A*cos(d*x + c)^2 + (3*A - B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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)]
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (145) = 290\).
Time = 2.11 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.46 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (9 \, A - 5 \, B\right )} \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {4 \, {\left (3 \, A - 2 \, B\right )} \log \left (\frac {{\left | -17179869184 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 34359738368 \, \sqrt {2} {\left | a \right |} + 51539607552 \, a \right |}}{{\left | -17179869184 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 34359738368 \, \sqrt {2} {\left | a \right |} + 51539607552 \, a \right |}}\right )}{\sqrt {-a} {\left | a \right |} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {16 \, \sqrt {2} {\left (3 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} A - A a\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} \sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{8 \, d} \]
-1/8*(sqrt(2)*(9*A - 5*B)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan (1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*a*sgn(cos(d*x + c))) - 4*(3*A - 2*B )*log(abs(-17179869184*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d* x + 1/2*c)^2 + a))^2 - 34359738368*sqrt(2)*abs(a) + 51539607552*a)/abs(-17 179869184*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 34359738368*sqrt(2)*abs(a) + 51539607552*a))/(sqrt(-a)*abs(a)*sg n(cos(d*x + c))) - 2*(sqrt(2)*A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos( d*x + c)))*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3 - 16*sqrt(2)*(3*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c )^2 + a))^2*A - A*a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d* x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2* d*x + 1/2*c)^2 + a))^2*a + a^2)*sqrt(-a)*sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]