3.1194 \(\int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=251 \[ \frac {4 a^2 (50 A+55 B+66 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^2 (7 A+8 B+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^2 (89 A+121 B+99 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{693 d}+\frac {4 a^2 (7 A+8 B+9 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {4 a^2 (50 A+55 B+66 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 (4 A+11 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{99 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}{11 d} \]

[Out]

4/15*a^2*(7*A+8*B+9*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d
+4/231*a^2*(50*A+55*B+66*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/
2))/d+4/45*a^2*(7*A+8*B+9*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/693*a^2*(89*A+121*B+99*C)*cos(d*x+c)^(5/2)*sin(d*
x+c)/d+2/11*A*cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+2/99*(4*A+11*B)*cos(d*x+c)^(5/2)*(a^2+a^2*cos(d
*x+c))*sin(d*x+c)/d+4/231*a^2*(50*A+55*B+66*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.60, antiderivative size = 251, normalized size of antiderivative = 1.00, 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, 2635, 2641, 2639} \[ \frac {4 a^2 (50 A+55 B+66 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^2 (7 A+8 B+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^2 (89 A+121 B+99 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{693 d}+\frac {4 a^2 (7 A+8 B+9 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {4 a^2 (50 A+55 B+66 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 (4 A+11 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{99 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(4*a^2*(7*A + 8*B + 9*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(50*A + 55*B + 66*C)*EllipticF[(c + d*x)/2
, 2])/(231*d) + (4*a^2*(50*A + 55*B + 66*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^2*(7*A + 8*B + 9*C
)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (2*a^2*(89*A + 121*B + 99*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(693
*d) + (2*A*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(11*d) + (2*(4*A + 11*B)*Cos[c + d*x]^(5/2)
*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(99*d)

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 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 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]

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \cos ^{\frac {11}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 \left (\frac {1}{2} a (5 A+11 C)+\frac {1}{2} a (4 A+11 B) \cos (c+d x)\right ) \, dx}{11 a}\\ &=\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (4 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {4 \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (65 A+55 B+99 C)+\frac {1}{4} a^2 (89 A+121 B+99 C) \cos (c+d x)\right ) \, dx}{99 a}\\ &=\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (4 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {4 \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {1}{4} a^3 (65 A+55 B+99 C)+\left (\frac {1}{4} a^3 (65 A+55 B+99 C)+\frac {1}{4} a^3 (89 A+121 B+99 C)\right ) \cos (c+d x)+\frac {1}{4} a^3 (89 A+121 B+99 C) \cos ^2(c+d x)\right ) \, dx}{99 a}\\ &=\frac {2 a^2 (89 A+121 B+99 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (4 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {8 \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {9}{4} a^3 (50 A+55 B+66 C)+\frac {77}{4} a^3 (7 A+8 B+9 C) \cos (c+d x)\right ) \, dx}{693 a}\\ &=\frac {2 a^2 (89 A+121 B+99 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (4 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {1}{9} \left (2 a^2 (7 A+8 B+9 C)\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{77} \left (2 a^2 (50 A+55 B+66 C)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {4 a^2 (50 A+55 B+66 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^2 (7 A+8 B+9 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a^2 (89 A+121 B+99 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (4 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {1}{15} \left (2 a^2 (7 A+8 B+9 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{231} \left (2 a^2 (50 A+55 B+66 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^2 (7 A+8 B+9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (50 A+55 B+66 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^2 (50 A+55 B+66 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^2 (7 A+8 B+9 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a^2 (89 A+121 B+99 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (4 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{99 d}\\ \end {align*}

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Mathematica [C]  time = 6.41, size = 1364, normalized size = 5.43 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(11/2)*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a^2*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/15*((7*A + 8*B + 9*C)*Cot[c])/d + ((941*
A + 1012*B + 1122*C)*Cos[d*x]*Sin[c])/(3696*d) + ((38*A + 37*B + 36*C)*Cos[2*d*x]*Sin[2*c])/(360*d) + ((101*A
+ 88*B + 44*C)*Cos[3*d*x]*Sin[3*c])/(2464*d) + ((2*A + B)*Cos[4*d*x]*Sin[4*c])/(144*d) + (A*Cos[5*d*x]*Sin[5*c
])/(352*d) + ((941*A + 1012*B + 1122*C)*Cos[c]*Sin[d*x])/(3696*d) + ((38*A + 37*B + 36*C)*Cos[2*c]*Sin[2*d*x])
/(360*d) + ((101*A + 88*B + 44*C)*Cos[3*c]*Sin[3*d*x])/(2464*d) + ((2*A + B)*Cos[4*c]*Sin[4*d*x])/(144*d) + (A
*Cos[5*c]*Sin[5*d*x])/(352*d)) - (50*A*(1 + Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*
x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*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]) - (5*B*(1 + Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2
]*Sec[c/2 + (d*x)/2]^4*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]]]])/(21*d*Sqrt[1 + Cot[c]^2]) - (2*C*(1 +
Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4
*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 - ArcT
an[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*Sqrt[1 + Cot[c]^2]) - (7*A*(1 + Cos[c + d*x])^2*Csc[c]
*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[T
an[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 + A
rcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(30*d) - (4*B*(1 + Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((Hypergeo
metricPFQ[{-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 + ArcT
an[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) - (3*C*(1 + Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, C
os[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Co
s[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]]))/(10*d))

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{4} + {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{3} + {\left (A + 2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{2} + {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right ) + A a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*a^2*cos(d*x + c)^5*sec(d*x + c)^4 + (B + 2*C)*a^2*cos(d*x + c)^5*sec(d*x + c)^3 + (A + 2*B + C)*a^
2*cos(d*x + c)^5*sec(d*x + c)^2 + (2*A + B)*a^2*cos(d*x + c)^5*sec(d*x + c) + A*a^2*cos(d*x + c)^5)*sqrt(cos(d
*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^2*(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)^2*cos(d*x + c)^(11/2), x)

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maple [A]  time = 5.85, size = 545, normalized size = 2.17 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{2} \left (10080 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-37520 A -6160 B \right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (57040 A +20240 B +3960 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-46192 A -26048 B -11484 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (22022 A +17248 B +12474 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-4563 A -4257 B -3861 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+750 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1617 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+825 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1848 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+990 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2079 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3465 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^2*(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^2*(10080*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-37520*A-6160*B)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(57040*A+20240*B+3960*C)*sin(1/2*d*x+1/2*c)
^8*cos(1/2*d*x+1/2*c)+(-46192*A-26048*B-11484*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(22022*A+17248*B+1247
4*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-4563*A-4257*B-3861*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+7
50*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))-1617*
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))+825*B*(s
in(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))-1848*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))+990*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))-2079*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 5.97, size = 404, normalized size = 1.61 \[ \frac {2\,C\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,A\,a^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,B\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,C\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(11/2)*(a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*C*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) - (2*A*a^2*cos(c + d*x)^(9/2)*
sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (4*A*a^2*cos(c + d*x)
^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (2*A*a^2*co
s(c + d*x)^(13/2)*sin(c + d*x)*hypergeom([1/2, 13/4], 17/4, cos(c + d*x)^2))/(13*d*(sin(c + d*x)^2)^(1/2)) - (
2*B*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/
2)) - (4*B*a^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)
^2)^(1/2)) - (2*B*a^2*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(si
n(c + d*x)^2)^(1/2)) - (4*C*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(
7*d*(sin(c + d*x)^2)^(1/2)) - (2*C*a^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x
)^2))/(9*d*(sin(c + d*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(11/2)*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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