\(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [541]

Optimal result

Integrand size = 35, antiderivative size = 234 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {(3 A-43 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \]

[Out]

1/4*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(3/2)+1/16*(3*A-11*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))
^(3/2)/sec(d*x+c)^(1/2)+2*B*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2
)/a^(5/2)/d+1/32*(3*A-43*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^(5/2)/d*2^(1/2)

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3040, 3056, 3061, 2861, 211, 2853, 222} \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(3 A-43 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

(2*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a^(5/2)*d
) + ((3*A - 43*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]])/(16*Sqrt[2]*a^(5/2)*d) + ((A - B)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2
)*Sec[c + d*x]^(3/2)) + ((3*A - 11*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, 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 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*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]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx \\ & = \frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} a (A-B)+4 a B \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 (3 A-11 B)+8 a^2 B \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = \frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left ((3 A-43 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}{32 a^2}+\frac {\left (B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{a^3} \\ & = \frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left ((3 A-43 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 )}{16 a d}-\frac {\left (2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^3 d} \\ & = \frac {2 B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {(3 A-43 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)}}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.85 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (-i \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (32 B \text {arcsinh}\left (e^{i (c+d x)}\right )-\sqrt {2} (3 A-43 B) \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-32 B \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )+\frac {1}{2} (3 A-11 B+(7 A-15 B) \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{8 d (a (1+\cos (c+d x)))^{5/2}} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((-I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*(32*B*ArcSinh[E^(I*(c + d*x))] - Sqrt[2]*(3*A - 43*B)*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2
*I)*(c + d*x))])] - 32*B*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/E^((I/2)*(c + d*x)) + ((3*A - 11*B + (7*A -
15*B)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/2))/(8*d
*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(468\) vs. \(2(195)=390\).

Time = 8.39 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.00

method	result	size
default	\(-\frac {{\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}^{2} \left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right ) \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (-2 \left (\csc ^{3}\left (d x +c \right )\right ) A \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{3}+2 \left (\csc ^{3}\left (d x +c \right )\right ) B \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{3}+5 A \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-32 B \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )-13 B \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-3 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+43 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{32 a^{3} d {\left (-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )}^{\frac {3}{2}} {\left (-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}^{\frac {5}{2}}}\)	\(469\)
parts	\(\frac {A \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (7 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \tan \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-6 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-3 \sec \left (d x +c \right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}+\frac {B \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (32 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )-15 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+64 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-11 \tan \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+43 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+32 \sec \left (d x +c \right ) \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+86 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+43 \sec \left (d x +c \right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}\)	\(486\)

[In]

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

[Out]

-1/32/a^3/d/(-(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(3/2)/(-csc(d*x+c)^2*(1-cos
(d*x+c))^2+1)^(5/2)*(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^2*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)*(a/(csc(d*x+c)^2*(1-
cos(d*x+c))^2+1))^(1/2)*(-2*csc(d*x+c)^3*A*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(1-cos(d*x+c))^3+2*csc(d*x
+c)^3*B*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(1-cos(d*x+c))^3+5*A*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)
*(csc(d*x+c)-cot(d*x+c))-32*B*2^(1/2)*arctan(2^(1/2)*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)/(csc(d*x+c)^2*(1
-cos(d*x+c))^2-1)*(csc(d*x+c)-cot(d*x+c)))-13*B*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(csc(d*x+c)-cot(d*x+c
))-3*A*arcsin(cot(d*x+c)-csc(d*x+c))+43*B*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 6.03 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\sqrt {2} {\left ({\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 64 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left ({\left (7 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

-1/32*(sqrt(2)*((3*A - 43*B)*cos(d*x + c)^3 + 3*(3*A - 43*B)*cos(d*x + c)^2 + 3*(3*A - 43*B)*cos(d*x + c) + 3*
A - 43*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 64*(B*c
os(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*
x + c))/(sqrt(a)*sin(d*x + c))) - 2*((7*A - 15*B)*cos(d*x + c)^2 + (3*A - 11*B)*cos(d*x + c))*sqrt(a*cos(d*x +
c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c
) + a^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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