3.1091 \(\int \frac {(a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=386 \[ -\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{15 d}+\frac {2 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{5 d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{15 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 B+4 a^3 b (A+3 C)+18 a^2 b^2 B+4 a b^3 (3 A+C)+b^4 B\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (3 A+5 C)+20 a^3 b B+30 a^2 b^2 (A-C)-20 a b^3 B-b^4 (5 A+3 C)\right )}{5 d}+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-2/5*(20*a^3*b*B-20*a*b^3*B+30*a^2*b^2*(A-C)-b^4*(5*A+3*C)+a^4*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2
*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(a^4*B+18*a^2*b^2*B+b^4*B+4*a*b^3*(3*A+C)+4*a^3*b*(A+3
*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d-2/15*b^2*(50*a*b*
B+b^2*(59*A-3*C)+3*a^2*(3*A+5*C))*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/15*(8*A*b+5*B*a)*(a+b*cos(d*x+c))^3*sin(d*x+
c)/d/cos(d*x+c)^(3/2)+2/5*A*(a+b*cos(d*x+c))^4*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/5*(16*A*b^2+15*a*b*B+a^2*(3*A+5
*C))*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/15*b*(105*a^2*b*B-5*b^3*B+4*a*b^2*(33*A-5*C)+6*a^3*(3*
A+5*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 1.29, antiderivative size = 386, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3047, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (A+3 C)+18 a^2 b^2 B+a^4 B+4 a b^3 (3 A+C)+b^4 B\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (30 a^2 b^2 (A-C)+a^4 (3 A+5 C)+20 a^3 b B-20 a b^3 B-b^4 (5 A+3 C)\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (3 a^2 (3 A+5 C)+50 a b B+b^2 (59 A-3 C)\right )}{15 d}+\frac {2 \sin (c+d x) \left (a^2 (3 A+5 C)+15 a b B+16 A b^2\right ) (a+b \cos (c+d x))^2}{5 d \sqrt {\cos (c+d x)}}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^3 (3 A+5 C)+105 a^2 b B+4 a b^2 (33 A-5 C)-5 b^3 B\right )}{15 d}+\frac {2 (5 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^4}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

(-2*(20*a^3*b*B - 20*a*b^3*B + 30*a^2*b^2*(A - C) - b^4*(5*A + 3*C) + a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2,
2])/(5*d) + (2*(a^4*B + 18*a^2*b^2*B + b^4*B + 4*a*b^3*(3*A + C) + 4*a^3*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2
])/(3*d) - (2*b*(105*a^2*b*B - 5*b^3*B + 4*a*b^2*(33*A - 5*C) + 6*a^3*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c +
d*x])/(15*d) - (2*b^2*(50*a*b*B + b^2*(59*A - 3*C) + 3*a^2*(3*A + 5*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*d
) + (2*(16*A*b^2 + 15*a*b*B + a^2*(3*A + 5*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) +
(2*(8*A*b + 5*a*B)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (2*A*(a + b*Cos[c + d*x])
^4*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3033

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

Rule 3047

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/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \cos (c+d x))^3 \left (\frac {1}{2} (8 A b+5 a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \cos (c+d x)-\frac {5}{2} b (A-C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {3}{4} \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right )+\frac {1}{4} \left (5 a^2 B+15 b^2 B+2 a b (A+15 C)\right ) \cos (c+d x)-\frac {5}{4} b (11 A b+5 a B-3 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8}{15} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{8} \left (192 A b^3+5 a^3 B+195 a b^2 B+a^2 (38 A b+90 b C)\right )-\frac {1}{8} \left (65 a^2 b B-15 b^3 B+a b^2 (101 A-45 C)+3 a^3 (3 A+5 C)\right ) \cos (c+d x)-\frac {5}{8} b \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {16}{75} \int \frac {\frac {5}{16} a \left (192 A b^3+5 a^3 B+195 a b^2 B+a^2 (38 A b+90 b C)\right )-\frac {15}{16} \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)-\frac {15}{16} b \left (105 a^2 b B-5 b^3 B+4 a b^2 (33 A-5 C)+6 a^3 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b \left (105 a^2 b B-5 b^3 B+4 a b^2 (33 A-5 C)+6 a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {32}{225} \int \frac {\frac {75}{32} \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right )-\frac {45}{32} \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b \left (105 a^2 b B-5 b^3 B+4 a b^2 (33 A-5 C)+6 a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{3} \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (-20 a^3 b B+20 a b^3 B-30 a^2 b^2 (A-C)+b^4 (5 A+3 C)-a^4 (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (20 a^3 b B-20 a b^3 B+30 a^2 b^2 (A-C)-b^4 (5 A+3 C)+a^4 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (a^4 B+18 a^2 b^2 B+b^4 B+4 a b^3 (3 A+C)+4 a^3 b (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 b \left (105 a^2 b B-5 b^3 B+4 a b^2 (33 A-5 C)+6 a^3 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}-\frac {2 b^2 \left (50 a b B+b^2 (59 A-3 C)+3 a^2 (3 A+5 C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 \left (16 A b^2+15 a b B+a^2 (3 A+5 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (8 A b+5 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 2.38, size = 316, normalized size = 0.82 \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^4 B+20 a^3 A b+60 a^3 b C+90 a^2 b^2 B+60 a A b^3+20 a b^3 C+5 b^4 B\right )+2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-9 a^4 A-15 a^4 C-60 a^3 b B-90 a^2 A b^2+90 a^2 b^2 C+60 a b^3 B+15 A b^4+9 b^4 C\right )}{15 d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2}{5} a^4 A \tan (c+d x) \sec ^2(c+d x)+\frac {2}{3} \sec ^2(c+d x) \left (a^4 B \sin (c+d x)+4 a^3 A b \sin (c+d x)\right )+\frac {2}{5} \sec (c+d x) \left (3 a^4 A \sin (c+d x)+5 a^4 C \sin (c+d x)+20 a^3 b B \sin (c+d x)+30 a^2 A b^2 \sin (c+d x)\right )+\frac {2}{3} b^3 (4 a C+b B) \sin (c+d x)+\frac {1}{5} b^4 C \sin (2 (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]

