Integrand size = 45, antiderivative size = 233 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A-7 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(19 A-15 B+3 C) \sin (c+d x)}{6 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-3 B+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}} \]
-1/2*(A-B+C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(3/2)+1/4*(11* A-7*B+3*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*se c(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)-1/6*( 19*A-15*B+3*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/6* (7*A-3*B+3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(1/2)
Time = 5.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.48 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {3 (11 A-7 B+3 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )+(-17 A+15 B-3 C-12 (A-B) \cos (c+d x)+2 A \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{6 a d \sqrt {\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]
(3*(11*A - 7*B + 3*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + (-17*A + 15*B - 3*C - 12*(A - B)*Cos[c + d*x] + 2*A*Cos[2*(c + d*x)])*Tan[(c + d* x)/2])/(6*a*d*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.36 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4753, 3042, 4572, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}
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.
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^{3/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{(a \sec (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sec ^2(c+d x)+B \sec (c+d x)+A}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-3 B+3 C)-4 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-3 B+3 C)-4 a (A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-3 B+3 C)-4 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {a^2 (19 A-15 B+3 C)-2 a^2 (7 A-3 B+3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A-15 B+3 C)-2 a^2 (7 A-3 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A-15 B+3 C)-2 a^2 (7 A-3 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A-7 B+3 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A-7 B+3 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a^2 (11 A-7 B+3 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (19 A-15 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (7 A-3 B+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A-15 B+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} a^{3/2} (11 A-7 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\right )\) |
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Se c[c + d*x])^(3/2),x]
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B + C)*Sin[c + d*x])/(d* Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((2*a*(7*A - 3*B + 3*C)*S in[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-3*Sqrt [2]*a^(3/2)*(11*A - 7*B + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^2*(19*A - 15*B + 3*C) *Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4* a^2))
3.13.81.3.1 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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; FreeQ[{a, b, d, e, f}, x] && 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[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -1]
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)]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(481\) vs. \(2(198)=396\).
Time = 0.87 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.07
| method | result | size |
| default | \(\frac {\left (8 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-24 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-33 A \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+24 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+21 B \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-9 C \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-38 A \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-33 A \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+30 B \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+21 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}-6 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-9 C \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{12 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\) | \(482\) |
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x,method=_RETURNVERBOSE)
1/12/a^2/d*(8*A*sin(d*x+c)*cos(d*x+c)^2*(-1/(1+cos(d*x+c)))^(1/2)-24*A*cos (d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)-33*A*cos(d*x+c)*2^(1/2)*arcta n(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+24*B*co s(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+21*B*2^(1/2)*cos(d*x+c)*arct an(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-9*C*co s(d*x+c)*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d *x+c)))^(1/2))-38*A*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)-33*A*2^(1/2)*arct an(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+30*B*s in(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)+21*B*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+ cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*2^(1/2)-6*C*sin(d*x+c)*(-1/(1+cos(d *x+c)))^(1/2)-9*C*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1 /(1+cos(d*x+c)))^(1/2)))*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos( d*x+c))^2/(-1/(1+cos(d*x+c)))^(1/2)
Time = 0.30 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A - 7 \, B + 3 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, A \cos \left (d x + c\right )^{2} - 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 19 \, A + 15 \, B - 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A - 7 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A - 7 \, B + 3 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - 2 \, {\left (4 \, A \cos \left (d x + c\right )^{2} - 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 19 \, A + 15 \, B - 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^(3/2),x, algorithm="fricas")
[1/24*(3*sqrt(2)*((11*A - 7*B + 3*C)*cos(d*x + c)^2 + 2*(11*A - 7*B + 3*C) *cos(d*x + c) + 11*A - 7*B + 3*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt( 2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin( d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(4*A*cos(d*x + c)^2 - 12*(A - B)*cos(d*x + c) - 19*A + 15*B - 3*C)*sqr t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(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), -1/12*(3*sqrt(2)*((11*A - 7*B + 3*C)*cos(d*x + c)^2 + 2*(11*A - 7*B + 3*C)*cos(d*x + c) + 11*A - 7*B + 3*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos( d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) - 2*(4*A*cos(d*x + c)^2 - 1 2*(A - B)*cos(d*x + c) - 19*A + 15*B - 3*C)*sqrt((a*cos(d*x + c) + a)/cos( d*x + c))*sqrt(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)]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 37260 vs. \(2 (198) = 396\).
Time = 0.97 (sec) , antiderivative size = 37260, normalized size of antiderivative = 159.91 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^(3/2),x, algorithm="maxima")
1/12*(3*(8*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2*sin(1/4* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 32*cos(3/4*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c)))^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 8*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2*sin(1 /4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 32*sin(3/4*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c)))^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2* d*x + 2*c))) + 8*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^3 + 4*(8*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2( sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - sin(2*d*x + 2*c) - 5*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) )*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 8*(4*cos(1/4*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - sin(2*d*x + 2*c) - 5*sin(1/2*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c))))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) ) - 7*(2*(2*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/4 *arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*co s(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*cos(1/4*arctan2(sin(...
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^(3/2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*se c(d*x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/co s(c + d*x))^(3/2),x)