3.1105 \(\int \frac{(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\sqrt{\cos (c+d x)}} \, dx\)

Optimal. Leaf size=246 \[ \frac{4 a^3 (21 A+11 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}-\frac{4 a^3 (27 A+17 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (63 A+73 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (27 A+17 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{4 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

[Out]

(-4*a^3*(27*A + 17*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(21*A + 11*C)*EllipticF[(c + d*x)/2, 2])/(21*
d) + (8*a^3*(21*A + 16*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (4*a^3*(27*A + 17*C)*Sin[c + d*x])/(15*d*
Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + (4*C*(a^2 + a^2*Cos
[c + d*x])^2*Sin[c + d*x])/(21*a*d*Cos[c + d*x]^(7/2)) + (2*(63*A + 73*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x
])/(315*d*Cos[c + d*x]^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.678175, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 10, 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, 2639, 2641} \[ \frac{4 a^3 (21 A+11 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{4 a^3 (27 A+17 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (63 A+73 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (27 A+17 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{4 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sqrt[Cos[c + d*x]],x]

[Out]

(-4*a^3*(27*A + 17*C)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(21*A + 11*C)*EllipticF[(c + d*x)/2, 2])/(21*
d) + (8*a^3*(21*A + 16*C)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (4*a^3*(27*A + 17*C)*Sin[c + d*x])/(15*d*
Sqrt[Cos[c + d*x]]) + (2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + (4*C*(a^2 + a^2*Cos
[c + d*x])^2*Sin[c + d*x])/(21*a*d*Cos[c + d*x]^(7/2)) + (2*(63*A + 73*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x
])/(315*d*Cos[c + d*x]^(5/2))

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 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]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx &=\int \frac{(a+a \cos (c+d x))^3 \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^3 \left (3 a C+\frac{1}{2} a (9 A+C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{4} a^2 (63 A+73 C)+\frac{1}{4} a^2 (63 A+13 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{63 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 \int \frac{(a+a \cos (c+d x)) \left (\frac{9}{2} a^3 (21 A+16 C)+\frac{3}{4} a^3 (63 A+23 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{8 \int \frac{\frac{9}{2} a^4 (21 A+16 C)+\left (\frac{9}{2} a^4 (21 A+16 C)+\frac{3}{4} a^4 (63 A+23 C)\right ) \cos (c+d x)+\frac{3}{4} a^4 (63 A+23 C) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{315 a}\\ &=\frac{8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{16 \int \frac{\frac{63}{8} a^4 (27 A+17 C)+\frac{45}{8} a^4 (21 A+11 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{945 a}\\ &=\frac{8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{21} \left (2 a^3 (21 A+11 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{15} \left (2 a^3 (27 A+17 C)\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^3 (21 A+11 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (27 A+17 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{1}{15} \left (2 a^3 (27 A+17 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{4 a^3 (27 A+17 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a^3 (21 A+11 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^3 (27 A+17 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 6.76038, size = 1135, normalized size = 4.61 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2))/Sqrt[Cos[c + d*x]],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)*(((27*A + 17*C)*Csc[c]
*Sec[c])/(15*d) + (C*Sec[c]*Sec[c + d*x]^5*Sin[d*x])/(18*d) + (Sec[c]*Sec[c + d*x]^4*(7*C*Sin[c] + 27*C*Sin[d*
x]))/(126*d) + (Sec[c]*Sec[c + d*x]^3*(135*C*Sin[c] + 63*A*Sin[d*x] + 238*C*Sin[d*x]))/(630*d) + (Sec[c]*Sec[c
+ d*x]*(105*A*Sin[c] + 110*C*Sin[c] + 378*A*Sin[d*x] + 238*C*Sin[d*x]))/(210*d) + (Sec[c]*Sec[c + d*x]^2*(63*
A*Sin[c] + 238*C*Sin[c] + 315*A*Sin[d*x] + 330*C*Sin[d*x]))/(630*d)))/(A + 2*C + 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 + 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 + A*Cos
[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (11*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 + 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]) + (9*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}
, 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]]))/(10*d*(A + 2*C + A*Cos[2
*c + 2*d*x])) + (17*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)
*((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]]))/(30*d*(A + 2*C + A*Cos[2*c + 2*d*x]))

________________________________________________________________________________________

Maple [B]  time = 9.904, size = 1246, 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)^(1/2),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*(1/8*A*(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))-1/5*(1/8*A+3/8*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)*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)^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)*Ellipt
icE(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)+1/8*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+sin(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*cos(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))))+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(cos(1/2*d*x+1/2*c),2^(1/2)))+(3/8*A+1/8*C)*(-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)))+3/8*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

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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((a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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}}{\sqrt{\cos \left (d x + c\right )}}, 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)^(1/2),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)/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((a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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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}}{\sqrt{\cos \left (d x + c\right )}}\,{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)^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/sqrt(cos(d*x + c)), x)