3.247 \(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=275 \[ \frac {2 a^3 (710 A+803 B) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a^3 (710 A+803 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

[Out]

2/11*a*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(9/2)+2/693*a^3*(194*A+209*B)*sin(d*x+c)/d/sec(d*x+c)^
(5/2)/(a+a*sec(d*x+c))^(1/2)+2/1155*a^3*(710*A+803*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+8/3
465*a^3*(710*A+803*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/3465*a^3*(710*A+803*B)*sin(d*x+c
)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a^2*(14*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 = 0.70, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {4017, 4015, 3805, 3804} \[ \frac {2 a^3 (710 A+803 B) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (14 A+11 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{99 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {16 a^3 (710 A+803 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3465 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{11 d \sec ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(194*A + 209*B)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(710*A + 803
*B)*Sin[c + d*x])/(1155*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a^3*(710*A + 803*B)*Sin[c + d*x])/
(3465*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^3*(710*A + 803*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x]
)/(3465*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(14*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 + 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]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2}{11} \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (14 A+11 B)+\frac {1}{2} a (6 A+11 B) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (14 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+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {4}{99} \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (194 A+209 B)+\frac {3}{4} a^2 (46 A+55 B) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (14 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+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {1}{231} \left (a^2 (710 A+803 B)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (14 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+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {\left (4 a^2 (710 A+803 B)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{1155}\\ &=\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (14 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+a \sec (c+d x))^{3/2} \sin (c+d x)}{11 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {\left (8 a^2 (710 A+803 B)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3465}\\ &=\frac {2 a^3 (194 A+209 B) \sin (c+d x)}{693 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (710 A+803 B) \sin (c+d x)}{1155 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (710 A+803 B) \sin (c+d x)}{3465 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^3 (710 A+803 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3465 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (14 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+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 = 4.25, size = 127, normalized size = 0.46 \[ \frac {2 a^3 \sin (c+d x) \left (8 (710 A+803 B) \sec ^5(c+d x)+4 (710 A+803 B) \sec ^4(c+d x)+3 (710 A+803 B) \sec ^3(c+d x)+5 (355 A+286 B) \sec ^2(c+d x)+35 (32 A+11 B) \sec (c+d x)+315 A\right )}{3465 d \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(315*A + 35*(32*A + 11*B)*Sec[c + d*x] + 5*(355*A + 286*B)*Sec[c + d*x]^2 + 3*(710*A + 803*B)*Sec[c + d
*x]^3 + 4*(710*A + 803*B)*Sec[c + d*x]^4 + 8*(710*A + 803*B)*Sec[c + d*x]^5)*Sin[c + d*x])/(3465*d*Sec[c + d*x
]^(9/2)*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.48, size = 162, normalized size = 0.59 \[ \frac {2 \, {\left (315 \, A a^{2} \cos \left (d x + c\right )^{6} + 35 \, {\left (32 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (355 \, A + 286 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (710 \, A + 803 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*A*a^2*cos(d*x + c)^6 + 35*(32*A + 11*B)*a^2*cos(d*x + c)^5 + 5*(355*A + 286*B)*a^2*cos(d*x + c)^4
+ 3*(710*A + 803*B)*a^2*cos(d*x + c)^3 + 4*(710*A + 803*B)*a^2*cos(d*x + c)^2 + 8*(710*A + 803*B)*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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 2.79, size = 165, normalized size = 0.60 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (315 A \left (\cos ^{5}\left (d x +c \right )\right )+1120 A \left (\cos ^{4}\left (d x +c \right )\right )+385 B \left (\cos ^{4}\left (d x +c \right )\right )+1775 A \left (\cos ^{3}\left (d x +c \right )\right )+1430 B \left (\cos ^{3}\left (d x +c \right )\right )+2130 A \left (\cos ^{2}\left (d x +c \right )\right )+2409 B \left (\cos ^{2}\left (d x +c \right )\right )+2840 A \cos \left (d x +c \right )+3212 B \cos \left (d x +c \right )+5680 A +6424 B \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{6}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} a^{2}}{3465 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(11/2),x)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(315*A*cos(d*x+c)^5+1120*A*cos(d*x+c)^4+385*B*cos(d*x+c)^4+1775*A*cos(d*x+c)^3+1430*
B*cos(d*x+c)^3+2130*A*cos(d*x+c)^2+2409*B*cos(d*x+c)^2+2840*A*cos(d*x+c)+3212*B*cos(d*x+c)+5680*A+6424*B)*(a*(
1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^6*(1/cos(d*x+c))^(11/2)/sin(d*x+c)*a^2

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maxima [B]  time = 1.51, size = 945, normalized size = 3.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/110880*(5*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*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) + 385*a^2*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) - 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(
8/11*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*arcta
n2(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))) + 3
465*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 + 1
1/2*c))))*A*sqrt(a) + 22*sqrt(2)*(8190*a^2*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) + 2100*a^2*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) +
756*a^2*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) + 225*a^2*cos(2/9*ar
ctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) - 8190*a^2*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))) - 2100*a^2*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))) - 756*a^2*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))) - 225*a^2*cos(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))
) + 70*a^2*sin(9/2*d*x + 9/2*c) + 225*a^2*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 756*a
^2*sin(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 2100*a^2*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c
), cos(9/2*d*x + 9/2*c))) + 8190*a^2*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*B*sqrt(a))/
d

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mupad [B]  time = 8.77, size = 392, normalized size = 1.43 \[ \frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {A\,a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{88\,d}+\frac {a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (5\,A+2\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{72\,d}+\frac {a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (13\,A+10\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{56\,d}+\frac {a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (19\,A+20\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{12\,d}+\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (23\,A+26\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{4\,d}+\frac {a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (25\,A+24\,B\right )\,\left (-2\,{\sin \left (\frac {11\,c}{4}+\frac {11\,d\,x}{4}\right )}^2+\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{40\,d}\right )}{2\,\sqrt {-\frac {1}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {c}{4}+\frac {d\,x}{4}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(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)*((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))/(8
8*d) + (a^2*sin((9*c)/2 + (9*d*x)/2)*(5*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^2*sin((7*c)/2 + (7*d*x)/2)*(13*A + 10*B)*(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 + 20*B)*(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 + 26*B)*(sin((11*c)/2 + (11*d*x)/2)*
1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(4*d) + (a^2*sin((5*c)/2 + (5*d*x)/2)*(25*A + 24*B)*(sin((11*c)/2 +
(11*d*x)/2)*1i - 2*sin((11*c)/4 + (11*d*x)/4)^2 + 1))/(40*d)))/(2*(-1/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(2*s
in(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))**(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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