\(\int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [220]

Optimal result

Integrand size = 25, antiderivative size = 201 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {38 a^3 \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {146 a^3 \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {584 a^3 \sin (c+d x)}{315 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {1168 a^3 \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

[Out]

38/63*a^3*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2)+146/105*a^3*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*
cos(d*x+c))^(1/2)+584/315*a^3*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+1168/315*a^3*sin(d*x+c)/d/c
os(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+2/9*a^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2841, 3059, 2851, 2850} \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {584 a^3 \sin (c+d x)}{315 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {146 a^3 \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {38 a^3 \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {1168 a^3 \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]

[In]

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

[Out]

(38*a^3*Sin[c + d*x])/(63*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d*x]]) + (146*a^3*Sin[c + d*x])/(105*d*Cos[c
+ d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (584*a^3*Sin[c + d*x])/(315*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c +
d*x]]) + (1168*a^3*Sin[c + d*x])/(315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*Sqrt[a + a*Cos[c
+ d*x]]*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2))

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 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 3059

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}-\frac {1}{9} (2 a) \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 \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {1}{21} \left (73 a^2\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {38 a^3 \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {146 a^3 \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {1}{105} \left (292 a^2\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {38 a^3 \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {146 a^3 \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {584 a^3 \sin (c+d x)}{315 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {1}{315} \left (584 a^2\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {38 a^3 \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {146 a^3 \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {584 a^3 \sin (c+d x)}{315 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {1168 a^3 \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.42 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} (727+698 \cos (c+d x)+803 \cos (2 (c+d x))+146 \cos (3 (c+d x))+146 \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{315 d \cos ^{\frac {9}{2}}(c+d x)} \]

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)/Cos[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)])*Tan[(c + d*x)/2])/(315*d*Cos[c + d*x]^(9/2))

Maple [A] (verified)

Time = 5.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.42

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (584 \left (\cos ^{4}\left (d x +c \right )\right )+292 \left (\cos ^{3}\left (d x +c \right )\right )+219 \left (\cos ^{2}\left (d x +c \right )\right )+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{315 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {9}{2}}}\)	\(85\)

[In]

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

[Out]

2/315/d*sin(d*x+c)*(584*cos(d*x+c)^4+292*cos(d*x+c)^3+219*cos(d*x+c)^2+130*cos(d*x+c)+35)*(a*(1+cos(d*x+c)))^(
1/2)/(1+cos(d*x+c))/cos(d*x+c)^(9/2)*a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.53 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \, {\left (584 \, a^{2} \cos \left (d x + c\right )^{4} + 292 \, a^{2} \cos \left (d x + c\right )^{3} + 219 \, a^{2} \cos \left (d x + c\right )^{2} + 130 \, a^{2} \cos \left (d x + c\right ) + 35 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)/cos(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)*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(5/2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.44 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {8 \, {\left (\frac {315 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {945 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1449 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1287 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {572 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {104 \, \sqrt {2} a^{\frac {5}{2}} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{315 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(5/2)/cos(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))

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 21.05 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.39 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (\frac {192\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5\,d}-\frac {16\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{3\,d}+\frac {1168\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{35\,d}+\frac {2336\,a^2\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{315\,d}\right )}{12\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+8\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+8\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )+2\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,9{}\mathrm {i}}{2}+\frac {d\,x\,9{}\mathrm {i}}{2}}\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )} \]

[In]

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

[Out]

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