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

Optimal. Leaf size=182 \[ \frac {163 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {163 a^3 \tan (c+d x)}{64 d \sqrt {a \cos (c+d x)+a}}+\frac {17 a^3 \tan (c+d x) \sec ^2(c+d x)}{24 d \sqrt {a \cos (c+d x)+a}}+\frac {163 a^3 \tan (c+d x) \sec (c+d x)}{96 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d} \]

[Out]

163/64*a^(5/2)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+163/64*a^3*tan(d*x+c)/d/(a+a*cos(d*x+c))^(
1/2)+163/96*a^3*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+17/24*a^3*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*
x+c))^(1/2)+1/4*a^2*sec(d*x+c)^3*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2762, 2980, 2772, 2773, 206} \[ \frac {163 a^3 \tan (c+d x)}{64 d \sqrt {a \cos (c+d x)+a}}+\frac {163 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {17 a^3 \tan (c+d x) \sec ^2(c+d x)}{24 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {163 a^3 \tan (c+d x) \sec (c+d x)}{96 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(163*a^(5/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (163*a^3*Tan[c + d*x])/(64*d*S
qrt[a + a*Cos[c + d*x]]) + (163*a^3*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a + a*Cos[c + d*x]]) + (17*a^3*Sec[c
+ d*x]^2*Tan[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c +
d*x])/(4*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 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 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*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 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]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \, dx &=\frac {a^2 \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{4} a \int \left (-\frac {17 a}{2}-\frac {13}{2} a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \sec ^4(c+d x) \, dx\\ &=\frac {17 a^3 \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{48} \left (163 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac {163 a^3 \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {17 a^3 \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{64} \left (163 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac {163 a^3 \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {163 a^3 \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {17 a^3 \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{128} \left (163 a^2\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac {163 a^3 \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {163 a^3 \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {17 a^3 \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (163 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}\\ &=\frac {163 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {163 a^3 \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {163 a^3 \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {17 a^3 \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 35.84, size = 2069, normalized size = 11.37 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-163/2048 + (163*I)/2048)*(1 + E^(I*c))*(Sqrt[2] - (1 - I)*E^((I/2)*c) + (16 - 16*I)*E^(((3*I)/2)*c + I*d*x)
+ (20 + 20*I)*Sqrt[2]*E^((2*I)*c + ((3*I)/2)*d*x) - (34 - 34*I)*E^(((5*I)/2)*c + (2*I)*d*x) - (20 + 20*I)*Sqr
t[2]*E^((3*I)*c + ((5*I)/2)*d*x) + (16 - 16*I)*E^(((7*I)/2)*c + (3*I)*d*x) + (4 + 4*I)*Sqrt[2]*E^((4*I)*c + ((
7*I)/2)*d*x) - (1 - I)*E^(((9*I)/2)*c + (4*I)*d*x) + (8*I)*E^((I/2)*(c + d*x)) - 16*Sqrt[2]*E^(I*(c + d*x)) -
(40*I)*E^(((3*I)/2)*(c + d*x)) + 34*Sqrt[2]*E^((2*I)*(c + d*x)) + (40*I)*E^(((5*I)/2)*(c + d*x)) - 16*Sqrt[2]*
E^((3*I)*(c + d*x)) - (8*I)*E^(((7*I)/2)*(c + d*x)) + Sqrt[2]*E^((4*I)*(c + d*x)) - (4 + 4*I)*Sqrt[2]*E^((I/2)
*(2*c + d*x)))*x*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5)/(((-1 - I) + Sqrt[2]*E^((I/2)*c))*(-1 + E^
(I*c))*(I - 2*Sqrt[2]*E^((I/2)*(c + d*x)) - (4*I)*E^(I*(c + d*x)) + 2*Sqrt[2]*E^(((3*I)/2)*(c + d*x)) + I*E^((
2*I)*(c + d*x)))^2) - (((163*I)/512)*ArcTan[(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[c/4 + (d*x)
/4])/(-Cos[c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*(a*(1 + Cos[c + d*x]))^(5/2)*Sec
[c/2 + (d*x)/2]^5)/(Sqrt[2]*d) - (((163*I)/512)*ArcTan[(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4] - Sqrt[2]*Sin[
c/4 + (d*x)/4])/(Cos[c/4 + (d*x)/4] + Sqrt[2]*Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])]*(a*(1 + Cos[c + d*x]))
^(5/2)*Sec[c/2 + (d*x)/2]^5)/(Sqrt[2]*d) - (163*(a*(1 + Cos[c + d*x]))^(5/2)*Log[2 - Sqrt[2]*Cos[c/2 + (d*x)/2
] - Sqrt[2]*Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^5)/(1024*Sqrt[2]*d) - (163*(a*(1 + Cos[c + d*x]))^(5/2)*Log
[2 + Sqrt[2]*Cos[c/2 + (d*x)/2] - Sqrt[2]*Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^5)/(1024*Sqrt[2]*d) - (((163*
I)/256)*ArcTan[((2*I)*Cos[c/2] - I*(-Sqrt[2] + 2*Sin[c/2])*Tan[(d*x)/4])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2
]]*(a*(1 + Cos[c + d*x]))^(5/2)*Cot[c/2]*Sec[c/2 + (d*x)/2]^5)/(d*Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]) + (1
63*(a*(1 + Cos[c + d*x]))^(5/2)*Csc[c/2]*Sec[c/2 + (d*x)/2]^5*(-(d*x*Cos[c/2]) + 2*Log[Sqrt[2] + 2*Cos[(d*x)/2
]*Sin[c/2] + 2*Cos[c/2]*Sin[(d*x)/2]]*Sin[c/2] + ((4*I)*Sqrt[2]*ArcTan[((2*I)*Cos[c/2] - I*(-Sqrt[2] + 2*Sin[c
/2])*Tan[(d*x)/4])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]]*Cos[c/2])/Sqrt[-2 + 4*Cos[c/2]^2 + 4*Sin[c/2]^2]))/
(256*Sqrt[2]*d*(4*Cos[c/2]^2 + 4*Sin[c/2]^2)) + ((a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2
])/(64*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^4) + ((a*(1 + Cos[c + d*x]))^(5/2)*Se
c[c/2 + (d*x)/2]^5*(23*Cos[c/2] - 17*Sin[c/2]))/(384*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (
d*x)/2])^3) + (43*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/(256*d*(Cos[c/2] - Sin[c/2])
*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + ((a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*(163*Cos[c/
2] - 77*Sin[c/2]))/(512*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + ((a*(1 + Cos[c +
d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*Sin[(d*x)/2])/(64*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d
*x)/2])^4) + ((a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*(-23*Cos[c/2] - 17*Sin[c/2]))/(384*d*(Cos[c/2]
+ Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3) + (43*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[c/2 + (d*x)/2
]^5*Sin[(d*x)/2])/(256*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + ((a*(1 + Cos[c +
d*x]))^(5/2)*Sec[c/2 + (d*x)/2]^5*(-163*Cos[c/2] - 77*Sin[c/2]))/(512*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x
)/2] + Sin[c/2 + (d*x)/2]))

