Integrand size = 35, antiderivative size = 200 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (75 A+112 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 A+16 C) \cos (c+d x) \sin (c+d x)}{32 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d} \]
1/64*a^(3/2)*(75*A+112*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2) )/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/64*a^2*(75*A+ 112*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/32*a^2*(13*A+16*C)*cos(d*x+c) *sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/8*a*A*cos(d*x+c)^2*sin(d*x+c)*(a+a* sec(d*x+c))^(1/2)/d
Time = 2.78 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.73 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left ((95 A+112 C+(62 A+32 C) \cos (c+d x)+20 A \cos (2 (c+d x))+4 A \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)} \sin (c+d x)+(75 A+112 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)\right )}{64 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
(a*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*((95*A + 112*C + (62*A + 32*C)* Cos[c + d*x] + 20*A*Cos[2*(c + d*x)] + 4*A*Cos[3*(c + d*x)])*Sqrt[1 - Sec[ c + d*x]]*Sin[c + d*x] + (75*A + 112*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Ta n[c + d*x]))/(64*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
Time = 1.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4575, 27, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4261, 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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \frac {1}{2} \cos ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (3 a A+a (3 A+8 C) \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) (\sec (c+d x) a+a)^{3/2} (3 a A+a (3 A+8 C) \sec (c+d x))dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a A+a (3 A+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {3}{2} \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \sec (c+d x) a^2\right )dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \sec (c+d x) a^2\right )dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {a^2 A \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {a^3 (13 A+16 C) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{4} a^2 (75 A+112 C) \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ((a^2*A *Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/d + ((a^3*(13*A + 1 6*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(75* A + 112*C)*((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x] ]])/d + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4)/2)/(8*a)
3.2.71.3.1 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_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[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*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[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 *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(383\) vs. \(2(176)=352\).
Time = 0.53 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.92
| method | result | size |
| default | \(\frac {a \left (16 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+40 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+50 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+112 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+32 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+75 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+112 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+112 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{64 d \left (\cos \left (d x +c \right )+1\right )}\) | \(384\) |
1/64*a/d*(16*A*cos(d*x+c)^4*sin(d*x+c)+40*A*cos(d*x+c)^3*sin(d*x+c)+75*A*( -cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos( d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+50*A*cos(d*x+c)^2*sin(d*x+c)+112* C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-c os(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+32*C*cos(d*x+c)^2*sin(d*x+c)+7 5*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/( -cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+75*A*cos(d*x+c)*sin(d*x+c)+112*C*(-cos( d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c )/(cos(d*x+c)+1))^(1/2))+112*C*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^( 1/2)/(cos(d*x+c)+1)
Time = 0.32 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.90 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 112 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (16 \, A a \cos \left (d x + c\right )^{4} + 40 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{128 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 112 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (16 \, A a \cos \left (d x + c\right )^{4} + 40 \, A a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/128*(((75*A + 112*C)*a*cos(d*x + c) + (75*A + 112*C)*a)*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)) + 2*(16*A *a*cos(d*x + c)^4 + 40*A*a*cos(d*x + c)^3 + 2*(25*A + 16*C)*a*cos(d*x + c) ^2 + (75*A + 112*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c) )*sin(d*x + c))/(d*cos(d*x + c) + d), -1/64*(((75*A + 112*C)*a*cos(d*x + c ) + (75*A + 112*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))) - (16*A*a*cos(d*x + c)^4 + 40*A*a* cos(d*x + c)^3 + 2*(25*A + 16*C)*a*cos(d*x + c)^2 + (75*A + 112*C)*a*cos(d *x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^4\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]