3.604 \(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {13}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=334 \[ \frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {16 a^3 (8368 A+9230 B+10439 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{45045 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)} \]

[Out]

2/143*a*(5*A+13*B)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(9/2)+2/13*A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+
c)/d/sec(d*x+c)^(11/2)+2/9009*a^3*(2224*A+2522*B+2717*C)*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+
2/15015*a^3*(8368*A+9230*B+10439*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+8/45045*a^3*(8368*A+9
230*B+10439*C)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/45045*a^3*(8368*A+9230*B+10439*C)*sin(d
*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/1287*a^2*(136*A+182*B+143*C)*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 = 0.94, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (136 A+182 B+143 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {16 a^3 (8368 A+9230 B+10439 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{45045 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 A+13 B) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d \sec ^{\frac {11}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(13/2),x]

[Out]

(2*a^3*(2224*A + 2522*B + 2717*C)*Sin[c + d*x])/(9009*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*
(8368*A + 9230*B + 10439*C)*Sin[c + d*x])/(15015*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (8*a^3*(8368
*A + 9230*B + 10439*C)*Sin[c + d*x])/(45045*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (16*a^3*(8368*A +
9230*B + 10439*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(45045*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(136*A + 182*B
+ 143*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(1287*d*Sec[c + d*x]^(7/2)) + (2*a*(5*A + 13*B)*(a + a*Sec[c
+ d*x])^(3/2)*Sin[c + d*x])/(143*d*Sec[c + d*x]^(9/2)) + (2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(13*d*S
ec[c + d*x]^(11/2))

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))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {13}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+13 B)+\frac {1}{2} a (6 A+13 C) \sec (c+d x)\right )}{\sec ^{\frac {11}{2}}(c+d x)} \, dx}{13 a}\\ &=\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (136 A+182 B+143 C)+\frac {1}{4} a^2 (96 A+78 B+143 C) \sec (c+d x)\right )}{\sec ^{\frac {9}{2}}(c+d x)} \, dx}{143 a}\\ &=\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (2224 A+2522 B+2717 C)+\frac {3}{8} a^3 (560 A+598 B+715 C) \sec (c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx}{1287 a}\\ &=\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}+\frac {\left (a^2 (8368 A+9230 B+10439 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{3003}\\ &=\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}+\frac {\left (4 a^2 (8368 A+9230 B+10439 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{15015}\\ &=\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}+\frac {\left (8 a^2 (8368 A+9230 B+10439 C)\right ) \int \frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{45045}\\ &=\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^3 (8368 A+9230 B+10439 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (136 A+182 B+143 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{1287 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {2 a (5 A+13 B) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{143 d \sec ^{\frac {9}{2}}(c+d x)}+\frac {2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{13 d \sec ^{\frac {11}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 1.61, size = 190, normalized size = 0.57 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} (4 (453146 A+454285 B+445588 C) \cos (c+d x)+(746519 A+676000 B+581152 C) \cos (2 (c+d x))+287060 A \cos (3 (c+d x))+94010 A \cos (4 (c+d x))+23940 A \cos (5 (c+d x))+3465 A \cos (6 (c+d x))+2798182 A+225550 B \cos (3 (c+d x))+58240 B \cos (4 (c+d x))+8190 B \cos (5 (c+d x))+2980640 B+148720 C \cos (3 (c+d x))+20020 C \cos (4 (c+d x))+3233516 C)}{720720 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(13/2),x]

[Out]

(a^2*(2798182*A + 2980640*B + 3233516*C + 4*(453146*A + 454285*B + 445588*C)*Cos[c + d*x] + (746519*A + 676000
*B + 581152*C)*Cos[2*(c + d*x)] + 287060*A*Cos[3*(c + d*x)] + 225550*B*Cos[3*(c + d*x)] + 148720*C*Cos[3*(c +
d*x)] + 94010*A*Cos[4*(c + d*x)] + 58240*B*Cos[4*(c + d*x)] + 20020*C*Cos[4*(c + d*x)] + 23940*A*Cos[5*(c + d*
x)] + 8190*B*Cos[5*(c + d*x)] + 3465*A*Cos[6*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(720720*
d*Sqrt[Sec[c + d*x]])

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fricas [A]  time = 0.44, size = 197, normalized size = 0.59 \[ \frac {2 \, {\left (3465 \, A a^{2} \cos \left (d x + c\right )^{7} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + 35 \, {\left (523 \, A + 416 \, B + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (4184 \, A + 4615 \, B + 3718 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]

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)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

2/45045*(3465*A*a^2*cos(d*x + c)^7 + 315*(38*A + 13*B)*a^2*cos(d*x + c)^6 + 35*(523*A + 416*B + 143*C)*a^2*cos
(d*x + c)^5 + 5*(4184*A + 4615*B + 3718*C)*a^2*cos(d*x + c)^4 + 3*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c)
^3 + 4*(8368*A + 9230*B + 10439*C)*a^2*cos(d*x + c)^2 + 8*(8368*A + 9230*B + 10439*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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {13}{2}}}\,{d x} \]

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)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(13/2), x)

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maple [A]  time = 2.63, size = 232, normalized size = 0.69 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (3465 A \left (\cos ^{6}\left (d x +c \right )\right )+11970 A \left (\cos ^{5}\left (d x +c \right )\right )+4095 B \left (\cos ^{5}\left (d x +c \right )\right )+18305 A \left (\cos ^{4}\left (d x +c \right )\right )+14560 B \left (\cos ^{4}\left (d x +c \right )\right )+5005 C \left (\cos ^{4}\left (d x +c \right )\right )+20920 A \left (\cos ^{3}\left (d x +c \right )\right )+23075 B \left (\cos ^{3}\left (d x +c \right )\right )+18590 C \left (\cos ^{3}\left (d x +c \right )\right )+25104 A \left (\cos ^{2}\left (d x +c \right )\right )+27690 B \left (\cos ^{2}\left (d x +c \right )\right )+31317 C \left (\cos ^{2}\left (d x +c \right )\right )+33472 A \cos \left (d x +c \right )+36920 B \cos \left (d x +c \right )+41756 C \cos \left (d x +c \right )+66944 A +73840 B +83512 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos ^{7}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {13}{2}} a^{2}}{45045 d \sin \left (d x +c \right )} \]

