3.12.81 \(\int \cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1181]

Optimal. Leaf size=93 \[ \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

2/5*(3*A+5*C)*(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))/d+2/3*B*(c
os(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))/d+2/5*A*cos(d*x+c)^(3/2)*s
in(d*x+c)/d+2/3*B*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4149, 3102, 2827, 2719, 2715, 2720} \begin {gather*} \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2719

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

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 2827

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 3102

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

Rule 4149

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.31, size = 72, normalized size = 0.77 \begin {gather*} \frac {2 \left (3 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 B F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} (5 B+3 A \cos (c+d x)) \sin (c+d x)\right )}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2] + 5*B*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*B + 3*A*Co
s[c + d*x])*Sin[c + d*x]))/(15*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(307\) vs. \(2(133)=266\).
time = 0.11, size = 308, normalized size = 3.31

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (24 A +20 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-6 A -10 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-9 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(308\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(
24*A+20*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*A-10*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-9*A*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+5*B*(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))-15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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)/(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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*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 = 0.52, size = 149, normalized size = 1.60 \begin {gather*} \frac {2 \, {\left (3 \, A \cos \left (d x + c\right ) + 5 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*(3*A*cos(d*x + c) + 5*B)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*I*sqrt(2)*B*weierstrassPInverse(-4, 0, co
s(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqr
t(2)*(-3*I*A - 5*I*C)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sq
rt(2)*(3*I*A + 5*I*C)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/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)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**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)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.60, size = 96, normalized size = 1.03 \begin {gather*} \frac {2\,B\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,A\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*cos(c + d*x)^(1/2)*sin(c
+ d*x))/(3*d) - (2*A*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c
+ d*x)^2)^(1/2))

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