3.563 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x) \, dx\)
Optimal. Leaf size=239 \[ \frac{2 a \left (3 a^2 B+9 a A b-2 b^2 B\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 \left (a^3 A+9 a^2 b B+9 a A b^2+b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 (a A-b B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{3 d \sqrt{\sec (c+d x)}} \]
[Out]
(-2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d
+ (2*(a^3*A + 9*a*A*b^2 + 9*a^2*b*B + b^3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])
/(3*d) + (2*a*(9*a*A*b + 3*a^2*B - 2*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a^2*(a*A - b*B)*Sec[c
+ d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*b*B*(b + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])
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Rubi [A] time = 0.565625, antiderivative size = 239, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used =
{2960, 4025, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 a \left (3 a^2 B+9 a A b-2 b^2 B\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 \left (a^3 A+9 a^2 b B+9 a A b^2+b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 (a A-b B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{3 d \sqrt{\sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2),x]
[Out]
(-2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d
+ (2*(a^3*A + 9*a*A*b^2 + 9*a^2*b*B + b^3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])
/(3*d) + (2*a*(9*a*A*b + 3*a^2*B - 2*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a^2*(a*A - b*B)*Sec[c
+ d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*b*B*(b + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])
Rule 2960
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 4025
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]
Rule 4076
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] && !LtQ[n, -1]
Rule 4047
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Rule 3771
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
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 4046
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
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]
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x) \, dx &=\int \frac{(b+a \sec (c+d x))^3 (B+A \sec (c+d x))}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{2}{3} \int \frac{(b+a \sec (c+d x)) \left (-\frac{1}{2} b (3 A b+7 a B)-\frac{1}{2} \left (6 a A b+3 a^2 B+b^2 B\right ) \sec (c+d x)-\frac{3}{2} a (a A-b B) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^2 (a A-b B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{4}{9} \int \frac{-\frac{3}{4} b^2 (3 A b+7 a B)-\frac{3}{4} \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) \sec (c+d x)-\frac{3}{4} a \left (9 a A b+3 a^2 B-2 b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^2 (a A-b B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{4}{9} \int \frac{-\frac{3}{4} b^2 (3 A b+7 a B)-\frac{3}{4} a \left (9 a A b+3 a^2 B-2 b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{3} \left (-a^3 A-9 a A b^2-9 a^2 b B-b^3 B\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a \left (9 a A b+3 a^2 B-2 b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 a^2 (a A-b B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{3} \left (\left (-a^3 A-9 a A b^2-9 a^2 b B-b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 a \left (9 a A b+3 a^2 B-2 b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 a^2 (a A-b B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\left (\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 a \left (9 a A b+3 a^2 B-2 b^2 B\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d}+\frac{2 a^2 (a A-b B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.86882, size = 166, normalized size = 0.69 \[ \frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (2 \left (a^3 A+9 a^2 b B+9 a A b^2+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{\sin (c+d x) \left (6 a^2 (a B+3 A b) \cos (c+d x)+2 a^3 A+b^3 B \cos (2 (c+d x))+b^3 B\right )}{\cos ^{\frac{3}{2}}(c+d x)}\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2),x]
[Out]
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-6*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2] +
2*(a^3*A + 9*a*A*b^2 + 9*a^2*b*B + b^3*B)*EllipticF[(c + d*x)/2, 2] + ((2*a^3*A + b^3*B + 6*a^2*(3*A*b + a*B)
*Cos[c + d*x] + b^3*B*Cos[2*(c + d*x)])*Sin[c + d*x])/Cos[c + d*x]^(3/2)))/(3*d)
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Maple [B] time = 10.385, size = 1212, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}
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))*sec(d*x+c)^(5/2),x)
[Out]
2/3*(-(-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*(8*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+2*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*sin(1/2*d*x+1/2*c)^2+18*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)
^2+18*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
2*b*sin(1/2*d*x+1/2*c)^2-6*A*(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))*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*b^3*sin(1/2*d*x+1/2*c)^2-36*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+18*B*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b*sin(1/2*d*x+1/
2*c)^2+2*B*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*b^3*sin(1/2*d*x+1/2*c)^2+6*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))*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*a^3*sin(1/2*d*x+1/2*c)^2-18*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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)^2-12*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^4-8*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3-9*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2-9*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b+3*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3+2*A*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+18*A*a^2*b*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-9*B*(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)
)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b-B*(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))*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3-3*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))*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*a^3+9*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))*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2+6*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*B*b^3*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., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
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))*sec(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{3} \cos \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{2} b + 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 )\right )} \sec \left (d x + c\right )^{\frac{5}{2}}, x\right ) \end{align*}
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))*sec(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((B*b^3*cos(d*x + c)^4 + A*a^3 + (3*B*a*b^2 + A*b^3)*cos(d*x + c)^3 + 3*(B*a^2*b + A*a*b^2)*cos(d*x +
c)^2 + (B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sec(d*x + c)^(5/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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))*sec(d*x+c)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
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))*sec(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(5/2), x)