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

Optimal. Leaf size=173 \[ \frac {5 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}-\frac {a (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d} \]

[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

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Rubi [A]  time = 0.37, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4087, 3917, 3915, 3774, 203, 3792} \[ \frac {a^3 (15 A+64 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (15 A-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {5 a^{5/2} A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {a (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d} \]

Antiderivative was successfully verified.

[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 203

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

Rule 3774

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 3792

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 3915

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 3917

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] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4087

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*} \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 (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 ) \operatorname {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 \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+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*}

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Mathematica [A]  time = 1.84, size = 145, normalized size = 0.84 \[ \frac {a^2 \tan (c+d x) \sec (c+d x) \sqrt {a (\sec (c+d x)+1)} \left (\sqrt {\sec (c+d x)-1} ((45 A+112 C) \cos (c+d x)+4 (15 A+43 C) \cos (2 (c+d x))+15 A \cos (3 (c+d x))+60 A+196 C)+300 A \cos ^2(c+d x) \tan ^{-1}\left (\sqrt {\sec (c+d x)-1}\right )\right )}{60 d (\cos (c+d x)+1) \sqrt {\sec (c+d x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(300*A*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Cos[c + d*x]^2 + (60*A + 196*C + (45*A + 112*C)*Cos[c + d*x] + 4*(
15*A + 43*C)*Cos[2*(c + d*x)] + 15*A*Cos[3*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sec[c + d*x]*Sqrt[a*(1 + Sec[c
+ d*x])]*Tan[c + d*x])/(60*d*(1 + Cos[c + d*x])*Sqrt[-1 + Sec[c + d*x]])

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fricas [A]  time = 0.47, size = 398, normalized size = 2.30 \[ \left [\frac {75 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {75 \, {\left (A a^{2} \cos \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 28 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

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+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)]

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giac [B]  time = 2.06, size = 484, normalized size = 2.80 \[ -\frac {75 \, A \sqrt {-a} a^{2} \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 ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 75 \, A \sqrt {-a} a^{2} \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 ) \mathrm {sgn}\left (\cos \left (d x + c\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 (15 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 60 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (30 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 80 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (15 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 32 \, \sqrt {2} 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}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\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} \]

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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/30*(75*A*sqrt(-a)*a^2*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)))*sgn(cos(d*x + c)) - 75*A*sqrt(-a)*a^2*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)))*sgn(cos(d*x + c)) + 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*(15*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 60*sqrt(2)
*C*a^5*sgn(cos(d*x + c)) - (30*sqrt(2)*A*a^5*sgn(cos(d*x + c)) + 80*sqrt(2)*C*a^5*sgn(cos(d*x + c)) - (15*sqrt
(2)*A*a^5*sgn(cos(d*x + c)) + 32*sqrt(2)*C*a^5*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)
^2)*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

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maple [B]  time = 2.08, size = 343, normalized size = 1.98 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (75 A \sqrt {2}\, \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+150 A \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+75 A \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )+120 A \left (\cos ^{4}\left (d x +c \right )\right )+120 A \left (\cos ^{3}\left (d x +c \right )\right )+688 C \left (\cos ^{3}\left (d x +c \right )\right )-240 A \left (\cos ^{2}\left (d x +c \right )\right )-464 C \left (\cos ^{2}\left (d x +c \right )\right )-176 C \cos \left (d x +c \right )-48 C \right ) a^{2}}{120 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}} \]

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+C*sec(d*x+c)^2),x)

[Out]

-1/120/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(75*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^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*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+150*A*2^(1/2)*sin(d*
x+c)*cos(d*x+c)*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*cos(d*x+c)
/(1+cos(d*x+c)))^(5/2)+75*A*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)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)+120*A*cos(d*x+c)^4+120*A*cos(d*x+c)^3+688*C*cos(d*x+c)^3
-240*A*cos(d*x+c)^2-464*C*cos(d*x+c)^2-176*C*cos(d*x+c)-48*C)/sin(d*x+c)/cos(d*x+c)^2*a^2

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maxima [B]  time = 0.64, size = 1384, normalized size = 8.00 \[ \text {result too large to display} \]

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+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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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+C*sec(d*x+c)**2),x)

[Out]

Timed out

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