\(\int \frac {(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [266]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^2 (28 A+33 C) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (112 A+143 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[Out]

2/11*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(9/2)+2/231*a^2*(28*A+33*C)*sin(d*x+c)/d/sec(d*x+c)^(5/2
)/(a+a*sec(d*x+c))^(1/2)+2/385*a^2*(112*A+143*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+8/1155*a
^2*(112*A+143*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/1155*a^2*(112*A+143*C)*sin(d*x+c)*sec
(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/33*a*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(7/2)

Rubi [A] (verified)

Time = 0.86 (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} \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^2 (112 A+143 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (28 A+33 C) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (112 A+143 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{1155 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^2*(28*A + 33*C)*Sin[c + d*x])/(231*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(112*A + 143*C
)*Sin[c + d*x])/(385*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(112*A + 143*C)*Sin[c + d*x])/(11
55*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(112*A + 143*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(
1155*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(33*d*Sec[c + d*x]^(7/2)) + (
2*A*(a + a*Sec[c + d*x])^(3/2)*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))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3 a A}{2}+\frac {1}{2} a (6 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{11 a} \\ & = \frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {3}{4} a^2 (28 A+33 C)+\frac {9}{4} a^2 (8 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a} \\ & = \frac {2 a^2 (28 A+33 C) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{77} (a (112 A+143 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (28 A+33 C) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{385} (4 a (112 A+143 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (28 A+33 C) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {(8 a (112 A+143 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{1155} \\ & = \frac {2 a^2 (28 A+33 C) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (112 A+143 C) \sin (c+d x)}{385 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (112 A+143 C) \sin (c+d x)}{1155 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (112 A+143 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{1155 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{33 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.45 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (105 A+245 A \sec (c+d x)+5 (56 A+33 C) \sec ^2(c+d x)+(336 A+429 C) \sec ^3(c+d x)+(448 A+572 C) \sec ^4(c+d x)+8 (112 A+143 C) \sec ^5(c+d x)\right ) \sin (c+d x)}{1155 d \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^2*(105*A + 245*A*Sec[c + d*x] + 5*(56*A + 33*C)*Sec[c + d*x]^2 + (336*A + 429*C)*Sec[c + d*x]^3 + (448*A
+ 572*C)*Sec[c + d*x]^4 + 8*(112*A + 143*C)*Sec[c + d*x]^5)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(9/2)*Sqrt[a*(1
+ Sec[c + d*x])])

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {2 a \left (105 A \cos \left (d x +c \right )^{5}+245 A \cos \left (d x +c \right )^{4}+280 A \cos \left (d x +c \right )^{3}+165 C \cos \left (d x +c \right )^{3}+336 A \cos \left (d x +c \right )^{2}+429 C \cos \left (d x +c \right )^{2}+448 A \cos \left (d x +c \right )+572 C \cos \left (d x +c \right )+896 A +1144 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{1155 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(134\)
parts	\(\frac {2 A a \left (15 \cos \left (d x +c \right )^{5}+35 \cos \left (d x +c \right )^{4}+40 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+64 \cos \left (d x +c \right )+128\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{165 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 C a \left (15 \cos \left (d x +c \right )^{3}+39 \cos \left (d x +c \right )^{2}+52 \cos \left (d x +c \right )+104\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{105 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(168\)

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

2/1155*a/d*(105*A*cos(d*x+c)^5+245*A*cos(d*x+c)^4+280*A*cos(d*x+c)^3+165*C*cos(d*x+c)^3+336*A*cos(d*x+c)^2+429
*C*cos(d*x+c)^2+448*A*cos(d*x+c)+572*C*cos(d*x+c)+896*A+1144*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 = 144, normalized size of antiderivative = 0.54 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \, {\left (105 \, A a \cos \left (d x + c\right )^{6} + 245 \, A a \cos \left (d x + c\right )^{5} + 5 \, {\left (56 \, A + 33 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{3} + 4 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (112 \, A + 143 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{1155 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/1155*(105*A*a*cos(d*x + c)^6 + 245*A*a*cos(d*x + c)^5 + 5*(56*A + 33*C)*a*cos(d*x + c)^4 + 3*(112*A + 143*C)
*a*cos(d*x + c)^3 + 4*(112*A + 143*C)*a*cos(d*x + c)^2 + 8*(112*A + 143*C)*a*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. \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(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. 794 vs. \(2 (230) = 460\).

Time = 0.55 (sec) , antiderivative size = 794, normalized size of antiderivative = 2.98 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

1/36960*(7*sqrt(2)*(3630*a*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 + 1
1/2*c) + 990*a*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) + 429*
a*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) + 165*a*cos(4/11*ar
ctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 55*a*cos(2/11*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) - 3630*a*cos(11/2*d*x + 11/2*c)*sin(10/11*arct
an2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 990*a*cos(11/2*d*x + 11/2*c)*sin(8/11*arctan2(sin(11/2*
d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 429*a*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))) - 165*a*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))) - 55*a*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)))
+ 30*a*sin(11/2*d*x + 11/2*c) + 55*a*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 165*a
*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 429*a*sin(5/11*arctan2(sin(11/2*d*x + 11/
2*c), cos(11/2*d*x + 11/2*c))) + 990*a*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 363
0*a*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) + 44*sqrt(2)*(735*a*cos(6/7*a
rctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 175*a*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) + 63*a*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) - 735*a*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))) - 175*a*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 63*a*c
os(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))) + 30*a*sin(7/2*d*x + 7/2*c) +
63*a*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 175*a*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c
), cos(7/2*d*x + 7/2*c))) + 735*a*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*C*sqrt(a))/d

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(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 = 22.82 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.41 \[ \int \frac {(a+a \sec (c+d x))^{3/2} \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 {a\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (9\,A+10\,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\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (7\,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\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,A+14\,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}+\frac {a\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (13\,A+12\,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 )}{40\,d}+\frac {A\,a\,\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 )}{24\,d}+\frac {A\,a\,\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}\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))^(3/2))/(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)*((a*sin((3*c)/2 + (3*d*x)/2)*(9*A + 10*C)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 +
1))/(12*d) + (a*sin((7*c)/2 + (7*d*x)/2)*(7*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*sin(c/2 + (d*x)/2)*(11*A + 14*C)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (
11*d*x)/4)^2 + 1))/(4*d) + (a*sin((5*c)/2 + (5*d*x)/2)*(13*A + 12*C)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((1
1*c)/4 + (11*d*x)/4)^2 + 1))/(40*d) + (A*a*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))/(24*d) + (A*a*sin((11*c)/2 + (11*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((1
1*c)/4 + (11*d*x)/4)^2 + 1))/(88*d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(2*sin(c/4 + (d*x)/4)^2 - 1))