Integrand size = 41, antiderivative size = 173 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {(9 A-5 B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-B+C) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \]
-(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+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/4*(9*A-5*B+C)*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+C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)
Time = 3.58 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\left (-4 (3 A-2 B) \arctan \left (\sqrt {-1+\sec (c+d x)}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}+\sqrt {2} (9 A-5 B+C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}+2 (3 A-B+C+2 A \cos (c+d x)) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{2 a d (-1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \]
Integrate[(Cos[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Se c[c + d*x])^(3/2),x]
-1/2*((-4*(3*A - 2*B)*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*S qrt[-1 + Sec[c + d*x]] + Sqrt[2]*(9*A - 5*B + C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] + 2*(3*A - B + C + 2*A*Cos[c + d*x])*Sin[(c + d*x)/2]^2)*Tan[(c + d*x)/2])/(a*d*(-1 + Cos[ c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.03 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4572, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (3 A-B+C)-a (3 A-3 B-C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (3 A-B+C)-a (3 A-3 B-C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (3 A-B+C)-a (3 A-3 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\int -\frac {2 a^2 (3 A-2 B)-a^2 (3 A-B+C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {2 a^2 (3 A-2 B)-a^2 (3 A-B+C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {2 a^2 (3 A-2 B)-a^2 (3 A-B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \sqrt {\sec (c+d x) a+a}dx-a^2 (9 A-5 B+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 a (3 A-2 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^2 (9 A-5 B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^2 (9 A-5 B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {4 a^2 (3 A-2 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^2 (9 A-5 B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (9 A-5 B+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {4 a^{3/2} (3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {2 a (3 A-B+C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^{3/2} (3 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (9 A-5 B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
-1/2*((A - B + C)*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 + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[ 2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a) + (2*a*(3*A - B + C)*Sin[c + d*x])/(d *Sqrt[a + a*Sec[c + d*x]]))/(4*a^2)
3.6.24.3.1 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*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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + 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*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(1025\) vs. \(2(148)=296\).
Time = 2.75 (sec) , antiderivative size = 1026, normalized size of antiderivative = 5.93
int(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,me thod=_RETURNVERBOSE)
-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)*((1 -cos(d*x+c))^2*csc(d*x+c)^2-1)^(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 +A*(1-cos(d*x+c))^5*csc(d*x+c)^5-4*B*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)*2^(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+C*(1-cos(d*x+c))^5*csc(d*x+c)^5-9*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)) *(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-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos( d*x+c))^2*csc(d*x+c)^2-C*((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))*(1-cos(d*x+c))^2* csc(d*x+c)^2+6*A*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)))*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)+4 *A*(1-cos(d*x+c))^3*csc(d*x+c)^3-4*B*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)*2^(1/2)-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)+5*B*ln(csc(d*x+c)-c ot(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)-C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c...
Time = 12.32 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.57 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right ) + 9 \, A - 5 \, B + C\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 + C\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 + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, A - 5 \, B + C\right )} \cos \left (d x + c\right ) + 9 \, A - 5 \, B + C\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 + C\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 ] \]
integrate(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="fricas")
[-1/8*(sqrt(2)*((9*A - 5*B + C)*cos(d*x + c)^2 + 2*(9*A - 5*B + C)*cos(d*x + c) + 9*A - 5*B + C)*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*c os(d*x + c)^2 + (3*A - B + C)*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 + C)*cos(d*x + c)^2 + 2*(9*A - 5*B + C)*cos (d*x + c) + 9*A - 5*B + C)*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))) - 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)*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 + C)*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)]
\[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)/(a*(sec(c + d*x) + 1))**(3/2), x)
\[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)/(a*sec(d*x + c) + a)^(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (148) = 296\).
Time = 2.47 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.50 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (9 \, A - 5 \, B + C\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 ) + \sqrt {2} C 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} \]
integrate(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x, algorithm="giac")
-1/8*(sqrt(2)*(9*A - 5*B + C)*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 (-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))/(sqrt(-a)*abs(a )*sgn(cos(d*x + c))) - 2*(sqrt(2)*A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn( cos(d*x + c)) + sqrt(2)*C*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. \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]