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

Optimal. Leaf size=329 \[ \frac {8 a b \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a b \left (96 a^2 C+891 A b^2+673 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3465 d}+\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{231 d}+\frac {2 \left (77 a^4 (3 A+C)+66 a^2 b^2 (7 A+5 C)+5 b^4 (11 A+9 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (64 a^4 C+9 a^2 b^2 (143 A+101 C)+15 b^4 (11 A+9 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{693 d}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4}{11 d}+\frac {16 a C \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3}{99 d} \]

[Out]

8/15*a*b*(3*a^2*(5*A+3*C)+b^2*(9*A+7*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+2/231*(77*a^4*(3*A+C)+66*a^2*b^2*(7*A+5*C)+5*b^4*(11*A+9*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3465*a*b*(891*A*b^2+96*C*a^2+673*C*b^2)*cos(d*x+c)
^(3/2)*sin(d*x+c)/d+2/693*(64*a^4*C+15*b^4*(11*A+9*C)+9*a^2*b^2*(143*A+101*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2
/231*(16*a^2*C+3*b^2*(11*A+9*C))*(a+b*cos(d*x+c))^2*sin(d*x+c)*cos(d*x+c)^(1/2)/d+16/99*a*C*(a+b*cos(d*x+c))^3
*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2/11*C*(a+b*cos(d*x+c))^4*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 1.09, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3050, 3049, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 \left (66 a^2 b^2 (7 A+5 C)+77 a^4 (3 A+C)+5 b^4 (11 A+9 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {8 a b \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a b \left (96 a^2 C+891 A b^2+673 b^2 C\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3465 d}+\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{231 d}+\frac {2 \left (9 a^2 b^2 (143 A+101 C)+64 a^4 C+15 b^4 (11 A+9 C)\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{693 d}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4}{11 d}+\frac {16 a C \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3}{99 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*b*(3*a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(77*a^4*(3*A + C) + 66*a^2
*b^2*(7*A + 5*C) + 5*b^4*(11*A + 9*C))*EllipticF[(c + d*x)/2, 2])/(231*d) + (2*(64*a^4*C + 15*b^4*(11*A + 9*C)
+ 9*a^2*b^2*(143*A + 101*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(693*d) + (4*a*b*(891*A*b^2 + 96*a^2*C + 673*b^
2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d) + (2*(16*a^2*C + 3*b^2*(11*A + 9*C))*Sqrt[Cos[c + d*x]]*(a + b*
Cos[c + d*x])^2*Sin[c + d*x])/(231*d) + (16*a*C*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(99*d)
+ (2*C*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(11*d)

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 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3050

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

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx &=\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac {2}{11} \int \frac {(a+b \cos (c+d x))^3 \left (\frac {1}{2} a (11 A+C)+\frac {1}{2} b (11 A+9 C) \cos (c+d x)+4 a C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {16 a C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac {4}{99} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{4} a^2 (99 A+17 C)+\frac {1}{2} a b (99 A+73 C) \cos (c+d x)+\frac {3}{4} \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac {16 a C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac {8}{693} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{8} a \left (9 b^2 (11 A+9 C)+a^2 (693 A+167 C)\right )+\frac {1}{8} b \left (45 b^2 (11 A+9 C)+a^2 (2079 A+1381 C)\right ) \cos (c+d x)+\frac {1}{4} a \left (891 A b^2+96 a^2 C+673 b^2 C\right ) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a b \left (891 A b^2+96 a^2 C+673 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac {16 a C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac {16 \int \frac {\frac {5}{16} a^2 \left (9 b^2 (11 A+9 C)+a^2 (693 A+167 C)\right )+\frac {231}{4} a b \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \cos (c+d x)+\frac {15}{16} \left (64 a^4 C+15 b^4 (11 A+9 C)+9 a^2 b^2 (143 A+101 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{3465}\\ &=\frac {2 \left (64 a^4 C+15 b^4 (11 A+9 C)+9 a^2 b^2 (143 A+101 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d}+\frac {4 a b \left (891 A b^2+96 a^2 C+673 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac {16 a C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac {32 \int \frac {\frac {45}{32} \left (77 a^4 (3 A+C)+66 a^2 b^2 (7 A+5 C)+5 b^4 (11 A+9 C)\right )+\frac {693}{8} a b \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{10395}\\ &=\frac {2 \left (64 a^4 C+15 b^4 (11 A+9 C)+9 a^2 b^2 (143 A+101 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d}+\frac {4 a b \left (891 A b^2+96 a^2 C+673 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac {16 a C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac {1}{15} \left (4 a b \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right )\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{231} \left (77 a^4 (3 A+C)+66 a^2 b^2 (7 A+5 C)+5 b^4 (11 A+9 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {8 a b \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (77 a^4 (3 A+C)+66 a^2 b^2 (7 A+5 C)+5 b^4 (11 A+9 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (64 a^4 C+15 b^4 (11 A+9 C)+9 a^2 b^2 (143 A+101 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{693 d}+\frac {4 a b \left (891 A b^2+96 a^2 C+673 b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac {2 \left (16 a^2 C+3 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac {16 a C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac {2 C \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^4 \sin (c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]  time = 2.12, size = 243, normalized size = 0.74 \[ \frac {14784 \left (3 a^3 b (5 A+3 C)+a b^3 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+240 \left (77 a^4 (3 A+C)+66 a^2 b^2 (7 A+5 C)+5 b^4 (11 A+9 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (616 a b \left (36 a^2 C+36 A b^2+43 b^2 C\right ) \cos (c+d x)+5 \left (1848 a^4 C+792 a^2 b^2 (14 A+13 C)+36 \left (66 a^2 b^2 C+11 A b^4+16 b^4 C\right ) \cos (2 (c+d x))+616 a b^3 C \cos (3 (c+d x))+3 b^4 (572 A+531 C)+63 b^4 C \cos (4 (c+d x))\right )\right )}{27720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14784*(3*a^3*b*(5*A + 3*C) + a*b^3*(9*A + 7*C))*EllipticE[(c + d*x)/2, 2] + 240*(77*a^4*(3*A + C) + 66*a^2*b^
2*(7*A + 5*C) + 5*b^4*(11*A + 9*C))*EllipticF[(c + d*x)/2, 2] + 2*Sqrt[Cos[c + d*x]]*(616*a*b*(36*A*b^2 + 36*a
^2*C + 43*b^2*C)*Cos[c + d*x] + 5*(1848*a^4*C + 792*a^2*b^2*(14*A + 13*C) + 3*b^4*(572*A + 531*C) + 36*(11*A*b
^4 + 66*a^2*b^2*C + 16*b^4*C)*Cos[2*(c + d*x)] + 616*a*b^3*C*Cos[3*(c + d*x)] + 63*b^4*C*Cos[4*(c + d*x)]))*Si
n[c + d*x])/(27720*d)

