3.1209 \(\int \cos ^{\frac{13}{2}}(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=310 \[ \frac{8 a^4 (100 A+113 B+132 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{231 d}+\frac{8 a^4 (185 A+208 B+247 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{4 a^4 (5255 A+6019 B+6721 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15015 d}+\frac{2 (13 A+17 B+11 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{9009 d}+\frac{8 a^4 (100 A+113 B+132 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d}+\frac{2 a (8 A+13 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{143 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^4}{13 d} \]
[Out]
(8*a^4*(185*A + 208*B + 247*C)*EllipticE[(c + d*x)/2, 2])/(195*d) + (8*a^4*(100*A + 113*B + 132*C)*EllipticF[(
c + d*x)/2, 2])/(231*d) + (8*a^4*(100*A + 113*B + 132*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^4*(52
55*A + 6019*B + 6721*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15015*d) + (2*a*(8*A + 13*B)*Cos[c + d*x]^(3/2)*(a +
a*Cos[c + d*x])^3*Sin[c + d*x])/(143*d) + (2*A*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(13*d)
+ (2*(13*A + 17*B + 11*C)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*d) + (4*(1355*A + 1
612*B + 1573*C)*Cos[c + d*x]^(3/2)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(9009*d)
________________________________________________________________________________________
Rubi [A] time = 0.909167, antiderivative size = 310, normalized size of antiderivative = 1.,
number of steps used = 11, number of rules used = 9, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.209, Rules used
= {4112, 3045, 2976, 2968, 3023, 2748, 2639, 2635, 2641} \[ \frac{8 a^4 (100 A+113 B+132 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{8 a^4 (185 A+208 B+247 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{4 a^4 (5255 A+6019 B+6721 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15015 d}+\frac{2 (13 A+17 B+11 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{9009 d}+\frac{8 a^4 (100 A+113 B+132 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d}+\frac{2 a (8 A+13 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{143 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^4}{13 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(13/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(8*a^4*(185*A + 208*B + 247*C)*EllipticE[(c + d*x)/2, 2])/(195*d) + (8*a^4*(100*A + 113*B + 132*C)*EllipticF[(
c + d*x)/2, 2])/(231*d) + (8*a^4*(100*A + 113*B + 132*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^4*(52
55*A + 6019*B + 6721*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15015*d) + (2*a*(8*A + 13*B)*Cos[c + d*x]^(3/2)*(a +
a*Cos[c + d*x])^3*Sin[c + d*x])/(143*d) + (2*A*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(13*d)
+ (2*(13*A + 17*B + 11*C)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*d) + (4*(1355*A + 1
612*B + 1573*C)*Cos[c + d*x]^(3/2)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(9009*d)
Rule 4112
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] && !IntegerQ[n] && IntegerQ[m]
Rule 3045
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*
sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
Rule 2976
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rubi steps
\begin{align*} \int \cos ^{\frac{13}{2}}(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^4 \left (\frac{1}{2} a (3 A+13 C)+\frac{1}{2} a (8 A+13 B) \cos (c+d x)\right ) \, dx}{13 a}\\ &=\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{4 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3 \left (\frac{1}{4} a^2 (57 A+39 B+143 C)+\frac{13}{4} a^2 (13 A+17 B+11 C) \cos (c+d x)\right ) \, dx}{143 a}\\ &=\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{8 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac{3}{4} a^3 (170 A+169 B+286 C)+\frac{1}{4} a^3 (1355 A+1612 B+1573 C) \cos (c+d x)\right ) \, dx}{1287 a}\\ &=\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{9009 d}+\frac{16 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac{15}{8} a^4 (509 A+559 B+715 C)+\frac{3}{8} a^4 (5255 A+6019 B+6721 C) \cos (c+d x)\right ) \, dx}{9009 a}\\ &=\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{9009 d}+\frac{16 \int \sqrt{\cos (c+d x)} \left (\frac{15}{8} a^5 (509 A+559 B+715 C)+\left (\frac{15}{8} a^5 (509 A+559 B+715 C)+\frac{3}{8} a^5 (5255 A+6019 B+6721 C)\right ) \cos (c+d x)+\frac{3}{8} a^5 (5255 A+6019 B+6721 C) \cos ^2(c+d x)\right ) \, dx}{9009 a}\\ &=\frac{4 a^4 (5255 A+6019 B+6721 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d}+\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{9009 d}+\frac{32 \int \sqrt{\cos (c+d x)} \left (\frac{231}{8} a^5 (185 A+208 B+247 C)+\frac{585}{8} a^5 (100 A+113 B+132 C) \cos (c+d x)\right ) \, dx}{45045 a}\\ &=\frac{4 a^4 (5255 A+6019 B+6721 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d}+\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{9009 d}+\frac{1}{77} \left (4 a^4 (100 A+113 B+132 C)\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{195} \left (4 a^4 (185 A+208 B+247 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{8 a^4 (185 A+208 B+247 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{8 a^4 (100 A+113 B+132 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{4 a^4 (5255 A+6019 B+6721 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d}+\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{9009 d}+\frac{1}{231} \left (4 a^4 (100 A+113 B+132 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{8 a^4 (185 A+208 B+247 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{8 a^4 (100 A+113 B+132 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{8 a^4 (100 A+113 B+132 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{4 a^4 (5255 A+6019 B+6721 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15015 d}+\frac{2 a (8 A+13 B) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{143 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^4 \sin (c+d x)}{13 d}+\frac{2 (13 A+17 B+11 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 d}+\frac{4 (1355 A+1612 B+1573 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{9009 d}\\ \end{align*}
