3.543 \(\int \frac{(A+B \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=317 \[ \frac{(579 A-199 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{192 a^3 d \sqrt{a \cos (c+d x)+a}}-\frac{(109 A-41 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{64 a^2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(1887 A-691 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{192 a^3 d \sqrt{a \cos (c+d x)+a}}+\frac{(1015 A-363 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(23 A-11 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{48 a d (a \cos (c+d x)+a)^{5/2}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}} \]

[Out]

((1015*A - 363*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(64*Sqrt[2]*a^(7/2)*d) - ((1887*A - 691*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(1
92*a^3*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)
) - ((23*A - 11*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(48*a*d*(a + a*Cos[c + d*x])^(5/2)) - ((109*A - 41*B)*Sec[
c + d*x]^(3/2)*Sin[c + d*x])/(64*a^2*d*(a + a*Cos[c + d*x])^(3/2)) + ((579*A - 199*B)*Sec[c + d*x]^(3/2)*Sin[c
+ d*x])/(192*a^3*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 1.14665, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2961, 2978, 2984, 12, 2782, 205} \[ \frac{(579 A-199 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{192 a^3 d \sqrt{a \cos (c+d x)+a}}-\frac{(109 A-41 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{64 a^2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(1887 A-691 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{192 a^3 d \sqrt{a \cos (c+d x)+a}}+\frac{(1015 A-363 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}-\frac{(23 A-11 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{48 a d (a \cos (c+d x)+a)^{5/2}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

((1015*A - 363*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(64*Sqrt[2]*a^(7/2)*d) - ((1887*A - 691*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(1
92*a^3*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)
) - ((23*A - 11*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(48*a*d*(a + a*Cos[c + d*x])^(5/2)) - ((109*A - 41*B)*Sec[
c + d*x]^(3/2)*Sin[c + d*x])/(64*a^2*d*(a + a*Cos[c + d*x])^(3/2)) + ((579*A - 199*B)*Sec[c + d*x]^(3/2)*Sin[c
+ d*x])/(192*a^3*d*Sqrt[a + a*Cos[c + d*x]])

Rule 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], 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[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*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 205

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

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{2} a (5 A-B)-4 a (A-B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{4} a^2 (63 A-19 B)-\frac{3}{2} a^2 (23 A-11 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac{(109 A-41 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{8} a^3 (579 A-199 B)-\frac{3}{2} a^3 (109 A-41 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac{(109 A-41 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(579 A-199 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{16} a^4 (1887 A-691 B)+\frac{3}{8} a^4 (579 A-199 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{72 a^7}\\ &=-\frac{(1887 A-691 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac{(109 A-41 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(579 A-199 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{9 a^5 (1015 A-363 B)}{32 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{36 a^8}\\ &=-\frac{(1887 A-691 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac{(109 A-41 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(579 A-199 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((1015 A-363 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac{(1887 A-691 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac{(109 A-41 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(579 A-199 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((1015 A-363 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{64 a^2 d}\\ &=\frac{(1015 A-363 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{64 \sqrt{2} a^{7/2} d}-\frac{(1887 A-691 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac{(23 A-11 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac{(109 A-41 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac{(579 A-199 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 5.5, size = 267, normalized size = 0.84 \[ \frac{\cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) (4 (9415 A-3579 B) \cos (c+d x)+8 (3069 A-1145 B) \cos (2 (c+d x))+10164 A \cos (3 (c+d x))+1887 A \cos (4 (c+d x))+21641 A-3748 B \cos (3 (c+d x))-691 B \cos (4 (c+d x))-8469 B)}{96 d}+\frac{i (1015 A-363 B) e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{d}\right )}{8 (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(Cos[(c + d*x)/2]^7*((I*(1015*A - 363*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c
+ d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/(d*E^((I/2)*(c + d*x))) - ((2
1641*A - 8469*B + 4*(9415*A - 3579*B)*Cos[c + d*x] + 8*(3069*A - 1145*B)*Cos[2*(c + d*x)] + 10164*A*Cos[3*(c +
d*x)] - 3748*B*Cos[3*(c + d*x)] + 1887*A*Cos[4*(c + d*x)] - 691*B*Cos[4*(c + d*x)])*Sec[(c + d*x)/2]^5*Sec[c
+ d*x]^(3/2)*Tan[(c + d*x)/2])/(96*d)))/(8*(a*(1 + Cos[c + d*x]))^(7/2))

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Maple [B]  time = 0.649, size = 729, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(7/2),x)

[Out]

-1/384/d*2^(1/2)/a^4*(-1+cos(d*x+c))*(-3045*A*cos(d*x+c)^4*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+c
os(d*x+c)))^(3/2)*sin(d*x+c)+1089*B*cos(d*x+c)^4*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c))
)^(3/2)*sin(d*x+c)-12180*A*sin(d*x+c)*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c
)))^(3/2)+4356*B*sin(d*x+c)*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-
18270*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+6534*B*ar
csin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+1887*A*cos(d*x+c)^5
*2^(1/2)-12180*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-69
1*B*cos(d*x+c)^5*2^(1/2)+4356*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*
x+c)))^(3/2)+3195*A*cos(d*x+c)^4*2^(1/2)-3045*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+c
os(d*x+c)))^(3/2)-1183*B*cos(d*x+c)^4*2^(1/2)+1089*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(3/2)-831*A*cos(d*x+c)^3*2^(1/2)+275*B*cos(d*x+c)^3*2^(1/2)-3355*A*cos(d*x+c)^2*2^(1/2)+1215*
B*cos(d*x+c)^2*2^(1/2)-1024*A*cos(d*x+c)*2^(1/2)+384*B*cos(d*x+c)*2^(1/2)+128*A*2^(1/2))*cos(d*x+c)*(1/cos(d*x
+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3/(1+cos(d*x+c))^2

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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+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.84682, size = 822, normalized size = 2.59 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left ({\left (1887 \, A - 691 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2541 \, A - 937 \, B\right )} \cos \left (d x + c\right )^{3} + 39 \,{\left (109 \, A - 41 \, B\right )} \cos \left (d x + c\right )^{2} + 128 \,{\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 128 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{384 \,{\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/384*(3*sqrt(2)*((1015*A - 363*B)*cos(d*x + c)^5 + 4*(1015*A - 363*B)*cos(d*x + c)^4 + 6*(1015*A - 363*B)*co
s(d*x + c)^3 + 4*(1015*A - 363*B)*cos(d*x + c)^2 + (1015*A - 363*B)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt(
a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((1887*A - 691*B)*cos(d*x + c)^4 + 2*(2541*
A - 937*B)*cos(d*x + c)^3 + 39*(109*A - 41*B)*cos(d*x + c)^2 + 128*(7*A - 3*B)*cos(d*x + c) - 128*A)*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(
d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

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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+B*cos(d*x+c))*sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(7/2), x)