\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx\) [476]

Optimal result

Integrand size = 33, antiderivative size = 193 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {3 (A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(5 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {3 (A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(5 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

1/3*(5*A-3*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d-(A-B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))-3*(A-B)*sin
(d*x+c)*sec(d*x+c)^(1/2)/a/d+3*(A-B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2
*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d+1/3*(5*A-3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*
c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4104, 3872, 3853, 3856, 2719, 2720} \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(5 A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}-\frac {3 (A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}+\frac {(5 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {3 (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[In]

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

[Out]

(3*(A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + ((5*A - 3*B)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) - (3*(A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*d
) + ((5*A - 3*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) - ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a +
a*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3039

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^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4104

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(
a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^{\frac {5}{2}}(c+d x) (B+A \sec (c+d x))}{a+a \sec (c+d x)} \, dx \\ & = -\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \sec ^{\frac {3}{2}}(c+d x) \left (-\frac {3}{2} a (A-B)+\frac {1}{2} a (5 A-3 B) \sec (c+d x)\right ) \, dx}{a^2} \\ & = -\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {(5 A-3 B) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a}-\frac {(3 (A-B)) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a} \\ & = -\frac {3 (A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(5 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {(5 A-3 B) \int \sqrt {\sec (c+d x)} \, dx}{6 a}+\frac {(3 (A-B)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a} \\ & = -\frac {3 (A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(5 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\left ((5 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}+\frac {\left (3 (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a} \\ & = \frac {3 (A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(5 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {3 (A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(5 A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.69 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.12 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-6 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+6 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+20 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-12 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-2 \sqrt {\sec (c+d x)} \left (18 (A-B) \cos (d x) \csc (c)-2 (5 A-3 B+2 A \sec (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{6 a d (1+\cos (c+d x))} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*((-6*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (6*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^
((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2
F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 20*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[S
ec[c + d*x]] - 12*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 2*Sqrt[Sec[c + d*x]]*(18
*(A - B)*Cos[d*x]*Csc[c] - 2*(5*A - 3*B + 2*A*Sec[c + d*x])*Tan[(c + d*x)/2])))/(6*a*d*(1 + Cos[c + d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(465\) vs. \(2(227)=454\).

Time = 8.75 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.41

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {\left (A -B \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}+2 A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{6 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )+\frac {\left (2 B -2 A \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(466\)

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((A-B)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)+2*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+
1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+(2*B-2*A)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x
+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(
sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d
*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.60 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left (\sqrt {2} {\left (-5 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-5 i \, A + 3 i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (5 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (5 i \, A - 3 i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (9 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/6*((sqrt(2)*(-5*I*A + 3*I*B)*cos(d*x + c)^2 + sqrt(2)*(-5*I*A + 3*I*B)*cos(d*x + c))*weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(5*I*A - 3*I*B)*cos(d*x + c)^2 + sqrt(2)*(5*I*A - 3*I*B)*cos(d*x
+ c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*(sqrt(2)*(-I*A + I*B)*cos(d*x + c)^2 + sq
rt(2)*(-I*A + I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x +
c))) - 9*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^2 + sqrt(2)*(I*A - I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weiers
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(9*(A - B)*cos(d*x + c)^2 + 2*(2*A - 3*B)*cos(d*x + c
) - 2*A)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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