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

Optimal. Leaf size=250 \[ -\frac{2 (A-7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (31 A-7 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}-\frac{2 (43 A-91 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos
[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(43*A - 91*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a +
a*Cos[c + d*x]]) + (2*(31*A - 7*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*(A
- 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(7/2)*Sin[c + d*x]
)/(7*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.837951, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2984, 12, 2782, 205} \[ -\frac{2 (A-7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (31 A-7 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}-\frac{2 (43 A-91 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos
[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(43*A - 91*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a +
a*Cos[c + d*x]]) + (2*(31*A - 7*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*(A
- 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(7/2)*Sin[c + d*x]
)/(7*d*Sqrt[a + a*Cos[c + d*x]])

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 2984

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(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]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{9}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (A-7 B)+3 a A \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{7 a}\\ &=-\frac{2 (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 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^2 (31 A-7 B)-a^2 (A-7 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 (31 A-7 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (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 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a^3 (43 A-91 B)+\frac{1}{4} a^3 (31 A-7 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{2 (43 A-91 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A-7 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (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 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (16 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{105 a^4 (A-B)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^4}\\ &=-\frac{2 (43 A-91 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A-7 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (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 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\left ((A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx\\ &=-\frac{2 (43 A-91 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A-7 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (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 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}-\frac{\left (2 a (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{\sqrt{2} (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}-\frac{2 (43 A-91 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A-7 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (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 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.79707, size = 250, normalized size = 1. \[ \frac{2 e^{-\frac{1}{2} i (c+d x)} \cos \left (\frac{1}{2} (c+d x)\right ) \left (105 i (A-B) \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-\frac{1}{2} \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right )+i \sin \left (\frac{1}{2} (c+d x)\right )\right ) (3 (47 A-119 B) \cos (c+d x)+(14 B-62 A) \cos (2 (c+d x))+43 A \cos (3 (c+d x))-122 A-91 B \cos (3 (c+d x))+14 B)\right )}{105 d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((105*I)*(A - B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*
x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - ((-122*A + 14*B + 3*(47*A - 119*
B)*Cos[c + d*x] + (-62*A + 14*B)*Cos[2*(c + d*x)] + 43*A*Cos[3*(c + d*x)] - 91*B*Cos[3*(c + d*x)])*Sec[c + d*x
]^(7/2)*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(c + d*x)/2])/2))/(105*d*E^((I/2)*(c + d*x))*Sqrt[a*(1 + C
os[c + d*x])])

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Maple [B]  time = 0.769, size = 657, normalized size = 2.6 \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))*sec(d*x+c)^(9/2)/(a+cos(d*x+c)*a)^(1/2),x)

[Out]

1/105/d*2^(1/2)/a*(105*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-105
*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+420*A*arcsin((-1+cos(d*x+
c))/sin(d*x+c))*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-420*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*
x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+630*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+
cos(d*x+c)))^(7/2)-630*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+420
*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-420*B*arcsin((-1+cos(d*x+c)
)/sin(d*x+c))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+105*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c
)/(1+cos(d*x+c)))^(7/2)-105*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+43*A*cos(d*
x+c)^3*2^(1/2)*sin(d*x+c)-91*B*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)-31*A*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)+7*B*cos(d*
x+c)^2*2^(1/2)*sin(d*x+c)+3*A*cos(d*x+c)*2^(1/2)*sin(d*x+c)-21*B*cos(d*x+c)*2^(1/2)*sin(d*x+c)-15*A*2^(1/2)*si
n(d*x+c))*cos(d*x+c)*sin(d*x+c)^6*(1/cos(d*x+c))^(9/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^3/(1+cos(d*x+c
))^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72449, size = 493, normalized size = 1.97 \begin{align*} -\frac{\frac{105 \, \sqrt{2}{\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{4} +{\left (A - B\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}} + \frac{2 \,{\left ({\left (43 \, A - 91 \, B\right )} \cos \left (d x + c\right )^{3} -{\left (31 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (A - 7 \, B\right )} \cos \left (d x + c\right ) - 15 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{105 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/105*(105*sqrt(2)*((A - B)*a*cos(d*x + c)^4 + (A - B)*a*cos(d*x + c)^3)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) +
a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*((43*A - 91*B)*cos(d*x + c)^3 - (31*A - 7*B)*cos(d*
x + c)^2 + 3*(A - 7*B)*cos(d*x + c) - 15*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos
(d*x + c)^4 + a*d*cos(d*x + c)^3)

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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+B*cos(d*x+c))*sec(d*x+c)**(9/2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{9}{2}}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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