\(\int \frac {1}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [722]

Optimal result

Integrand size = 23, antiderivative size = 223 \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {a \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}-\frac {a \left (a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^2 (a+b)^2 d}+\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \sec (c+d x))} \] Output:

-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/(a^2-b^2)/d+(a^2-2*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^2/(a^2-b^2)/d-a*(a^2-3*b^2)*cos(d*x+c)^(1/2)*E 
llipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/(a-b)/b^2 
/(a+b)^2/d+a*sec(d*x+c)^(1/2)*sin(d*x+c)/(a^2-b^2)/d/(b+a*sec(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 3.57 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\cos (2 (c+d x)) \csc (c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (-b (-a+b) (a+b \cos (c+d x)) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-\left (a^2-3 b^2\right ) (a+b \cos (c+d x)) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+a b \left (a \tan ^2(c+d x)-(a+b \cos (c+d x)) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{(a-b) b^2 (a+b) d (b+a \sec (c+d x)) \left (-2+\sec ^2(c+d x)\right )} \] Input:

Integrate[1/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)),x]
 

Output:

(Cos[2*(c + d*x)]*Csc[c + d*x]*Sec[c + d*x]^(3/2)*(-(b*(-a + b)*(a + b*Cos 
[c + d*x])*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sq 
rt[-Tan[c + d*x]^2]) - (a^2 - 3*b^2)*(a + b*Cos[c + d*x])*EllipticPi[-(a/b 
), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2 
] + a*b*(a*Tan[c + d*x]^2 - (a + b*Cos[c + d*x])*EllipticE[ArcSin[Sqrt[Sec 
[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/((a - b)*b^2* 
(a + b)*d*(b + a*Sec[c + d*x])*(-2 + Sec[c + d*x]^2))
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.95, 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, 4330, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4330

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 4336

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

Int[1/((a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)),x]
 

Output:

-1/2*(((2*a*b*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d* 
x]])/d - (2*(a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/d)/b^2 + (2*a*(a^2 - 3*b^2)*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))/(a^2 
- b^2) + (a*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(b + a*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 4330

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

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 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. \(793\) vs. \(2(214)=428\).

Time = 7.20 (sec) , antiderivative size = 794, normalized size of antiderivative = 3.56

method	result	size
default	\(\text {Expression too large to display}\)	\(794\)

Input:

int(1/(a+cos(d*x+c)*b)^2/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(1/(a+b*cos(d*x+c))**2/sec(d*x+c)**(3/2),x)
                                                                                    
                                                                                    
 

Output:

Integral(1/((a + b*cos(c + d*x))**2*sec(c + d*x)**(3/2)), x)
 

Maxima [F]

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

integrate(1/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2)), x)
 

Giac [F]

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

integrate(1/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^2*sec(d*x + c)^(3/2)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(1/(a+b*cos(d*x+c))^2/sec(d*x+c)^(3/2),x)
 

Output:

int(sqrt(sec(c + d*x))/(cos(c + d*x)**2*sec(c + d*x)**2*b**2 + 2*cos(c + d 
*x)*sec(c + d*x)**2*a*b + sec(c + d*x)**2*a**2),x)