3.1085 \(\int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=294 \[ \frac {2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}+\frac {2 \sin (c+d x) \left (21 a^3 B+21 a^2 b (3 A+5 C)+98 a b^2 B+24 A b^3\right )}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (7 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

-2/5*(3*a^3*B+15*a*b^2*B+5*b^3*(A-C)+3*a^2*b*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(21*a^2*b*B+21*b^3*B+21*a*b^2*(A+3*C)+a^3*(5*A+7*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+2/105*a*(24*A*b^2+63*a*b*B+5*a^2*(5*A+7*
C))*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/35*(6*A*b+7*B*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/7*A*(a
+b*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/35*(24*A*b^3+21*a^3*B+98*a*b^2*B+21*a^2*b*(3*A+5*C))*sin(d*x+
c)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.85, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3047, 3031, 3021, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 b (3 A+5 C)+3 a^3 B+15 a b^2 B+5 b^3 (A-C)\right )}{5 d}+\frac {2 a \sin (c+d x) \left (5 a^2 (5 A+7 C)+63 a b B+24 A b^2\right )}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) \left (21 a^2 b (3 A+5 C)+21 a^3 B+98 a b^2 B+24 A b^3\right )}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (7 a B+6 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

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]^(9/2),x]

[Out]

(-2*(3*a^3*B + 15*a*b^2*B + 5*b^3*(A - C) + 3*a^2*b*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(21*a^2
*b*B + 21*b^3*B + 21*a*b^2*(A + 3*C) + a^3*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*(24*A*b^2 + 6
3*a*b*B + 5*a^2*(5*A + 7*C))*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (2*(24*A*b^3 + 21*a^3*B + 98*a*b^2*B +
21*a^2*b*(3*A + 5*C))*Sin[c + d*x])/(35*d*Sqrt[Cos[c + d*x]]) + (2*(6*A*b + 7*a*B)*(a + b*Cos[c + d*x])^2*Sin
[c + d*x])/(35*d*Cos[c + d*x]^(5/2)) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2))

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 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 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 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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{2} (6 A b+7 a B)+\frac {1}{2} (5 a A+7 b B+7 a C) \cos (c+d x)-\frac {1}{2} b (A-7 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4}{35} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right )+\frac {1}{4} \left (38 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \cos (c+d x)-\frac {1}{4} b (11 A b+7 a B-35 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {8}{105} \int \frac {-\frac {3}{8} \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right )-\frac {5}{8} \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) \cos (c+d x)+\frac {3}{8} b^2 (11 A b+7 a B-35 b C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right ) \sin (c+d x)}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {16}{105} \int \frac {-\frac {5}{16} \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right )+\frac {21}{16} \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right ) \sin (c+d x)}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {1}{5} \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} \left (-21 a^2 b B-21 b^3 B-21 a b^2 (A+3 C)-a^3 (5 A+7 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 \left (3 a^3 B+15 a b^2 B+5 b^3 (A-C)+3 a^2 b (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (21 a^2 b B+21 b^3 B+21 a b^2 (A+3 C)+a^3 (5 A+7 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a \left (24 A b^2+63 a b B+5 a^2 (5 A+7 C)\right ) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (24 A b^3+21 a^3 B+98 a b^2 B+21 a^2 b (3 A+5 C)\right ) \sin (c+d x)}{35 d \sqrt {\cos (c+d x)}}+\frac {2 (6 A b+7 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 4.96, size = 251, normalized size = 0.85 \[ \frac {2 \left (\frac {15 a^3 A \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {5 a \sin (c+d x) \left (a^2 (5 A+7 C)+21 a b B+21 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 a^2 (a B+3 A b) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+5 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (5 A+7 C)+21 a^2 b B+21 a b^2 (A+3 C)+21 b^3 B\right )-21 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 b^3 (A-C)\right )+\frac {21 \sin (c+d x) \left (3 a^3 B+3 a^2 b (3 A+5 C)+15 a b^2 B+5 A b^3\right )}{\sqrt {\cos (c+d x)}}\right )}{105 d} \]

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]^(9/2),x]

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{5} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {9}{2}}}, x\right ) \]

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)^(9/2),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)^(9/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

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)^(9/2),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)^(9/2), x)

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maple [B]  time = 10.06, size = 1205, normalized size = 4.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),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
))-EllipticE(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*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*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*a*(3*A*b^2+3*B*a*b
+C*a^2)*(-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*b*(A*b^2+3*B*a*b+3*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)+2*A*a^3*(-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)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(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)))-2/5*
a^2*(3*A*b+B*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*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))/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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

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)^(9/2),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)^(9/2), x)

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mupad [B]  time = 7.09, size = 442, normalized size = 1.50 \[ \frac {2\,\left (C\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,C\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {\frac {2\,A\,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\,A\,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\,A\,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\,A\,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\,B\,b^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,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 {2\,C\,a^3\,\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 {6\,B\,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\,B\,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}}+\frac {6\,C\,a^2\,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}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^(9/2),x)

[Out]

(2*(C*b^3*ellipticE(c/2 + (d*x)/2, 2) + 3*C*a*b^2*ellipticF(c/2 + (d*x)/2, 2)))/d + ((2*A*a^3*sin(c + d*x)*hyp
ergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + 2*A*b^3*cos(c + d*x)^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4,
cos(c + d*x)^2) + 2*A*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*A*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*cos(c + d*x)^(7/2)*(1 - cos(
c + d*x)^2)^(1/2)) + (2*B*b^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*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)) + (2*C*a^3*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)) + (6*B*a*b^2*sin(c + d*x)*hypergeo
m([-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*B*a^2*b*sin(c + d*x)*h
ypergeom([-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)) + (6*C*a^2*b*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))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2),x)

[Out]

Timed out

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