3.340 \(\int \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \, dx\)

Optimal. Leaf size=115 \[ \frac{2 a \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(16*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a*Sec[c + d*x]^(3/2)*Sin[c + d*x])
/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.22103, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4222, 2772, 2771} \[ \frac{2 a \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2),x]

[Out]

(16*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a*Sec[c + d*x]^(3/2)*Sin[c + d*x])
/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]])

Rule 4222

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rule 2772

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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

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]

Rubi steps

\begin{align*} \int \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{1}{5} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{8 a \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{1}{15} \left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{16 a \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{8 a \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.128528, size = 61, normalized size = 0.53 \[ \frac{2 (4 \cos (c+d x)+4 \cos (2 (c+d x))+7) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)}}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2),x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*(7 + 4*Cos[c + d*x] + 4*Cos[2*(c + d*x)])*Sec[c + d*x]^(5/2)*Tan[(c + d*x)/2])/(
15*d)

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Maple [A]  time = 0.46, size = 72, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 16\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\cos \left ( dx+c \right ) -6 \right ) \cos \left ( dx+c \right ) }{15\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{7}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)*(a+cos(d*x+c)*a)^(1/2),x)

[Out]

-2/15/d*(8*cos(d*x+c)^3-4*cos(d*x+c)^2-cos(d*x+c)-3)*cos(d*x+c)*(1/cos(d*x+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/
sin(d*x+c)

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Maxima [B]  time = 1.60247, size = 320, normalized size = 2.78 \begin{align*} \frac{2 \,{\left (\frac{15 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{25 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{17 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{7 \, \sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}{\left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, d{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{7}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{7}{2}}{\left (\frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/15*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 17*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x +
c)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)
^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1))

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Fricas [A]  time = 1.61622, size = 193, normalized size = 1.68 \begin{align*} \frac{2 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*sqrt(a*cos(d*x + c) + a)*(8*cos(d*x + c)^2 + 4*cos(d*x + c) + 3)*sin(d*x + c)/((d*cos(d*x + c)^3 + d*cos(
d*x + c)^2)*sqrt(cos(d*x + c)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(7/2)*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)