\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [531]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (warning: unable to verify)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 35, antiderivative size = 223 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(11 A-7 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/2*(A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/6*(7*A-3*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/
(a+a*cos(d*x+c))^(1/2)+1/4*(11*A-7*B)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^
(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)-1/6*(19*A-15*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+
a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3040, 3057, 3063, 12, 2861, 211} \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(11 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(7 A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{6 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(19 A-15 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 a d \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

((11*A - 7*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((19*A - 15*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((7*A -
3*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*a*d*Sqrt[a + a*Cos[c + d*x]])

Rule 12

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

Rule 211

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

Rule 2861

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 3040

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 3057

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 3063

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])

Rubi steps \begin{align*} \text {integral}& = \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))^{3/2}} \, dx \\ & = -\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{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 {1}{2} a (7 A-3 B)-2 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a^2 (19 A-15 B)+\frac {1}{2} a^2 (7 A-3 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{3 a^3} \\ & = -\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {3 a^3 (11 A-7 B)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{3 a^4} \\ & = -\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((11 A-7 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}{4 a} \\ & = -\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((11 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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 )}{2 d} \\ & = \frac {(11 A-7 B) \arctan \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)}}{2 \sqrt {2} a^{3/2} d}-\frac {(19 A-15 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(7 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.59 (sec) , antiderivative size = 981, normalized size of antiderivative = 4.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 \cos ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (-\frac {(A-B) \left (1-2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{12 \left (1+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}}+\frac {(A-B) \left (1+2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{12 \left (1-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}}-\frac {1}{2} (A-B) \left (5 \arctan \left (\frac {1-2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right )+\frac {1+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\left (1-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+\frac {3 \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{1-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )+\frac {1}{2} (A-B) \left (5 \arctan \left (\frac {1+2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right )+\frac {1-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{\left (1+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+\frac {3 \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}{1+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )+\frac {(A+3 B) \csc ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-12 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{2};-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )-12 \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (4-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+3 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+7 \sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \left (15-20 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+8 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (\left (3-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-3 \text {arctanh}\left (\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{126 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2}}\right )}{d (a (1+\cos (c+d x)))^{3/2}} \]

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(-1/12*((A -
B)*(1 - 2*Sin[c/2 + (d*x)/2]))/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A - B)*(1 + 2
*Sin[c/2 + (d*x)/2]))/(12*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - ((A - B)*(5*ArcTan[(1
 - 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 + Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2]
)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/2 + ((A
- B)*(5*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 - Sin[c/2 + (d*x)/2])/((1 + S
in[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*
x)/2])))/2 + ((A + 3*B)*Csc[c/2 + (d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2},
-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeometric2F1[2, 7/2, 9/2,
 -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c/2 + (d*x)/2]^2 + 3*Si
n[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 - 2*Sin[c/2 + (d*x)/2]^2
)^3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 + (d*x)/2]^2)*Sqrt[-(Sin[c/2 + (d*
x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]
*(1 - 2*Sin[c/2 + (d*x)/2]^2))))/(126*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))))/(d*(a*(1 + Cos[c + d*x]))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(416\) vs. \(2(188)=376\).

Time = 10.33 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.87

method result size
default \(-\frac {\left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (33 A \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-21 B \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+19 A \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+66 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-15 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-42 B \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+12 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+33 A \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-12 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-21 B \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-4 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{12 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2}}\) \(417\)
parts \(-\frac {A \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (33 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+19 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+66 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+12 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+33 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{12 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}+\frac {B \left (7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}+14 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+4 \sqrt {2}\, \sin \left (d x +c \right )+7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}\) \(439\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/12/a^2/d*sec(d*x+c)^(5/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^2*(33*A*cos(d*x+c)^4*arcsin(cot(d*x+c)-cs
c(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-21*B*cos(d*x+c)^4*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)+19*A*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+66*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsi
n(cot(d*x+c)-csc(d*x+c))-15*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)-42*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*arcsin(cot(d*x+c)-csc(d*x+c))+12*A*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+33*A*cos(d*x+c)^2*arcsin(cot(d*x+c)-csc
(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-12*B*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)-21*B*cos(d*x+c)^2*arcsin(cot(d
*x+c)-csc(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-4*A*sin(d*x+c)*cos(d*x+c)*2^(1/2))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} {\left ({\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 4 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(2)*((11*A - 7*B)*cos(d*x + c)^3 + 2*(11*A - 7*B)*cos(d*x + c)^2 + (11*A - 7*B)*cos(d*x + c))*sqr
t(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((19*A - 15*B)*cos
(d*x + c)^2 + 12*(A - B)*cos(d*x + c) - 4*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*
cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(3/2), x)

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(3/2), x)