3.6.4 \(\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [504]

Optimal. Leaf size=184 \[ \frac {a^{5/2} (5 A+2 B) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-10 B-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]

[Out]

a^(5/2)*(5*A+2*B)*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+70*B+64*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2
)-1/15*a^2*(15*A-10*B-16*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.28, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4171, 4002, 4000, 3859, 209, 3877} \begin {gather*} \frac {a^{5/2} (5 A+2 B) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (15 A+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (15 A-10 B-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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(5*A + 2*B)*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 + 70*B + 64*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(15*A - 10*
B - 16*C)*Sqrt[a + a*Sec[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 4171

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - 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 - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 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*} \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\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 {1}{2} a (5 A+2 B)-\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 {5}{4} a^2 (5 A+2 B)-\frac {1}{4} a^2 (15 A-10 B-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-10 B-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 {15}{8} a^3 (5 A+2 B)+\frac {1}{8} a^3 (15 A+70 B+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-10 B-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 (a^2 (5 A+2 B)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{30} \left (a^2 (15 A+70 B+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+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-10 B-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 (a^3 (5 A+2 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 {a^{5/2} (5 A+2 B) \tan ^{-1}\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+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-10 B-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]
time = 1.92, size = 209, normalized size = 1.14 \begin {gather*} \frac {\cos (c+d x) (a (1+\sec (c+d x)))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {60 (5 A+2 B) \text {ArcTan}\left (\sqrt {-1+\sec (c+d x)}\right ) \sin (c+d x)}{\sqrt {-1+\sec (c+d x)} (1+\sec (c+d x))^3}+\frac {\cos (c+d x) (60 A+160 B+196 C+(45 A+40 B+112 C) \cos (c+d x)+4 (15 A+40 B+43 C) \cos (2 (c+d x))+15 A \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{(1+\cos (c+d x))^2}\right )}{30 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]*(a*(1 + Sec[c + d*x]))^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((60*(5*A + 2*B)*ArcTan[Sqr
t[-1 + Sec[c + d*x]]]*Sin[c + d*x])/(Sqrt[-1 + Sec[c + d*x]]*(1 + Sec[c + d*x])^3) + (Cos[c + d*x]*(60*A + 160
*B + 196*C + (45*A + 40*B + 112*C)*Cos[c + d*x] + 4*(15*A + 40*B + 43*C)*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*
x)])*Tan[(c + d*x)/2])/(1 + Cos[c + d*x])^2))/(30*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(603\) vs. \(2(164)=328\).
time = 41.16, size = 604, normalized size = 3.28

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (75 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+30 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+150 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+60 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}+75 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+30 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )+120 A \left (\cos ^{4}\left (d x +c \right )\right )+120 A \left (\cos ^{3}\left (d x +c \right )\right )+640 B \left (\cos ^{3}\left (d x +c \right )\right )+688 C \left (\cos ^{3}\left (d x +c \right )\right )-240 A \left (\cos ^{2}\left (d x +c \right )\right )-560 B \left (\cos ^{2}\left (d x +c \right )\right )-464 C \left (\cos ^{2}\left (d x +c \right )\right )-80 B \cos \left (d x +c \right )-176 C \cos \left (d x +c \right )-48 C \right ) a^{2}}{120 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}\)	\(604\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(75*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+30*B*cos(d*x+c)^2*si
n(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(
d*x+c)*2^(1/2))*2^(1/2)+150*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+60*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+7
5*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/c
os(d*x+c)*2^(1/2))*sin(d*x+c)+30*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)+120*A*cos(d*x+c)^4+120*A*cos(d*x+c)^3+640*B*cos(d
*x+c)^3+688*C*cos(d*x+c)^3-240*A*cos(d*x+c)^2-560*B*cos(d*x+c)^2-464*C*cos(d*x+c)^2-80*B*cos(d*x+c)-176*C*cos(
d*x+c)-48*C)/cos(d*x+c)^2/sin(d*x+c)*a^2

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 2780 vs. \(2 (164) = 328\).
time = 0.73, size = 2780, normalized size = 15.11 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(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(s
in(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
), cos(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
), cos(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*arctan
2(sin(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)*arcta
n2(-(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))*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)*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) + (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) + 2*(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*ar
ctan2(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)*(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)*(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((...

________________________________________________________________________________________

Fricas [A]
time = 2.61, size = 440, normalized size = 2.39 \begin {gather*} \left [\frac {15 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (5 \, A + 2 \, B\right )} 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 + 40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B + 14 \, C\right )} 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 {15 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (5 \, A + 2 \, B\right )} 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 + 40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B + 14 \, C\right )} 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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*((5*A + 2*B)*a^2*cos(d*x + c)^3 + (5*A + 2*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*(15*A*a^2*cos(d*x + c)^3 + 2*(15*A + 40*B + 43*C)*a^2*cos(d*x + c)^2 + 2*(5*B + 14*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*(15*((5*A + 2*B)*a^2*cos(d*x + c)^3 + (5*A + 2*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))) - (15*A*a^2*cos(d*x + c)^3 + 2*(15*A + 40*B + 43*C)
*a^2*cos(d*x + c)^2 + 2*(5*B + 14*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)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (164) = 328\).
time = 2.21, size = 559, normalized size = 3.04 \begin {gather*} -\frac {15 \, {\left (5 \, A \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right ) - 15 \, {\left (5 \, A \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right ) + \frac {60 \, {\left (3 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} A \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}} - \frac {4 \, {\left ({\left (\sqrt {2} {\left (15 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 32 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \sqrt {2} {\left (3 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*(5*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 2*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*
d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3))) - 15*(5*A*sqrt(-a)*a^2*sgn(cos(d*x
+ c)) + 2*B*sqrt(-a)*a^2*sgn(cos(d*x + c)))*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/
2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 60*(3*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/
2*c)^2 + a))^2*A*sqrt(-a)*a^3*sgn(cos(d*x + c)) - sqrt(2)*A*sqrt(-a)*a^4*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*
x + 1/2*c)^2 + a))^2*a + a^2) - 4*((sqrt(2)*(15*A*a^5*sgn(cos(d*x + c)) + 35*B*a^5*sgn(cos(d*x + c)) + 32*C*a^
5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 10*sqrt(2)*(3*A*a^5*sgn(cos(d*x + c)) + 8*B*a^5*sgn(cos(d*x + c)
) + 8*C*a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(A*a^5*sgn(cos(d*x + c)) + 3*B*a^5*sgn(cos
(d*x + c)) + 4*C*a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/
2*d*x + 1/2*c)^2 + a)))/d

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \cos \left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________