3.1.90 \(\int \cos ^3(c+d x) (b \cos (c+d x))^{5/2} \, dx\) [90]

Optimal. Leaf size=125 \[ \frac {30 b^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 d \sqrt {b \cos (c+d x)}}+\frac {30 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{77 d}+\frac {18 (b \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}+\frac {2 (b \cos (c+d x))^{9/2} \sin (c+d x)}{11 b^2 d} \]

[Out]

18/77*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/d+2/11*(b*cos(d*x+c))^(9/2)*sin(d*x+c)/b^2/d+30/77*b^3*(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)/d/(b*cos(d*x+c))^(1/2)+3
0/77*b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.06, antiderivative size = 125, 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, 2715, 2721, 2720} \begin {gather*} \frac {30 b^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{77 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x) (b \cos (c+d x))^{9/2}}{11 b^2 d}+\frac {30 b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{77 d}+\frac {18 \sin (c+d x) (b \cos (c+d x))^{5/2}}{77 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.09, size = 83, normalized size = 0.66 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} \left (240 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} (290 \sin (c+d x)+57 \sin (3 (c+d x))+7 \sin (5 (c+d x)))\right )}{616 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(240*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(290*Sin[c + d*x] + 57*Sin[3*(c +
d*x)] + 7*Sin[5*(c + d*x)])))/(616*d*Cos[c + d*x]^(5/2))

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Maple [A]
time = 0.08, size = 236, normalized size = 1.89

method	result	size
default	\(-\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \left (448 \left (\cos ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1568 \left (\cos ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2384 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2040 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1084 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-370 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+62 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{77 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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^3*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 108, normalized size = 0.86 \begin {gather*} \frac {-15 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (7 \, b^{2} \cos \left (d x + c\right )^{4} + 9 \, b^{2} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{77 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/77*(-15*I*sqrt(2)*b^(5/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*I*sqrt(2)*b^(5/2)*w
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(7*b^2*cos(d*x + c)^4 + 9*b^2*cos(d*x + c)^2 + 15
*b^2)*sqrt(b*cos(d*x + c))*sin(d*x + c))/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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