3.397 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=207 \[ -\frac{a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (31 A+42 B+24 C) \tan (c+d x) \sqrt{a \cos (c+d x)+a}}{24 d}+\frac{a^{5/2} (25 A+38 B+40 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}+\frac{a (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d} \]

[Out]

(a^(5/2)*(25*A + 38*B + 40*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) - (a^3*(49*A + 5
4*B - 24*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(31*A + 42*B + 24*C)*Sqrt[a + a*Cos[c + d*x]]
*Tan[c + d*x])/(24*d) + (a*(5*A + 6*B)*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(12*d) + (A*(a +
a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.726982, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3043, 2975, 2981, 2773, 206} \[ -\frac{a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (31 A+42 B+24 C) \tan (c+d x) \sqrt{a \cos (c+d x)+a}}{24 d}+\frac{a^{5/2} (25 A+38 B+40 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}+\frac{a (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(a^(5/2)*(25*A + 38*B + 40*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) - (a^3*(49*A + 5
4*B - 24*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(31*A + 42*B + 24*C)*Sqrt[a + a*Cos[c + d*x]]
*Tan[c + d*x])/(24*d) + (a*(5*A + 6*B)*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(12*d) + (A*(a +
a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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 -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (5 A+6 B)-\frac{1}{2} a (A-6 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac{a (5 A+6 B) (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int (a+a \cos (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) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac{a^2 (31 A+42 B+24 C) \sqrt{a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac{a (5 A+6 B) (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{\int \sqrt{a+a \cos (c+d x)} \left (\frac{3}{8} a^3 (25 A+38 B+40 C)-\frac{1}{8} a^3 (49 A+54 B-24 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac{a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (31 A+42 B+24 C) \sqrt{a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac{a (5 A+6 B) (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{16} \left (a^2 (25 A+38 B+40 C)\right ) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=-\frac{a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (31 A+42 B+24 C) \sqrt{a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac{a (5 A+6 B) (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (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 \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 d}\\ &=\frac{a^{5/2} (25 A+38 B+40 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 d}-\frac{a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (31 A+42 B+24 C) \sqrt{a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac{a (5 A+6 B) (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac{A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.51506, size = 156, normalized size = 0.75 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) (4 (17 A+6 (B+3 C)) \cos (c+d x)+3 (25 A+22 B+8 C) \cos (2 (c+d x))+91 A+66 B+24 C \cos (3 (c+d x))+24 C)+3 \sqrt{2} (25 A+38 B+40 C) \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(3*Sqrt[2]*(25*A + 38*B + 40*C)*ArcTanh[Sqrt[2
]*Sin[(c + d*x)/2]]*Cos[c + d*x]^3 + (91*A + 66*B + 24*C + 4*(17*A + 6*(B + 3*C))*Cos[c + d*x] + 3*(25*A + 22*
B + 8*C)*Cos[2*(c + d*x)] + 24*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

________________________________________________________________________________________

Maple [B]  time = 0.257, size = 1925, normalized size = 9.3 \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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

1/6*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*(32*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2
)*a^(1/2)+25*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+25*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+38*B*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+38*B*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+40*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+40*C*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)*sin(1/2*d*x+1/2*c)^6+12*(50*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+44*B*a^(1/2)*2^(1/2)*(a*sin(1/
2*d*x+1/2*c)^2)^(1/2)+112*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+75*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+75*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+1
14*B*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+114*B*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+120*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+120*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)*sin(1/2*d*x+1/2*c)^4+(-736*A*2^(1/2)*(
a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-450*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-450*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-576*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)-684*B*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-684*B*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-768*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-720*
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-720*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)^2+234*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+75*
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+75*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+156*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+114*B*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+114*B*l
n(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+144*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+120*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+120*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)/(2*cos(1/2
*d*x+1/2*c)+2^(1/2))^3/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^3/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 3.11268, size = 606, normalized size = 2.93 \begin{align*} \frac{3 \,{\left ({\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} +{\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\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 (48 \, C a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (25 \, A + 22 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/96*(3*((25*A + 38*B + 40*C)*a^2*cos(d*x + c)^4 + (25*A + 38*B + 40*C)*a^2*cos(d*x + c)^3)*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*(48*C*a^2*cos(d*x + c)^3 + 3*(25*A + 22*B + 8*C)*a^2*cos(d*x + c)^2 + 2*
(17*A + 6*B)*a^2*cos(d*x + c) + 8*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x
+ c)^3)

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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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 3.84894, size = 1210, normalized size = 5.85 \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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/48*(96*sqrt(2)*C*a^3*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + 3*(25*A*a^(5/2) + 38*B*a^(5/2
) + 40*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*(25*A*a^(5/2) + 38*B*a^(5/2) + 40*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)*(75*(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^(7/2) + 114*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10
*B*a^(7/2) + 72*(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^(7/2) - 1125*(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^(9/2) - 1710*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*a^(9/2) - 888*(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^(9/2) + 6174*(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^(1
1/2) + 6804*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(11/2) + 3024*(sqrt(a)*t
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^(11/2) - 4314*(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^(13/2) - 4284*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1
/2*c)^2 + a))^4*B*a^(13/2) - 1776*(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^(1
3/2) + 807*(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^(15/2) + 858*(sqrt(a)*tan
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(15/2) + 360*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(15/2) - 49*A*a^(17/2) - 54*B*a^(17/2) - 24*C*a^(17/2))/((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)^3)/d