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

Optimal. Leaf size=250 \[ \frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{(163 A-75 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((163*A - 75*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt
[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - ((17*A - 9*B)*Si
n[c + d*x])/(16*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((95*A - 39*B)*Sin[c + d*x])/(48*a^2*d*Co
s[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((299*A - 147*B)*Sin[c + d*x])/(48*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[
a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.751786, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2978, 2984, 12, 2782, 205} \[ \frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{(163 A-75 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((163*A - 75*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt
[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - ((17*A - 9*B)*Si
n[c + d*x])/(16*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((95*A - 39*B)*Sin[c + d*x])/(48*a^2*d*Co
s[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((299*A - 147*B)*Sin[c + d*x])/(48*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[
a + a*Cos[c + d*x]])

Rule 2978

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

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)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (11 A-3 B)-3 a (A-B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (95 A-39 B)-a^2 (17 A-9 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{8} a^3 (299 A-147 B)+\frac{1}{4} a^3 (95 A-39 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{3 a^4 (163 A-75 B)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{6 a^6}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{(163 A-75 B) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(163 A-75 B) \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 )}{16 a d}\\ &=\frac{(163 A-75 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 )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}-\frac{(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.33274, size = 239, normalized size = 0.96 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) ((1537 A-825 B) \cos (c+d x)+2 (503 A-255 B) \cos (2 (c+d x))+299 A \cos (3 (c+d x))+878 A-147 B \cos (3 (c+d x))-510 B)}{8 \cos ^{\frac{3}{2}}(c+d x)}+\frac{3 i (163 A-75 B) e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{12 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((3*I)*(163*A - 75*B)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]
*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[1 + E^((2*I)*(c + d*x))] - ((878
*A - 510*B + (1537*A - 825*B)*Cos[c + d*x] + 2*(503*A - 255*B)*Cos[2*(c + d*x)] + 299*A*Cos[3*(c + d*x)] - 147
*B*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*Tan[(c + d*x)/2])/(8*Cos[c + d*x]^(3/2))))/(12*d*(a*(1 + Cos[c + d*x])
)^(5/2))

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Maple [B]  time = 0.527, size = 571, normalized size = 2.3 \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))/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(5/2),x)

[Out]

-1/96/d*(a*(1+cos(d*x+c)))^(1/2)*(489*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcs
in((-1+cos(d*x+c))/sin(d*x+c))-225*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin(
(-1+cos(d*x+c))/sin(d*x+c))+1467*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*(cos(d*x
+c)/(1+cos(d*x+c)))^(3/2)-675*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(3/2)+1467*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+
cos(d*x+c)))^(3/2)-675*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d
*x+c)))^(3/2)+489*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-22
5*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-598*A*cos(d*x+c)^4
+294*B*cos(d*x+c)^4-408*A*cos(d*x+c)^3+216*B*cos(d*x+c)^3+686*A*cos(d*x+c)^2-318*B*cos(d*x+c)^2+384*A*cos(d*x+
c)-192*B*cos(d*x+c)-64*A)/a^3/sin(d*x+c)/(1+cos(d*x+c))^2/cos(d*x+c)^(3/2)

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Maxima [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))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.44291, size = 728, normalized size = 2.91 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \,{\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \,{\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \,{\left ({\left (299 \, A - 147 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (503 \, A - 255 \, B\right )} \cos \left (d x + c\right )^{2} + 32 \,{\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 32 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{96 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/96*(3*sqrt(2)*((163*A - 75*B)*cos(d*x + c)^5 + 3*(163*A - 75*B)*cos(d*x + c)^4 + 3*(163*A - 75*B)*cos(d*x +
c)^3 + (163*A - 75*B)*cos(d*x + c)^2)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x
+ c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*((299*A - 147*B)*cos(d*x + c)^3 + (503*A - 255*B)
*cos(d*x + c)^2 + 32*(5*A - 3*B)*cos(d*x + c) - 32*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)
)/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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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))/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2)), x)