\(\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [177]

Optimal result

Integrand size = 33, antiderivative size = 173 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {5 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

5*a^(5/2)*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+A*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d-1/5*a*(5
*A-2*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+1/15*a^3*(15*A+64*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/15*a^2*
(15*A-16*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4172, 4002, 4000, 3859, 209, 3877} \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {5 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}-\frac {a (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d} \]

[In]

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

[Out]

(5*a^(5/2)*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (A*(a + a*Sec[c + d*x])^(5/2)*Sin[c
+ d*x])/d + (a^3*(15*A + 64*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(15*A - 16*C)*Sqrt[a + a*S
ec[c + d*x]]*Tan[c + d*x])/(15*d) - (a*(5*A - 2*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 4172

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {\int (a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}-\frac {1}{2} a (5 A-2 C) \sec (c+d x)\right ) \, dx}{a} \\ & = \frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2 \int (a+a \sec (c+d x))^{3/2} \left (\frac {25 a^2 A}{4}-\frac {1}{4} a^2 (15 A-16 C) \sec (c+d x)\right ) \, dx}{5 a} \\ & = \frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a^2 (15 A-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4 \int \sqrt {a+a \sec (c+d x)} \left (\frac {75 a^3 A}{8}+\frac {1}{8} a^3 (15 A+64 C) \sec (c+d x)\right ) \, dx}{15 a} \\ & = \frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a^2 (15 A-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{2} \left (5 a^2 A\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{30} \left (a^2 (15 A+64 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac {\left (5 a^3 A\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 {5 a^{5/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (75 A \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {1-\sec (c+d x)} \left (15 A \sin (c+d x)+2 \left (15 A+43 C+14 C \sec (c+d x)+3 C \sec ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(a^3*(75*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[1 - Sec[c + d*x]]*(15*A*Sin[c + d*x] + 2*(15*A
+ 43*C + 14*C*Sec[c + d*x] + 3*C*Sec[c + d*x]^2)*Tan[c + d*x])))/(15*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[
c + d*x])])

Maple [A] (verified)

Time = 51.03 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+75 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+15 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+30 A \sin \left (d x +c \right )+86 C \sin \left (d x +c \right )+28 C \tan \left (d x +c \right )+6 C \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\)	\(217\)

[In]

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

[Out]

1/15*a^2/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(75*A*(-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))*cos(d*x+c)+75*A*(-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))+15*A*cos(d*x+c)*sin(d*x+c)+30*A*sin(d*x+c)+86*C
*sin(d*x+c)+28*C*tan(d*x+c)+6*C*sec(d*x+c)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.30 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {75 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {75 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/30*(75*(A*a^2*cos(d*x + c)^3 + A*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*(15*A*
a^2*cos(d*x + c)^3 + 2*(15*A + 43*C)*a^2*cos(d*x + c)^2 + 28*C*a^2*cos(d*x + c) + 6*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), -1/15*(75*(A*a^2*cos(d*x + c)^3 + A*
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))
) - (15*A*a^2*cos(d*x + c)^3 + 2*(15*A + 43*C)*a^2*cos(d*x + c)^2 + 28*C*a^2*cos(d*x + c) + 6*C*a^2)*sqrt((a*c
os(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. \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))**(5/2)*(A+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. 1384 vs. \(2 (153) = 306\).

Time = 0.43 (sec) , antiderivative size = 1384, normalized size of antiderivative = 8.00 \[ \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/4*(18*(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)*((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x + 2*c) + 4*a^2*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c) + 1)) + (a^2*cos(2*d*x + 2*c)^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos
(2*d*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (4*a
^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x + 2*c) + 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c) + 1)) - ((a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c)^2 + a^2*cos(d*x + c) + (a^2*cos(d*x + c) -
a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*cos(d*x + c) - a^2)*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1)))*sqrt(a) + 5*((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(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*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(d*x + c)*cos(1/2
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(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)*(cos(1/2*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (c
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(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), c
os(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))*A/((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]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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