3.127 \(\int \frac {\cos ^8(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=128 \[ \frac {2 \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^7 d}+\frac {18 \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 b^5 d}+\frac {30 \sin (c+d x) \sqrt {b \cos (c+d x)}}{77 b^3 d}+\frac {30 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

18/77*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^5/d+2/11*(b*cos(d*x+c))^(9/2)*sin(d*x+c)/b^7/d+30/77*(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))*cos(d*x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/
2)+30/77*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/b^3/d

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Rubi [A]  time = 0.08, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2635, 2642, 2641} \[ \frac {2 \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^7 d}+\frac {18 \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 b^5 d}+\frac {30 \sin (c+d x) \sqrt {b \cos (c+d x)}}{77 b^3 d}+\frac {30 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 b^2 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^8/(b*Cos[c + d*x])^(5/2),x]

[Out]

(30*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*b^2*d*Sqrt[b*Cos[c + d*x]]) + (30*Sqrt[b*Cos[c + d*x]]*S
in[c + d*x])/(77*b^3*d) + (18*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*b^5*d) + (2*(b*Cos[c + d*x])^(9/2)*Sin[
c + d*x])/(11*b^7*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^8(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx &=\frac {\int (b \cos (c+d x))^{11/2} \, dx}{b^8}\\ &=\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^7 d}+\frac {9 \int (b \cos (c+d x))^{7/2} \, dx}{11 b^6}\\ &=\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^5 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^7 d}+\frac {45 \int (b \cos (c+d x))^{3/2} \, dx}{77 b^4}\\ &=\frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b^3 d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^5 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^7 d}+\frac {15 \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx}{77 b^2}\\ &=\frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b^3 d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^5 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^7 d}+\frac {\left (15 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {30 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 b^2 d \sqrt {b \cos (c+d x)}}+\frac {30 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 b^3 d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 b^5 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^7 d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 76, normalized size = 0.59 \[ \frac {347 \sin (2 (c+d x))+64 \sin (4 (c+d x))+7 \sin (6 (c+d x))+480 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{1232 b^2 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^8/(b*Cos[c + d*x])^(5/2),x]

[Out]

(480*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 347*Sin[2*(c + d*x)] + 64*Sin[4*(c + d*x)] + 7*Sin[6*(c +
d*x)])/(1232*b^2*d*Sqrt[b*Cos[c + d*x]])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{5}}{b^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c))*cos(d*x + c)^5/b^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{8}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^8/(b*cos(d*x + c))^(5/2), x)

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maple [A]  time = 0.19, size = 236, normalized size = 1.84 \[ -\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (448 \left (\cos ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1568 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2384 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2040 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1084 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-370 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+62 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{77 b^{2} \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8/(b*cos(d*x+c))^(5/2),x)

[Out]

-2/77*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/b^2*(448*cos(1/2*d*x+1/2*c)^13-1568*cos(1/2*d*
x+1/2*c)^11+2384*cos(1/2*d*x+1/2*c)^9-2040*cos(1/2*d*x+1/2*c)^7+1084*cos(1/2*d*x+1/2*c)^5-370*cos(1/2*d*x+1/2*
c)^3+15*(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))+6
2*cos(1/2*d*x+1/2*c))/(-b*(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)/(b*(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 \[ \int \frac {\cos \left (d x + c\right )^{8}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^8/(b*cos(d*x + c))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^8}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^8/(b*cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^8/(b*cos(c + d*x))^(5/2), 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)**8/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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