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

Optimal. Leaf size=155 \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {4 \sin (c+d x) \sqrt {\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

-1/10*(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+1/6*(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*sin(d*x+c)*cos(d*x+c)^
(1/2)/d/(a+a*cos(d*x+c))^3+4/15*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2+1/10*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 = 0.30, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2765, 2978, 2748, 2641, 2639} \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {4 \sin (c+d x) \sqrt {\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-EllipticE[(c + d*x)/2, 2]/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(6*a^3*d) - (Sqrt[Cos[c + d*x]]*Sin[c + d*x]
)/(5*d*(a + a*Cos[c + d*x])^3) + (4*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (Sqrt[C
os[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 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 {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {\int \frac {\frac {a}{2}-\frac {7}{2} a \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {4 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {\int \frac {-\frac {a^2}{2}-2 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {4 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\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 {5 a^3}{4}+\frac {3}{4} a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {4 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {\int \sqrt {\cos (c+d x)} \, dx}{20 a^3}+\frac {\int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=-\frac {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 )}{6 a^3 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {4 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\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 = 3.95, size = 334, normalized size = 2.15 \[ \frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\cos (c+d x)} \left (14 \cos \left (\frac {1}{2} (c-d x)\right )+16 \cos \left (\frac {1}{2} (3 c+d x)\right )+20 \cos \left (\frac {1}{2} (c+3 d x)\right )-5 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+3 \cos \left (\frac {1}{2} (3 c+5 d x)\right )\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {4 i \sqrt {2} e^{-i (c+d x)} \left (3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+3 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{15 a^3 (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*(((-4*I)*Sqrt[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c
+ d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1
+ E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*E^(I*(c + d*x))*(-1 + E^((
2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (Sqrt[Cos[c + d*x]]*(14*Cos[(c - d*x)/2] + 16*Cos[
(3*c + d*x)/2] + 20*Cos[(c + 3*d*x)/2] - 5*Cos[(5*c + 3*d*x)/2] + 3*Cos[(3*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Se
c[(c + d*x)/2]^5)/(8*d)))/(15*a^3*(1 + Cos[c + d*x])^3)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos(d*x + c)^(3/2)/(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 \[ \int \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.52, size = 270, normalized size = 1.74 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3\right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*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+6*(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))-2*cos(1/2*d*x+1/2*c)^6-24*cos(1/2*d*x+1/2*c)^4+17*cos(1/2*d*x+1/2*c)^2-3)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*s
in(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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