3.503 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x) \, dx\)

Optimal. Leaf size=181 \[ \frac{2 a^2 (8 A+7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (52 A+63 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (52 A+63 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{7 d} \]

[Out]

(4*a^2*(52*A + 63*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(52*A + 63*B)*
Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(8*A + 7*B)*Sec[c + d*x]^(5/2)*Sin[
c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(
7*d)

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Rubi [A]  time = 0.560834, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2975, 2980, 2772, 2771} \[ \frac{2 a^2 (8 A+7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (52 A+63 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a^2 (52 A+63 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(4*a^2*(52*A + 63*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(52*A + 63*B)*
Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(8*A + 7*B)*Sec[c + d*x]^(5/2)*Sin[
c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(
7*d)

Rule 2961

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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{7} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (8 A+7 B)+\frac{1}{2} a (4 A+7 B) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (8 A+7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{35} \left (a (52 A+63 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (52 A+63 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (8 A+7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{105} \left (2 a (52 A+63 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 (52 A+63 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (52 A+63 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (8 A+7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.544828, size = 102, normalized size = 0.56 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} (3 (78 A+77 B) \cos (c+d x)+(52 A+63 B) \cos (2 (c+d x))+52 A \cos (3 (c+d x))+82 A+63 B \cos (3 (c+d x))+63 B)}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(82*A + 63*B + 3*(78*A + 77*B)*Cos[c + d*x] + (52*A + 63*B)*Cos[2*(c + d*x)] + 5
2*A*Cos[3*(c + d*x)] + 63*B*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/2)*Tan[(c + d*x)/2])/(105*d)

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Maple [A]  time = 0.652, size = 117, normalized size = 0.7 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 104\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+126\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+52\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+39\,A\cos \left ( dx+c \right ) +21\,B\cos \left ( dx+c \right ) +15\,A \right ) \cos \left ( dx+c \right ) }{105\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{9}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x)

[Out]

-2/105/d*a*(-1+cos(d*x+c))*(104*A*cos(d*x+c)^3+126*B*cos(d*x+c)^3+52*A*cos(d*x+c)^2+63*B*cos(d*x+c)^2+39*A*cos
(d*x+c)+21*B*cos(d*x+c)+15*A)*cos(d*x+c)*(1/cos(d*x+c))^(9/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)

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Maxima [B]  time = 2.50775, size = 711, normalized size = 3.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

4/105*((105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 273*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 171*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*
x + c) + 1)^7 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1))
+ 21*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 15*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 17*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/
((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x + c)^2/(c
os(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d

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Fricas [A]  time = 1.46307, size = 288, normalized size = 1.59 \begin{align*} \frac{2 \,{\left (2 \,{\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{3} +{\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/105*(2*(52*A + 63*B)*a*cos(d*x + c)^3 + (52*A + 63*B)*a*cos(d*x + c)^2 + 3*(13*A + 7*B)*a*cos(d*x + c) + 15*
A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x + c)))

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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+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(9/2),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 (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{\frac{9}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^(9/2), x)