3.353 \(\int (a+a \cos (c+d x))^{5/2} \sec ^{\frac {11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=201 \[ \frac {38 a^3 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {146 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {584 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {1168 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d} \]

[Out]

584/315*a^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+146/105*a^3*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+
a*cos(d*x+c))^(1/2)+38/63*a^3*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/9*a^2*sec(d*x+c)^(9/2)*si
n(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+1168/315*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 = 0.41, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {2 a^2 \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d}+\frac {38 a^3 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {146 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {584 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {1168 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(11/2),x]

[Out]

(1168*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (584*a^3*Sec[c + d*x]^(3/2)*Sin[
c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (146*a^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c
+ d*x]]) + (38*a^3*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*Sqrt[a + a*Cos[c
+ d*x]]*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*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 {11}{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 {11}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac {1}{9} \left (2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\left (-\frac {19 a}{2}-\frac {15}{2} a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {38 a^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{21} \left (73 a^2 \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 {146 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {38 a^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{105} \left (292 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 {584 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {146 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {38 a^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{315} \left (584 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 {1168 a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {584 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}+\frac {146 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {38 a^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 5.38, size = 84, normalized size = 0.42 \[ \frac {a^2 (698 \cos (c+d x)+803 \cos (2 (c+d x))+146 \cos (3 (c+d x))+146 \cos (4 (c+d x))+727) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a (\cos (c+d x)+1)}}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(11/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(727 + 698*Cos[c + d*x] + 803*Cos[2*(c + d*x)] + 146*Cos[3*(c + d*x)] + 146*Co
s[4*(c + d*x)])*Sec[c + d*x]^(9/2)*Tan[(c + d*x)/2])/(315*d)

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fricas [A]  time = 0.83, size = 107, normalized size = 0.53 \[ \frac {2 \, {\left (584 \, a^{2} \cos \left (d x + c\right )^{4} + 292 \, a^{2} \cos \left (d x + c\right )^{3} + 219 \, a^{2} \cos \left (d x + c\right )^{2} + 130 \, a^{2} \cos \left (d x + c\right ) + 35 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )} \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/315*(584*a^2*cos(d*x + c)^4 + 292*a^2*cos(d*x + c)^3 + 219*a^2*cos(d*x + c)^2 + 130*a^2*cos(d*x + c) + 35*a^
2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^5 + d*cos(d*x + c)^4)*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)^(11/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.21, size = 95, normalized size = 0.47 \[ -\frac {2 \left (584 \left (\cos ^{5}\left (d x +c \right )\right )-292 \left (\cos ^{4}\left (d x +c \right )\right )-73 \left (\cos ^{3}\left (d x +c \right )\right )-89 \left (\cos ^{2}\left (d x +c \right )\right )-95 \cos \left (d x +c \right )-35\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} a^{2}}{315 d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x)

[Out]

-2/315/d*(584*cos(d*x+c)^5-292*cos(d*x+c)^4-73*cos(d*x+c)^3-89*cos(d*x+c)^2-95*cos(d*x+c)-35)*cos(d*x+c)*(a*(1
+cos(d*x+c)))^(1/2)*(1/cos(d*x+c))^(11/2)/sin(d*x+c)*a^2

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maxima [A]  time = 1.02, size = 289, normalized size = 1.44 \[ \frac {8 \, {\left (\frac {315 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {945 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1449 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1287 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {572 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {104 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{315 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

8/315*(315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 945*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 1449*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1287*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d
*x + c) + 1)^7 + 572*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 104*sqrt(2)*a^(5/2)*sin(d*x + c)^11
/(cos(d*x + c) + 1)^11)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(
11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(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 = 4.85, size = 306, normalized size = 1.52 \[ \frac {\sqrt {\frac {1}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {192\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{5\,d}-\frac {16\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{3\,d}+\frac {1168\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{35\,d}+\frac {2336\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{315\,d}\right )}{12\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+8\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+8\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )+2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(11/2)*(a + a*cos(c + d*x))^(5/2),x)

[Out]

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((192*a^2*exp((c*9i)/2 + (d*x*9i)/2)*sin(c/2 + (d*x
)/2)*(a + a*cos(c + d*x))^(1/2))/(5*d) - (16*a^2*exp((c*9i)/2 + (d*x*9i)/2)*sin((3*c)/2 + (3*d*x)/2)*(a + a*co
s(c + d*x))^(1/2))/(3*d) + (1168*a^2*exp((c*9i)/2 + (d*x*9i)/2)*sin((5*c)/2 + (5*d*x)/2)*(a + a*cos(c + d*x))^
(1/2))/(35*d) + (2336*a^2*exp((c*9i)/2 + (d*x*9i)/2)*sin((9*c)/2 + (9*d*x)/2)*(a + a*cos(c + d*x))^(1/2))/(315
*d)))/(12*exp((c*9i)/2 + (d*x*9i)/2)*cos(c/2 + (d*x)/2) + 8*exp((c*9i)/2 + (d*x*9i)/2)*cos((3*c)/2 + (3*d*x)/2
) + 8*exp((c*9i)/2 + (d*x*9i)/2)*cos((5*c)/2 + (5*d*x)/2) + 2*exp((c*9i)/2 + (d*x*9i)/2)*cos((7*c)/2 + (7*d*x)
/2) + 2*exp((c*9i)/2 + (d*x*9i)/2)*cos((9*c)/2 + (9*d*x)/2))

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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)**(11/2),x)

[Out]

Timed out

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