3.354 \(\int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \, dx\)

Optimal. Leaf size=161 \[ \frac {6 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}+\frac {46 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d \sqrt {a \cos (c+d x)+a}}+\frac {92 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{21 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d} \]

[Out]

46/21*a^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+6/7*a^3*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(
d*x+c))^(1/2)+2/7*a^2*sec(d*x+c)^(7/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+92/21*a^3*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 = 0.35, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4222, 2762, 2980, 2772, 2771} \[ \frac {2 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{7 d}+\frac {6 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}}+\frac {46 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d \sqrt {a \cos (c+d x)+a}}+\frac {92 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{21 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(9/2),x]

[Out]

(92*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(21*d*Sqrt[a + a*Cos[c + d*x]]) + (46*a^3*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(21*d*Sqrt[a + a*Cos[c + d*x]]) + (6*a^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]])
+ (2*a^2*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d)

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

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 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), 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, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

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 (a+a \cos (c+d x))^{5/2} \sec ^{\frac {9}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {1}{7} \left (2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\left (-\frac {15 a}{2}-\frac {11}{2} a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {6 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (23 a^2 \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 {46 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d \sqrt {a+a \cos (c+d x)}}+\frac {6 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{21} \left (46 a^2 \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 {92 a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{21 d \sqrt {a+a \cos (c+d x)}}+\frac {46 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d \sqrt {a+a \cos (c+d x)}}+\frac {6 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 5.36, size = 74, normalized size = 0.46 \[ \frac {a^2 (93 \cos (c+d x)+23 \cos (2 (c+d x))+23 \cos (3 (c+d x))+29) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (\cos (c+d x)+1)}}{21 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(9/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(29 + 93*Cos[c + d*x] + 23*Cos[2*(c + d*x)] + 23*Cos[3*(c + d*x)])*Sec[c + d*x
]^(7/2)*Tan[(c + d*x)/2])/(21*d)

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fricas [A]  time = 1.05, size = 94, normalized size = 0.58 \[ \frac {2 \, {\left (46 \, a^{2} \cos \left (d x + c\right )^{3} + 23 \, a^{2} \cos \left (d x + c\right )^{2} + 12 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{21 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/21*(46*a^2*cos(d*x + c)^3 + 23*a^2*cos(d*x + c)^2 + 12*a^2*cos(d*x + c) + 3*a^2)*sqrt(a*cos(d*x + c) + a)*si
n(d*x + c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x + c)))

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giac [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((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.20, size = 85, normalized size = 0.53 \[ -\frac {2 \left (46 \left (\cos ^{4}\left (d x +c \right )\right )-23 \left (\cos ^{3}\left (d x +c \right )\right )-11 \left (\cos ^{2}\left (d x +c \right )\right )-9 \cos \left (d x +c \right )-3\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} a^{2}}{21 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(9/2),x)

[Out]

-2/21/d*(46*cos(d*x+c)^4-23*cos(d*x+c)^3-11*cos(d*x+c)^2-9*cos(d*x+c)-3)*cos(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*(
1/cos(d*x+c))^(9/2)/sin(d*x+c)*a^2

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maxima [A]  time = 1.04, size = 243, normalized size = 1.51 \[ \frac {8 \, {\left (\frac {21 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {56 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {36 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {8 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{21 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

8/21*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c)
+ 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/(
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)*(2*sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1))

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mupad [B]  time = 4.21, size = 227, normalized size = 1.41 \[ \frac {-\frac {35\,a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{2}+35\,a^2\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}+\frac {23\,a^2\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {2\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{2}}{\frac {63\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {63\,d\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{8}+\frac {21\,d\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{8}+\frac {21\,d\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{8}} \]

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))^(5/2),x)

[Out]

(35*a^2*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) - (35*a^2*sin(c/2 + (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))/2 + (23*a^2*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))/2)/((63*d*cos(c/2 + (d*x)/2))/8 + (63*d*cos((3*c)/2 + (3*d*x)/2))/8 + (21*d*cos((5*c)/2 +
(5*d*x)/2))/8 + (21*d*cos((7*c)/2 + (7*d*x)/2))/8)

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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((a+a*cos(d*x+c))**(5/2)*sec(d*x+c)**(9/2),x)

[Out]

Timed out

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