\(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx\) [643]

Optimal result

Integrand size = 25, antiderivative size = 473 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=-\frac {8 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^5 (a-b) (a+b)^{3/2} d}+\frac {2 \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 (a-b) (a+b)^{3/2} d}+\frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/3*b^2*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(3/2)+4/3*b^2*(5*a^2-3*b^2)*sin(d*x+c)/a^2/
(a^2-b^2)^2/d/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(1/2)+2/3*(a^4-13*a^2*b^2+8*b^4)*sin(d*x+c)*(a+b*cos(d*x+c))^(
1/2)/a^3/(a^2-b^2)^2/d/cos(d*x+c)^(3/2)-8/3*b*(2*a^4-7*a^2*b^2+4*b^4)*cot(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1
/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))
^(1/2)/a^5/(a-b)/(a+b)^(3/2)/d+2/3*(a^4+9*a^3*b+16*a^2*b^2-12*a*b^3-16*b^4)*cot(d*x+c)*EllipticF((a+b*cos(d*x+
c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/
(a-b))^(1/2)/a^4/(a-b)/(a+b)^(3/2)/d

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 473, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2881, 3134, 3077, 2895, 3073} \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}-\frac {8 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a^5 d (a-b) (a+b)^{3/2}}+\frac {2 \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{3 a^4 d (a-b) (a+b)^{3/2}}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^(5/2)),x]

[Out]

(-8*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos
[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(
3*a^5*(a - b)*(a + b)^(3/2)*d) + (2*(a^4 + 9*a^3*b + 16*a^2*b^2 - 12*a*b^3 - 16*b^4)*Cot[c + d*x]*EllipticF[Ar
cSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]
))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(3*a^4*(a - b)*(a + b)^(3/2)*d) + (2*b^2*Sin[c + d*x])/(3*a*
(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^(3/2)) + (4*b^2*(5*a^2 - 3*b^2)*Sin[c + d*x])/(3*a^2*(a^
2 - b^2)^2*d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Cos[c + d*x]]) + (2*(a^4 - 13*a^2*b^2 + 8*b^4)*Sqrt[a + b*Cos[c + d
*x]]*Sin[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2))

Rule 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 2895

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(
Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqrt[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]
*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2]], -(a + b)/(a - b)], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 3073

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A*(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e +
f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e +
f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ
[A, B] && PosQ[(c + d)/b]

Rule 3077

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*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] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]

Rule 3134

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {2 \int \frac {\frac {3}{2} \left (a^2-2 b^2\right )-\frac {3}{2} a b \cos (c+d x)+2 b^2 \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {4 \int \frac {\frac {3}{4} \left (a^4-13 a^2 b^2+8 b^4\right )-\frac {1}{2} a b \left (3 a^2-b^2\right ) \cos (c+d x)+b^2 \left (5 a^2-3 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {-\frac {3}{2} b \left (2 a^4-7 a^2 b^2+4 b^4\right )+\frac {3}{8} a \left (a^4+7 a^2 b^2-4 b^4\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{9 a^3 \left (a^2-b^2\right )^2} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{3 a^3 (a-b) (a+b)^2}-\frac {\left (4 b \left (2 a^4-7 a^2 b^2+4 b^4\right )\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {8 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^5 (a-b) (a+b)^{3/2} d}+\frac {2 \left (a^4+9 a^3 b+16 a^2 b^2-12 a b^3-16 b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 (a-b) (a+b)^{3/2} d}+\frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}+\frac {4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.44 (sec) , antiderivative size = 1351, normalized size of antiderivative = 2.86 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\frac {-\frac {4 a \left (a^6+15 a^4 b^2-32 a^2 b^4+16 b^6\right ) \sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a+b) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-4 a \left (8 a^5 b-28 a^3 b^3+16 a b^5\right ) \left (\frac {\sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a+b) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{b \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )+2 \left (8 a^4 b^2-28 a^2 b^4+16 b^6\right ) \left (\frac {i \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {a+b \cos (c+d x)} E\left (i \text {arcsinh}\left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos (c+d x)}}\right )|-\frac {2 a}{-a-b}\right ) \sec (c+d x)}{b \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt {\frac {(a+b \cos (c+d x)) \sec (c+d x)}{a+b}}}+\frac {2 a \left (\frac {a \sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{(a+b) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {a \sqrt {\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{-a+b}} \sqrt {-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\frac {\sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}}}{\sqrt {2}}\right ),-\frac {2 a}{-a+b}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{b \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )}{b}+\frac {\sqrt {a+b \cos (c+d x)} \sin (c+d x)}{b \sqrt {\cos (c+d x)}}\right )}{3 a^4 (a-b)^2 (a+b)^2 d}+\frac {\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2 b^4 \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {8 \left (3 a^2 b^4 \sin (c+d x)-2 b^6 \sin (c+d x)\right )}{3 a^4 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {16 b \tan (c+d x)}{3 a^4}+\frac {2 \sec (c+d x) \tan (c+d x)}{3 a^3}\right )}{d} \]

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^(5/2)),x]

[Out]

((-4*a*(a^6 + 15*a^4*b^2 - 32*a^2*b^4 + 16*b^6)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Co
s[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[Ar
cSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)
*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(8*a^5*b - 28*a^3*b^3 + 16*a*b^5)*((Sqrt[((a + b)*Cot[(c +
d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c +
d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a
)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c
+ d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c
+ d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[
2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(8*a^4*b^2 - 28
*a^2*b^4 + 16*b^6)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos
[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x]
)*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*
Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[(
(a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c
+ d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*
x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), A
rcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt
[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c + d*x]]))
)/(3*a^4*(a - b)^2*(a + b)^2*d) + (Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((2*b^4*Sin[c + d*x])/(3*a^3*(a
^2 - b^2)*(a + b*Cos[c + d*x])^2) + (8*(3*a^2*b^4*Sin[c + d*x] - 2*b^6*Sin[c + d*x]))/(3*a^4*(a^2 - b^2)^2*(a
+ b*Cos[c + d*x])) - (16*b*Tan[c + d*x])/(3*a^4) + (2*Sec[c + d*x]*Tan[c + d*x])/(3*a^3)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5223\) vs. \(2(435)=870\).

Time = 15.91 (sec) , antiderivative size = 5224, normalized size of antiderivative = 11.04

method	result	size
default	\(\text {Expression too large to display}\)	\(5224\)

[In]

int(1/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*b)^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*cos(d*x + c)^6 + 3*a*b^2*cos(d*x + c)^5 + 3*a^2*b*co
s(d*x + c)^4 + a^3*cos(d*x + c)^3), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2)), x)

Giac [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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