3.506 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=207 \[ \frac {a^{5/2} (25 A+38 B+40 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {a (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d} \]
[Out]
1/8*a^(5/2)*(25*A+38*B+40*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/12*a*(5*A+6*B)*cos(d*x+c)*(
a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/24*a^3*(49*A+54*B-
24*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/4*a^2*(3*A+2*B-8*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
________________________________________________________________________________________
Rubi [A] time = 0.66, antiderivative size = 207, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used =
{4086, 4017, 4018, 4015, 3774, 203} \[ \frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (3 A+2 B-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {a^{5/2} (25 A+38 B+40 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 d}+\frac {a (5 A+6 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{3/2}}{12 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^{5/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*(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)*(25*A + 38*B + 40*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*d) + (a^3*(49*A + 54
*B - 24*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(3*A + 2*B - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin
[c + d*x])/(4*d) + (a*(5*A + 6*B)*Cos[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (A*Cos[c + d*
x]^2*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*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 4015
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 4017
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]
Rule 4018
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] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Rule 4086
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 ^3(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 \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+6 B)-\frac {1}{2} a (A-6 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac {a (5 A+6 B) \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (31 A+42 B+24 C)-\frac {3}{4} a^2 (3 A+2 B-8 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=-\frac {a^2 (3 A+2 B-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a (5 A+6 B) \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {\int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (49 A+54 B-24 C)+\frac {1}{8} a^3 (13 A+30 B+72 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (3 A+2 B-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a (5 A+6 B) \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac {1}{16} \left (a^2 (25 A+38 B+40 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (3 A+2 B-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a (5 A+6 B) \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}-\frac {\left (a^3 (25 A+38 B+40 C)\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 )}{8 d}\\ &=\frac {a^{5/2} (25 A+38 B+40 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (3 A+2 B-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a (5 A+6 B) \cos (c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 1.80, size = 140, normalized size = 0.68 \[ \frac {a^2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \left (\sqrt {\sec (c+d x)-1} (3 (27 A+22 B+8 C) \cos (c+d x)+(17 A+6 B) \cos (2 (c+d x))+2 A \cos (3 (c+d x))+17 A+6 B+48 C)+3 (25 A+38 B+40 C) \tan ^{-1}\left (\sqrt {\sec (c+d x)-1}\right )\right )}{24 d \sqrt {\sec (c+d x)-1}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^2*(3*(25*A + 38*B + 40*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]] + (17*A + 6*B + 48*C + 3*(27*A + 22*B + 8*C)*Cos[
c + d*x] + (17*A + 6*B)*Cos[2*(c + d*x)] + 2*A*Cos[3*(c + d*x)])*Sqrt[-1 + Sec[c + d*x]])*Sqrt[a*(1 + Sec[c +
d*x])]*Tan[(c + d*x)/2])/(24*d*Sqrt[-1 + Sec[c + d*x]])
________________________________________________________________________________________
fricas [A] time = 0.62, size = 410, normalized size = 1.98 \[ \left [\frac {3 \, {\left ({\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (8 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (25 \, A + 22 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/48*(3*((25*A + 38*B + 40*C)*a^2*cos(d*x + c) + (25*A + 38*B + 40*C)*a^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*(8*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25*A + 22*B + 8*C)*a^2*cos(d*x
+ c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/24*(3*((25*A +
38*B + 40*C)*a^2*cos(d*x + c) + (25*A + 38*B + 40*C)*a^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))) - (8*A*a^2*cos(d*x + c)^3 + 2*(17*A + 6*B)*a^2*cos(d*x + c)^2 + 3*(25
*A + 22*B + 8*C)*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x
+ c) + d)]
________________________________________________________________________________________
giac [B] time = 3.42, size = 1319, normalized size = 6.37 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
-1/48*(96*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*C*a^3*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*
d*x + 1/2*c)^2 - a) + 3*(25*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 38*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 40*C*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))) - 3*(25*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 38*B*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 40*C*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))) + 4*(75*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A
*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 114*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^10*B*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 72*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a))^10*C*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 1125*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 1710*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 888*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 6174*sqrt(2)*(sqrt(-a)*tan(1
/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 6804*sqrt(2)*(sqrt
(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 3024*sqr
t(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^5*sgn(cos(d*x + c))
- 4314*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^6*sgn(cos(
d*x + c)) - 4284*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^
6*sgn(cos(d*x + c)) - 1776*sqrt(2)*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*s
qrt(-a)*a^6*sgn(cos(d*x + c)) + 807*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^7*sgn(cos(d*x + c)) + 858*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*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 360*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*C*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 49*sqrt(2)*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 54*sq
rt(2)*B*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 24*sqrt(2)*C*sqrt(-a)*a^8*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)^3)/d
________________________________________________________________________________________
maple [B] time = 1.80, size = 846, normalized size = 4.09 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-1/192/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)+114*B*cos(d*x+c)^2*s
in(d*x+c)*2^(1/2)*(-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))+120*C*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)*arct
anh(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)+228*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-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))+240*C*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)*s
in(d*x+c)+114*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*C*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/co
s(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)+64*A*cos(d*x+c)^6+208*A*cos(d*x+c)^5
+96*B*cos(d*x+c)^5+328*A*cos(d*x+c)^4+432*B*cos(d*x+c)^4+192*C*cos(d*x+c)^4-600*A*cos(d*x+c)^3-528*B*cos(d*x+c
)^3+192*C*cos(d*x+c)^3-384*C*cos(d*x+c)^2)/sin(d*x+c)/cos(d*x+c)^2*a^2
________________________________________________________________________________________
maxima [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)^3*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^3*(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)^3*(a + a/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
________________________________________________________________________________________
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)**3*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
________________________________________________________________________________________