3.182 \(\int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=290 \[ \frac {a^{5/2} (1015 A+1304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{512 d}+\frac {a^3 (1015 A+1304 C) \sin (c+d x)}{512 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (109 A+136 C) \sin (c+d x) \cos ^2(c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (1015 A+1304 C) \sin (c+d x) \cos (c+d x)}{768 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (23 A+24 C) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{96 d}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^{5/2}}{6 d}+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d} \]

[Out]

1/512*a^(5/2)*(1015*A+1304*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/12*a*A*cos(d*x+c)^4*(a+a*s
ec(d*x+c))^(3/2)*sin(d*x+c)/d+1/6*A*cos(d*x+c)^5*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/512*a^3*(1015*A+1304*C)
*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/768*a^3*(1015*A+1304*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+
1/192*a^3*(109*A+136*C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96*a^2*(23*A+24*C)*cos(d*x+c)^3*sin
(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.86, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4087, 4017, 4015, 3805, 3774, 203} \[ \frac {a^3 (1015 A+1304 C) \sin (c+d x)}{512 d \sqrt {a \sec (c+d x)+a}}+\frac {a^{5/2} (1015 A+1304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{512 d}+\frac {a^2 (23 A+24 C) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{96 d}+\frac {a^3 (109 A+136 C) \sin (c+d x) \cos ^2(c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (1015 A+1304 C) \sin (c+d x) \cos (c+d x)}{768 d \sqrt {a \sec (c+d x)+a}}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^{5/2}}{6 d}+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^(5/2)*(1015*A + 1304*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(512*d) + (a^3*(1015*A + 1
304*C)*Sin[c + d*x])/(512*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(1015*A + 1304*C)*Cos[c + d*x]*Sin[c + d*x])/(768
*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(109*A + 136*C)*Cos[c + d*x]^2*Sin[c + d*x])/(192*d*Sqrt[a + a*Sec[c + d*x
]]) + (a^2*(23*A + 24*C)*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(96*d) + (a*A*Cos[c + d*x]^4*(a
+ a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (A*Cos[c + d*x]^5*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(6*
d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3805

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 4015

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (5 A+12 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {5}{4} a^2 (23 A+24 C)+\frac {15}{4} a^2 (5 A+8 C) \sec (c+d x)\right ) \, dx}{30 a}\\ &=\frac {a^2 (23 A+24 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{96 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\int \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {15}{8} a^3 (109 A+136 C)+\frac {5}{8} a^3 (235 A+312 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^3 (109 A+136 C) \cos ^2(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (23 A+24 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{96 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {1}{384} \left (a^2 (1015 A+1304 C)\right ) \int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (1015 A+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (109 A+136 C) \cos ^2(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (23 A+24 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{96 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {1}{512} \left (a^2 (1015 A+1304 C)\right ) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (1015 A+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (109 A+136 C) \cos ^2(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (23 A+24 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{96 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac {\left (a^2 (1015 A+1304 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx}{1024}\\ &=\frac {a^3 (1015 A+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (109 A+136 C) \cos ^2(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (23 A+24 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{96 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}-\frac {\left (a^3 (1015 A+1304 C)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{512 d}\\ &=\frac {a^{5/2} (1015 A+1304 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{512 d}+\frac {a^3 (1015 A+1304 C) \sin (c+d x)}{512 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (1015 A+1304 C) \cos (c+d x) \sin (c+d x)}{768 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (109 A+136 C) \cos ^2(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (23 A+24 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{96 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 2.23, size = 182, normalized size = 0.63 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (3 \sqrt {2} (1015 A+1304 C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+\left (\sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) ((3234 A+2896 C) \cos (c+d x)+4 (315 A+184 C) \cos (2 (c+d x))+428 A \cos (3 (c+d x))+112 A \cos (4 (c+d x))+16 A \cos (5 (c+d x))+4193 A+96 C \cos (3 (c+d x))+4648 C)\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(1015*A + 1304*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]
*Sqrt[Cos[c + d*x]] + (4193*A + 4648*C + (3234*A + 2896*C)*Cos[c + d*x] + 4*(315*A + 184*C)*Cos[2*(c + d*x)] +
428*A*Cos[3*(c + d*x)] + 96*C*Cos[3*(c + d*x)] + 112*A*Cos[4*(c + d*x)] + 16*A*Cos[5*(c + d*x)])*(-Sin[(c + d
*x)/2] + Sin[(3*(c + d*x))/2])))/(3072*d)

