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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.577069, antiderivative size = 244, 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, 4026, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 a \left (3 a^2 A+15 a b B+14 A b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 \left (3 a^2 A b+a^3 B+9 a b^2 B+3 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (5 a B+9 A b) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 a A \sin (c+d x) \sqrt{\sec (c+d x)} (a \sec (c+d x)+b)^2}{5 d} \]

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 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 4026

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 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{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*}

Mathematica [A]  time = 1.55486, size = 192, normalized size = 0.79 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (5 \left (3 a^2 A b+a^3 B+9 a b^2 B+3 A b^3\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-3 \left (3 a^3 A+15 a^2 b B+15 a A b^2-5 b^3 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{a \sin (c+d x) \left (9 \left (a^2 A+5 a b B+5 A b^2\right ) \cos (2 (c+d x))+15 \left (a^2 A+3 a b B+3 A b^2\right )+10 a (a B+3 A b) \cos (c+d x)\right )}{2 \cos ^{\frac{5}{2}}(c+d x)}\right )}{15 d} \]

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)

________________________________________________________________________________________

Maple [B]  time = 10.882, size = 997, normalized size = 4.1 \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)^(7/2),x)

[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*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
))-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*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*a*b*(A*b+B*a)*(-(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/
2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)-2/5*A*a^3/(8*sin(1/2*d*x+1/2*c)^6-1
2*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*(12*EllipticE(cos(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)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*c
os(1/2*d*x+1/2*c)-12*EllipticE(cos(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)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-8*sin(1/2*d*x+1/2*c)^2*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)+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)/(cos(1/2*d*x+1/2*c)^2-1/2)^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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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{7}{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)^(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)

________________________________________________________________________________________

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{7}{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)^(7/2),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)^(7/2), x)

________________________________________________________________________________________

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)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{7}{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)^(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)