\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [420]

Optimal result

Integrand size = 25, antiderivative size = 189 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}} \]

[Out]

-arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2)*sec(
d*x+c)^(1/2)/d/a^(1/2)+2/5*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+26/15*sin(d*x+c)/d/cos(d*x+c)^
(1/2)/(a+a*sec(d*x+c))^(1/2)-2/15*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4349, 3908, 4107, 4098, 3893, 212} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}} \]

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos[c +
d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (26*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])
- (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5
*d*Sqrt[a + a*Sec[c + d*x]])

Rule 212

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

Rule 3893

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*b*(d/
(a*f)), Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3908

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[Cot[e +
f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[1/(2*b*d*n), Int[(d*Csc[e + f*x])^(n + 1)
*((a + b*(2*n + 1)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b
^2, 0] && LtQ[n, 0] && IntegerQ[2*n]

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4107

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx \\ & = \frac {2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {a-4 a \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{5 a} \\ & = -\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {13 a^2}{2}+a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{15 a^2} \\ & = \frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx \\ & = \frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = -\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left (15 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^{\frac {5}{2}}(c+d x)+2 \sqrt {1-\sec (c+d x)} \left (3-\sec (c+d x)+13 \sec ^2(c+d x)\right )\right ) \sin (c+d x)}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(Cos[c + d*x]^(3/2)*(15*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^(5/2)
+ 2*Sqrt[1 - Sec[c + d*x]]*(3 - Sec[c + d*x] + 13*Sec[c + d*x]^2))*Sin[c + d*x])/(15*d*Sqrt[1 - Sec[c + d*x]]
*Sqrt[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.03

method	result	size
default	\(-\frac {\left (15 \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+15 \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-6 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )-26 \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{15 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(194\)

[In]

int(cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/d/a*(15*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(-1/(cos(d*x+c)+
1))^(1/2)*cos(d*x+c)+15*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(-1/(c
os(d*x+c)+1))^(1/2)-6*cos(d*x+c)^2*sin(d*x+c)+2*cos(d*x+c)*sin(d*x+c)-26*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)*
cos(d*x+c)^(1/2)/(cos(d*x+c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/30*(4*(3*cos(d*x + c)^2 - cos(d*x + c) + 13)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin
(d*x + c) + 15*sqrt(2)*(a*cos(d*x + c) + a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sq
rt(a))/(a*d*cos(d*x + c) + a*d), 1/15*(15*sqrt(2)*(a*cos(d*x + c) + a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d
*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) + 2*(3*cos(d*x + c)^2 - cos(d*x + c) +
13)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)]

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (156) = 312\).

Time = 0.37 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (60 \, \cos \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 \, \cos \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 60 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {4}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 5 \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) \sin \left (\frac {2}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 30 \, \log \left (\cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 1\right ) + 30 \, \log \left (\cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 1\right ) + 6 \, \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 60 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )}}{60 \, \sqrt {a} d} \]

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/60*sqrt(2)*(60*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 5*cos(2/5
*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 60*cos(5/2*d*x + 5/2*c)*sin(4/5*a
rctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 5*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*d*x + 5/2
*c), cos(5/2*d*x + 5/2*c))) - 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*
arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x +
5/2*c))) + 1) + 30*log(cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arctan2(sin(5
/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 - 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 1
) + 6*sin(5/2*d*x + 5/2*c) - 5*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 60*sin(1/5*arcta
n2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/(sqrt(a)*d)

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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