[Out]

(2*(-9*a^4*A - 90*a^2*A*b^2 + 15*A*b^4 - 60*a^3*b*B + 60*a*b^3*B - 15*a^4*C + 90*a^2*b^2*C + 9*b^4*C)*Elliptic
E[(c + d*x)/2, 2] + 2*(20*a^3*A*b + 60*a*A*b^3 + 5*a^4*B + 90*a^2*b^2*B + 5*b^4*B + 60*a^3*b*C + 20*a*b^3*C)*E
llipticF[(c + d*x)/2, 2])/(15*d) + (Sqrt[Cos[c + d*x]]*((2*b^3*(b*B + 4*a*C)*Sin[c + d*x])/3 + (2*Sec[c + d*x]
^2*(4*a^3*A*b*Sin[c + d*x] + a^4*B*Sin[c + d*x]))/3 + (2*Sec[c + d*x]*(3*a^4*A*Sin[c + d*x] + 30*a^2*A*b^2*Sin
[c + d*x] + 20*a^3*b*B*Sin[c + d*x] + 5*a^4*C*Sin[c + d*x]))/5 + (b^4*C*Sin[2*(c + d*x)])/5 + (2*a^4*A*Sec[c +
d*x]^2*Tan[c + d*x])/5))/d

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{4} \cos \left (d x + c\right )^{6} + {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{5} + A a^{4} + {\left (6 \, C a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, C a^{3} b + 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 4 \, B a^{3} b + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((C*b^4*cos(d*x + c)^6 + (4*C*a*b^3 + B*b^4)*cos(d*x + c)^5 + A*a^4 + (6*C*a^2*b^2 + 4*B*a*b^3 + A*b^4
)*cos(d*x + c)^4 + 2*(2*C*a^3*b + 3*B*a^2*b^2 + 2*A*a*b^3)*cos(d*x + c)^3 + (C*a^4 + 4*B*a^3*b + 6*A*a^2*b^2)*
cos(d*x + c)^2 + (B*a^4 + 4*A*a^3*b)*cos(d*x + c))/cos(d*x + c)^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^4/cos(d*x + c)^(7/2), x)

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maple [B]  time = 11.14, size = 1884, normalized size = 4.88 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5*C*b^4*(-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^6+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)+1/3*(4*B*b^4+16*C*a*b^3-12*C*b^4)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+2*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x
+1/2*c))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(2*A*b^4+8*B*a*b^3-4*B*b^4+12*C*a^2*b^2-16*C*a*b
^3+6*C*b^4)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*
x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+8*a*A*b^3*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*A*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*a^2*b^2*B*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-8*B*a*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2
)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*b^4*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+8*a^3*b*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*C*a^2*b^2*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+8*C*a*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*C*b^4*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-2/5*A*a^4/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)
^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+
1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*
d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(
1/2*d*x+1/2*c),2^(1/2))-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)+2*a^3*(4*A*b+B*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/
2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*a^2*(6*A*b^2+4*B*a*b+C*a^2)*(-(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1
/2*c)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^4/cos(d*x + c)^(7/2), x)

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mupad [B]  time = 6.01, size = 524, normalized size = 1.36 \[ \frac {2\,\left (B\,b^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+12\,B\,a\,b^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+B\,b^4\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+18\,B\,a^2\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,b^4\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,A\,a\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {8\,C\,a^3\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {12\,C\,a^2\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^4\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,A\,a^3\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {8\,B\,a^3\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {12\,A\,a^2\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(7/2),x)

[Out]

(2*(B*b^4*ellipticF(c/2 + (d*x)/2, 2) + 12*B*a*b^3*ellipticE(c/2 + (d*x)/2, 2) + B*b^4*cos(c + d*x)^(1/2)*sin(
c + d*x) + 18*B*a^2*b^2*ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*b^4*ellipticE(c/2 + (d*x)/2, 2))/d + (8*A*a
*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (8*C*a^3*b*ellipticF(c/2 + (d*x)/2, 2))/d + (4*C*a*b^3*((2*cos(c + d*x)^
(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (12*C*a^2*b^2*ellipticE(c/2 + (d*x)/2, 2))/d +
(2*A*a^4*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^
(1/2)) + (2*B*a^4*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d
*x)^2)^(1/2)) + (2*C*a^4*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(
c + d*x)^2)^(1/2)) - (2*C*b^4*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*
d*(sin(c + d*x)^2)^(1/2)) + (8*A*a^3*b*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c +
d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (8*B*a^3*b*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*c
os(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (12*A*a^2*b^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x
)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))

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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+b*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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