\(\int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [508]

Optimal result

Integrand size = 43, antiderivative size = 261 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (283 A+326 B+400 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B+400 C) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (787 A+950 B+1040 C) \cos (c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (79 A+110 B+80 C) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \]

[Out]

1/128*a^(5/2)*(283*A+326*B+400*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/8*a*(A+2*B)*cos(d*x+c)
^3*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/128*a^3*(283*A
+326*B+400*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/960*a^3*(787*A+950*B+1040*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*
sec(d*x+c))^(1/2)+1/240*a^2*(79*A+110*B+80*C)*cos(d*x+c)^2*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4171, 4102, 4100, 3890, 3859, 209} \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (283 A+326 B+400 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{128 d}+\frac {a^3 (283 A+326 B+400 C) \sin (c+d x)}{128 d \sqrt {a \sec (c+d x)+a}}+\frac {a^3 (787 A+950 B+1040 C) \sin (c+d x) \cos (c+d x)}{960 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (79 A+110 B+80 C) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{240 d}+\frac {a (A+2 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{8 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^{5/2}}{5 d} \]

[In]

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

[Out]

(a^(5/2)*(283*A + 326*B + 400*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(128*d) + (a^3*(283*
A + 326*B + 400*C)*Sin[c + d*x])/(128*d*Sqrt[a + a*Sec[c + d*x]]) + (a^3*(787*A + 950*B + 1040*C)*Cos[c + d*x]
*Sin[c + d*x])/(960*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(79*A + 110*B + 80*C)*Cos[c + d*x]^2*Sqrt[a + a*Sec[c +
d*x]]*Sin[c + d*x])/(240*d) + (a*(A + 2*B)*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(8*d) + (A
*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2)*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 4171

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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*} \text {integral}& = \frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {5}{2} a (A+2 B)+\frac {1}{2} a (3 A+10 C) \sec (c+d x)\right ) \, dx}{5 a} \\ & = \frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (79 A+110 B+80 C)+\frac {1}{4} a^2 (39 A+30 B+80 C) \sec (c+d x)\right ) \, dx}{20 a} \\ & = \frac {a^2 (79 A+110 B+80 C) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {\int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (787 A+950 B+1040 C)+\frac {3}{8} a^3 (157 A+170 B+240 C) \sec (c+d x)\right ) \, dx}{60 a} \\ & = \frac {a^3 (787 A+950 B+1040 C) \cos (c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (79 A+110 B+80 C) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{128} \left (a^2 (283 A+326 B+400 C)\right ) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a^3 (283 A+326 B+400 C) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (787 A+950 B+1040 C) \cos (c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (79 A+110 B+80 C) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{256} \left (a^2 (283 A+326 B+400 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a^3 (283 A+326 B+400 C) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (787 A+950 B+1040 C) \cos (c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (79 A+110 B+80 C) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac {\left (a^3 (283 A+326 B+400 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 A+326 B+400 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B+400 C) \sin (c+d x)}{128 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (787 A+950 B+1040 C) \cos (c+d x) \sin (c+d x)}{960 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (79 A+110 B+80 C) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{240 d}+\frac {a (A+2 B) \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^{5/2} \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 = 2.52 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.14 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (77805 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+85050 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+92400 C \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+34953 A \sqrt {1-\sec (c+d x)}+29106 B \sqrt {1-\sec (c+d x)}+17360 C \sqrt {1-\sec (c+d x)}+111087 A \cos (c+d x) \sqrt {1-\sec (c+d x)}+106974 B \cos (c+d x) \sqrt {1-\sec (c+d x)}+89040 C \cos (c+d x) \sqrt {1-\sec (c+d x)}+37959 A \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+29358 B \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+17360 C \cos (2 (c+d x)) \sqrt {1-\sec (c+d x)}+11454 A \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+7308 B \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}-1120 C \cos (3 (c+d x)) \sqrt {1-\sec (c+d x)}+3006 A \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+252 B \cos (4 (c+d x)) \sqrt {1-\sec (c+d x)}+216 A \cos (5 (c+d x)) \sqrt {1-\sec (c+d x)}+107520 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+64512 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},5,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}+46080 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},6,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{40320 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