Mathematica [C] time = 6.51246, size = 1416, normalized size = 4.57 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(13/2)*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
a^4*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*(-((185*A + 208*B + 247*C)*Cot[c])/(390*d) +
((3764*A + 4087*B + 4488*C)*Cos[d*x]*Sin[c])/(14784*d) + ((15625*A + 15392*B + 13208*C)*Cos[2*d*x]*Sin[2*c])/
(149760*d) + ((404*A + 321*B + 176*C)*Cos[3*d*x]*Sin[3*c])/(9856*d) + ((98*A + 52*B + 13*C)*Cos[4*d*x]*Sin[4*c
])/(7488*d) + ((4*A + B)*Cos[5*d*x]*Sin[5*c])/(1408*d) + (A*Cos[6*d*x]*Sin[6*c])/(3328*d) + ((3764*A + 4087*B
+ 4488*C)*Cos[c]*Sin[d*x])/(14784*d) + ((15625*A + 15392*B + 13208*C)*Cos[2*c]*Sin[2*d*x])/(149760*d) + ((404*
A + 321*B + 176*C)*Cos[3*c]*Sin[3*d*x])/(9856*d) + ((98*A + 52*B + 13*C)*Cos[4*c]*Sin[4*d*x])/(7488*d) + ((4*A
+ B)*Cos[5*c]*Sin[5*d*x])/(1408*d) + (A*Cos[6*c]*Sin[6*d*x])/(3328*d)) - (50*A*(1 + Cos[c + d*x])^4*Csc[c]*Hy
pergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*Sec[d*x - ArcTan[Cot[c]]]
*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin
[d*x - ArcTan[Cot[c]]]])/(231*d*Sqrt[1 + Cot[c]^2]) - (113*B*(1 + Cos[c + d*x])^4*Csc[c]*HypergeometricPFQ[{1/
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x -
ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c
]]]])/(462*d*Sqrt[1 + Cot[c]^2]) - (2*C*(1 + Cos[c + d*x])^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d
*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^8*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqr
t[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*Sqrt[1 + C
ot[c]^2]) - (37*A*(1 + Cos[c + d*x])^4*Csc[c]*Sec[c/2 + (d*x)/2]^8*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Co
s[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos
[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[
d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(
Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(156*d) - (4*B*(1 + Cos[c +
d*x])^4*Csc[c]*Sec[c/2 + (d*x)/2]^8*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[
d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Co
s[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/S
qrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[
c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(15*d) - (19*C*(1 + Cos[c + d*x])^4*Csc[c]*Sec[c/2 + (d*x)/
2]^8*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(
Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*
Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2
*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sq
rt[1 + Tan[c]^2]]))/(60*d))
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Maple [A] time = 2.398, size = 576, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-8/45045*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(-110880*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^14+(594720*A+65520*B)*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)+(-1345120*A-323960*B-40040*C)*sin(1/2*d
*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(1667840*A+659620*B+183040*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-123749
0*A-713518*B-336622*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(572110*A+448448*B+322322*C)*sin(1/2*d*x+1/2*c)
^4*cos(1/2*d*x+1/2*c)+(-117945*A-110097*B-97383*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+19500*A*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-42735*A*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+22035*B*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-48048*B*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+25740*C*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-57057*C*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{4} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right )^{6} +{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right )^{5} +{\left (A + 4 \, B + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right )^{4} + 2 \,{\left (2 \, A + 3 \, B + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right )^{3} +{\left (6 \, A + 4 \, B + C\right )} a^{4} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right )^{2} +{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{6} \sec \left (d x + c\right ) + A a^{4} \cos \left (d x + c\right )^{6}\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*a^4*cos(d*x + c)^6*sec(d*x + c)^6 + (B + 4*C)*a^4*cos(d*x + c)^6*sec(d*x + c)^5 + (A + 4*B + 6*C)*
a^4*cos(d*x + c)^6*sec(d*x + c)^4 + 2*(2*A + 3*B + 2*C)*a^4*cos(d*x + c)^6*sec(d*x + c)^3 + (6*A + 4*B + C)*a^
4*cos(d*x + c)^6*sec(d*x + c)^2 + (4*A + B)*a^4*cos(d*x + c)^6*sec(d*x + c) + A*a^4*cos(d*x + c)^6)*sqrt(cos(d
*x + c)), x)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**(13/2)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{4} \cos \left (d x + c\right )^{\frac{13}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^4*cos(d*x + c)^(13/2), x)