3.874 \(\int \frac {(a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=255 \[ \frac {2 a \left (5 a^2 B+21 a b C+18 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (7 a C+11 b B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (5 a^3 B+21 a^2 b C+21 a b^2 B+21 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (3 a^3 C+9 a^2 b B+15 a b^2 C+5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (3 a^3 C+9 a^2 b B+15 a b^2 C+5 b^3 B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-2/5*(9*B*a^2*b+5*B*b^3+3*C*a^3+15*C*a*b^2)*(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+2/21*(5*B*a^3+21*B*a*b^2+21*C*a^2*b+21*C*b^3)*(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+2/35*a^2*(11*B*b+7*C*a)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*a*(5*
B*a^2+18*B*b^2+21*C*a*b)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*a*B*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(7/2
)+2/5*(9*B*a^2*b+5*B*b^3+3*C*a^3+15*C*a*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3029, 2989, 3031, 3021, 2748, 2636, 2639, 2641} \[ \frac {2 \left (21 a^2 b C+5 a^3 B+21 a b^2 B+21 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (9 a^2 b B+3 a^3 C+15 a b^2 C+5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (5 a^2 B+21 a b C+18 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (9 a^2 b B+3 a^3 C+15 a b^2 C+5 b^3 B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 (7 a C+11 b B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

(-2*(9*a^2*b*B + 5*b^3*B + 3*a^3*C + 15*a*b^2*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(5*a^3*B + 21*a*b^2*B +
21*a^2*b*C + 21*b^3*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a^2*(11*b*B + 7*a*C)*Sin[c + d*x])/(35*d*Cos[c
+ d*x]^(5/2)) + (2*a*(5*a^2*B + 18*b^2*B + 21*a*b*C)*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*(9*a^2*b*B +
5*b^3*B + 3*a^3*C + 15*a*b^2*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a*B*(a + b*Cos[c + d*x])^2*Sin[c
+ d*x])/(7*d*Cos[c + d*x]^(7/2))

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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 2989

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

Rule 3021

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

Rule 3029

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

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]

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx &=\int \frac {(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} a (11 b B+7 a C)+\frac {1}{2} \left (5 a^2 B+7 b^2 B+14 a b C\right ) \cos (c+d x)+\frac {1}{2} b (a B+7 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (11 b B+7 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} a \left (5 a^2 B+18 b^2 B+21 a b C\right )-\frac {7}{4} \left (9 a^2 b B+5 b^3 B+3 a^3 C+15 a b^2 C\right ) \cos (c+d x)-\frac {5}{4} b^2 (a B+7 b C) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (11 b B+7 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2 B+18 b^2 B+21 a b C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {8}{105} \int \frac {-\frac {21}{8} \left (9 a^2 b B+5 b^3 B+3 a^3 C+15 a b^2 C\right )-\frac {5}{8} \left (5 a^3 B+21 a b^2 B+21 a^2 b C+21 b^3 C\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (11 b B+7 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2 B+18 b^2 B+21 a b C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (-9 a^2 b B-5 b^3 B-3 a^3 C-15 a b^2 C\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{21} \left (-5 a^3 B-21 a b^2 B-21 a^2 b C-21 b^3 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (5 a^3 B+21 a b^2 B+21 a^2 b C+21 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a^2 (11 b B+7 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2 B+18 b^2 B+21 a b C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (9 a^2 b B+5 b^3 B+3 a^3 C+15 a b^2 C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (9 a^2 b B+5 b^3 B+3 a^3 C+15 a b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (9 a^2 b B+5 b^3 B+3 a^3 C+15 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (5 a^3 B+21 a b^2 B+21 a^2 b C+21 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a^2 (11 b B+7 a C) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (5 a^2 B+18 b^2 B+21 a b C\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (9 a^2 b B+5 b^3 B+3 a^3 C+15 a b^2 C\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.69, size = 221, normalized size = 0.87 \[ \frac {2 \left (\frac {15 a^3 B \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {5 a \left (5 a^2 B+21 a b C+21 b^2 B\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 a^2 (a C+3 b B) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+5 \left (5 a^3 B+21 a^2 b C+21 a b^2 B+21 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-21 \left (3 a^3 C+9 a^2 b B+15 a b^2 C+5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {21 \left (3 a^3 C+9 a^2 b B+15 a b^2 C+5 b^3 B\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]

[Out]

(2*(-21*(9*a^2*b*B + 5*b^3*B + 3*a^3*C + 15*a*b^2*C)*EllipticE[(c + d*x)/2, 2] + 5*(5*a^3*B + 21*a*b^2*B + 21*
a^2*b*C + 21*b^3*C)*EllipticF[(c + d*x)/2, 2] + (15*a^3*B*Sin[c + d*x])/Cos[c + d*x]^(7/2) + (21*a^2*(3*b*B +
a*C)*Sin[c + d*x])/Cos[c + d*x]^(5/2) + (5*a*(5*a^2*B + 21*b^2*B + 21*a*b*C)*Sin[c + d*x])/Cos[c + d*x]^(3/2)
+ (21*(9*a^2*b*B + 5*b^3*B + 3*a^3*C + 15*a*b^2*C)*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(105*d)

________________________________________________________________________________________

fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{4} + B a^{3} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (C a^{2} b + B a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {9}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

integral((C*b^3*cos(d*x + c)^4 + B*a^3 + (3*C*a*b^2 + B*b^3)*cos(d*x + c)^3 + 3*(C*a^2*b + B*a*b^2)*cos(d*x +
c)^2 + (C*a^3 + 3*B*a^2*b)*cos(d*x + c))/cos(d*x + c)^(9/2), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(11/2), x)

________________________________________________________________________________________

maple [B]  time = 8.70, size = 944, normalized size = 3.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^3*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)
)-2/5*a^2*(3*B*b+C*a)/(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*s
in(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*B*(-
1/56*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)^4-5/4
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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+5/21*(
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^2*(B*b+3*C*a)*(-(-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)+6*a*b*(B*b+C*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))))/sin(1/2*d*x+1/2
*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(11/2), x)

________________________________________________________________________________________

mupad [B]  time = 5.20, size = 311, normalized size = 1.22 \[ \frac {\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+2\,B\,b^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+2\,B\,a\,b^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+\frac {6\,B\,a^2\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5}}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,C\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,C\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/cos(c + d*x)^(11/2),x)

[Out]

((2*B*a^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + 2*B*b^3*cos(c + d*x)^3*sin(c + d*x)*h
ypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 2*B*a*b^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4,
cos(c + d*x)^2) + (6*B*a^2*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/5)/(d*co
s(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*C*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a^3*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)) + (6*C*a*b^2*si
n(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*a
^2*b*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))

________________________________________________________________________________________

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))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________