3.2.17 \(\int \frac {\cos ^5(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx\) [117]

Optimal. Leaf size=100 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b^2 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^4 d} \]

[Out]

2/7*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^4/d+10/21*(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/d/(b*cos(d*x+c))^(1/2)+10/21*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/b^2/d

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Rubi [A]
time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2715, 2721, 2720} \begin {gather*} \frac {2 \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^4 d}+\frac {10 \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 b^2 d}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*b*d*Sqrt[b*Cos[c + d*x]]) + (10*Sqrt[b*Cos[c + d*x]]*Sin
[c + d*x])/(21*b^2*d) + (2*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b^4*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 2715

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] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

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

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx &=\frac {\int (b \cos (c+d x))^{7/2} \, dx}{b^5}\\ &=\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^4 d}+\frac {5 \int (b \cos (c+d x))^{3/2} \, dx}{7 b^3}\\ &=\frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b^2 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^4 d}+\frac {5 \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx}{21 b}\\ &=\frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b^2 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^4 d}+\frac {\left (5 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b \sqrt {b \cos (c+d x)}}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {10 \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b^2 d}+\frac {2 (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^4 d}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 66, normalized size = 0.66 \begin {gather*} \frac {40 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+26 \sin (2 (c+d x))+3 \sin (4 (c+d x))}{84 b d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(40*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 26*Sin[2*(c + d*x)] + 3*Sin[4*(c + d*x)])/(84*b*d*Sqrt[b*Co
s[c + d*x]])

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Maple [A]
time = 0.08, size = 210, normalized size = 2.10

method	result	size
default	\(-\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 b \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\)	\(210\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(b*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/21*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/b*(48*cos(1/2*d*x+1/2*c)^9-120*cos(1/2*d*x+1/2
*c)^7+128*cos(1/2*d*x+1/2*c)^5-72*cos(1/2*d*x+1/2*c)^3+5*(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))+16*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 91, normalized size = 0.91 \begin {gather*} \frac {2 \, \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 5 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{21 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(2*sqrt(b*cos(d*x + c))*(3*cos(d*x + c)^2 + 5)*sin(d*x + c) - 5*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x +
c)))/(b^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^5}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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