3.1082 \(\int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=279 \[ \frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (-6 a^2 (7 A-3 C)+21 a b B+b^2 (7 A+5 C)\right )}{21 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (21 a^3 B+21 a^2 b (3 A+C)+21 a b^2 B+b^3 (7 A+5 C)\right )}{21 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 (A-C)+15 a^2 b B+3 a b^2 (5 A+3 C)+3 b^3 B\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (35 a A-11 a C-7 b B)}{35 d}-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}} \]

[Out]

2/5*(15*a^2*b*B+3*b^3*B-5*a^3*(A-C)+3*a*b^2*(5*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(21*a^3*B+21*a*b^2*B+21*a^2*b*(3*A+C)+b^3*(7*A+5*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-2/35*b^2*(35*A*a-7*B*b-11*C*a)*cos(d*x+c)
^(3/2)*sin(d*x+c)/d+2*A*(a+b*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/21*b*(21*a*b*B-6*a^2*(7*A-3*C)+b^2*
(7*A+5*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/d-2/7*b*(7*A-C)*(a+b*cos(d*x+c))^2*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.83, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3047, 3049, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (21 a^2 b (3 A+C)+21 a^3 B+21 a b^2 B+b^3 (7 A+5 C)\right )}{21 d}+\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 (A-C)+15 a^2 b B+3 a b^2 (5 A+3 C)+3 b^3 B\right )}{5 d}+\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (-6 a^2 (7 A-3 C)+21 a b B+b^2 (7 A+5 C)\right )}{21 d}-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (35 a A-11 a C-7 b B)}{35 d}-\frac {2 b (7 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}{7 d}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{d \sqrt {\cos (c+d x)}} \]

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]^(3/2),x]

[Out]

(2*(15*a^2*b*B + 3*b^3*B - 5*a^3*(A - C) + 3*a*b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(21*a^3*
B + 21*a*b^2*B + 21*a^2*b*(3*A + C) + b^3*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*b*(21*a*b*B - 6*
a^2*(7*A - 3*C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) - (2*b^2*(35*a*A - 7*b*B - 11*a*C)*
Cos[c + d*x]^(3/2)*Sin[c + d*x])/(35*d) - (2*b*(7*A - C)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2*Sin[c + d*x
])/(7*d) + (2*A*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(d*Sqrt[Cos[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]

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]

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])))

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 {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+2 \int \frac {(a+b \cos (c+d x))^2 \left (\frac {1}{2} (6 A b+a B)+\frac {1}{2} (b B-a (A-C)) \cos (c+d x)-\frac {1}{2} b (7 A-C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b (7 A-C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {4}{7} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} a (35 A b+7 a B+b C)+\frac {1}{4} \left (14 a b B-7 a^2 (A-C)+b^2 (7 A+5 C)\right ) \cos (c+d x)-\frac {1}{4} b (35 a A-7 b B-11 a C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 (35 a A-7 b B-11 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}-\frac {2 b (7 A-C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {8}{35} \int \frac {\frac {5}{8} a^2 (35 A b+7 a B+b C)+\frac {7}{8} \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \cos (c+d x)+\frac {5}{8} b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b^2 (35 a A-7 b B-11 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}-\frac {2 b (7 A-C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {16}{105} \int \frac {\frac {5}{16} \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right )+\frac {21}{16} \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b^2 (35 a A-7 b B-11 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}-\frac {2 b (7 A-C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {1}{5} \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (15 a^2 b B+3 b^3 B-5 a^3 (A-C)+3 a b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (21 a^3 B+21 a b^2 B+21 a^2 b (3 A+C)+b^3 (7 A+5 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \left (21 a b B-6 a^2 (7 A-3 C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 b^2 (35 a A-7 b B-11 a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}-\frac {2 b (7 A-C) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.86, size = 212, normalized size = 0.76 \[ \frac {20 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (21 a^3 B+21 a^2 b (3 A+C)+21 a b^2 B+b^3 (7 A+5 C)\right )-84 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 (A-C)-15 a^2 b B-3 a b^2 (5 A+3 C)-3 b^3 B\right )+\frac {\sin (c+d x) \left (420 a^3 A+5 b \cos (c+d x) \left (84 a^2 C+84 a b B+28 A b^2+29 b^2 C\right )+42 b^2 (3 a C+b B) \cos (2 (c+d x))+126 a b^2 C+42 b^3 B+15 b^3 C \cos (3 (c+d x))\right )}{\sqrt {\cos (c+d x)}}}{210 d} \]

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]^(3/2),x]

[Out]

(-84*(-15*a^2*b*B - 3*b^3*B + 5*a^3*(A - C) - 3*a*b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 20*(21*a^3*B +
21*a*b^2*B + 21*a^2*b*(3*A + C) + b^3*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2] + ((420*a^3*A + 42*b^3*B + 126*a*
b^2*C + 5*b*(28*A*b^2 + 84*a*b*B + 84*a^2*C + 29*b^2*C)*Cos[c + d*x] + 42*b^2*(b*B + 3*a*C)*Cos[2*(c + d*x)] +
15*b^3*C*Cos[3*(c + d*x)])*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(210*d)

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

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)^(3/2),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)^(3/2), x)

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

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)^(3/2),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)^(3/2), x)

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maple [B]  time = 3.56, size = 1278, normalized size = 4.58 \[ \text {result too large to display} \]

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)^(3/2),x)

[Out]

-2/105*(240*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-24*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(7*B*b+21*C*a+15*C*b)*sin(1/2*d*x+1/2*c)^6*cos(1/
2*d*x+1/2*c)+28*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(5*A*b^2+15*B*a*b+6*B*b^2+15*C*a^2+18*C
*a*b+10*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
(105*A*a^3+35*A*b^3+105*B*a*b^2+21*B*b^3+105*C*a^2*b+63*C*a*b^2+40*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2
*c)+315*A*a^2*b*(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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+35*A*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
+105*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))*a^3-315*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))*a*b^2+1
05*a^3*B*(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))*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+105*B*a*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-315
*B*(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*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)*a^2*b-63*B*(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*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)*b^3+105*C*
a^2*b*(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))*(-2*
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+25*b^3*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-105*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)^(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^3-189*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)^(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^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*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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),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)^(3/2), x)

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mupad [B]  time = 3.40, size = 398, normalized size = 1.43 \[ \frac {2\,\left (C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+C\,a^2\,b\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {A\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,A\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {3\,B\,a\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,C\,a\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

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)^(3/2),x)

[Out]

(2*(C*a^3*ellipticE(c/2 + (d*x)/2, 2) + C*a^2*b*ellipticF(c/2 + (d*x)/2, 2) + C*a^2*b*cos(c + d*x)^(1/2)*sin(c
+ d*x)))/d + (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 + (2*B*a^3
*ellipticF(c/2 + (d*x)/2, 2))/d + (6*A*a*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (6*A*a^2*b*ellipticF(c/2 + (d*x)
/2, 2))/d + (6*B*a^2*b*ellipticE(c/2 + (d*x)/2, 2))/d + (3*B*a*b^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2
*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*A*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*co
s(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) - (2*B*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)) - (2*C*b^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4
], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (6*C*a*b^2*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))

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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)**(3/2),x)

[Out]

Timed out

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