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

Optimal. Leaf size=166 \[ -\frac{5 (A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-7 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(4 A-7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{5 (A-2 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.389269, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2977, 2748, 2639, 2635, 2641} \[ -\frac{5 (A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-7 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(4 A-7 B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{5 (A-2 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 2635

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

Rubi steps

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

Mathematica [C]  time = 6.682, size = 1218, normalized size = 7.34 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((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 - (((7*I)/2)*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 + (10*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]) - (20*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]]*((-4*(2*A - 3*B + 2*A*Cos[c] - 4*B*Cos[c])*Csc[c])/d + (8*B
*Cos[d*x]*Sin[c])/(3*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(2*A*Sin[(d*x)/2] - 3*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*B*Cos[c]*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

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Maple [B]  time = 3.277, size = 435, 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(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^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)*(-16*B*cos(1/2*d*x+1/2*c)^8+24*A*cos(1/2*d*x+1/2*c
)^6+10*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+24*A*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))-12*B*cos(1/2*d*x+1/2*c)^6-20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3-42*B*cos(1/2*d*x+1/2*c)^3*(si
n(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))-38*A*cos(1/2
*d*x+1/2*c)^4+48*B*cos(1/2*d*x+1/2*c)^4+15*A*cos(1/2*d*x+1/2*c)^2-21*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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**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 )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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