3.1.80 \(\int \frac {\cos ^4(c+d x) (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) [80]

Optimal. Leaf size=115 \[ \frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b^3 d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^5 d} \]

[Out]

2/7*C*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^5/d+2/21*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell
ipticF(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)+2/21*(7*A+5*C)*sin(d*x+c)*(b*co
s(d*x+c))^(1/2)/b^3/d

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Rubi [A]
time = 0.06, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {16, 3093, 2715, 2721, 2720} \begin {gather*} \frac {2 (7 A+5 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 b^3 d}+\frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

(2*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*b^2*d*Sqrt[b*Cos[c + d*x]]) + (2*(7*A + 5*C)*
Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(21*b^3*d) + (2*C*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b^5*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]

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.27, size = 80, normalized size = 0.70 \begin {gather*} \frac {4 (7 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(14 A+13 C+3 C \cos (2 (c+d x))) \sin (2 (c+d x))}{42 b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

(4*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (14*A + 13*C + 3*C*Cos[2*(c + d*x)])*Sin[2*(c +
d*x)])/(42*b^2*d*Sqrt[b*Cos[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(295\) vs. \(2(127)=254\).
time = 0.39, size = 296, normalized size = 2.57

method	result	size
default	\(-\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (28 A +56 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-14 A -16 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+7 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{21 b^{2} \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\)	\(296\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),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^2*(48*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8-72*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(28*A+56*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-14*A-16
*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+7*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+5*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^4/(b*cos(d*x + c))^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 109, normalized size = 0.95 \begin {gather*} \frac {\sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + 7 \, A + 5 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{21 \, b^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/21*(sqrt(2)*(-7*I*A - 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(7*
I*A + 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*C*cos(d*x + c)^2 + 7*A +
5*C)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b^3*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)**4*(A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(5/2),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)^4*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^4/(b*cos(d*x + c))^(5/2), 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 )}^4\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\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)^4*(A + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^4*(A + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(5/2), x)

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