\(\int \frac {(a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [275]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^3 (568 A+759 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[Out]

10/99*a*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(7/2)+2/11*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d/sec(
d*x+c)^(9/2)+2/693*a^3*(232*A+297*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/693*a^3*(568*A+759
*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4/693*a^3*(568*A+759*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d
/(a+a*sec(d*x+c))^(1/2)+2/231*a^2*(32*A+33*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4172, 4102, 4100, 3890, 3889} \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {4 a^3 (568 A+759 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{693 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (32 A+33 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {10 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^3*(232*A + 297*C)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(568*A + 759
*C)*Sin[c + d*x])/(693*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^3*(568*A + 759*C)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(32*A + 33*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x
])/(231*d*Sec[c + d*x]^(5/2)) + (10*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)) + (
2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

Rule 3889

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[-2*a*(Co
t[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, 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 4172

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*} \text {integral}& = \frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (4 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{11 a} \\ & = \frac {10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3}{4} a^2 (32 A+33 C)+\frac {1}{4} a^2 (56 A+99 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a} \\ & = \frac {2 a^2 (32 A+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {5}{8} a^3 (232 A+297 C)+\frac {1}{8} a^3 (776 A+1089 C) \sec (c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{693 a} \\ & = \frac {2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{231} \left (a^2 (568 A+759 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{693} \left (2 a^2 (568 A+759 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^3 (568 A+759 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (32 A+33 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.58 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.45 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (63 A+224 A \sec (c+d x)+(355 A+99 C) \sec ^2(c+d x)+6 (71 A+66 C) \sec ^3(c+d x)+(568 A+759 C) \sec ^4(c+d x)+2 (568 A+759 C) \sec ^5(c+d x)\right ) \sin (c+d x)}{693 d \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^3*(63*A + 224*A*Sec[c + d*x] + (355*A + 99*C)*Sec[c + d*x]^2 + 6*(71*A + 66*C)*Sec[c + d*x]^3 + (568*A +
759*C)*Sec[c + d*x]^4 + 2*(568*A + 759*C)*Sec[c + d*x]^5)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(9/2)*Sqrt[a*(1 +
Sec[c + d*x])])

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.51

method	result	size
default	\(\frac {2 a^{2} \left (63 A \cos \left (d x +c \right )^{5}+224 A \cos \left (d x +c \right )^{4}+355 A \cos \left (d x +c \right )^{3}+99 C \cos \left (d x +c \right )^{3}+426 A \cos \left (d x +c \right )^{2}+396 C \cos \left (d x +c \right )^{2}+568 A \cos \left (d x +c \right )+759 C \cos \left (d x +c \right )+1136 A +1518 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{693 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(136\)
parts	\(\frac {2 A \,a^{2} \left (63 \cos \left (d x +c \right )^{5}+224 \cos \left (d x +c \right )^{4}+355 \cos \left (d x +c \right )^{3}+426 \cos \left (d x +c \right )^{2}+568 \cos \left (d x +c \right )+1136\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{693 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C \,a^{2} \left (3 \cos \left (d x +c \right )^{3}+12 \cos \left (d x +c \right )^{2}+23 \cos \left (d x +c \right )+46\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{21 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(172\)

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

2/693*a^2/d*(63*A*cos(d*x+c)^5+224*A*cos(d*x+c)^4+355*A*cos(d*x+c)^3+99*C*cos(d*x+c)^3+426*A*cos(d*x+c)^2+396*
C*cos(d*x+c)^2+568*A*cos(d*x+c)+759*C*cos(d*x+c)+1136*A+1518*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*
x+c)^(3/2)*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.58 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \, {\left (63 \, A a^{2} \cos \left (d x + c\right )^{6} + 224 \, A a^{2} \cos \left (d x + c\right )^{5} + {\left (355 \, A + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 6 \, {\left (71 \, A + 66 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/693*(63*A*a^2*cos(d*x + c)^6 + 224*A*a^2*cos(d*x + c)^5 + (355*A + 99*C)*a^2*cos(d*x + c)^4 + 6*(71*A + 66*C
)*a^2*cos(d*x + c)^3 + (568*A + 759*C)*a^2*cos(d*x + c)^2 + 2*(568*A + 759*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)*sqrt(cos(d*x + c)))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 845 vs. \(2 (230) = 460\).

Time = 0.50 (sec) , antiderivative size = 845, normalized size of antiderivative = 3.18 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

1/22176*(sqrt(2)*(31878*a^2*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x +
11/2*c) + 8778*a^2*cos(8/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) +
3465*a^2*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 1287*a^2*c
os(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 385*a^2*cos(2/11*arc
tan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 31878*a^2*cos(11/2*d*x + 11/2*c
)*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 8778*a^2*cos(11/2*d*x + 11/2*c)*sin(8/1
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 3465*a^2*cos(11/2*d*x + 11/2*c)*sin(6/11*arctan2(
sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 1287*a^2*cos(11/2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d
*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 385*a^2*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2*c
), cos(11/2*d*x + 11/2*c))) + 126*a^2*sin(11/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c)
, cos(11/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3465
*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2*d*
x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2
*c))))*A*sqrt(a) + 132*sqrt(2)*(315*a^2*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d
*x + 7/2*c) + 77*a^2*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*a^
2*cos(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/
2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan
2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 21*a^2*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2
*c), cos(7/2*d*x + 7/2*c))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/
2*d*x + 7/2*c))) + 77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arcta
n2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*C*sqrt(a))/d

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 22.89 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.45 \[ \int \frac {(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {5\,A\,a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{72\,d}+\frac {A\,a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{88\,d}+\frac {a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (5\,A+4\,C\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{8\,d}+\frac {a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (13\,A+4\,C\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{56\,d}+\frac {a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (19\,A+22\,C\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{12\,d}+\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (23\,A+30\,C\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{4\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \]

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(11/2),x)

[Out]

((a - a/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin((11*c)/2 + (11*d*x)/2)*1i + 2*sin((11*c)/4 + (11*d*x)/4)^2 -
1)*((5*A*a^2*sin((9*c)/2 + (9*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(7
2*d) + (A*a^2*sin((11*c)/2 + (11*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))
/(88*d) + (a^2*sin((5*c)/2 + (5*d*x)/2)*(5*A + 4*C)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)
/4)^2 + 1))/(8*d) + (a^2*sin((7*c)/2 + (7*d*x)/2)*(13*A + 4*C)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4
+ (11*d*x)/4)^2 + 1))/(56*d) + (a^2*sin((3*c)/2 + (3*d*x)/2)*(19*A + 22*C)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2
*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(12*d) + (a^2*sin(c/2 + (d*x)/2)*(23*A + 30*C)*(sin((11*c)/2 + (11*d*x)/2)
*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(4*d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(2*sin(c/4 + (d*
x)/4)^2 - 1))