3.513 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x) \, dx\)

Optimal. Leaf size=181 \[ \frac{2 a^2 (10 A+7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{35 d}+\frac{2 a^3 (10 A+11 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (230 A+301 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d} \]

[Out]

(2*a^3*(230*A + 301*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(10*A + 11*B
)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(10*A + 7*B)*Sqrt[a + a*Cos[c + d*
x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(7*d)

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Rubi [A]  time = 0.676108, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2961, 2975, 2980, 2771} \[ \frac{2 a^2 (10 A+7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{35 d}+\frac{2 a^3 (10 A+11 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (230 A+301 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(2*a^3*(230*A + 301*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(10*A + 11*B
)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(10*A + 7*B)*Sqrt[a + a*Cos[c + d*
x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*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 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))^{5/2} (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))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a A (a+a \cos (c+d x))^{3/2} \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{(a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (10 A+7 B)+\frac{1}{2} a (2 A+7 B) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (10 A+7 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{35} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{7}{4} a^2 (10 A+11 B)+\frac{1}{4} a^2 (30 A+49 B) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (10 A+11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (10 A+7 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{105} \left (a^2 (230 A+301 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{2 a^3 (230 A+301 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (10 A+11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (10 A+7 B) \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.700292, size = 104, normalized size = 0.57 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} ((930 A+987 B) \cos (c+d x)+2 (115 A+98 B) \cos (2 (c+d x))+230 A \cos (3 (c+d x))+290 A+301 B \cos (3 (c+d x))+196 B)}{210 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(290*A + 196*B + (930*A + 987*B)*Cos[c + d*x] + 2*(115*A + 98*B)*Cos[2*(c + d*
x)] + 230*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/2)*Tan[(c + d*x)/2])/(210*d)

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Maple [A]  time = 0.655, size = 119, normalized size = 0.7 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 230\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+301\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+115\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+98\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+60\,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)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x)

[Out]

-2/105/d*a^2*(-1+cos(d*x+c))*(230*A*cos(d*x+c)^3+301*B*cos(d*x+c)^3+115*A*cos(d*x+c)^2+98*B*cos(d*x+c)^2+60*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 = 2.33033, size = 659, normalized size = 3.64 \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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

8/105*(5*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + 7*(15*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(
d*x + c) + 1) - 50*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/
(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/
2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d
*x + c) + 1)^4 + 1)))/d

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Fricas [A]  time = 1.44344, size = 300, normalized size = 1.66 \begin{align*} \frac{2 \,{\left ({\left (230 \, A + 301 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (115 \, A + 98 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/105*((230*A + 301*B)*a^2*cos(d*x + c)^3 + (115*A + 98*B)*a^2*cos(d*x + c)^2 + 3*(20*A + 7*B)*a^2*cos(d*x + c
) + 15*A*a^2)*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))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Timed out