3.1083 \(\int \frac {(a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=271 \[ -\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{3 d}+\frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^3 B+15 a^2 b (A-C)-15 a b^2 B-b^3 (5 A+3 C)\right )}{5 d}-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{15 d}+\frac {2 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-2/5*(5*a^3*B-15*a*b^2*B+15*a^2*b*(A-C)-b^3*(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+2/3*(9*a^2*b*B+b^3*B+3*a*b^2*(3*A+C)+a^3*(A+3*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-2/15*b^2*(35*A*b+15*B*a-3*C*b)*cos(d*x+c)^(3/2)*s
in(d*x+c)/d+2/3*A*(a+b*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2*(2*A*b+B*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)
/d/cos(d*x+c)^(1/2)-2/3*b*(6*a^2*B-b^2*B+3*a*b*(5*A-C))*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.86, antiderivative size = 271, 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, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )}{3 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (15 a^2 b (A-C)+5 a^3 B-15 a b^2 B-b^3 (5 A+3 C)\right )}{5 d}-\frac {2 b \sin (c+d x) \sqrt {\cos (c+d x)} \left (6 a^2 B+3 a b (5 A-C)-b^2 B\right )}{3 d}-\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (15 a B+35 A b-3 b C)}{15 d}+\frac {2 (a B+2 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

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]^(5/2),x]

[Out]

(-2*(5*a^3*B - 15*a*b^2*B + 15*a^2*b*(A - C) - b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(9*a^2*b
*B + b^3*B + 3*a*b^2*(3*A + C) + a^3*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*d) - (2*b*(6*a^2*B - b^2*B + 3*a
*b*(5*A - C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) - (2*b^2*(35*A*b + 15*a*B - 3*b*C)*Cos[c + d*x]^(3/2)*Sin
[c + d*x])/(15*d) + (2*(2*A*b + a*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*A*(a + b
*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))

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 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]

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 {5}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+b \cos (c+d x))^2 \left (\frac {3}{2} (2 A b+a B)+\frac {1}{2} (3 b B+a (A+3 C)) \cos (c+d x)-\frac {1}{2} b (5 A-3 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 (2 A b+a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4}{3} \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} \left (24 A b^2+15 a b B+a^2 (A+3 C)\right )-\frac {1}{4} \left (10 a A b+3 a^2 B-3 b^2 B-6 a b C\right ) \cos (c+d x)-\frac {1}{4} b (35 A b+15 a B-3 b C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 (35 A b+15 a B-3 b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (2 A b+a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8}{15} \int \frac {\frac {5}{8} a \left (24 A b^2+15 a b B+a^2 (A+3 C)\right )-\frac {3}{8} \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \cos (c+d x)-\frac {15}{8} b \left (6 a^2 B-b^2 B+3 a b (5 A-C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b \left (6 a^2 B-b^2 B+3 a b (5 A-C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (35 A b+15 a B-3 b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (2 A b+a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16}{45} \int \frac {\frac {15}{16} \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right )-\frac {9}{16} \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b \left (6 a^2 B-b^2 B+3 a b (5 A-C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (35 A b+15 a B-3 b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (2 A b+a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (-5 a^3 B+15 a b^2 B-15 a^2 b (A-C)+b^3 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {2 \left (5 a^3 B-15 a b^2 B+15 a^2 b (A-C)-b^3 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a b^2 (3 A+C)+a^3 (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 b \left (6 a^2 B-b^2 B+3 a b (5 A-C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (35 A b+15 a B-3 b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 (2 A b+a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 2.40, size = 186, normalized size = 0.69 \[ \frac {10 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^3 (A+3 C)+9 a^2 b B+3 a b^2 (3 A+C)+b^3 B\right )+2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-15 a^3 B-45 a^2 b (A-C)+45 a b^2 B+3 b^3 (5 A+3 C)\right )+\frac {5 \left (2 a^3 A \tan (c+d x)+b^2 (3 a C+b B) \sin (2 (c+d x))\right )+6 \sin (c+d x) \left (5 a^2 (a B+3 A b)+b^3 C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}}}{15 d} \]

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]^(5/2),x]

[Out]

(2*(-15*a^3*B + 45*a*b^2*B - 45*a^2*b*(A - C) + 3*b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 10*(9*a^2*b*B +
b^3*B + 3*a*b^2*(3*A + C) + a^3*(A + 3*C))*EllipticF[(c + d*x)/2, 2] + (6*(5*a^2*(3*A*b + a*B) + b^3*C*Cos[c
+ d*x]^2)*Sin[c + d*x] + 5*(b^2*(b*B + 3*a*C)*Sin[2*(c + d*x)] + 2*a^3*A*Tan[c + d*x]))/Sqrt[Cos[c + d*x]])/(1
5*d)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{5} + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]

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)^(5/2),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)^(5/2), x)

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

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)^(5/2),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)^(5/2), x)

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maple [B]  time = 8.30, size = 1837, normalized size = 6.78 \[ \text {Expression too large to display} \]

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)^(5/2),x)

[Out]

2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+
1)/sin(1/2*d*x+1/2*c)^3*(6*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-45*a^2*b*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))+72*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*
d*x+1/2*c)^6-90*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^2*b*sin(1/2*d*x+1/2*c)^2+90*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)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2*sin(1/2*d*x+1/2*c)^2+90*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^2*b*sin(1/2*d*x+1/2*c)^2+90*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))*a^2*b*sin(1/2*d*x+1/2*c)^2-90*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2*sin(1/2*
d*x+1/2*c)^2+30*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))*a*b^2*sin(1/2*d*x+1/2*c)^2+45*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-15*C*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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-40*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+10*A*a^3*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^2+10*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-48*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*
d*x+1/2*c)^8+40*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4
-45*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^2*
b-45*A*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
))+45*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*
b^2-36*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+30*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+10*A*(si
n(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))*a^3*sin(1/2*d
*x+1/2*c)^2-30*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))*b^3*sin(1/2*d*x+1/2*c)^2+10*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(c
os(1/2*d*x+1/2*c),2^(1/2))*b^3*sin(1/2*d*x+1/2*c)^2+30*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*sin(1/2*d*x+1/2*c)^2+30*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*sin(1/2*d*x+1/2*c)^2-18*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))*b^3*sin(1/2*d*x+1/2*c)^
2+9*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))*b^3+
15*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))*b^3-5
*A*a^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))-15*
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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-5*b^
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))-15*C*a
^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))+120*C*a
*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-180*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-120*C*a*b^2*c
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+90*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+30*C*a*b^2*cos(1/2*d
*x+1/2*c)*sin(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)/(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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),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)^(5/2), x)

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mupad [B]  time = 4.07, size = 379, normalized size = 1.40 \[ \frac {2\,\left (A\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^3+3\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\,b^2\right )}{d}+\frac {B\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {3\,C\,a\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,A\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

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)^(5/2),x)

[Out]

(2*(A*b^3*ellipticE(c/2 + (d*x)/2, 2) + 3*A*a*b^2*ellipticF(c/2 + (d*x)/2, 2)))/d + (B*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*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*B*a
*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (6*B*a^2*b*ellipticF(c/2 + (d*x)/2, 2))/d + (6*C*a^2*b*ellipticE(c/2 + (
d*x)/2, 2))/d + (3*C*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([-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
)) + (2*B*a^3*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)) - (2*C*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)) + (6*A*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)*(s
in(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)**(5/2),x)

[Out]

Timed out

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