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

Optimal. Leaf size=284 \[ \frac {2 a^2 (336 A+374 B+429 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (84 A+110 B+99 C) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (336 A+374 B+429 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (3 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[Out]

2/11*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(9/2)+2/693*a^2*(84*A+110*B+99*C)*sin(d*x+c)/d/sec(d*x+c
)^(5/2)/(a+a*sec(d*x+c))^(1/2)+2/1155*a^2*(336*A+374*B+429*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(
1/2)+8/3465*a^2*(336*A+374*B+429*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/3465*a^2*(336*A+37
4*B+429*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a*(3*A+11*B)*sin(d*x+c)*(a+a*sec(d*x+c))^
(1/2)/d/sec(d*x+c)^(7/2)

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Rubi [A]  time = 0.74, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4086, 4017, 4015, 3805, 3804} \[ \frac {2 a^2 (336 A+374 B+429 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (84 A+110 B+99 C) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a^2 (336 A+374 B+429 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (3 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(84*A + 110*B + 99*C)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(336*A
+ 374*B + 429*C)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a^2*(336*A + 374*B +
429*C)*Sin[c + d*x])/(3465*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^2*(336*A + 374*B + 429*C)*S
qrt[Sec[c + d*x]]*Sin[c + d*x])/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(3*A + 11*B)*Sqrt[a + a*Sec[c + d*x]]
*Sin[c + d*x])/(99*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 3804

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 3805

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 4015

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 4017

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 4086

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*} \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx &=\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 {1}{2} a (3 A+11 B)+\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 (3 A+11 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 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 {1}{4} a^2 (84 A+110 B+99 C)+\frac {3}{4} a^2 (24 A+22 B+33 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac {2 a^2 (84 A+110 B+99 C) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+11 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 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}{231} (a (336 A+374 B+429 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (84 A+110 B+99 C) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (336 A+374 B+429 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+11 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 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 a (336 A+374 B+429 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{1155}\\ &=\frac {2 a^2 (84 A+110 B+99 C) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (336 A+374 B+429 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+11 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 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 (336 A+374 B+429 C)) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3465}\\ &=\frac {2 a^2 (84 A+110 B+99 C) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (336 A+374 B+429 C) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^2 (336 A+374 B+429 C) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (336 A+374 B+429 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (3 A+11 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{99 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*}

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Mathematica [A]  time = 2.27, size = 158, normalized size = 0.56 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} ((34734 A+44 (799 B+759 C)) \cos (c+d x)+8 (1743 A+1507 B+1287 C) \cos (2 (c+d x))+4935 A \cos (3 (c+d x))+1470 A \cos (4 (c+d x))+315 A \cos (5 (c+d x))+55482 A+3740 B \cos (3 (c+d x))+770 B \cos (4 (c+d x))+59158 B+1980 C \cos (3 (c+d x))+65208 C)}{27720 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(55482*A + 59158*B + 65208*C + (34734*A + 44*(799*B + 759*C))*Cos[c + d*x] + 8*(1743*A + 1507*B + 1287*C)*C
os[2*(c + d*x)] + 4935*A*Cos[3*(c + d*x)] + 3740*B*Cos[3*(c + d*x)] + 1980*C*Cos[3*(c + d*x)] + 1470*A*Cos[4*(
c + d*x)] + 770*B*Cos[4*(c + d*x)] + 315*A*Cos[5*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(277
20*d*Sqrt[Sec[c + d*x]])

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fricas [A]  time = 0.44, size = 162, normalized size = 0.57 \[ \frac {2 \, {\left (315 \, A a \cos \left (d x + c\right )^{6} + 35 \, {\left (21 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{5} + 5 \, {\left (168 \, A + 187 \, B + 99 \, C\right )} a \cos \left (d x + c\right )^{4} + 3 \, {\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )^{3} + 4 \, {\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (336 \, A + 374 \, B + 429 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*A*a*cos(d*x + c)^6 + 35*(21*A + 11*B)*a*cos(d*x + c)^5 + 5*(168*A + 187*B + 99*C)*a*cos(d*x + c)^4
+ 3*(336*A + 374*B + 429*C)*a*cos(d*x + c)^3 + 4*(336*A + 374*B + 429*C)*a*cos(d*x + c)^2 + 8*(336*A + 374*B
+ 429*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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(11/2), x)