________________________________________________________________________________________

fricas [A]  time = 1.47, size = 196, normalized size = 1.08 \[ \frac {489 \, {\left (a^{2} \cos \left (d x + c\right )^{5} + a^{2} \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (489 \, a^{2} \cos \left (d x + c\right )^{3} + 326 \, a^{2} \cos \left (d x + c\right )^{2} + 184 \, a^{2} \cos \left (d x + c\right ) + 48 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(489*(a^2*cos(d*x + c)^5 + a^2*cos(d*x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sq
rt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(
489*a^2*cos(d*x + c)^3 + 326*a^2*cos(d*x + c)^2 + 184*a^2*cos(d*x + c) + 48*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)

________________________________________________________________________________________

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)^5,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.60, size = 872, normalized size = 4.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(7824*a*(ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+ln(-4/(-2*cos(1/2*d*x+1/2*c)+
2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a)))*sin(1/2*d*x+1/2*c
)^8-7824*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+2*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*
(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^6+130
4*(11*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+9*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1
/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+9*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(
1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^4+(-3912*
ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/
2*c)+2*a))*a-3912*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2
)*cos(1/2*d*x+1/2*c)+2*a))*a-9212*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))*sin(1/2*d*x+1/2*c)^2+489*ln(
-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c
)+2*a))*a+489*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*co
s(1/2*d*x+1/2*c)+2*a))*a+2094*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^4
/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^4/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

________________________________________________________________________________________

maxima [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)^5,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)**5,x)

[Out]

Timed out

________________________________________________________________________________________