3.510 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac{15}{2}}(c+d x) \, dx\)

Optimal. Leaf size=322 \[ \frac{2 a^2 (16 A+13 B) \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{143 d}+\frac{2 a^3 (280 A+299 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{1287 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{9009 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{15015 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{45045 d \sqrt{a \cos (c+d x)+a}}+\frac{32 a^3 (4184 A+4615 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{45045 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{13}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d} \]

[Out]

(32*a^3*(4184*A + 4615*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(45045*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^3*(4184*
A + 4615*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(45045*d*Sqrt[a + a*Cos[c + d*x]]) + (4*a^3*(4184*A + 4615*B)*Sec
[c + d*x]^(5/2)*Sin[c + d*x])/(15015*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(4184*A + 4615*B)*Sec[c + d*x]^(7/2)
*Sin[c + d*x])/(9009*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(280*A + 299*B)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(12
87*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(16*A + 13*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(11/2)*Sin[c + d*x
])/(143*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(13/2)*Sin[c + d*x])/(13*d)

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Rubi [A]  time = 0.938889, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2975, 2980, 2772, 2771} \[ \frac{2 a^2 (16 A+13 B) \sin (c+d x) \sec ^{\frac{11}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{143 d}+\frac{2 a^3 (280 A+299 B) \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{1287 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{9009 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{15015 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a^3 (4184 A+4615 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{45045 d \sqrt{a \cos (c+d x)+a}}+\frac{32 a^3 (4184 A+4615 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{45045 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{13}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{13 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(15/2),x]

[Out]

(32*a^3*(4184*A + 4615*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(45045*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a^3*(4184*
A + 4615*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(45045*d*Sqrt[a + a*Cos[c + d*x]]) + (4*a^3*(4184*A + 4615*B)*Sec
[c + d*x]^(5/2)*Sin[c + d*x])/(15015*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(4184*A + 4615*B)*Sec[c + d*x]^(7/2)
*Sin[c + d*x])/(9009*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(280*A + 299*B)*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(12
87*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(16*A + 13*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(11/2)*Sin[c + d*x
])/(143*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(13/2)*Sin[c + d*x])/(13*d)

Rule 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac{15}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{15}{2}}(c+d x)} \, dx\\ &=\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac{1}{13} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (16 A+13 B)+\frac{1}{2} a (8 A+13 B) \cos (c+d x)\right )}{\cos ^{\frac{13}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (16 A+13 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac{1}{143} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{4} a^2 (280 A+299 B)+\frac{1}{4} a^2 (216 A+247 B) \cos (c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (280 A+299 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (16 A+13 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac{\left (a^2 (4184 A+4615 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{1287}\\ &=\frac{2 a^3 (4184 A+4615 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (280 A+299 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (16 A+13 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac{\left (2 a^2 (4184 A+4615 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{3003}\\ &=\frac{4 a^3 (4184 A+4615 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (4184 A+4615 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (280 A+299 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (16 A+13 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac{\left (8 a^2 (4184 A+4615 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{15015}\\ &=\frac{16 a^3 (4184 A+4615 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt{a+a \cos (c+d x)}}+\frac{4 a^3 (4184 A+4615 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (4184 A+4615 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (280 A+299 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (16 A+13 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}+\frac{\left (16 a^2 (4184 A+4615 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{45045}\\ &=\frac{32 a^3 (4184 A+4615 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{45045 d \sqrt{a+a \cos (c+d x)}}+\frac{16 a^3 (4184 A+4615 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45045 d \sqrt{a+a \cos (c+d x)}}+\frac{4 a^3 (4184 A+4615 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{15015 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (4184 A+4615 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9009 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (280 A+299 B) \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{1287 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (16 A+13 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{11}{2}}(c+d x) \sin (c+d x)}{143 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{13}{2}}(c+d x) \sin (c+d x)}{13 d}\\ \end{align*}

