3.4.39 \(\int \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \, dx\) [339]

Optimal. Leaf size=153 \[ \frac {32 a \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

16/35*a*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+12/35*a*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*
x+c))^(1/2)+2/7*a*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+32/35*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 = 0.18, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4307, 2851, 2850} \begin {gather*} \frac {2 a \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}+\frac {12 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}+\frac {32 a \sin (c+d x) \sqrt {\sec (c+d x)}}{35 d \sqrt {a \cos (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2),x]

[Out]

(32*a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (16*a*Sec[c + d*x]^(3/2)*Sin[c + d*x]
)/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (12*a*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (
2*a*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]])

Rule 2850

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 2851

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 4307

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 {9}{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 {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{7} \left (6 \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 {12 a \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{35} \left (24 \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 {16 a \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{35} \left (16 \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 {32 a \sqrt {\sec (c+d x)} \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {16 a \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {12 a \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 71, normalized size = 0.46 \begin {gather*} \frac {2 \sqrt {a (1+\cos (c+d x))} (9+18 \cos (c+d x)+4 \cos (2 (c+d x))+4 \cos (3 (c+d x))) \sec ^{\frac {7}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{35 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2),x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*(9 + 18*Cos[c + d*x] + 4*Cos[2*(c + d*x)] + 4*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/
2)*Tan[(c + d*x)/2])/(35*d)

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Maple [A]
time = 0.87, size = 82, normalized size = 0.54

method	result	size
default	\(-\frac {2 \left (16 \left (\cos ^{4}\left (d x +c \right )\right )-8 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-5\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{35 d \sin \left (d x +c \right )}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/35/d*(16*cos(d*x+c)^4-8*cos(d*x+c)^3-2*cos(d*x+c)^2-cos(d*x+c)-5)*cos(d*x+c)*(1/cos(d*x+c))^(9/2)*(a*(1+cos
(d*x+c)))^(1/2)/sin(d*x+c)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (129) = 258\).
time = 0.53, size = 283, normalized size = 1.85 \begin {gather*} \frac {2 \, {\left (\frac {35 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {70 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {84 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {58 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {9 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{35 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/35*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/(
d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(4*sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x
+ c)^8/(cos(d*x + c) + 1)^8 + 1))

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Fricas [A]
time = 0.39, size = 81, normalized size = 0.53 \begin {gather*} \frac {2 \, {\left (16 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 5\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{35 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*cos(d*x + c)^3 + 8*cos(d*x + c)^2 + 6*cos(d*x + c) + 5)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos
(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x + c)))

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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(sec(d*x+c)**(9/2)*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 0.54, size = 143, normalized size = 0.93 \begin {gather*} \frac {4 \, \sqrt {2} {\left ({\left ({\left ({\left (7 \, {\left (5 \, {\left (\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} + 267\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 3684\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1869\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 350\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 35\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 )}{35 \, {\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 {7}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

4/35*sqrt(2)*((((7*(5*(tan(1/4*d*x + 1/4*c)^2 - 10)*tan(1/4*d*x + 1/4*c)^2 + 267)*tan(1/4*d*x + 1/4*c)^2 - 368
4)*tan(1/4*d*x + 1/4*c)^2 + 1869)*tan(1/4*d*x + 1/4*c)^2 - 350)*tan(1/4*d*x + 1/4*c)^2 + 35)*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)^(7/2)*d)

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Mupad [B]
time = 5.19, size = 163, normalized size = 1.07 \begin {gather*} \frac {14\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}+4\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{\frac {105\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {105\,d\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{8}+\frac {35\,d\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{8}+\frac {35\,d\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(1/2),x)

[Out]

(14*sin((3*c)/2 + (3*d*x)/2)*(a + a*cos(c + d*x))^(1/2)*((2*exp(c*1i + d*x*1i))/(exp(c*2i + d*x*2i) + 1))^(1/2
) + 4*sin((7*c)/2 + (7*d*x)/2)*(a + a*cos(c + d*x))^(1/2)*((2*exp(c*1i + d*x*1i))/(exp(c*2i + d*x*2i) + 1))^(1
/2))/((105*d*cos(c/2 + (d*x)/2))/8 + (105*d*cos((3*c)/2 + (3*d*x)/2))/8 + (35*d*cos((5*c)/2 + (5*d*x)/2))/8 +
(35*d*cos((7*c)/2 + (7*d*x)/2))/8)

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