3.1202 \(\int \cos ^{\frac{11}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=267 \[ \frac{4 a^3 (105 A+121 B+143 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{231 d}+\frac{4 a^3 (15 A+17 B+21 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{1155 d}+\frac{2 (105 A+143 B+99 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{693 d}+\frac{4 a^3 (105 A+121 B+143 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d}+\frac{2 (6 A+11 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 a d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{11 d} \]
[Out]
(4*a^3*(15*A + 17*B + 21*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(105*A + 121*B + 143*C)*EllipticF[(c +
d*x)/2, 2])/(231*d) + (4*a^3*(105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^3*(210*A
+ 253*B + 264*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d) + (2*A*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(11*d) + (2*(6*A + 11*B)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*a*d) + (2
*(105*A + 143*B + 99*C)*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(693*d)
________________________________________________________________________________________
Rubi [A] time = 0.731935, antiderivative size = 267, normalized size of antiderivative = 1.,
number of steps used = 10, 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} \[ \frac{4 a^3 (105 A+121 B+143 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (15 A+17 B+21 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^3 (210 A+253 B+264 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{1155 d}+\frac{2 (105 A+143 B+99 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{693 d}+\frac{4 a^3 (105 A+121 B+143 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d}+\frac{2 (6 A+11 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{99 a d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^3}{11 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(4*a^3*(15*A + 17*B + 21*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(105*A + 121*B + 143*C)*EllipticF[(c +
d*x)/2, 2])/(231*d) + (4*a^3*(105*A + 121*B + 143*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^3*(210*A
+ 253*B + 264*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1155*d) + (2*A*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3*Si
n[c + d*x])/(11*d) + (2*(6*A + 11*B)*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(99*a*d) + (2
*(105*A + 143*B + 99*C)*Cos[c + d*x]^(3/2)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(693*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{11}{2}}(c+d x) (a+a \sec (c+d x))^3 \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))^3 \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))^3 \sin (c+d x)}{11 d}+\frac{2 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3 \left (\frac{1}{2} a (3 A+11 C)+\frac{1}{2} a (6 A+11 B) \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{4 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \left (\frac{3}{4} a^2 (15 A+11 B+33 C)+\frac{1}{4} a^2 (105 A+143 B+99 C) \cos (c+d x)\right ) \, dx}{99 a}\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (105 A+143 B+99 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{8 \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x)) \left (\frac{15}{4} a^3 (21 A+22 B+33 C)+\frac{3}{4} a^3 (210 A+253 B+264 C) \cos (c+d x)\right ) \, dx}{693 a}\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (105 A+143 B+99 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{8 \int \sqrt{\cos (c+d x)} \left (\frac{15}{4} a^4 (21 A+22 B+33 C)+\left (\frac{15}{4} a^4 (21 A+22 B+33 C)+\frac{3}{4} a^4 (210 A+253 B+264 C)\right ) \cos (c+d x)+\frac{3}{4} a^4 (210 A+253 B+264 C) \cos ^2(c+d x)\right ) \, dx}{693 a}\\ &=\frac{4 a^3 (210 A+253 B+264 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (105 A+143 B+99 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{16 \int \sqrt{\cos (c+d x)} \left (\frac{231}{8} a^4 (15 A+17 B+21 C)+\frac{45}{8} a^4 (105 A+121 B+143 C) \cos (c+d x)\right ) \, dx}{3465 a}\\ &=\frac{4 a^3 (210 A+253 B+264 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (105 A+143 B+99 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{1}{15} \left (2 a^3 (15 A+17 B+21 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{77} \left (2 a^3 (105 A+121 B+143 C)\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{4 a^3 (15 A+17 B+21 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^3 (105 A+121 B+143 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{4 a^3 (210 A+253 B+264 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (105 A+143 B+99 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}+\frac{1}{231} \left (2 a^3 (105 A+121 B+143 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (15 A+17 B+21 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^3 (105 A+121 B+143 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (105 A+121 B+143 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{4 a^3 (210 A+253 B+264 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{1155 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d}+\frac{2 (6 A+11 B) \cos ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{99 a d}+\frac{2 (105 A+143 B+99 C) \cos ^{\frac{3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{693 d}\\ \end{align*}
Mathematica [C] time = 6.46387, size = 1364, normalized size = 5.11 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
a^3*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-((15*A + 17*B + 21*C)*Cot[c])/(30*d) + ((1
953*A + 2134*B + 2354*C)*Cos[d*x]*Sin[c])/(7392*d) + ((75*A + 73*B + 54*C)*Cos[2*d*x]*Sin[2*c])/(720*d) + ((18
9*A + 132*B + 44*C)*Cos[3*d*x]*Sin[3*c])/(4928*d) + ((3*A + B)*Cos[4*d*x]*Sin[4*c])/(288*d) + (A*Cos[5*d*x]*Si
n[5*c])/(704*d) + ((1953*A + 2134*B + 2354*C)*Cos[c]*Sin[d*x])/(7392*d) + ((75*A + 73*B + 54*C)*Cos[2*c]*Sin[2
*d*x])/(720*d) + ((189*A + 132*B + 44*C)*Cos[3*c]*Sin[3*d*x])/(4928*d) + ((3*A + B)*Cos[4*c]*Sin[4*d*x])/(288*
d) + (A*Cos[5*c]*Sin[5*d*x])/(704*d)) - (5*A*(1 + Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4},
Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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]]]])/(22*d*Sqrt
[1 + Cot[c]^2]) - (11*B*(1 + Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[
c]]]^2]*Sec[c/2 + (d*x)/2]^6*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]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (13
*C*(1 + Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*
x)/2]^6*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]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (A*(1 + Cos[c + d*x])^3*
Csc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + Ar
cTan[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]]))/(4*d) - (17*B*(1 + Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((Hy
pergeometricPFQ[{-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]]]*Sqrt[1 + Ta
n[c]^2]]))/(60*d) - (7*C*(1 + Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((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]]]*Sqrt[1 + Tan[c]^2]]))/(20*d))
________________________________________________________________________________________
Maple [A] time = 2.18, size = 545, normalized size = 2. \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)^(11/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-4/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(10080*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-43680*A-6160*B)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(77280*A+24200*B+3960*C)*sin(1/2*d*x+1/2*c)
^8*cos(1/2*d*x+1/2*c)+(-72240*A-37532*B-14256*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(39270*A+29722*B+1986
6*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-8820*A-8118*B-6864*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1
575*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))-3465
*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))+1815*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))-3927*B*(si
n(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))+2145*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))-4851*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
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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)^(11/2)*(a+a*sec(d*x+c))^3*(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^{3} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{5} +{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{3} +{\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right ) + A a^{3} \cos \left (d x + c\right )^{5}\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)^(11/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*a^3*cos(d*x + c)^5*sec(d*x + c)^5 + (B + 3*C)*a^3*cos(d*x + c)^5*sec(d*x + c)^4 + (A + 3*B + 3*C)*
a^3*cos(d*x + c)^5*sec(d*x + c)^3 + (3*A + 3*B + C)*a^3*cos(d*x + c)^5*sec(d*x + c)^2 + (3*A + B)*a^3*cos(d*x
+ c)^5*sec(d*x + c) + A*a^3*cos(d*x + c)^5)*sqrt(cos(d*x + c)), x)
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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)**(11/2)*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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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 )}^{3} \cos \left (d x + c\right )^{\frac{11}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^3*(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)^3*cos(d*x + c)^(11/2), x)