3.1262 \(\int \cos ^{\frac {13}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

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

[Out]

2/143*a*(5*A+13*B)*cos(d*x+c)^(9/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/13*A*cos(d*x+c)^(11/2)*(a+a*sec(d*x+
c))^(5/2)*sin(d*x+c)/d+2/15015*a^3*(8368*A+9230*B+10439*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2
)+2/9009*a^3*(2224*A+2522*B+2717*C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/45045*a^3*(8368*A+
9230*B+10439*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+8/45045*a^3*(8368*A+9230*B+10439*C)*sin(d
*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/1287*a^2*(136*A+182*B+143*C)*cos(d*x+c)^(7/2)*sin(d*x+c)*(a+
a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 1.10, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4265, 4086, 4017, 4015, 3805, 3804} \[ \frac {2 a^2 (136 A+182 B+143 C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{1287 d}+\frac {2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{9009 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{15015 d \sqrt {a \sec (c+d x)+a}}+\frac {8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{45045 d \sqrt {a \sec (c+d x)+a}}+\frac {16 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {2 a (5 A+13 B) \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{143 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {11}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

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

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Mathematica [A]  time = 2.49, size = 190, normalized size = 0.57 \[ \frac {a^2 \sqrt {\cos (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} (4 (453146 A+454285 B+445588 C) \cos (c+d x)+(746519 A+676000 B+581152 C) \cos (2 (c+d x))+287060 A \cos (3 (c+d x))+94010 A \cos (4 (c+d x))+23940 A \cos (5 (c+d x))+3465 A \cos (6 (c+d x))+2798182 A+225550 B \cos (3 (c+d x))+58240 B \cos (4 (c+d x))+8190 B \cos (5 (c+d x))+2980640 B+148720 C \cos (3 (c+d x))+20020 C \cos (4 (c+d x))+3233516 C)}{720720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[Cos[c + d*x]]*(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)] + 1
48720*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)] + 23
940*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)

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fricas [A]  time = 0.44, size = 189, normalized size = 0.57 \[ \frac {2 \, {\left (3465 \, A a^{2} \cos \left (d x + c\right )^{6} + 315 \, {\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 35 \, {\left (523 \, A + 416 \, B + 143 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \, {\left (4184 \, A + 4615 \, B + 3718 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 4 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 2.04, size = 222, normalized size = 0.66 \[ -\frac {2 a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (3465 A \left (\cos ^{6}\left (d x +c \right )\right )+11970 A \left (\cos ^{5}\left (d x +c \right )\right )+4095 B \left (\cos ^{5}\left (d x +c \right )\right )+18305 A \left (\cos ^{4}\left (d x +c \right )\right )+14560 B \left (\cos ^{4}\left (d x +c \right )\right )+5005 C \left (\cos ^{4}\left (d x +c \right )\right )+20920 A \left (\cos ^{3}\left (d x +c \right )\right )+23075 B \left (\cos ^{3}\left (d x +c \right )\right )+18590 C \left (\cos ^{3}\left (d x +c \right )\right )+25104 A \left (\cos ^{2}\left (d x +c \right )\right )+27690 B \left (\cos ^{2}\left (d x +c \right )\right )+31317 C \left (\cos ^{2}\left (d x +c \right )\right )+33472 A \cos \left (d x +c \right )+36920 B \cos \left (d x +c \right )+41756 C \cos \left (d x +c \right )+66944 A +73840 B +83512 C \right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{45045 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45045/d*a^2*(-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+25
104*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)*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)

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maxima [B]  time = 0.86, size = 1135, normalized size = 3.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(13/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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) + 1144*sqrt(2)*(225*a^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 378*a^2*sin(5/4*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2100*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4095*a^2
*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 63*(65*a^2*sin(4*d*x + 4*c) + 6*a^2*sin(2*d*x + 2*c))*
cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 7*(585*a^2*cos(4*d*x + 4*c) + 54*a^2*cos(2*d*x + 2*c) +
 5*a^2)*sin(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(a) + 130*(770*sqrt(2)*a^2*sin(9/4*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1287*sqrt(2)*a^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
+ 6930*sqrt(2)*a^2*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 8778*sqrt(2)*a^2*sin(3/4*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c))) + 63756*sqrt(2)*a^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3
3*(266*sqrt(2)*a^2*sin(4*d*x + 4*c) + 39*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(11/4*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c))) + 3*(2926*sqrt(2)*a^2*cos(4*d*x + 4*c) + 429*sqrt(2)*a^2*cos(2*d*x + 2*c) + 42*sqrt(2)*a^2)*sin
(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B*sqrt(a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{13/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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