3.153 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=136 \[ \frac {(2 A-5 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(A-4 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(2 A-5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-(A-4*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+1/3*(2*A-
5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+1/3*(A-B)*cos
(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2+1/3*(2*A-5*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d/(1+cos(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2977, 2748, 2641, 2639} \[ \frac {(2 A-5 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(A-4 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(2 A-5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}+\frac {(A-B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 6.66, size = 1184, normalized size = 8.71 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*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)]*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 + ((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))*Si
n[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)]*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 - (4*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - A
rcTan[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*Sqrt[
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[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*Sqrt[1 + Cot[c
]^2]) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((-4*(-A + 2*B + 2*B*Cos[c])*Csc[c])/d + (4*Sec[c/2]*Sec[c/2
+ (d*x)/2]*(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) - (2*(A - B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2

________________________________________________________________________________________

fricas [F]  time = 1.21, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 1.17, size = 421, normalized size = 3.10 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 B \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-20 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-A +B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2*d*x+1/2*c)^6+4*A*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+6*A*co
s(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))-24*B*cos(1/2*d*x+1/2*c)^6-10*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3-24*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-20*A*cos(1/2*d*x+1/2*c)^4+38*B*cos(1/2*d
*x+1/2*c)^4+9*A*cos(1/2*d*x+1/2*c)^2-15*B*cos(1/2*d*x+1/2*c)^2-A+B)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________