Mathematica [A]  time = 0.890915, size = 171, normalized size = 0.53 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{13}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} (35 (5552 A+5083 B) \cos (c+d x)+14 (15167 A+15925 B) \cos (2 (c+d x))+62760 A \cos (3 (c+d x))+62760 A \cos (4 (c+d x))+8368 A \cos (5 (c+d x))+8368 A \cos (6 (c+d x))+171806 A+69225 B \cos (3 (c+d x))+69225 B \cos (4 (c+d x))+9230 B \cos (5 (c+d x))+9230 B \cos (6 (c+d x))+162955 B)}{90090 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(15/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(171806*A + 162955*B + 35*(5552*A + 5083*B)*Cos[c + d*x] + 14*(15167*A + 15925
*B)*Cos[2*(c + d*x)] + 62760*A*Cos[3*(c + d*x)] + 69225*B*Cos[3*(c + d*x)] + 62760*A*Cos[4*(c + d*x)] + 69225*
B*Cos[4*(c + d*x)] + 8368*A*Cos[5*(c + d*x)] + 9230*B*Cos[5*(c + d*x)] + 8368*A*Cos[6*(c + d*x)] + 9230*B*Cos[
6*(c + d*x)])*Sec[c + d*x]^(13/2)*Tan[(c + d*x)/2])/(90090*d)

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Maple [A]  time = 0.651, size = 185, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 66944\,A \left ( \cos \left ( dx+c \right ) \right ) ^{6}+73840\,B \left ( \cos \left ( dx+c \right ) \right ) ^{6}+33472\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+36920\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+25104\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+27690\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+20920\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+23075\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+18305\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+14560\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+11970\,A\cos \left ( dx+c \right ) +4095\,B\cos \left ( dx+c \right ) +3465\,A \right ) \cos \left ( dx+c \right ) }{45045\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{15}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x)

[Out]

-2/45045/d*a^2*(-1+cos(d*x+c))*(66944*A*cos(d*x+c)^6+73840*B*cos(d*x+c)^6+33472*A*cos(d*x+c)^5+36920*B*cos(d*x
+c)^5+25104*A*cos(d*x+c)^4+27690*B*cos(d*x+c)^4+20920*A*cos(d*x+c)^3+23075*B*cos(d*x+c)^3+18305*A*cos(d*x+c)^2
+14560*B*cos(d*x+c)^2+11970*A*cos(d*x+c)+4095*B*cos(d*x+c)+3465*A)*cos(d*x+c)*(1/cos(d*x+c))^(15/2)*(a*(1+cos(
d*x+c)))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.19606, size = 1030, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, algorithm="maxima")

[Out]

8/45045*((45045*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 165165*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 414414*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 604890*sqrt(2)*a^(5/2)*sin(d*x +
c)^7/(cos(d*x + c) + 1)^7 + 522665*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 289185*sqrt(2)*a^(5/
2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 88980*sqrt(2)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 11864
*sqrt(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*
x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*sin(d*x + c
)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 1)) + 65*(693*sqrt(2)*a^(5/2)*sin(d*x + c)/
(cos(d*x + c) + 1) - 3003*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6930*sqrt(2)*a^(5/2)*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 10098*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9053*sqrt(2)*a^(5/2)*
sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 4875*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 1500*sqrt(2
)*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - 200*sqrt(2)*a^(5/2)*sin(d*x + c)^15/(cos(d*x + c) + 1)^15)*B
*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(15/2)*(-sin(d*x + c)/(cos
(d*x + c) + 1) + 1)^(15/2)*(5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1
0*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c)
+ 1)^10 + 1)))/d

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Fricas [A]  time = 1.51198, size = 487, normalized size = 1.51 \begin{align*} \frac{2 \,{\left (16 \,{\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} + 8 \,{\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} + 6 \,{\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 5 \,{\left (4184 \, A + 4615 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 35 \,{\left (523 \, A + 416 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 315 \,{\left (38 \, A + 13 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3465 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45045 \,{\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, algorithm="fricas")

[Out]

2/45045*(16*(4184*A + 4615*B)*a^2*cos(d*x + c)^6 + 8*(4184*A + 4615*B)*a^2*cos(d*x + c)^5 + 6*(4184*A + 4615*B
)*a^2*cos(d*x + c)^4 + 5*(4184*A + 4615*B)*a^2*cos(d*x + c)^3 + 35*(523*A + 416*B)*a^2*cos(d*x + c)^2 + 315*(3
8*A + 13*B)*a^2*cos(d*x + c) + 3465*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^7 + d*cos(d*
x + c)^6)*sqrt(cos(d*x + c)))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(15/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(15/2),x, algorithm="giac")

[Out]

Timed out