3.1205 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=226 \[ \frac{4 a^3 (3 A+5 (B+C)) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{4 a^3 (9 A+5 B-5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (6 A-5 B-20 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 (3 A-15 B-35 C) \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac{2 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{a d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
[Out]
(4*a^3*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2]
)/(3*d) + (4*a^3*(6*A - 5*B - 20*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*C*(a + a*Cos[c + d*x])^3*Sin[
c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(B + 2*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d
*x]]) + (2*(3*A - 15*B - 35*C)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d)
________________________________________________________________________________________
Rubi [A] time = 0.733771, antiderivative size = 226, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 9, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.209, Rules used =
{4112, 3043, 2975, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac{4 a^3 (3 A+5 (B+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 (9 A+5 B-5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (6 A-5 B-20 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 d}+\frac{2 (3 A-15 B-35 C) \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac{2 (B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{a d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(4*a^3*(9*A + 5*B - 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(3*A + 5*(B + C))*EllipticF[(c + d*x)/2, 2]
)/(3*d) + (4*a^3*(6*A - 5*B - 20*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*C*(a + a*Cos[c + d*x])^3*Sin[
c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(B + 2*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d
*x]]) + (2*(3*A - 15*B - 35*C)*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*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 3043
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Rule 2975
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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 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 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 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]
Rubi steps
\begin{align*} \int \cos ^{\frac{5}{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 \frac{(a+a \cos (c+d x))^3 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^3 \left (\frac{3}{2} a (B+2 C)+\frac{1}{2} a (3 A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{4 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{4} a^2 (3 A+15 B+25 C)+\frac{1}{4} a^2 (3 A-15 B-35 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{3 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 (3 A-15 B-35 C) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{8 \int \frac{(a+a \cos (c+d x)) \left (\frac{3}{4} a^3 (3 A+10 B+15 C)+\frac{3}{4} a^3 (6 A-5 B-20 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 (3 A-15 B-35 C) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{8 \int \frac{\frac{3}{4} a^4 (3 A+10 B+15 C)+\left (\frac{3}{4} a^4 (6 A-5 B-20 C)+\frac{3}{4} a^4 (3 A+10 B+15 C)\right ) \cos (c+d x)+\frac{3}{4} a^4 (6 A-5 B-20 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{4 a^3 (6 A-5 B-20 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 (3 A-15 B-35 C) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{16 \int \frac{\frac{15}{8} a^4 (3 A+5 (B+C))+\frac{9}{8} a^4 (9 A+5 B-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{45 a}\\ &=\frac{4 a^3 (6 A-5 B-20 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 (3 A-15 B-35 C) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{5} \left (2 a^3 (9 A+5 B-5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^3 (3 A+5 (B+C))\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^3 (9 A+5 B-5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (3 A+5 (B+C)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 (6 A-5 B-20 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (B+2 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt{\cos (c+d x)}}+\frac{2 (3 A-15 B-35 C) \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}\\ \end{align*}
Mathematica [C] time = 6.81363, size = 1672, normalized size = 7.4 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(Cos[c + d*x]^(11/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-((1
8*A + 5*B - 25*C + 18*A*Cos[2*c] + 15*B*Cos[2*c] + 5*C*Cos[2*c])*Csc[c]*Sec[c])/(20*d) + ((3*A + B)*Cos[d*x]*S
in[c])/(6*d) + (A*Cos[2*d*x]*Sin[2*c])/(20*d) + ((3*A + B)*Cos[c]*Sin[d*x])/(6*d) + (C*Sec[c]*Sec[c + d*x]^2*S
in[d*x])/(6*d) + (Sec[c]*Sec[c + d*x]*(C*Sin[c] + 3*B*Sin[d*x] + 9*C*Sin[d*x]))/(6*d) + (A*Cos[2*c]*Sin[2*d*x]
)/(20*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]) - (A*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4
, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] +
C*Sec[c + d*x]^2)*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]]]])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c
+ 2*d*x])*Sqrt[1 + Cot[c]^2]) - (5*B*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Arc
Tan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*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]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]
^2]) - (5*C*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 +
(d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcT
an[Cot[c]]]])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (9*A*Cos[c + d*x]^5
*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPF
Q[{-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 + ArcT
an[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqr
t[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*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) - (B*Cos[c + d*x]^5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*S
ec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcT
an[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTa
n[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 + S
in[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos
[2*c + 2*d*x])) + (C*Cos[c + d*x]^5*Csc[c]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C
*Sec[c + d*x]^2)*((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 + ArcT
an[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]]))/(2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))
________________________________________________________________________________________
Maple [B] time = 7.306, size = 950, normalized size = 4.2 \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)^(5/2)*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c
)^2+1)/sin(1/2*d*x+1/2*c)^3*(-24*A*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+96*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d
*x+1/2*c)+20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+30*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))*sin(1/2*d*x+1/2*c)^2-54*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-78*A*cos(1/2*d*x+1/2*c)
*sin(1/2*d*x+1/2*c)^4+50*B*(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))*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-30*B*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-50*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+50*C*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*
c)^2+30*C*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
sin(1/2*d*x+1/2*c)^2-90*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-15*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))+27*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))+18*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-25*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))+15*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))+20*B*sin(1/2*d*x+1/
2*c)^2*cos(1/2*d*x+1/2*c)-25*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))-15*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))+50*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*C)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(
1/2)/(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)^(5/2)*(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 )^{2} \sec \left (d x + c\right )^{5} +{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} +{\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A a^{3} \cos \left (d x + c\right )^{2}\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)^(5/2)*(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)^2*sec(d*x + c)^5 + (B + 3*C)*a^3*cos(d*x + c)^2*sec(d*x + c)^4 + (A + 3*B + 3*C)*
a^3*cos(d*x + c)^2*sec(d*x + c)^3 + (3*A + 3*B + C)*a^3*cos(d*x + c)^2*sec(d*x + c)^2 + (3*A + B)*a^3*cos(d*x
+ c)^2*sec(d*x + c) + A*a^3*cos(d*x + c)^2)*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)**(5/2)*(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{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(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)^(5/2), x)