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fricas [A]  time = 0.56, size = 488, normalized size = 1.68 \[ \left [\frac {3 \, {\left ({\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (1015 \, A + 1304 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (256 \, A a^{2} \cos \left (d x + c\right )^{6} + 896 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (203 \, A + 184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3072 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (1015 \, A + 1304 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (256 \, A a^{2} \cos \left (d x + c\right )^{6} + 896 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (203 \, A + 184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{1536 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/3072*(3*((1015*A + 1304*C)*a^2*cos(d*x + c) + (1015*A + 1304*C)*a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*s
qrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c)
+ 1)) + 2*(256*A*a^2*cos(d*x + c)^6 + 896*A*a^2*cos(d*x + c)^5 + 48*(29*A + 8*C)*a^2*cos(d*x + c)^4 + 8*(203*A
+ 184*C)*a^2*cos(d*x + c)^3 + 2*(1015*A + 1304*C)*a^2*cos(d*x + c)^2 + 3*(1015*A + 1304*C)*a^2*cos(d*x + c))*
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/1536*(3*((1015*A + 1304*C)*a^2*
cos(d*x + c) + (1015*A + 1304*C)*a^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqr
t(a)*sin(d*x + c))) - (256*A*a^2*cos(d*x + c)^6 + 896*A*a^2*cos(d*x + c)^5 + 48*(29*A + 8*C)*a^2*cos(d*x + c)^
4 + 8*(203*A + 184*C)*a^2*cos(d*x + c)^3 + 2*(1015*A + 1304*C)*a^2*cos(d*x + c)^2 + 3*(1015*A + 1304*C)*a^2*co
s(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

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giac [B]  time = 6.26, size = 1613, normalized size = 5.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/3072*(3*(1015*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 1304*C*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*t
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 3*(1015*A*sqrt(-a)*a^2*sg
n(cos(d*x + c)) + 1304*C*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(3045*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a))^22*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 3912*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^22*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 100485*sqrt(2)*(sqrt(-a)*tan(1/2*d
*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^20*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 129096*sqrt(2)*(sqrt(
-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^20*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 1303699*
sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*A*sqrt(-a)*a^5*sgn(cos(d*x +
c)) + 1693560*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^18*C*sqrt(-a)*a^5*
sgn(cos(d*x + c)) - 9936699*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^16*A
*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 11951544*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^16*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 38257266*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^14*A*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 48800976*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 83779026*sqrt(2)*(sqrt(-a)*tan(
1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 106200016*sqrt(2
)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*a^8*sgn(cos(d*x + c)) +
74917446*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(-a)*a^9*sgn(c
os(d*x + c)) + 94661616*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*sqr
t(-a)*a^9*sgn(cos(d*x + c)) - 30850806*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^8*A*sqrt(-a)*a^10*sgn(cos(d*x + c)) - 39751536*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2
*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a^10*sgn(cos(d*x + c)) + 7187801*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^11*sgn(cos(d*x + c)) + 9070440*sqrt(2)*(sqrt(-a)*tan(1/2*d*
x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^11*sgn(cos(d*x + c)) - 929817*sqrt(2)*(sqrt(-
a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^12*sgn(cos(d*x + c)) - 1176936*s
qrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a^12*sgn(cos(d*x + c
)) + 64887*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^13*sgn
(cos(d*x + c)) + 82200*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt(
-a)*a^13*sgn(cos(d*x + c)) - 1887*sqrt(2)*A*sqrt(-a)*a^14*sgn(cos(d*x + c)) - 2392*sqrt(2)*C*sqrt(-a)*a^14*sgn
(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^6)/d

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maple [B]  time = 2.02, size = 1118, normalized size = 3.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x)

[Out]

1/98304/d*(3045*A*sin(d*x+c)*cos(d*x+c)^5*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+3912*C*sin(d*x+c)*cos(d*x+c)^5*2^(1/2)*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+15225
*A*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+19560*C*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+30450*A*sin(d*x+c)*
cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)/cos(d*x+c)*2^(1/2))+39120*C*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arc
tanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+30450*A*sin(d*x+c)*cos(d*x+c)^2*2
^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d
*x+c)*2^(1/2))+39120*C*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+15225*A*sin(d*x+c)*cos(d*x+c)*2^(1/2)*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+1
9560*C*sin(d*x+c)*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+3045*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*2^(1/2)*arctanh(1/2*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+3912*C*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(11/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+
c)-16384*A*cos(d*x+c)^12-40960*A*cos(d*x+c)^11-31744*A*cos(d*x+c)^10-24576*C*cos(d*x+c)^10-14848*A*cos(d*x+c)^
9-69632*C*cos(d*x+c)^9-25984*A*cos(d*x+c)^8-72704*C*cos(d*x+c)^8-64960*A*cos(d*x+c)^7-83456*C*cos(d*x+c)^7+194
880*A*cos(d*x+c)^6+250368*C*cos(d*x+c)^6)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^5*a^2

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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)^6*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^6\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^6*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), x)

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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)**6*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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