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)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x)

[Out]

-2/45045/d*(-1+cos(d*x+c))*(3465*A*cos(d*x+c)^6+11970*A*cos(d*x+c)^5+4095*B*cos(d*x+c)^5+18305*A*cos(d*x+c)^4+
14560*B*cos(d*x+c)^4+5005*C*cos(d*x+c)^4+20920*A*cos(d*x+c)^3+23075*B*cos(d*x+c)^3+18590*C*cos(d*x+c)^3+25104*
A*cos(d*x+c)^2+27690*B*cos(d*x+c)^2+31317*C*cos(d*x+c)^2+33472*A*cos(d*x+c)+36920*B*cos(d*x+c)+41756*C*cos(d*x
+c)+66944*A+73840*B+83512*C)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^7*(1/cos(d*x+c))^(13/2)/sin(d*x+c)
*a^2

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maxima [B]  time = 0.88, size = 1562, normalized size = 4.68 \[ \text {result too large to display} \]

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)+C*sec(d*x+c)^2)/sec(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

1/2882880*(sqrt(2)*(3783780*a^2*cos(12/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*
x + 13/2*c) + 1066065*a^2*cos(10/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13
/2*c) + 459459*a^2*cos(8/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) +
193050*a^2*cos(6/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) + 70070*a^
2*cos(4/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) + 20475*a^2*cos(2/1
3*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))*sin(13/2*d*x + 13/2*c) - 3783780*a^2*cos(13/2*d*x +
13/2*c)*sin(12/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 1066065*a^2*cos(13/2*d*x + 13/2*
c)*sin(10/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 459459*a^2*cos(13/2*d*x + 13/2*c)*sin(
8/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 193050*a^2*cos(13/2*d*x + 13/2*c)*sin(6/13*arc
tan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 70070*a^2*cos(13/2*d*x + 13/2*c)*sin(4/13*arctan2(sin(
13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 20475*a^2*cos(13/2*d*x + 13/2*c)*sin(2/13*arctan2(sin(13/2*d*x
+ 13/2*c), cos(13/2*d*x + 13/2*c))) + 6930*a^2*sin(13/2*d*x + 13/2*c) + 20475*a^2*sin(11/13*arctan2(sin(13/2*d
*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 70070*a^2*sin(9/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/
2*c))) + 193050*a^2*sin(7/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 459459*a^2*sin(5/13*ar
ctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 1066065*a^2*sin(3/13*arctan2(sin(13/2*d*x + 13/2*c),
cos(13/2*d*x + 13/2*c))) + 3783780*a^2*sin(1/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))))*A*sq
rt(a) + 130*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))))*B*sqrt(a) + 572*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*a
rctan2(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), co
s(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))))*C*sqrt(a))
/d

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mupad [B]  time = 12.74, size = 458, normalized size = 1.37 \[ \frac {\sqrt {a-\frac {a}{2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1}}\,\left (2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {A\,a^2\,\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{208\,d}+\frac {a^2\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (7\,A+5\,B+2\,C\right )}{72\,d}+\frac {a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (15\,A+13\,B+10\,C\right )}{56\,d}+\frac {a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (21\,A+23\,B+26\,C\right )}{4\,d}+\frac {a^2\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (51\,A+50\,B+48\,C\right )}{80\,d}+\frac {a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )\,\left (71\,A+76\,B+80\,C\right )}{48\,d}+\frac {a^2\,\sin \left (\frac {11\,c}{2}+\frac {11\,d\,x}{2}\right )\,\left (5\,A+2\,B\right )\,\left (-2\,{\sin \left (\frac {13\,c}{4}+\frac {13\,d\,x}{4}\right )}^2+\sin \left (\frac {13\,c}{2}+\frac {13\,d\,x}{2}\right )\,1{}\mathrm {i}+1\right )}{176\,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))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(13/2),x)

[Out]

((a - a/(2*sin(c/2 + (d*x)/2)^2 - 1))^(1/2)*(sin((13*c)/2 + (13*d*x)/2)*1i + 2*sin((13*c)/4 + (13*d*x)/4)^2 -
1)*((A*a^2*sin((13*c)/2 + (13*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1))/(2
08*d) + (a^2*sin((9*c)/2 + (9*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1)*(7*
A + 5*B + 2*C))/(72*d) + (a^2*sin((7*c)/2 + (7*d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d
*x)/4)^2 + 1)*(15*A + 13*B + 10*C))/(56*d) + (a^2*sin(c/2 + (d*x)/2)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((1
3*c)/4 + (13*d*x)/4)^2 + 1)*(21*A + 23*B + 26*C))/(4*d) + (a^2*sin((5*c)/2 + (5*d*x)/2)*(sin((13*c)/2 + (13*d*
x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1)*(51*A + 50*B + 48*C))/(80*d) + (a^2*sin((3*c)/2 + (3*d*x)/2)*(s
in((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1)*(71*A + 76*B + 80*C))/(48*d) + (a^2*sin((11
*c)/2 + (11*d*x)/2)*(5*A + 2*B)*(sin((13*c)/2 + (13*d*x)/2)*1i - 2*sin((13*c)/4 + (13*d*x)/4)^2 + 1))/(176*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))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(13/2),x)

[Out]

Timed out

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