3.6.62 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [562]

Optimal. Leaf size=244 \[ -\frac {2 \left (3 a^3 A+15 a A b^2+15 a^2 b B-5 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 B\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a \left (3 a^2 A+14 A b^2+15 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (9 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d} \]

[Out]

2/15*a^2*(9*A*b+5*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*(3*A*a^2+14*A*b^2+15*B*a*b)*sin(d*x+c)*sec(d*x+c)^(
1/2)/d+2/5*a*A*(b+a*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(3*A*a^3+15*A*a*b^2+15*B*a^2*b-5*B*b^3)*(c
os(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))*cos(d*x+c)^(1/2)*sec(d*x+c
)^(1/2)/d+2/3*(3*A*a^2*b+3*A*b^3+B*a^3+9*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.37, antiderivative size = 244, normalized size of antiderivative = 1.00, 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 = {3039, 4111, 4161, 4132, 3856, 2720, 4131, 2719} \begin {gather*} \frac {2 a \left (3 a^2 A+15 a b B+14 A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a^2 (5 a B+9 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 \left (a^3 B+3 a^2 A b+9 a b^2 B+3 A b^3\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^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+b)^2}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]

[Out]

(-2*(3*a^3*A + 15*a*A*b^2 + 15*a^2*b*B - 5*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*
x]])/(5*d) + (2*(3*a^2*A*b + 3*A*b^3 + a^3*B + 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Se
c[c + d*x]])/(3*d) + (2*a*(3*a^2*A + 14*A*b^2 + 15*a*b*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a^2*(9*A
*b + 5*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*a*A*Sqrt[Sec[c + d*x]]*(b + a*Sec[c + d*x])^2*Sin[c +
d*x])/(5*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3039

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 3856

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 4111

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m +
n))), x] + Dist[1/(m + n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*Simp[a^2*A*(m + n) + a*b*B*n +
(a*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1))*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] &&
!(IGtQ[n, 1] &&  !IntegerQ[m])

Rule 4131

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 4132

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 4161

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]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx &=\int \frac {(b+a \sec (c+d x))^3 (B+A \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2}{5} \int \frac {(b+a \sec (c+d x)) \left (-\frac {1}{2} b (a A-5 b B)+\frac {1}{2} \left (3 a^2 A+5 b (A b+2 a B)\right ) \sec (c+d x)+\frac {1}{2} a (9 A b+5 a B) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a^2 (9 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {-\frac {3}{4} b^2 (a A-5 b B)+\frac {5}{4} \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 B\right ) \sec (c+d x)+\frac {3}{4} a \left (3 a^2 A+14 A b^2+15 a b B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a^2 (9 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {-\frac {3}{4} b^2 (a A-5 b B)+\frac {3}{4} a \left (3 a^2 A+14 A b^2+15 a b B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 B\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a \left (3 a^2 A+14 A b^2+15 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (9 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (-3 a^3 A-15 a A b^2-15 a^2 b B+5 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 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 (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 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 (3 a^2 A+14 A b^2+15 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (9 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \left (\left (-3 a^3 A-15 a A b^2-15 a^2 b B+5 b^3 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^3 A+15 a A b^2+15 a^2 b B-5 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 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 (3 a^2 A+14 A b^2+15 a b B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^2 (9 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 1.73, size = 192, normalized size = 0.79 \begin {gather*} \frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 \left (3 a^3 A+15 a A b^2+15 a^2 b B-5 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (3 a^2 A b+3 A b^3+a^3 B+9 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {a \left (15 \left (a^2 A+3 A b^2+3 a b B\right )+10 a (3 A b+a B) \cos (c+d x)+9 \left (a^2 A+5 A b^2+5 a b B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}\right )}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(969\) vs. \(2(272)=544\).
time = 1.00, size = 970, normalized size = 3.98

method	result	size
default	\(\text {Expression too large to display}\)	\(970\)

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)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*A*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/
2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*B*a*b^2*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*a^2*(3*A*b+B*a)*(-
1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+6*a*b*(A*b+B*a)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)))+2/5*A*a^3/(8*sin(1/2*d*x
+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c)*sin(1/
2*d*x+1/2*c)^6-12*EllipticE(cos(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)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*EllipticE(cos(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)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*c
os(1/2*d*x+1/2*c)-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)*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*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(7/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 326, normalized size = 1.34 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (i \, B a^{3} + 3 i \, A a^{2} b + 9 i \, B a b^{2} + 3 i \, A b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{3} - 3 i \, A a^{2} b - 9 i \, B a b^{2} - 3 i \, A b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (3 i \, A a^{3} + 15 i \, B a^{2} b + 15 i \, A a b^{2} - 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, A a^{3} - 15 i \, B a^{2} b - 15 i \, A a b^{2} + 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, A a^{3} + 9 \, {\left (A a^{3} + 5 \, B a^{2} b + 5 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

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)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(5*sqrt(2)*(I*B*a^3 + 3*I*A*a^2*b + 9*I*B*a*b^2 + 3*I*A*b^3)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, c
os(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^3 - 3*I*A*a^2*b - 9*I*B*a*b^2 - 3*I*A*b^3)*cos(d*x + c)^2*we
ierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(3*I*A*a^3 + 15*I*B*a^2*b + 15*I*A*a*b^2 -
5*I*B*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) +
3*sqrt(2)*(-3*I*A*a^3 - 15*I*B*a^2*b - 15*I*A*a*b^2 + 5*I*B*b^3)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^3 + 9*(A*a^3 + 5*B*a^2*b + 5*A*a*b^2)*cos(d*x
+ c)^2 + 5*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(7/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^3,x)

[Out]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^3, x)

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