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

Optimal. Leaf size=201 \[ \frac{5 (2 A-B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(7 A-4 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x) (\cos (c+d x)+1)}+\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(7 A-4 B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[Out]

((7*A - 4*B)*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (5*(2*A - B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + (5*(2*A
- B)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)) - ((7*A - 4*B)*Sin[c + d*x])/(a^2*d*Sqrt[Cos[c + d*x]]) - ((7*
A - 4*B)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)*(1 + Cos[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Cos[c + d
*x]^(3/2)*(a + a*Cos[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.363733, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2978, 2748, 2636, 2641, 2639} \[ \frac{5 (2 A-B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(7 A-4 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x) (\cos (c+d x)+1)}+\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(7 A-4 B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A - 4*B)*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (5*(2*A - B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + (5*(2*A
- B)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)) - ((7*A - 4*B)*Sin[c + d*x])/(a^2*d*Sqrt[Cos[c + d*x]]) - ((7*
A - 4*B)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)*(1 + Cos[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Cos[c + d
*x]^(3/2)*(a + a*Cos[c + d*x])^2)

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

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))^2} \, dx &=-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a (3 A-B)-\frac{5}{2} a (A-B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x) (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{15}{2} a^2 (2 A-B)-\frac{3}{2} a^2 (7 A-4 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x) (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac{(7 A-4 B) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}+\frac{(5 (2 A-B)) \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{2 a^2}\\ &=\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(7 A-4 B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x) (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac{(7 A-4 B) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}+\frac{(5 (2 A-B)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{(7 A-4 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{5 (2 A-B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{(7 A-4 B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x) (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 7.19355, size = 1258, normalized size = 6.26 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((7*I)/2)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2
*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*
d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(
-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqr
t[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c
] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*Co
s[c + d*x])^2 - ((2*I)*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4,
7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*S
in[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Co
s[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin
[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*
d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])
))/(a + a*Cos[c + d*x])^2 - (20*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x -
ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot
[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqr
t[1 + Cot[c]^2]) + (10*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[C
ot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Si
n[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot
[c]^2]) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((-2*(4*A - 2*B + 3*A*Cos[c] - 2*B*Cos[c])*Csc[c/2]*Sec[c/2
]*Sec[c])/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/2] - 2*B*Sin[(d*x)/2]))/d - (2*Sec[c/2]*Sec[c/2 +
(d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(3*d) + (8*A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (8*Sec[c]*S
ec[c + d*x]*(A*Sin[c] - 6*A*Sin[d*x] + 3*B*Sin[d*x]))/(3*d) - (2*(A - B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d))
)/(a + a*Cos[c + d*x])^2

________________________________________________________________________________________

Maple [B]  time = 11.656, size = 750, normalized size = 3.7 \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)^2,x)

[Out]

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*((4*A-2*B)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin
(1/2*d*x+1/2*c)^2)^(1/2)+(-8*A+4*B)*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^
2-1)+4*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1
/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+1/3*(A-B)*(2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*EllipticF(c
os(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6
+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*
d*x+1/2*c)/(-1+sin(1/2*d*x+1/2*c)^2))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{5} + 2 \, a^{2} \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}}, x\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))^2,x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(a^2*cos(d*x + c)^5 + 2*a^2*cos(d*x + c)^4 + a^2*cos(d*x + c)
^3), x)

________________________________________________________________________________________

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________

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 )}^{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))^2,x, algorithm="giac")

[Out]

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