Integrand size = 43, antiderivative size = 215 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (163 A+200 B+304 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (299 A+392 B+432 C) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {a (5 A+8 B) \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d} \]
1/64*a^(5/2)*(163*A+200*B+304*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c) )^(1/2))/d+1/24*a*(5*A+8*B)*cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c) /d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/192*a^3*(299*A +392*B+432*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/32*a^2*(17*A+24*B+16*C )*cos(d*x+c)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.03 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.27 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left (6075 A \sqrt {1-\sec (c+d x)} \sin (c+d x)+6600 B \sqrt {1-\sec (c+d x)} \sin (c+d x)-720 C \sqrt {1-\sec (c+d x)} \sin (c+d x)+2088 A \cos ^2(c+d x) \sqrt {1-\sec (c+d x)} \sin (c+d x)-320 B \cos ^2(c+d x) \sqrt {1-\sec (c+d x)} \sin (c+d x)+144 A \cos ^3(c+d x) \sqrt {1-\sec (c+d x)} \sin (c+d x)+2025 A \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))+1240 B \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))-2160 C \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))+6075 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+6600 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+5040 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+23040 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)+7680 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)+4608 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan (c+d x)\right )}{2880 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
(a^2*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6075*A*Sqrt[1 - Sec[c + d*x] ]*Sin[c + d*x] + 6600*B*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] - 720*C*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 2088*A*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x ]]*Sin[c + d*x] - 320*B*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 144*A*Cos[c + d*x]^3*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 2025*A*Sqrt[ 1 - Sec[c + d*x]]*Sin[2*(c + d*x)] + 1240*B*Sqrt[1 - Sec[c + d*x]]*Sin[2*( c + d*x)] - 2160*C*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)] + 6075*A*ArcTan h[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 6600*B*ArcTanh[Sqrt[1 - Sec[c + d *x]]]*Tan[c + d*x] + 5040*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 23040*C*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x] + 7680*B*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d* x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x] + 4608*A*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]]*Tan[c + d*x]))/(2880*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
Time = 1.40 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4574, 27, 3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \frac {1}{2} \cos ^3(c+d x) (\sec (c+d x) a+a)^{5/2} (a (5 A+8 B)+a (A+8 C) \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) (\sec (c+d x) a+a)^{5/2} (a (5 A+8 B)+a (A+8 C) \sec (c+d x))dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+8 B)+a (A+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {1}{2} \cos ^2(c+d x) (\sec (c+d x) a+a)^{3/2} \left (3 (17 A+24 B+16 C) a^2+(11 A+8 B+48 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \int \cos ^2(c+d x) (\sec (c+d x) a+a)^{3/2} \left (3 (17 A+24 B+16 C) a^2+(11 A+8 B+48 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (17 A+24 B+16 C) a^2+(11 A+8 B+48 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((299 A+392 B+432 C) a^3+(95 A+104 B+240 C) \sec (c+d x) a^3\right )dx+\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a} \left ((299 A+392 B+432 C) a^3+(95 A+104 B+240 C) \sec (c+d x) a^3\right )dx+\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((299 A+392 B+432 C) a^3+(95 A+104 B+240 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (163 A+200 B+304 C) \int \sqrt {\sec (c+d x) a+a}dx+\frac {a^4 (299 A+392 B+432 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (163 A+200 B+304 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^4 (299 A+392 B+432 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {a^4 (299 A+392 B+432 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 a^4 (163 A+200 B+304 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {a^2 (5 A+8 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}+\frac {1}{6} \left (\frac {3 a^3 (17 A+24 B+16 C) \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{7/2} (163 A+200 B+304 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^4 (299 A+392 B+432 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{8 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\) |
(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d) + ((a^2*( 5*A + 8*B)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ((3*a^3*(17*A + 24*B + 16*C)*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((3*a^(7/2)*(163*A + 200*B + 304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^4*(299*A + 392*B + 432*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/4)/6)/(8*a)
3.6.7.3.1 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_)]*(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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), 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 - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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. \(562\) vs. \(2(191)=382\).
Time = 1.03 (sec) , antiderivative size = 563, normalized size of antiderivative = 2.62
\[\frac {a^{2} \left (48 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+184 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+64 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+489 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 )+326 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+600 B \,\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 ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+272 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+912 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 )+96 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+489 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 )+489 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+600 B \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 )+600 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+912 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 )+528 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 )}}{192 d \left (\cos \left (d x +c \right )+1\right )}\]
1/192*a^2/d*(48*A*cos(d*x+c)^4*sin(d*x+c)+184*A*cos(d*x+c)^3*sin(d*x+c)+64 *B*sin(d*x+c)*cos(d*x+c)^3+489*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan h(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c) +326*A*cos(d*x+c)^2*sin(d*x+c)+600*B*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)/(cos(d*x+c)+1))^(1/2)*cos(d* x+c)+272*B*sin(d*x+c)*cos(d*x+c)^2+912*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))*co s(d*x+c)+96*C*cos(d*x+c)^2*sin(d*x+c)+489*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)) +489*A*cos(d*x+c)*sin(d*x+c)+600*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct anh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+600*B*co s(d*x+c)*sin(d*x+c)+912*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))+528*C*cos(d*x+c)* sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.44 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.07 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 200 \, B + 176 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A + 200 \, B + 176 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="fricas")
[1/384*(3*((163*A + 200*B + 304*C)*a^2*cos(d*x + c) + (163*A + 200*B + 304 *C)*a^2)*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*(48*A*a^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B + 48*C)*a^2*cos(d*x + c)^2 + 3*(163*A + 200*B + 1 76*C)*a^2*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), -1/192*(3*((163*A + 200*B + 304*C)*a^2*cos(d*x + c) + (163*A + 200*B + 304*C)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a )/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a^2*cos(d*x + c)^4 + 8*(23*A + 8*B)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B + 48*C)*a^2*c os(d*x + c)^2 + 3*(163*A + 200*B + 176*C)*a^2*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. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="maxima")
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \,d x } \]
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c) ^2),x, algorithm="giac")
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]