\(\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [378]

Optimal result

Integrand size = 40, antiderivative size = 142 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^{5/2} B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (35 B+32 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 B+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

2*a^(5/2)*B*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/5*a*C*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/
15*a^3*(35*B+32*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/15*a^2*(5*B+8*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4157, 4002, 4000, 3859, 209, 3877} \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^{5/2} B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^3 (35 B+32 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a^2 (5 B+8 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {2 a C \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d} \]

[In]

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

[Out]

(2*a^(5/2)*B*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^3*(35*B + 32*C)*Tan[c + d*x])/(
15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(5*B + 8*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*a*C*(a
+ a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 209

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

Rule 3859

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

Rule 3877

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

Rule 4000

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

Rule 4002

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

Rule 4157

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

Rubi steps \begin{align*} \text {integral}& = \int (a+a \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx \\ & = \frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2}{5} \int (a+a \sec (c+d x))^{3/2} \left (\frac {5 a B}{2}+\frac {1}{2} a (5 B+8 C) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (5 B+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \sqrt {a+a \sec (c+d x)} \left (\frac {15 a^2 B}{4}+\frac {1}{4} a^2 (35 B+32 C) \sec (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (5 B+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\left (a^2 B\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{15} \left (a^2 (35 B+32 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a^3 (35 B+32 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 B+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac {\left (2 a^3 B\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {2 a^{5/2} B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 (35 B+32 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (5 B+8 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 a C (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.82 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.90 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (1+\sec (c+d x))} \left (30 \sqrt {2} B \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {5}{2}}(c+d x)+2 (40 B+49 C+2 (5 B+14 C) \cos (c+d x)+(40 B+43 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{30 d} \]

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^2*Sqrt[a*(1 + Sec[c + d*x])]*(30*Sqrt[2]*B*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]
*Cos[c + d*x]^(5/2) + 2*(40*B + 49*C + 2*(5*B + 14*C)*Cos[c + d*x] + (40*B + 43*C)*Cos[2*(c + d*x)])*Sin[(c +
d*x)/2]))/(30*d)

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+15 B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+40 B \sin \left (d x +c \right )+43 C \sin \left (d x +c \right )+5 B \tan \left (d x +c \right )+14 C \tan \left (d x +c \right )+3 C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\)	\(211\)

[In]

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

[Out]

2/15*a^2/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(15*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d
*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+15*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh
(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+40*B*sin(d*x+c)+43*C*sin(d*x+c)+5*B*tan(d*x+c)+
14*C*tan(d*x+c)+3*C*sec(d*x+c)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.66 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {15 \, {\left (B a^{2} \cos \left (d x + c\right )^{3} + B a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left ({\left (40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} \cos \left (d x + c\right )^{3} + B a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/15*(15*(B*a^2*cos(d*x + c)^3 + B*a^2*cos(d*x + c)^2)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*((40*B
+ 43*C)*a^2*cos(d*x + c)^2 + (5*B + 14*C)*a^2*cos(d*x + c) + 3*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), -2/15*(15*(B*a^2*cos(d*x + c)^3 + B*a^2*cos(d*x + c)^2)*
sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((40*B + 43*C)*a
^2*cos(d*x + c)^2 + (5*B + 14*C)*a^2*cos(d*x + c) + 3*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x +
c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1396 vs. \(2 (124) = 248\).

Time = 0.41 (sec) , antiderivative size = 1396, normalized size of antiderivative = 9.83 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/6*(30*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/
4)*((12*a^2*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 3*a^2*sin(2*d*x + 2*c) - 4
*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (12*a^2*sin(2*d*x + 2*c)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c))) + 3*a^2*cos(2*d*x + 2*c) - a^2 + 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*((a^2*cos(2*d*x + 2*c)^
2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(c
os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) +
sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
) + 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2
*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2
*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/
4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2
*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +
2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*sqrt(a))*B/((cos(
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*d)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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