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maple [A]  time = 2.58, size = 197, normalized size = 0.69 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (315 A \left (\cos ^{5}\left (d x +c \right )\right )+735 A \left (\cos ^{4}\left (d x +c \right )\right )+385 B \left (\cos ^{4}\left (d x +c \right )\right )+840 A \left (\cos ^{3}\left (d x +c \right )\right )+935 B \left (\cos ^{3}\left (d x +c \right )\right )+495 C \left (\cos ^{3}\left (d x +c \right )\right )+1008 A \left (\cos ^{2}\left (d x +c \right )\right )+1122 B \left (\cos ^{2}\left (d x +c \right )\right )+1287 C \left (\cos ^{2}\left (d x +c \right )\right )+1344 A \cos \left (d x +c \right )+1496 B \cos \left (d x +c \right )+1716 C \cos \left (d x +c \right )+2688 A +2992 B +3432 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{6}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} a}{3465 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465/d*(-1+cos(d*x+c))*(315*A*cos(d*x+c)^5+735*A*cos(d*x+c)^4+385*B*cos(d*x+c)^4+840*A*cos(d*x+c)^3+935*B*c
os(d*x+c)^3+495*C*cos(d*x+c)^3+1008*A*cos(d*x+c)^2+1122*B*cos(d*x+c)^2+1287*C*cos(d*x+c)^2+1344*A*cos(d*x+c)+1
496*B*cos(d*x+c)+1716*C*cos(d*x+c)+2688*A+2992*B+3432*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^6*(1/c
os(d*x+c))^(11/2)/sin(d*x+c)*a

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maxima [B]  time = 0.86, size = 1188, normalized size = 4.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/110880*(21*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 +
11/2*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) + 42
9*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*
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) + 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*ar
ctan2(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 + 1
1/2*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))) + 3
630*a*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))*A*sqrt(a) + 22*sqrt(2)*(3780*a*cos(8/
9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 1050*a*cos(2/3*arctan2(sin(9/2*d
*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 378*a*cos(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2
*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 135*a*cos(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(
9/2*d*x + 9/2*c) - 3780*a*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) -
1050*a*cos(9/2*d*x + 9/2*c)*sin(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 378*a*cos(9/2*d*x +
9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 135*a*cos(9/2*d*x + 9/2*c)*sin(2/9*arct
an2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 70*a*sin(9/2*d*x + 9/2*c) + 135*a*sin(7/9*arctan2(sin(9/2*d
*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 378*a*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 105
0*a*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 3780*a*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c)
, cos(9/2*d*x + 9/2*c))))*B*sqrt(a) + 132*sqrt(2)*(735*a*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) + 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*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))) + 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*ar
ctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*C*sqrt(a))/d

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mupad [B]  time = 11.49, size = 392, normalized size = 1.38 \[ \frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {a\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (3\,A+2\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{72\,d}+\frac {A\,a\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{88\,d}+\frac {a\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (7\,A+6\,B+4\,C\right )}{56\,d}+\frac {a\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (9\,A+10\,B+10\,C\right )}{12\,d}+\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (11\,A+12\,B+14\,C\right )}{4\,d}+\frac {a\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (13\,A+12\,B+12\,C\right )}{40\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^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((9*c)/2 + (9*d*x)/2)*(3*A + 2*B)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 +
1))/(72*d) + (A*a*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*sin((7*c)/2 + (7*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1)
*(7*A + 6*B + 4*C))/(56*d) + (a*sin((3*c)/2 + (3*d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11*c)/4 + (11
*d*x)/4)^2 + 1)*(9*A + 10*B + 10*C))/(12*d) + (a*sin(c/2 + (d*x)/2)*(sin((11*c)/2 + (11*d*x)/2)*1i - 2*sin((11
*c)/4 + (11*d*x)/4)^2 + 1)*(11*A + 12*B + 14*C))/(4*d) + (a*sin((5*c)/2 + (5*d*x)/2)*(sin((11*c)/2 + (11*d*x)/
2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1)*(13*A + 12*B + 12*C))/(40*d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1)
)^(1/2)*(2*sin(c/4 + (d*x)/4)^2 - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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