(a^2*(77805*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 85050*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 92400*C*ArcTanh[Sqrt
[1 - Sec[c + d*x]]] + 34953*A*Sqrt[1 - Sec[c + d*x]] + 29106*B*Sqrt[1 - Sec[c + d*x]] + 17360*C*Sqrt[1 - Sec[c
+ d*x]] + 111087*A*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 106974*B*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 89040
*C*Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 37959*A*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 29358*B*Cos[2*(c +
d*x)]*Sqrt[1 - Sec[c + d*x]] + 17360*C*Cos[2*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 11454*A*Cos[3*(c + d*x)]*Sqrt
[1 - Sec[c + d*x]] + 7308*B*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] - 1120*C*Cos[3*(c + d*x)]*Sqrt[1 - Sec[c +
d*x]] + 3006*A*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 252*B*Cos[4*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 216*
A*Cos[5*(c + d*x)]*Sqrt[1 - Sec[c + d*x]] + 107520*C*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 -
Sec[c + d*x]] + 64512*B*Hypergeometric2F1[1/2, 5, 3/2, 1 - Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]] + 46080*A*Hyp
ergeometric2F1[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])
/(40320*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(613\) vs. \(2(233)=466\).

Time = 0.52 (sec) , antiderivative size = 614, normalized size of antiderivative = 2.35

\[\frac {a^{2} \left (384 A \cos \left (d x +c \right )^{5} \sin \left (d x +c \right )+1392 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+480 B \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+2264 A \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+1840 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+640 C \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+4245 A \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 )+2830 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+4890 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 )+3260 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+6000 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 )+2720 C \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+4245 A \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 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+4890 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 )+4890 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+6000 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 )+6000 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)^5*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/1920*a^2/d*(384*A*cos(d*x+c)^5*sin(d*x+c)+1392*A*cos(d*x+c)^4*sin(d*x+c)+480*B*cos(d*x+c)^4*sin(d*x+c)+2264*
A*cos(d*x+c)^3*sin(d*x+c)+1840*B*sin(d*x+c)*cos(d*x+c)^3+640*C*cos(d*x+c)^3*sin(d*x+c)+4245*A*(-cos(d*x+c)/(co
s(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)+2830*A*cos
(d*x+c)^2*sin(d*x+c)+4890*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)+3260*B*sin(d*x+c)*cos(d*x+c)^2+6000*C*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arc
tanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+2720*C*cos(d*x+c)^2*sin(d*x+c)+4
245*A*(-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*A*cos(d*x+c)*sin(d*x+c)+4890*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-co
s(d*x+c)/(cos(d*x+c)+1))^(1/2))+4890*B*cos(d*x+c)*sin(d*x+c)+6000*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))+6000*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 = 0.44 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.88 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {15 \, {\left ({\left (283 \, A + 326 \, B + 400 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, A + 326 \, B + 400 \, 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 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, A + 230 \, B + 80 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, A + 326 \, B + 272 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, A + 326 \, B + 400 \, 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 \, A + 326 \, B + 400 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (283 \, A + 326 \, B + 400 \, 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 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (283 \, A + 230 \, B + 80 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (283 \, A + 326 \, B + 272 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (283 \, A + 326 \, B + 400 \, 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)^5*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/3840*(15*((283*A + 326*B + 400*C)*a^2*cos(d*x + c) + (283*A + 326*B + 400*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*A*a^2*cos(d*x + c)^5 + 48*(29*A + 10*B)*a^2*cos(d*x + c)^4 + 8*(283*A + 230*B + 80
*C)*a^2*cos(d*x + c)^3 + 10*(283*A + 326*B + 272*C)*a^2*cos(d*x + c)^2 + 15*(283*A + 326*B + 400*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*A + 326*
B + 400*C)*a^2*cos(d*x + c) + (283*A + 326*B + 400*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*A*a^2*cos(d*x + c)^5 + 48*(29*A + 10*B)*a^2*cos(d*x + c)^4 + 8
*(283*A + 230*B + 80*C)*a^2*cos(d*x + c)^3 + 10*(283*A + 326*B + 272*C)*a^2*cos(d*x + c)^2 + 15*(283*A + 326*B
+ 400*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. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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