∫ ∘ __________ a + a cos(c + dx) cos5

Optimal. Leaf size=77 \[ \frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

2/3*a*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+4/3*a*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c)

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Rubi [A]
time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2851, 2850} \begin {gather*} \frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}} \end {gather*}

(2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim

p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b,

c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp

[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]

+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n

+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &

\begin {align*} \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2}{3} \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 51, normalized size = 0.66 \begin {gather*} \frac {2 \sqrt {a (1+\cos (c+d x))} (1+2 \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

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

(2*Sqrt[a*(1 + Cos[c + d*x])]*(1 + 2*Cos[c + d*x])*Tan[(c + d*x)/2])/(3*d*Cos[c + d*x]^(3/2))

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method	result	size

default	\(-\frac {2 \left (2 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-1\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(54\)

int((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

-2/3/d*(2*cos(d*x+c)^2-cos(d*x+c)-1)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(3/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (65) = 130\).
time = 0.50, size = 190, normalized size = 2.47 \begin {gather*} \frac {2 \, {\left (\frac {3 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} \end {gather*}

integrate((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

2/3*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3

+ sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/(d*(sin(d*

x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d*x +

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Fricas [A]
time = 0.44, size = 61, normalized size = 0.79 \begin {gather*} \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

integrate((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

2/3*sqrt(a*cos(d*x + c) + a)*(2*cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

integrate((a+a*cos(d*x+c))**(1/2)/cos(d*x+c)**(5/2),x)

Integral(sqrt(a*(cos(c + d*x) + 1))/cos(c + d*x)**(5/2), x)

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Giac [A]
time = 0.55, size = 87, normalized size = 1.13 \begin {gather*} \frac {4 \, \sqrt {2} {\left ({\left (3 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 10\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 3\right )} \sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )}{3 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {3}{2}} d} \end {gather*}

integrate((a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="giac")

4/3*sqrt(2)*((3*tan(1/4*d*x + 1/4*c)^2 - 10)*tan(1/4*d*x + 1/4*c)^2 + 3)*sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))*tan

(1/4*d*x + 1/4*c)/((tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1)^(3/2)*d)

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Mupad [B]
time = 1.27, size = 82, normalized size = 1.06 \begin {gather*} \frac {4\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (\sin \left (c+d\,x\right )+\sin \left (2\,c+2\,d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (3\,\cos \left (c+d\,x\right )+2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )+2\right )} \end {gather*}

(4*(a*(cos(c + d*x) + 1))^(1/2)*(sin(c + d*x) + sin(2*c + 2*d*x) + sin(3*c + 3*d*x)))/(3*d*cos(c + d*x)^(1/2)*

(3*cos(c + d*x) + 2*cos(2*c + 2*d*x) + cos(3*c + 3*d*x) + 2))

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3.3.3 \(\int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [203]