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]

8/15*a*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*a*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c
))^(1/2)+16/15*a*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [A]  time = 1.03, size = 71, normalized size = 0.62 \[ \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}} \]

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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giac [A]  time = 0.59, size = 116, normalized size = 1.01 \[ \frac {4 \, \sqrt {2} {\left ({\left ({\left (5 \, {\left (3 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 20\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 282\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 100\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 15\right )} \sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )}{15 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {5}{2}} d} \]

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]

4/15*sqrt(2)*(((5*(3*tan(1/4*d*x + 1/4*c)^2 - 20)*tan(1/4*d*x + 1/4*c)^2 + 282)*tan(1/4*d*x + 1/4*c)^2 - 100)*
tan(1/4*d*x + 1/4*c)^2 + 15)*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)^(5/2)*d)

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maple [A]  time = 0.21, size = 72, normalized size = 0.63 \[ -\frac {2 \left (8 \left (\cos ^{3}\left (d x +c \right )\right )-4 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-3\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)*(a+a*cos(d*x+c))^(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 = 0.98, size = 237, normalized size = 2.06 \[ \frac {2 \, {\left (\frac {15 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {17 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \]

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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mupad [B]  time = 1.84, size = 134, normalized size = 1.17 \[ \frac {8\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (7\,\sin \left (c+d\,x\right )+4\,\sin \left (2\,c+2\,d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )+2\,\sin \left (4\,c+4\,d\,x\right )+2\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*(a*(cos(c + d*x) + 1))^(1/2)*(1/cos(c + d*x))^(1/2)*(7*sin(c + d*x) + 4*sin(2*c + 2*d*x) + 9*sin(3*c + 3*d*
x) + 2*sin(4*c + 4*d*x) + 2*sin(5*c + 5*d*x)))/(15*d*(10*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 5*cos(3*c + 3*d*x
) + 2*cos(4*c + 4*d*x) + cos(5*c + 5*d*x) + 6))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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