3.192 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=155 \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac {9 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {9 \sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{5 a d (a \cos (c+d x)+a)^2} \]

[Out]

-9/10*(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^3/d+1/2*(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^3/d-1/5*cos(d*x+c)^(3/2)*sin(d
*x+c)/d/(a+a*cos(d*x+c))^3-2/5*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2+9/10*sin(d*x+c)*cos(d*x+c)^(
1/2)/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.31, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2765, 2977, 2978, 2748, 2641, 2639} \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac {9 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {9 \sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{5 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^3,x]

[Out]

(-9*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(2*a^3*d) - (Cos[c + d*x]^(3/2)*Sin[c +
d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*a*d*(a + a*Cos[c + d*x])^2) + (9*S
qrt[Cos[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[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]

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 2765

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

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 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])

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 a}{2}-\frac {9}{2} a \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {\int \frac {3 a^2-\frac {21}{2} a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {\int \frac {-\frac {15 a^3}{4}+\frac {27}{4} a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{4 a^3}-\frac {9 \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac {9 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 6.41, size = 705, normalized size = 4.55 \[ -\frac {2 \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin (c) \left (-\sqrt {\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{d \sqrt {\cot ^2(c)+1} (a \cos (c+d x)+a)^3}-\frac {9 i \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {2 e^{2 i d x} \sqrt {e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt {i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{3 i d \cos (c) \left (1+e^{2 i d x}\right )-3 d \sin (c) \left (-1+e^{2 i d x}\right )}-\frac {2 \sqrt {e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt {i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{d \sin (c) \left (-1+e^{2 i d x}\right )-i d \cos (c) \left (1+e^{2 i d x}\right )}\right )}{10 (a \cos (c+d x)+a)^3}+\frac {\sqrt {\cos (c+d x)} \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {2 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{5 d}+\frac {2 \tan \left (\frac {c}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{5 d}-\frac {12 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{5 d}-\frac {12 \tan \left (\frac {c}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{5 d}+\frac {36 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right )}{5 d}+\frac {36 \csc (c)}{5 d}\right )}{(a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^3,x]

[Out]

(((-9*I)/10)*Cos[c/2 + (d*x)/2]^6*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])^3 - (2*Cos[c/2 + (d*x)/2]^6*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]*S
in[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2])
+ (Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((36*Csc[c])/(5*d) + (36*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/
(5*d) - (12*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])
/(5*d) - (12*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(5*d) + (2*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d
*x])^3

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^3, x)

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maple [A]  time = 0.48, size = 270, normalized size = 1.74 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (36 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-66 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right )}{20 a^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x)

[Out]

-1/20*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(36*cos(1/2*d*x+1/2*c)^8+10*(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)^5+18*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))-66*cos(1/2*d*x+1/2*c)^6+38*cos(1/2*d*x+1/2*c)^4-9*cos(1/2*d*x+1/2*c)^2+1)/a^3/(-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)^5/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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(5/2)/(a + a*cos(c + d*x))^3,x)

[Out]

int(cos(c + d*x)^(5/2)/(a + a*cos(c + d*x))^3, x)

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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)**(5/2)/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

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