3.101 \(\int (a+a \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)
Optimal. Leaf size=290 \[ \frac{a^3 (1015 A+1304 C) \tan (c+d x)}{512 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{5/2} (1015 A+1304 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{512 d}+\frac{a^2 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{96 d}+\frac{a^3 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{192 d \sqrt{a \cos (c+d x)+a}}+\frac{a^3 (1015 A+1304 C) \tan (c+d x) \sec (c+d x)}{768 d \sqrt{a \cos (c+d x)+a}}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d} \]
[Out]
(a^(5/2)*(1015*A + 1304*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(512*d) + (a^3*(1015*A +
1304*C)*Tan[c + d*x])/(512*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1015*A + 1304*C)*Sec[c + d*x]*Tan[c + d*x])/(76
8*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(109*A + 136*C)*Sec[c + d*x]^2*Tan[c + d*x])/(192*d*Sqrt[a + a*Cos[c + d*
x]]) + (a^2*(23*A + 24*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(96*d) + (a*A*(a + a*Cos[c + d
*x])^(3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(12*d) + (A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^5*Tan[c + d*x])/(6
*d)
________________________________________________________________________________________
Rubi [A] time = 0.906508, antiderivative size = 290, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{3044, 2975, 2980, 2772, 2773, 206} \[ \frac{a^3 (1015 A+1304 C) \tan (c+d x)}{512 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{5/2} (1015 A+1304 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{512 d}+\frac{a^2 (23 A+24 C) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{96 d}+\frac{a^3 (109 A+136 C) \tan (c+d x) \sec ^2(c+d x)}{192 d \sqrt{a \cos (c+d x)+a}}+\frac{a^3 (1015 A+1304 C) \tan (c+d x) \sec (c+d x)}{768 d \sqrt{a \cos (c+d x)+a}}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]
[Out]
(a^(5/2)*(1015*A + 1304*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(512*d) + (a^3*(1015*A +
1304*C)*Tan[c + d*x])/(512*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1015*A + 1304*C)*Sec[c + d*x]*Tan[c + d*x])/(76
8*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(109*A + 136*C)*Sec[c + d*x]^2*Tan[c + d*x])/(192*d*Sqrt[a + a*Cos[c + d*
x]]) + (a^2*(23*A + 24*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(96*d) + (a*A*(a + a*Cos[c + d
*x])^(3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(12*d) + (A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^5*Tan[c + d*x])/(6
*d)
Rule 3044
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])
Rule 2975
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])
Rule 2980
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Rule 2772
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Rule 2773
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int (a+a \cos (c+d x))^{5/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (5 A+12 C) \cos (c+d x)\right ) \sec ^6(c+d x) \, dx}{6 a}\\ &=\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int (a+a \cos (c+d x))^{3/2} \left (\frac{5}{4} a^2 (23 A+24 C)+\frac{15}{4} a^2 (5 A+8 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{30 a}\\ &=\frac{a^2 (23 A+24 C) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int \sqrt{a+a \cos (c+d x)} \left (\frac{15}{8} a^3 (109 A+136 C)+\frac{5}{8} a^3 (235 A+312 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{120 a}\\ &=\frac{a^3 (109 A+136 C) \sec ^2(c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (23 A+24 C) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{384} \left (a^2 (1015 A+1304 C)\right ) \int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (1015 A+1304 C) \sec (c+d x) \tan (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (109 A+136 C) \sec ^2(c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (23 A+24 C) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{512} \left (a^2 (1015 A+1304 C)\right ) \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (1015 A+1304 C) \tan (c+d x)}{512 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1015 A+1304 C) \sec (c+d x) \tan (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (109 A+136 C) \sec ^2(c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (23 A+24 C) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\left (a^2 (1015 A+1304 C)\right ) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{1024}\\ &=\frac{a^3 (1015 A+1304 C) \tan (c+d x)}{512 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1015 A+1304 C) \sec (c+d x) \tan (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (109 A+136 C) \sec ^2(c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (23 A+24 C) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{\left (a^3 (1015 A+1304 C)\right ) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{512 d}\\ &=\frac{a^{5/2} (1015 A+1304 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{512 d}+\frac{a^3 (1015 A+1304 C) \tan (c+d x)}{512 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1015 A+1304 C) \sec (c+d x) \tan (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (109 A+136 C) \sec ^2(c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (23 A+24 C) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{96 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}
Mathematica [A] time = 2.7184, size = 198, normalized size = 0.68 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) (14 (4591 A+4056 C) \cos (c+d x)+16 (1711 A+1496 C) \cos (2 (c+d x))+21721 A \cos (3 (c+d x))+4060 A \cos (4 (c+d x))+3045 A \cos (5 (c+d x))+27412 A+25448 C \cos (3 (c+d x))+5216 C \cos (4 (c+d x))+3912 C \cos (5 (c+d x))+18720 C)+24 \sqrt{2} (1015 A+1304 C) \cos ^6(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{24576 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^6*(24*Sqrt[2]*(1015*A + 1304*C)*ArcTanh[Sqrt[2]*
Sin[(c + d*x)/2]]*Cos[c + d*x]^6 + (27412*A + 18720*C + 14*(4591*A + 4056*C)*Cos[c + d*x] + 16*(1711*A + 1496*
C)*Cos[2*(c + d*x)] + 21721*A*Cos[3*(c + d*x)] + 25448*C*Cos[3*(c + d*x)] + 4060*A*Cos[4*(c + d*x)] + 5216*C*C
os[4*(c + d*x)] + 3045*A*Cos[5*(c + d*x)] + 3912*C*Cos[5*(c + d*x)])*Sin[(c + d*x)/2]))/(24576*d)
________________________________________________________________________________________
Maple [B] time = 0.13, size = 2271, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)
[Out]
1/48*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(192*a*(1015*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(
1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+1015*A*ln(4/(2*cos(1/
2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))+1304*
C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+
1/2*c)+2*a))+1304*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1
