\(\int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [758]

Optimal result

Integrand size = 25, antiderivative size = 325 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^3 \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 (a+2 b) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 b^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 7.80 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (\left (a b^2+\left (a^2-2 b^2\right ) (a+b \cos (c+d x))\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)} \sin (c+d x)-\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (a^3+a^2 b-2 a b^2-2 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )-2 a \left (a^2-a b-2 b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+\left (a^2-2 b^2\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)} \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4710, 3042, 3281, 27, 3042, 3477, 3042, 3295, 3473}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*(a - b)*Sqrt[a + b]*(a^2 - 2*b^ 
2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqr 
t[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)])/(a^2*d) - (2*(a - b)*Sqrt[a + b]*( 
a + 2*b)*Cot[c + d*x]*EllipticF[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)])/(a*d))/(a*(a^2 - b^2)) + (2* 
b^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] + Simp[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)*Si 
n[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 3295

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]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473

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 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[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 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(868\) vs. \(2(291)=582\).

Time = 23.49 (sec) , antiderivative size = 869, normalized size of antiderivative = 2.67

method	result	size
default	\(\text {Expression too large to display}\)	\(869\)

Input:

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

Output:

-2/d*sec(d*x+c)^(3/2)*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)*((2*csc(d*x+c)^3*( 
1-cos(d*x+c))^3+2*cot(d*x+c)-2*csc(d*x+c))*b^3+(csc(d*x+c)^3*(1-cos(d*x+c) 
)^3-cot(d*x+c)+csc(d*x+c))*a^3+(-csc(d*x+c)^3*(1-cos(d*x+c))^3-cot(d*x+c)+ 
csc(d*x+c))*b*a^2-2*csc(d*x+c)^3*a*b^2*(1-cos(d*x+c))^3-2*EllipticF(cot(d* 
x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)+2*EllipticF(cot(d*x+c)-csc 
(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+ 
cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)+4*EllipticF(cot(d*x+c)-csc(d*x+c 
),(-(a-b)/(a+b))^(1/2))*a*b^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d* 
x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)+2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a 
-b)/(a+b))^(1/2))*a^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/ 
(cos(d*x+c)+1)/(a+b))^(1/2)+2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b 
))^(1/2))*a^2*b*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d 
*x+c)+1)/(a+b))^(1/2)-4*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/ 
2))*a*b^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+ 
1)/(a+b))^(1/2)-4*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b^ 
3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b) 
)^(1/2))*(a+cos(d*x+c)*b)^(1/2)/(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-csc(d*x+c 
)^2*b*(1-cos(d*x+c))^2+a+b)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(a+b)/(a-b)/ 
a^2
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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