\(\int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [717]

Optimal result

Integrand size = 23, antiderivative size = 172 \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 a \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^3 d}-\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}} \] Output:

-2*a*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/ 
2)/b^2/d+2/3*(3*a^2+b^2)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^ 
(1/2))*sec(d*x+c)^(1/2)/b^3/d-2*a^3*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d* 
x+1/2*c),2*b/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/b^3/(a+b)/d+2/3*sin(d*x+c)/b/ 
d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 21.66 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\cot (c+d x) \left (-b^2 \sqrt {\sec (c+d x)}+6 a b \sec ^{\frac {3}{2}}(c+d x)-6 a b \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)+b^2 \cos (3 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)-12 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+4 (3 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-12 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{6 b^3 d} \] Input:

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

Output:

-1/6*(Cot[c + d*x]*(-(b^2*Sqrt[Sec[c + d*x]]) + 6*a*b*Sec[c + d*x]^(3/2) - 
 6*a*b*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) + b^2*Cos[3*(c + d*x)]*Sec[c + 
d*x]^(3/2) - 12*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c 
+ d*x]^2] + 4*(3*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[- 
Tan[c + d*x]^2] - 12*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1 
]*Sqrt[-Tan[c + d*x]^2]))/(b^3*d)
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.739, Rules used = {3042, 3717, 3042, 4340, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3717

\(\displaystyle \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )}dx\)

\(\Big \downarrow \) 4340

\(\displaystyle \frac {2 \int -\frac {-a \sec ^2(c+d x)-b \sec (c+d x)+3 a}{2 \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{3 b}+\frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-a \sec ^2(c+d x)-b \sec (c+d x)+3 a}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-a \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \csc \left (c+d x+\frac {\pi }{2}\right )+3 a}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)}dx}{b^2}+\frac {\int \frac {3 a b-\left (3 a^2+b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\int \frac {3 a b+\left (-3 a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {3 a b \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\left (3 a^2+b^2\right ) \int \sqrt {\sec (c+d x)}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {3 a b \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (3 a^2+b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {3 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {3 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{3 b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{3 b}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b^2}+\frac {\frac {6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}+\frac {\frac {6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{3 b}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}-\frac {\frac {6 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 d (a+b)}+\frac {\frac {6 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{3 b}\)

Input:

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

Output:

-1/3*(((6*a*b*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d* 
x]])/d - (2*(3*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/d)/b^2 + (6*a^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + 
b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*(a + b)*d))/b + (2*Sin[c + d* 
x])/(3*b*d*Sqrt[Sec[c + d*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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 3120

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3717

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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 4274

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

rule 4336

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]]   Int[ 
1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4340

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n)), x] - Sim 
p[1/(a*d*n)   Int[((d*Csc[e + f*x])^(n + 1)/(a + b*Csc[e + f*x]))*Simp[b*n 
- a*(n + 1)*Csc[e + f*x] - b*(n + 1)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]
 

rule 4594

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2)   Int[(d*Csc[e + 
f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2   Int[(a*A - (A*b - a 
*B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, 
B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(551\) vs. \(2(159)=318\).

Time = 7.47 (sec) , antiderivative size = 552, normalized size of antiderivative = 3.21

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \,b^{2}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b^{3}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{2}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}+3 a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 a^{2} b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+b^{2} a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-3 a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )\right )}{3 b^{3} \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(552\)

Input:

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

Output:

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4*b^3-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2+2*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*b^3+3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* 
d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*a^2*b*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2* 
d*x+1/2*c),2^(1/2))+b^2*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-b^3*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2 
^(1/2))+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El 
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3*(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))*a*b^ 
2-3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli 
pticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/b^3/(a-b)/(-2*sin(1/2*d*x+1 
/2*c)^4+sin(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 [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} b +\sec \left (d x +c \right )^{3} a}d x \] Input:

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

Output:

int(sqrt(sec(c + d*x))/(cos(c + d*x)*sec(c + d*x)**3*b + sec(c + d*x)**3*a 
),x)