\(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [178]

Optimal result

Integrand size = 35, antiderivative size = 188 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d} \]

[Out]

1/4*a^(5/2)*(19*A+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-1/6*a*(3*A-4*C)*(a+a*sec(d*x+c))^(3
/2)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/12*a^3*(27*A-56*C)*sin(d*x+c)/d/(a+a*s
ec(d*x+c))^(1/2)-1/2*a^2*(A-8*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4172, 4103, 4100, 3859, 209} \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (A-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}-\frac {a (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d} \]

[In]

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

[Out]

(a^(5/2)*(19*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(27*A - 56*C)*Sin[
c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) - (a*(
3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*Sin[c +
d*x])/(2*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 4100

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

Rule 4103

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

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 \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}-\frac {1}{2} a (3 A-4 C) \sec (c+d x)\right ) \, dx}{2 a} \\ & = -\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (21 A-8 C)-\frac {3}{4} a^2 (A-8 C) \sec (c+d x)\right ) \, dx}{3 a} \\ & = -\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {2 \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (27 A-56 C)+\frac {5}{8} a^3 (3 A+8 C) \sec (c+d x)\right ) \, dx}{3 a} \\ & = \frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {1}{8} \left (a^2 (19 A+8 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}-\frac {\left (a^3 (19 A+8 C)\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 )}{4 d} \\ & = \frac {a^{5/2} (19 A+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \left (6 \sqrt {2} (19 A+8 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (33 A+16 C+(9 A+128 C) \cos (c+d x)+33 A \cos (2 (c+d x))+3 A \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \]

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*(19*A + 8*C)*ArcSin[Sqrt[2]*Sin[(c +
d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(33*A + 16*C + (9*A + 128*C)*Cos[c + d*x] + 33*A*Cos[2*(c + d*x)] + 3*A*Cos[3*
(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(337\) vs. \(2(164)=328\).

Time = 52.42 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.80

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (57 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+6 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+24 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+57 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+33 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+24 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+64 C \sin \left (d x +c \right )+8 C \tan \left (d x +c \right )\right )}{12 d \left (\cos \left (d x +c \right )+1\right )}\)	\(338\)

[In]

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

[Out]

1/12*a^2/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(57*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)+6*A*cos(d*x+c)^2*sin(d*x+c)+24*C*(-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)+57*A*(-c
os(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))+33*A*cos
(d*x+c)*sin(d*x+c)+24*C*(-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))+64*C*sin(d*x+c)+8*C*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.14 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 33 \, A a^{2} \cos \left (d x + c\right )^{2} + 64 \, C a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {3 \, {\left ({\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 33 \, A a^{2} \cos \left (d x + c\right )^{2} + 64 \, C a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]

[In]

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

[Out]

[1/24*(3*((19*A + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 8*C)*a^2*cos(d*x + c))*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*(6*A*a^2*cos(d*x + c)^3 + 33*A*a^2*cos(d*x + c)^2 + 64*C*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), -1/12*(3*((19*A + 8*C)*a^2*cos(
d*x + c)^2 + (19*A + 8*C)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c
)/(sqrt(a)*sin(d*x + c))) - (6*A*a^2*cos(d*x + c)^3 + 33*A*a^2*cos(d*x + c)^2 + 64*C*a^2*cos(d*x + c) + 8*C*a^
2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[In]

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

[Out]

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