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{4} \cos \left (d x + c\right )^{6} + 4 \, C a b^{3} \cos \left (d x + c\right )^{5} + 4 \, A a^{3} b \cos \left (d x + c\right ) + A a^{4} + {\left (6 \, C a^{2} b^{2} + A b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (C a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{4} + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}}{\sqrt {\cos \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.16, size = 924, normalized size = 2.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x)

[Out]

-2/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(20160*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-49280*C*a*b^3-50400*C*b^4)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(7920*A*b^4+47520*C*a^2*b^2+9856
0*C*a*b^3+56880*C*b^4)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-22176*A*a*b^3-11880*A*b^4-22176*C*a^3*b-71280
*C*a^2*b^2-91168*C*a*b^3-34920*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(27720*A*a^2*b^2+22176*A*a*b^3+9
240*A*b^4+4620*C*a^4+22176*C*a^3*b+55440*C*a^2*b^2+41888*C*a*b^3+13860*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)+(-13860*A*a^2*b^2-5544*A*a*b^3-2640*A*b^4-2310*C*a^4-5544*C*a^3*b-15840*C*a^2*b^2-7392*C*a*b^3-2790*C*
b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+3465*A*a^4*(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))+6930*A*a^2*b^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))+825*A*b^4*(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))-13860*A*(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))*a^3*b-8316*A*(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))*a*b^3+1155*a^4*C*(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))+4950*C*a^2*b^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))+675*C*b^4*(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))-8316*C*(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))*a^3*b-6468*C*(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))*a*b^3)/(-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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.19, size = 400, normalized size = 1.22 \[ \frac {2\,\left (A\,a^4\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+4\,A\,a^3\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+2\,A\,a^2\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+2\,A\,a^2\,b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {C\,a^4\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,A\,b^4\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^4\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,A\,a\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,C\,a^3\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {8\,C\,a\,b^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,C\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^4)/cos(c + d*x)^(1/2),x)

[Out]

(2*(A*a^4*ellipticF(c/2 + (d*x)/2, 2) + 4*A*a^3*b*ellipticE(c/2 + (d*x)/2, 2) + 2*A*a^2*b^2*ellipticF(c/2 + (d
*x)/2, 2) + 2*A*a^2*b^2*cos(c + d*x)^(1/2)*sin(c + d*x)))/d + (C*a^4*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 +
(2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (2*A*b^4*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, c
os(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (2*C*b^4*cos(c + d*x)^(13/2)*sin(c + d*x)*hypergeom([1/2, 13/4]
, 17/4, cos(c + d*x)^2))/(13*d*(sin(c + d*x)^2)^(1/2)) - (8*A*a*b^3*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*C*a^3*b*cos(c + d*x)^(7/2)*sin(c + d*x)*h
ypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (8*C*a*b^3*cos(c + d*x)^(11/2)*sin(
c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2)) - (4*C*a^2*b^2*cos(c + d*
x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(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+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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