/2*d*x+1/2*c)^2)^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^12-192*(1015*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)
+1304*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+3045*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+3045*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2
^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+3912*C*ln(-4/(-2*
cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))
*a+3912*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^10+16*(34510*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+44336*C
*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+45675*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)
*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+45675*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2
))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+58680*C*ln(-4/(-2*cos(
1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+5
8680*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)
^2)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^8-96*(6699*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+8504*C*2^(1/
2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+5075*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+5075*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^
(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+6520*C*ln(-4/(-2*cos(1/2*d*x+1
/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+6520*C*ln(
4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+
2*a))*a)*sin(1/2*d*x+1/2*c)^6+12*(32596*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+39712*C*2^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+15225*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+15225*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*
cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+19560*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)
+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+19560*C*ln(4/(2
*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)
)*a)*sin(1/2*d*x+1/2*c)^4-4*(31897*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+35176*C*2^(1/2)*(a*sin(1/2
*d*x+1/2*c)^2)^(1/2)*a^(1/2)+9135*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+9135*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2
*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+11736*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2
))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+11736*C*ln(4/(2*cos(1/
2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a)*si
n(1/2*d*x+1/2*c)^2+18486*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+3045*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^
(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+3045*A*ln(-4/(-2*c
os(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*
a+16752*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+3912*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)
*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+3912*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)
+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)/(2*cos(1/2*d*x
+1/2*c)+2^(1/2))^6/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^6/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 2.29953, size = 711, normalized size = 2.45 \begin{align*} \frac{3 \,{\left ({\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{7} +{\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{6}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left (3 \,{\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + 2 \,{\left (1015 \, A + 1304 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \,{\left (203 \, A + 184 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 48 \,{\left (29 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 896 \, A a^{2} \cos \left (d x + c\right ) + 256 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{6144 \,{\left (d \cos \left (d x + c\right )^{7} + d \cos \left (d x + c\right )^{6}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")
[Out]
1/6144*(3*((1015*A + 1304*C)*a^2*cos(d*x + c)^7 + (1015*A + 1304*C)*a^2*cos(d*x + c)^6)*sqrt(a)*log((a*cos(d*x
+ c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(
d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(1015*A + 1304*C)*a^2*cos(d*x + c)^5 + 2*(1015*A + 1304*C)*a^2*cos(d*x +
c)^4 + 8*(203*A + 184*C)*a^2*cos(d*x + c)^3 + 48*(29*A + 8*C)*a^2*cos(d*x + c)^2 + 896*A*a^2*cos(d*x + c) + 25
6*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^7 + d*cos(d*x + c)^6)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 3.81402, size = 1526, normalized size = 5.26 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")
[Out]
1/3072*(3*(1015*A*a^(5/2) + 1304*C*a^(5/2))*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))) - 3*(1015*A*a^(5/2) + 1304*C*a^(5/2))*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))) + 4*sqrt(2)*(3045*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^22*A*a^(7/2) + 3912*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x +
1/2*c)^2 + a))^22*C*a^(7/2) - 100485*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^20*A
*a^(9/2) - 129096*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^20*C*a^(9/2) + 1303699*(
sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*A*a^(11/2) + 1693560*(sqrt(a)*tan(1/2*d*
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*C*a^(11/2) - 9936699*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(
a*tan(1/2*d*x + 1/2*c)^2 + a))^16*A*a^(13/2) - 11951544*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1
/2*c)^2 + a))^16*C*a^(13/2) + 38257266*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*
A*a^(15/2) + 48800976*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*a^(15/2) - 8377
9026*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(17/2) - 106200016*(sqrt(a)*ta
n(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*a^(17/2) + 74917446*(sqrt(a)*tan(1/2*d*x + 1/2*c
) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(19/2) + 94661616*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/
2*d*x + 1/2*c)^2 + a))^10*C*a^(19/2) - 30850806*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^8*A*a^(21/2) - 39751536*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*a^(21/2)
+ 7187801*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(23/2) + 9070440*(sqrt(a)
*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^(23/2) - 929817*(sqrt(a)*tan(1/2*d*x + 1/2*c
) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^(25/2) - 1176936*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*
d*x + 1/2*c)^2 + a))^4*C*a^(25/2) + 64887*(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^(27/2) + 82200*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(27/2) - 1887*A
*a^(29/2) - 2392*C*a^(29/2))/((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)^6)/d