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

Optimal. Leaf size=172 \[ \frac{2 a^2 (4 A+5 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (104 A+175 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{6 a A \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

[Out]

(2*a^2*(4*A + 5*C)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(104*A + 175*C)*S
in[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (6*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x
])/(35*d*Cos[c + d*x]^(5/2)) + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2))

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Rubi [A]  time = 0.523874, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {3044, 2975, 2980, 2771} \[ \frac{2 a^2 (4 A+5 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (104 A+175 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{6 a A \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{7 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(4*A + 5*C)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(104*A + 175*C)*S
in[c + d*x])/(105*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (6*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x
])/(35*d*Cos[c + d*x]^(5/2)) + (2*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2))

Rule 3044

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

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 \frac{(a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{3 a A}{2}+\frac{1}{2} a (2 A+7 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac{6 a A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{7}{4} a^2 (4 A+5 C)+\frac{1}{4} a^2 (16 A+35 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac{2 a^2 (4 A+5 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{6 a A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{1}{105} (a (104 A+175 C)) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (4 A+5 C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (104 A+175 C) \sin (c+d x)}{105 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{6 a A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{35 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{7 d \cos ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.53584, size = 102, normalized size = 0.59 \[ \frac{a \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((468 A+525 C) \cos (c+d x)+2 (52 A+35 C) \cos (2 (c+d x))+104 A \cos (3 (c+d x))+164 A+175 C \cos (3 (c+d x))+70 C)}{210 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(164*A + 70*C + (468*A + 525*C)*Cos[c + d*x] + 2*(52*A + 35*C)*Cos[2*(c + d*x)]
+ 104*A*Cos[3*(c + d*x)] + 175*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(210*d*Cos[c + d*x]^(7/2))

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Maple [A]  time = 0.104, size = 100, normalized size = 0.6 \begin{align*} -{\frac{2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 104\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+175\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+52\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+35\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+39\,A\cos \left ( dx+c \right ) +15\,A \right ) }{105\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/d*a*(-1+cos(d*x+c))*(104*A*cos(d*x+c)^3+175*C*cos(d*x+c)^3+52*A*cos(d*x+c)^2+35*C*cos(d*x+c)^2+39*A*cos
(d*x+c)+15*A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(7/2)

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Maxima [B]  time = 1.89633, size = 525, normalized size = 3.05 \begin{align*} \frac{4 \,{\left (\frac{35 \,{\left (\frac{3 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{5 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{2 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} C}{{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{5}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{5}{2}}} + \frac{{\left (\frac{105 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{245 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{273 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{171 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{38 \, \sqrt{2} a^{\frac{3}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} A{\left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{9}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{9}{2}}{\left (\frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}\right )}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/105*(35*(3*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*C/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)
*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)) + (105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sq
rt(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)))/d

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Fricas [A]  time = 1.49635, size = 271, normalized size = 1.58 \begin{align*} \frac{2 \,{\left ({\left (104 \, A + 175 \, C\right )} a \cos \left (d x + c\right )^{3} +{\left (52 \, A + 35 \, C\right )} a \cos \left (d x + c\right )^{2} + 39 \, A a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*((104*A + 175*C)*a*cos(d*x + c)^3 + (52*A + 35*C)*a*cos(d*x + c)^2 + 39*A*a*cos(d*x + c) + 15*A*a)*sqrt(
a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

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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+C*cos(d*x+c)**2)/cos(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 \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}{\cos \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+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

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