3.1084 \(\int \frac{(a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=273 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}+\frac{2 a \sin (c+d x) \left (3 a^2 (3 A+5 C)+35 a b B+24 A b^2\right )}{15 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 \sin (c+d x) \sqrt{\cos (c+d x)} (5 a B+9 A b-5 b C)}{15 d}+\frac{2 (5 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^3}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-2*(15*a^2*b*B - 5*b^3*B + 15*a*b^2*(A - C) + a^3*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(a^3*B +
9*a*b^2*B + b^3*(3*A + C) + 3*a^2*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*(24*A*b^2 + 35*a*b*B +
3*a^2*(3*A + 5*C))*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) - (2*b^2*(9*A*b + 5*a*B - 5*b*C)*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(15*d) + (2*(6*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (
2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
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Rubi [A] time = 0.823653, antiderivative size = 273, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used =
{3047, 3031, 3023, 2748, 2641, 2639} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )}{3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )}{5 d}+\frac{2 a \sin (c+d x) \left (3 a^2 (3 A+5 C)+35 a b B+24 A b^2\right )}{15 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 \sin (c+d x) \sqrt{\cos (c+d x)} (5 a B+9 A b-5 b C)}{15 d}+\frac{2 (5 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^3}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
[Out]
(-2*(15*a^2*b*B - 5*b^3*B + 15*a*b^2*(A - C) + a^3*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(a^3*B +
9*a*b^2*B + b^3*(3*A + C) + 3*a^2*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*(24*A*b^2 + 35*a*b*B +
3*a^2*(3*A + 5*C))*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) - (2*b^2*(9*A*b + 5*a*B - 5*b*C)*Sqrt[Cos[c + d*x]
]*Sin[c + d*x])/(15*d) + (2*(6*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (
2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
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]
Rule 3031
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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*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] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]
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 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 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 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]
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^3 \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))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \cos (c+d x))^2 \left (\frac{1}{2} (6 A b+5 a B)+\frac{1}{2} (3 a A+5 b B+5 a C) \cos (c+d x)-\frac{1}{2} b (3 A-5 C) \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4}{15} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{4} \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right )+\frac{1}{4} \left (5 a^2 B+15 b^2 B+6 a b (A+5 C)\right ) \cos (c+d x)-\frac{3}{4} b (9 A b+5 a B-5 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{8}{15} \int \frac{\frac{1}{8} \left (-24 A b^3-5 a^3 B-50 a b^2 B-15 a^2 b (A+3 C)\right )+\frac{3}{8} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \cos (c+d x)+\frac{3}{8} b^2 (9 A b+5 a B-5 b C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (9 A b+5 a B-5 b C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{16}{45} \int \frac{-\frac{15}{16} \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right )+\frac{9}{16} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (9 A b+5 a B-5 b C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{1}{3} \left (-a^3 B-9 a b^2 B-b^3 (3 A+C)-3 a^2 b (A+3 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (15 a^2 b B-5 b^3 B+15 a b^2 (A-C)+a^3 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (a^3 B+9 a b^2 B+b^3 (3 A+C)+3 a^2 b (A+3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a \left (24 A b^2+35 a b B+3 a^2 (3 A+5 C)\right ) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (9 A b+5 a B-5 b C) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (6 A b+5 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 2.13405, size = 248, normalized size = 0.91 \[ \frac{10 \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (A+3 C)+a^3 B+9 a b^2 B+b^3 (3 A+C)\right )-6 \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^3 (3 A+5 C)+15 a^2 b B+15 a b^2 (A-C)-5 b^3 B\right )+30 a^2 A b \sin (c+d x)+9 a^3 A \sin (2 (c+d x))+6 a^3 A \tan (c+d x)+45 a^2 b B \sin (2 (c+d x))+10 a^3 B \sin (c+d x)+15 a^3 C \sin (2 (c+d x))+45 a A b^2 \sin (2 (c+d x))+10 b^3 C \sin (c+d x) \cos ^2(c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
[Out]
(-6*(15*a^2*b*B - 5*b^3*B + 15*a*b^2*(A - C) + a^3*(3*A + 5*C))*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] +
10*(a^3*B + 9*a*b^2*B + b^3*(3*A + C) + 3*a^2*b*(A + 3*C))*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 30*
a^2*A*b*Sin[c + d*x] + 10*a^3*B*Sin[c + d*x] + 10*b^3*C*Cos[c + d*x]^2*Sin[c + d*x] + 9*a^3*A*Sin[2*(c + d*x)]
+ 45*a*A*b^2*Sin[2*(c + d*x)] + 45*a^2*b*B*Sin[2*(c + d*x)] + 15*a^3*C*Sin[2*(c + d*x)] + 6*a^3*A*Tan[c + d*x
])/(15*d*Cos[c + d*x]^(3/2))
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Maple [B] time = 3.658, size = 1419, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^3*(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)*(4/3*C*b^3*(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*Ell
ipticE(cos(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)-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*B*b^3+6*C*a*b^2-4*C*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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*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))+6*a*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))-2*b^3*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))+6*a^2*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))-6*C*a*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)*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2))+2*C*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^2*(3*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)))-2/5*A*a^3/(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))*(sin(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)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(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)*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*EllipticE(cos(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)-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*A*b^2+3*B*a*b+C*a^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)*(-2*sin(1/2*d*
x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(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., 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 (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(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)^3/cos(d*x + c)^(7/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{3} \cos \left (d x + c\right )^{5} +{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{7}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(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^3*cos(d*x + c)^5 + (3*C*a*b^2 + B*b^3)*cos(d*x + c)^4 + A*a^3 + (3*C*a^2*b + 3*B*a*b^2 + A*b^3)*
cos(d*x + c)^3 + (C*a^3 + 3*B*a^2*b + 3*A*a*b^2)*cos(d*x + c)^2 + (B*a^3 + 3*A*a^2*b)*cos(d*x + c))/cos(d*x +
c)^(7/2), x)
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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+b*cos(d*x+c))**3*(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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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 (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(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)^3/cos(d*x + c)^(7/2), x)