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

Optimal. Leaf size=202 \[ -\frac {2 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 \left (3 a^3 C+9 a^2 b B+3 a b^2 C+b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (5 a^3 B-15 a^2 b C-15 a b^2 B-3 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3029, 2989, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 \left (9 a^2 b B+3 a^3 C+3 a b^2 C+b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}-\frac {2 \left (-15 a^2 b C+5 a^3 B-15 a b^2 B-3 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}} \]

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]^(5/2),x]

[Out]

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

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 {5}{2}}(c+d x)} \, dx &=\int \frac {(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+2 \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} a (5 b B+a C)-\frac {1}{2} \left (a^2 B-b^2 B-2 a b C\right ) \cos (c+d x)-\frac {1}{2} b (5 a B-b C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 (5 a B-b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {4}{5} \int \frac {\frac {5}{4} a^2 (5 b B+a C)-\frac {1}{4} \left (5 a^3 B-15 a b^2 B-15 a^2 b C-3 b^3 C\right ) \cos (c+d x)-\frac {5}{4} b \left (6 a^2 B-b^2 B-3 a b 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 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (5 a B-b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {8}{15} \int \frac {\frac {5}{8} \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right )-\frac {3}{8} \left (5 a^3 B-15 a b^2 B-15 a^2 b C-3 b^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 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (5 a B-b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {1}{3} \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 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 C+3 b^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 C-3 b^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^3 C+3 a b^2 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 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (5 a B-b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.18, size = 150, normalized size = 0.74 \[ \frac {\frac {\sin (c+d x) \left (3 \left (10 a^3 B+b^3 C \cos (2 (c+d x))+b^3 C\right )+10 b^2 (3 a C+b B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}+10 \left (3 a^3 C+9 a^2 b B+3 a b^2 C+b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (-30 a^3 B+90 a^2 b C+90 a b^2 B+18 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d} \]

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]^(5/2),x]

[Out]

((-30*a^3*B + 90*a*b^2*B + 90*a^2*b*C + 18*b^3*C)*EllipticE[(c + d*x)/2, 2] + 10*(9*a^2*b*B + b^3*B + 3*a^3*C
+ 3*a*b^2*C)*EllipticF[(c + d*x)/2, 2] + ((10*b^2*(b*B + 3*a*C)*Cos[c + d*x] + 3*(10*a^3*B + b^3*C + b^3*C*Cos
[2*(c + d*x)]))*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(15*d)

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

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)^(5/2),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)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(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))*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)

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maple [B]  time = 2.78, size = 867, normalized size = 4.29 \[ \text {result too large to display} \]

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)^(5/2),x)

[Out]

-2/15*(-24*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)^6+
4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(5*B*b+15*C*a+6*C*b)*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)*(15*B*a^3+5*B*b^3+15*C*a*b^2+3*C*b^3)*sin(1/2
*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+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)*Elli
pticF(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)+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))*(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2)+15*B*(-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^3-45*B*(-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+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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+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)*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)-45*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^2*b-9*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))*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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),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)^(5/2), x)

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mupad [B]  time = 2.75, size = 248, normalized size = 1.23 \[ \frac {B\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {3\,C\,a\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

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)^(5/2),x)

[Out]

(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*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))

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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*(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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