\(\int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [383]

Optimal result

Integrand size = 42, antiderivative size = 254 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (283 B+326 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (283 B+326 C) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (283 B+326 C) \cos (c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (157 B+170 C) \cos ^2(c+d x) \sin (c+d x)}{240 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (13 B+10 C) \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{40 d}+\frac {a B \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4157, 4102, 4100, 3890, 3859, 209} \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (283 B+326 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a^3 (283 B+326 C) \sin (c+d x)}{128 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (157 B+170 C) \sin (c+d x) \cos ^2(c+d x)}{240 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (283 B+326 C) \sin (c+d x) \cos (c+d x)}{192 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (13 B+10 C) \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{40 d}+\frac {a B \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[In]

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

[Out]

(a^(5/2)*(283*B + 326*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(128*d) + (a^3*(283*B + 326*
C)*Sin[c + d*x])/(128*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(283*B + 326*C)*Cos[c + d*x]*Sin[c + d*x])/(192*d*Sqr
t[a + a*Sec[c + d*x]]) + (a^3*(157*B + 170*C)*Cos[c + d*x]^2*Sin[c + d*x])/(240*d*Sqrt[a + a*Sec[c + d*x]]) +
(a^2*(13*B + 10*C)*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(40*d) + (a*B*Cos[c + d*x]^4*(a + a*S
ec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

Rule 209

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

Rule 3859

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 3890

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 4100

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 4102

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 4157

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.17 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.64 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (25935 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+28350 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+11651 B \sqrt {1-\sec (c+d x)}+9702 C \sqrt {1-\sec (c+d x)}+37029 B \cos (c+d x) \sqrt {1-\sec (c+d x)}+35658 C \cos (c+d x) \sqrt {1-\sec (c+d x)}+12653 B \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+9786 C \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+3818 B \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+2436 C \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+1002 B \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+84 C \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+72 B \cos (5 (c+d x)) \sqrt {1-\sec (c+d x)}+21504 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+15360 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},6,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{13440 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

(a^2*(25935*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 28350*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 11651*B*Sqrt[1 - Sec
[c + d*x]] + 9702*C*Sqrt[1 - Sec[c + d*x]] + 37029*B*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 35658*C*Cos[c + d*x
]*Sqrt[1 - Sec[c + d*x]] + 12653*B*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 9786*C*Cos[2*(c + d*x)]*Sqrt[1 -
Sec[c + d*x]] + 3818*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 2436*C*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]
] + 1002*B*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 84*C*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 72*B*Cos[5
*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 21504*C*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c +
d*x]] + 15360*B*Hypergeometric2F1[1/2, 6, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])*Sqrt[a*(1 + Sec[c +
d*x])]*Sin[c + d*x])/(13440*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.72

\[\frac {a^{2} \left (384 B \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )+1392 B \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+480 C \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+2264 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+1840 C \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+4245 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2830 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+4890 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+3260 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+4245 B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4245 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+4890 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4890 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{1920 d \left (\cos \left (d x +c \right )+1\right )}\]

[In]

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

[Out]

1/1920*a^2/d*(384*B*cos(d*x+c)^5*sin(d*x+c)+1392*B*cos(d*x+c)^4*sin(d*x+c)+480*C*cos(d*x+c)^4*sin(d*x+c)+2264*
B*sin(d*x+c)*cos(d*x+c)^3+1840*C*cos(d*x+c)^3*sin(d*x+c)+4245*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)
/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2830*B*sin(d*x+c)*cos(d*x+c)^2+4890*C*(-
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x
+c)+3260*C*cos(d*x+c)^2*sin(d*x+c)+4245*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)
/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4245*B*cos(d*x+c)*sin(d*x+c)+4890*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ar
ctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4890*C*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*
x+c)))^(1/2)/(cos(d*x+c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.81 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {15 \, {\left ({\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, B + 326 \, 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 (384 \, B a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, B + 326 \, 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 )}{3840 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {15 \, {\left ({\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, B + 326 \, 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 (384 \, B a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, B + 326 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, B + 326 \, 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 )}{1920 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

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

[Out]

[1/3840*(15*((283*B + 326*C)*a^2*cos(d*x + c) + (283*B + 326*C)*a^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt
(-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*(384*B*a^2*cos(d*x + c)^5 + 48*(29*B + 10*C)*a^2*cos(d*x + c)^4 + 8*(283*B + 230*C)*a^2*cos(d*x + c)^3
+ 10*(283*B + 326*C)*a^2*cos(d*x + c)^2 + 15*(283*B + 326*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), -1/1920*(15*((283*B + 326*C)*a^2*cos(d*x + c) + (283*B + 326*C)*a
^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (384*B*a^2*c
os(d*x + c)^5 + 48*(29*B + 10*C)*a^2*cos(d*x + c)^4 + 8*(283*B + 230*C)*a^2*cos(d*x + c)^3 + 10*(283*B + 326*C
)*a^2*cos(d*x + c)^2 + 15*(283*B + 326*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)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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