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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.989349, antiderivative size = 334, normalized size of antiderivative = 1., 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 = {3043, 2975, 2980, 2772, 2771} \[ \frac{8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (136 A+182 B+143 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{1287 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{16 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a (5 A+13 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{13 d \cos ^{\frac{13}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3043

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

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 \frac{(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{15}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (5 A+13 B)+\frac{1}{2} a (6 A+13 C) \cos (c+d x)\right )}{\cos ^{\frac{13}{2}}(c+d x)} \, dx}{13 a}\\ &=\frac{2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \cos (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) \cos (c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx}{143 a}\\ &=\frac{2 a^2 (136 A+182 B+143 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1287 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+a \cos (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) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{1287 a}\\ &=\frac{2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (136 A+182 B+143 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1287 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}+\frac{\left (a^2 (8368 A+9230 B+10439 C)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{3003}\\ &=\frac{2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (136 A+182 B+143 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1287 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}+\frac{\left (4 a^2 (8368 A+9230 B+10439 C)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{15015}\\ &=\frac{2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (136 A+182 B+143 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1287 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}+\frac{\left (8 a^2 (8368 A+9230 B+10439 C)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{45045}\\ &=\frac{2 a^3 (2224 A+2522 B+2717 C) \sin (c+d x)}{9009 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{15015 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{8 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{16 a^3 (8368 A+9230 B+10439 C) \sin (c+d x)}{45045 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (136 A+182 B+143 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{1287 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2 a (5 A+13 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{143 d \cos ^{\frac{11}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{13 d \cos ^{\frac{13}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 1.21634, size = 224, normalized size = 0.67 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} (70 (5552 A+5083 B+4576 C) \cos (c+d x)+14 (30334 A+31850 B+32747 C) \cos (2 (c+d x))+125520 A \cos (3 (c+d x))+125520 A \cos (4 (c+d x))+16736 A \cos (5 (c+d x))+16736 A \cos (6 (c+d x))+343612 A+138450 B \cos (3 (c+d x))+138450 B \cos (4 (c+d x))+18460 B \cos (5 (c+d x))+18460 B \cos (6 (c+d x))+325910 B+141570 C \cos (3 (c+d x))+156585 C \cos (4 (c+d x))+20878 C \cos (5 (c+d x))+20878 C \cos (6 (c+d x))+322751 C)}{180180 d \cos ^{\frac{13}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(343612*A + 325910*B + 322751*C + 70*(5552*A + 5083*B + 4576*C)*Cos[c + d*x] +
14*(30334*A + 31850*B + 32747*C)*Cos[2*(c + d*x)] + 125520*A*Cos[3*(c + d*x)] + 138450*B*Cos[3*(c + d*x)] + 1
41570*C*Cos[3*(c + d*x)] + 125520*A*Cos[4*(c + d*x)] + 138450*B*Cos[4*(c + d*x)] + 156585*C*Cos[4*(c + d*x)] +
16736*A*Cos[5*(c + d*x)] + 18460*B*Cos[5*(c + d*x)] + 20878*C*Cos[5*(c + d*x)] + 16736*A*Cos[6*(c + d*x)] + 1
8460*B*Cos[6*(c + d*x)] + 20878*C*Cos[6*(c + d*x)])*Tan[(c + d*x)/2])/(180180*d*Cos[c + d*x]^(13/2))

________________________________________________________________________________________

Maple [A]  time = 0.11, size = 232, normalized size = 0.7 \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}+83512\,C \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}+41756\,C \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}+31317\,C \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}+18590\,C \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}+5005\,C \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 ) }{45045\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(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+83512*C*cos(d*x+c)^6+33472*A*cos(d*x
+c)^5+36920*B*cos(d*x+c)^5+41756*C*cos(d*x+c)^5+25104*A*cos(d*x+c)^4+27690*B*cos(d*x+c)^4+31317*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+18305*A*cos(d*x+c)^2+14560*B*cos(d*x+c)^2+5005
*C*cos(d*x+c)^2+11970*A*cos(d*x+c)+4095*B*cos(d*x+c)+3465*A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(1
3/2)

________________________________________________________________________________________

Maxima [B]  time = 1.80642, size = 1355, normalized size = 4.06 \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)+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x, algorithm="maxima")

[Out]

8/45045*(143*(315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 945*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 + 1449*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1287*sqrt(2)*a^(5/2)*sin(d*x + c)^7
/(cos(d*x + c) + 1)^7 + 572*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 104*sqrt(2)*a^(5/2)*sin(d*x
+ c)^11/(cos(d*x + c) + 1)^11)*C*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1)
+ 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)) + 65*(693*sqrt(2)*a^(5/2)*sin(d*x + c)
/(cos(d*x + c) + 1) - 2310*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 4620*sqrt(2)*a^(5/2)*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 - 5478*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3575*sqrt(2)*a^(5/2)*
sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 1300*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 200*sqrt(2)
*a^(5/2)*sin(d*x + c)^13/(cos(d*x + c) + 1)^13)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(
cos(d*x + c) + 1) + 1)^(13/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(13/2)*(4*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x
+ c) + 1)^8 + 1)) + (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*s
qrt(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)))/d

________________________________________________________________________________________

Fricas [A]  time = 1.94444, size = 547, normalized size = 1.64 \begin{align*} \frac{2 \,{\left (8 \,{\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + 4 \,{\left (8368 \, A + 9230 \, B + 10439 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 3 \,{\left (8368 \, A + 9230 \, B + 10439 \, 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} + 35 \,{\left (523 \, A + 416 \, B + 143 \, C\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} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{45045 \,{\left (d \cos \left (d x + c\right )^{8} + d \cos \left (d x + c\right )^{7}\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)+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)+C*cos(d*x+c)**2)/cos(d*x+c)**(15/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{15}{2}}}\,{d x} \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)+C*cos(d*x+c)^2)/cos(d*x+c)^(15/2),x, algorithm="giac")

[Out]

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