\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\) [99]

Optimal result

Integrand size = 33, antiderivative size = 254 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^{5/2} (283 A+326 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d} \]

[Out]

1/128*a^(5/2)*(283*A+326*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/5*a*A*(a+a*cos(d*x+c))^(3/2
)*sec(d*x+c)^4*tan(d*x+c)/d+1/128*a^3*(283*A+326*B)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/192*a^3*(283*A+326*B
)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/240*a^3*(157*A+170*B)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*
x+c))^(1/2)+1/40*a^2*(13*A+10*B)*sec(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3054, 3059, 2851, 2852, 212} \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^{5/2} (283 A+326 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{128 d}+\frac {a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (157 A+170 B) \tan (c+d x) \sec ^2(c+d x)}{240 d \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (283 A+326 B) \tan (c+d x) \sec (c+d x)}{192 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (13 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{40 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

[In]

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

[Out]

(a^(5/2)*(283*A + 326*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(128*d) + (a^3*(283*A + 326
*B)*Tan[c + d*x])/(128*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(283*A + 326*B)*Sec[c + d*x]*Tan[c + d*x])/(192*d*Sq
rt[a + a*Cos[c + d*x]]) + (a^3*(157*A + 170*B)*Sec[c + d*x]^2*Tan[c + d*x])/(240*d*Sqrt[a + a*Cos[c + d*x]]) +
(a^2*(13*A + 10*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(40*d) + (a*A*(a + a*Cos[c + d*x])^(
3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 212

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

Rule 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]
+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3054

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
*Sin[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 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (13 A+10 B)+\frac {5}{2} a (A+2 B) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int \sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (157 A+170 B)+\frac {5}{4} a^2 (21 A+26 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{96} \left (a^2 (283 A+326 B)\right ) \int \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \, dx \\ & = \frac {a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{128} \left (a^2 (283 A+326 B)\right ) \int \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \, dx \\ & = \frac {a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{256} \left (a^2 (283 A+326 B)\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx \\ & = \frac {a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (a^3 (283 A+326 B)\right ) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d} \\ & = \frac {a^{5/2} (283 A+326 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+10 B) \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.69 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (60 \sqrt {2} (283 A+326 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5(c+d x)+(24863 A+22030 B+36 (781 A+650 B) \cos (c+d x)+4 (6509 A+6730 B) \cos (2 (c+d x))+5660 A \cos (3 (c+d x))+6520 B \cos (3 (c+d x))+4245 A \cos (4 (c+d x))+4890 B \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15360 d} \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^5*(60*Sqrt[2]*(283*A + 326*B)*ArcTanh[Sqrt[2]*Si
n[(c + d*x)/2]]*Cos[c + d*x]^5 + (24863*A + 22030*B + 36*(781*A + 650*B)*Cos[c + d*x] + 4*(6509*A + 6730*B)*Co
s[2*(c + d*x)] + 5660*A*Cos[3*(c + d*x)] + 6520*B*Cos[3*(c + d*x)] + 4245*A*Cos[4*(c + d*x)] + 4890*B*Cos[4*(c
+ d*x)])*Sin[(c + d*x)/2]))/(15360*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1974\) vs. \(2(226)=452\).

Time = 2.58 (sec) , antiderivative size = 1975, normalized size of antiderivative = 7.78

\[\text {Expression too large to display}\]

[In]

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

[Out]

1/120*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-480*a*(283*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1
/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))+283*A*ln(-4/(2*cos(1/2
*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))+326*B*
ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/
2)+2*a))+326*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c
)^2)^(1/2)*a^(1/2)-2*a)))*sin(1/2*d*x+1/2*c)^10+240*(566*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+652*
B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1415*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2
*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+1415*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))
*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+1630*B*ln(4/(2*cos(1/2*d
*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+1630*B
*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(
1/2)-2*a))*a)*sin(1/2*d*x+1/2*c)^8-80*(3962*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+4564*B*2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+4245*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+4245*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*
cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+4890*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^
(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+4890*B*ln(-4/(2*co
s(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a
)*sin(1/2*d*x+1/2*c)^6+8*(36224*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+40960*B*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)*a^(1/2)+21225*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+21225*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d
*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+24450*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(
2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+24450*B*ln(-4/(2*cos(1/2*d
*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1
/2*d*x+1/2*c)^4-10*(12556*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+13400*B*2^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)*a^(1/2)+4245*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/
2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+4245*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c
)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+4890*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a
*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+4890*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-
2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)*sin(1/2*d*x+1/2
*c)^2+4245*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)+2*a))*a+4245*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+22230*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+4890*B*ln(4/
(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*
a))*a+4890*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^
2)^(1/2)*a^(1/2)-2*a))*a+20940*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2)
)^5/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^5/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.99 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {15 \, {\left ({\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (15 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (283 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right ) + 384 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{7680 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

[In]

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

[Out]

1/7680*(15*((283*A + 326*B)*a^2*cos(d*x + c)^6 + (283*A + 326*B)*a^2*cos(d*x + c)^5)*sqrt(a)*log((a*cos(d*x +
c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x
+ c)^3 + cos(d*x + c)^2)) + 4*(15*(283*A + 326*B)*a^2*cos(d*x + c)^4 + 10*(283*A + 326*B)*a^2*cos(d*x + c)^3
+ 8*(283*A + 230*B)*a^2*cos(d*x + c)^2 + 48*(29*A + 10*B)*a^2*cos(d*x + c) + 384*A*a^2)*sqrt(a*cos(d*x + c) +
a)*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**6,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.79 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.48 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=-\frac {\sqrt {2} {\left (15 \, \sqrt {2} {\left (283 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 326 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (67920 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 78240 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 158480 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 182560 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144896 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 163840 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 62780 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 67000 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11115 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10470 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}\right )} \sqrt {a}}{7680 \, d} \]

[In]

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

[Out]

-1/7680*sqrt(2)*(15*sqrt(2)*(283*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 326*B*a^2*sgn(cos(1/2*d*x + 1/2*c)))*log(ab
s(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(67920*A*a^2*sgn(cos(1/2*d
*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 + 78240*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 - 158480*A*
a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 - 182560*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/
2*c)^7 + 144896*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 163840*B*a^2*sgn(cos(1/2*d*x + 1/2*c)
)*sin(1/2*d*x + 1/2*c)^5 - 62780*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 67000*B*a^2*sgn(cos(
1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 11115*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 10470*
B*a^2*sgn(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 - 1)^5)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

[In]

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

[Out]

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