3.1106 \(\int \frac{(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=279 \[ \frac{4 a^3 (143 A+105 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{231 d}-\frac{4 a^3 (7 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (143 A+105 C) \sin (c+d x)}{231 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8 a^3 (44 A+35 C) \sin (c+d x)}{385 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (33 A+35 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{231 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^3 (7 A+5 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{4 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{11 d \cos ^{\frac{11}{2}}(c+d x)} \]
[Out]
(-4*a^3*(7*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(143*A + 105*C)*EllipticF[(c + d*x)/2, 2])/(231*
d) + (8*a^3*(44*A + 35*C)*Sin[c + d*x])/(385*d*Cos[c + d*x]^(5/2)) + (4*a^3*(143*A + 105*C)*Sin[c + d*x])/(231
*d*Cos[c + d*x]^(3/2)) + (4*a^3*(7*A + 5*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])
^3*Sin[c + d*x])/(11*d*Cos[c + d*x]^(11/2)) + (4*C*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(33*a*d*Cos[c + d*
x]^(9/2)) + (2*(33*A + 35*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(231*d*Cos[c + d*x]^(7/2))
________________________________________________________________________________________
Rubi [A] time = 0.713054, antiderivative size = 279, normalized size of antiderivative = 1.,
number of steps used = 11, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used
= {4114, 3044, 2975, 2968, 3021, 2748, 2636, 2641, 2639} \[ \frac{4 a^3 (143 A+105 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}-\frac{4 a^3 (7 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (143 A+105 C) \sin (c+d x)}{231 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8 a^3 (44 A+35 C) \sin (c+d x)}{385 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 (33 A+35 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{231 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^3 (7 A+5 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{4 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{11 d \cos ^{\frac{11}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
[Out]
(-4*a^3*(7*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(143*A + 105*C)*EllipticF[(c + d*x)/2, 2])/(231*
d) + (8*a^3*(44*A + 35*C)*Sin[c + d*x])/(385*d*Cos[c + d*x]^(5/2)) + (4*a^3*(143*A + 105*C)*Sin[c + d*x])/(231
*d*Cos[c + d*x]^(3/2)) + (4*a^3*(7*A + 5*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])
^3*Sin[c + d*x])/(11*d*Cos[c + d*x]^(11/2)) + (4*C*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(33*a*d*Cos[c + d*
x]^(9/2)) + (2*(33*A + 35*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(231*d*Cos[c + d*x]^(7/2))
Rule 4114
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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 + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Rule 3044
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 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 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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 2636
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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]
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 \frac{(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{(a+a \cos (c+d x))^3 \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac{13}{2}}(c+d x)} \, dx\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^3 \left (3 a C+\frac{1}{2} a (11 A+3 C) \cos (c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{3}{4} a^2 (33 A+35 C)+\frac{9}{4} a^2 (11 A+5 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (33 A+35 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{231 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{8 \int \frac{(a+a \cos (c+d x)) \left (\frac{9}{2} a^3 (44 A+35 C)+\frac{45}{4} a^3 (11 A+7 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{693 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (33 A+35 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{231 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{8 \int \frac{\frac{9}{2} a^4 (44 A+35 C)+\left (\frac{45}{4} a^4 (11 A+7 C)+\frac{9}{2} a^4 (44 A+35 C)\right ) \cos (c+d x)+\frac{45}{4} a^4 (11 A+7 C) \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{693 a}\\ &=\frac{8 a^3 (44 A+35 C) \sin (c+d x)}{385 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (33 A+35 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{231 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{16 \int \frac{\frac{45}{8} a^4 (143 A+105 C)+\frac{693}{8} a^4 (7 A+5 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{3465 a}\\ &=\frac{8 a^3 (44 A+35 C) \sin (c+d x)}{385 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (33 A+35 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{231 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{1}{5} \left (2 a^3 (7 A+5 C)\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{77} \left (2 a^3 (143 A+105 C)\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{8 a^3 (44 A+35 C) \sin (c+d x)}{385 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (143 A+105 C) \sin (c+d x)}{231 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (7 A+5 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (33 A+35 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{231 d \cos ^{\frac{7}{2}}(c+d x)}-\frac{1}{5} \left (2 a^3 (7 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{231} \left (2 a^3 (143 A+105 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{4 a^3 (7 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^3 (143 A+105 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{8 a^3 (44 A+35 C) \sin (c+d x)}{385 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (143 A+105 C) \sin (c+d x)}{231 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (7 A+5 C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{11 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{33 a d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 (33 A+35 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{231 d \cos ^{\frac{7}{2}}(c+d x)}\\ \end{align*}
Mathematica [C] time = 6.87191, size = 1179, normalized size = 4.23 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]
[Out]
(Cos[c + d*x]^(11/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*(((7*A + 5*C)*Csc[c]*S
ec[c])/(5*d) + (C*Sec[c]*Sec[c + d*x]^6*Sin[d*x])/(22*d) + (Sec[c]*Sec[c + d*x]^5*(3*C*Sin[c] + 11*C*Sin[d*x])
)/(66*d) + (Sec[c]*Sec[c + d*x]^4*(77*C*Sin[c] + 33*A*Sin[d*x] + 126*C*Sin[d*x]))/(462*d) + (Sec[c]*Sec[c + d*
x]^3*(165*A*Sin[c] + 630*C*Sin[c] + 693*A*Sin[d*x] + 770*C*Sin[d*x]))/(2310*d) + (Sec[c]*Sec[c + d*x]^2*(693*A
*Sin[c] + 770*C*Sin[c] + 1430*A*Sin[d*x] + 1050*C*Sin[d*x]))/(2310*d) + (Sec[c]*Sec[c + d*x]*(715*A*Sin[c] + 5
25*C*Sin[c] + 1617*A*Sin[d*x] + 1155*C*Sin[d*x]))/(1155*d)))/(A + 2*C + A*Cos[2*c + 2*d*x]) - (13*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 + 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]]]])/(21*d*(A + 2*C + 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 - Ar
cTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + 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]]]])/(11*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) + (7*A*Cos[c + d*x]^5*Csc[c]*
Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*((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]]))/(10*d*(A + 2*C + 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 + C*Sec[c + d*x]^2)*((Hype
rgeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Co
s[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Ta
n[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 + A*Cos[2*c + 2*d*x]))
________________________________________________________________________________________
Maple [B] time = 10.798, size = 1408, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x)
[Out]
-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-1/5*(3/8*A+1/8*C)/(8*sin(1/2*d*x+1/2*c)^6-
12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticE(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)^4-24*sin(1/2*d*x+1/2*c)^6*
cos(1/2*d*x+1/2*c)-12*(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+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+3/8*C*(-1/144*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^
4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*co
s(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1
)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))+(1/8*A+3/8*C)*(-1/56*cos(1/2*d*
x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/
2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2)))+1/8*C*(-1/352*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)/(cos(1/2*d*x+1/2*c)^2-1/2)^6-9/616*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/
(cos(1/2*d*x+1/2*c)^2-1/2)^4-15/154*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(c
os(1/2*d*x+1/2*c)^2-1/2)^2+15/77*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*
x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+3/8*A*(-1/6*cos(1/2*d*x+1/2*c)*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2)))+1/8*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)*(-2*sin(1/2*d*x+1/2*
c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/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((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C a^{3} \sec \left (d x + c\right )^{5} + 3 \, C a^{3} \sec \left (d x + c\right )^{4} +{\left (A + 3 \, C\right )} a^{3} \sec \left (d x + c\right )^{3} +{\left (3 \, A + C\right )} a^{3} \sec \left (d x + c\right )^{2} + 3 \, A a^{3} \sec \left (d x + c\right ) + A a^{3}}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
integral((C*a^3*sec(d*x + c)^5 + 3*C*a^3*sec(d*x + c)^4 + (A + 3*C)*a^3*sec(d*x + c)^3 + (3*A + C)*a^3*sec(d*x
+ c)^2 + 3*A*a^3*sec(d*x + c) + A*a^3)/cos(d*x + c)^(3/2), 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((a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/cos(d*x + c)^(3/2), x)