3.7.12 \(\int \frac {(a+b \cos (c+d x))^{3/2}}{\sqrt {\cos (c+d x)}} \, dx\) [612]

3.7.12.1 Optimal result
3.7.12.2 Mathematica [A] (verified)
3.7.12.3 Rubi [A] (verified)
3.7.12.4 Maple [B] (verified)
3.7.12.5 Fricas [F]
3.7.12.6 Sympy [F]
3.7.12.7 Maxima [F]
3.7.12.8 Giac [F]
3.7.12.9 Mupad [F(-1)]

3.7.12.1 Optimal result

Integrand size = 25, antiderivative size = 375 \[ \int \frac {(a+b \cos (c+d x))^{3/2}}{\sqrt {\cos (c+d x)}} \, dx=-\frac {(a-b) b \sqrt {a+b} \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d}+\frac {\sqrt {a+b} (2 a+b) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d}-\frac {3 a \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d}+\frac {b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

output
b*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)-(a-b)*b*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+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a- 
b))^(1/2)/a/d+(2*a+b)*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+b)^(1/2)*(a*(1-sec(d*x+c))/( 
a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/d-3*a*cot(d*x+c)*EllipticPi((a+ 
b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1 
/2))*(a+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^( 
1/2)/d
 
3.7.12.2 Mathematica [A] (verified)

Time = 5.20 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b \cos (c+d x))^{3/2}}{\sqrt {\cos (c+d x)}} \, dx=-\frac {\sqrt {\cos (c+d x)} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 b (a+b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right )+4 a (a-2 b) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+12 a b \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+b \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {a+b \cos (c+d x)} \left (-1+\tan ^4\left (\frac {1}{2} (c+d x)\right )\right )} \]

input
Integrate[(a + b*Cos[c + d*x])^(3/2)/Sqrt[Cos[c + d*x]],x]
 
output
-((Sqrt[Cos[c + d*x]]*Sec[(c + d*x)/2]^2*(2*b*(a + b)*Sqrt[Cos[c + d*x]/(1 
 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]* 
EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 4*a*(a - 2*b)*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)] + 12 
*a*b*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + 
 b)*(1 + Cos[c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b) 
/(a + b)] + b*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c 
+ d*x)/2]))/(d*Sqrt[a + b*Cos[c + d*x]]*(-1 + Tan[(c + d*x)/2]^4)))
 
3.7.12.3 Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3300, 27, 3042, 3532, 3042, 3288, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3300

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3532

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

\(\displaystyle \frac {1}{2} \left (-\int \frac {a b-2 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (a (2 a+b) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

input
Int[(a + b*Cos[c + d*x])^(3/2)/Sqrt[Cos[c + d*x]],x]
 
output
((-2*(a - b)*b*Sqrt[a + b]*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)])/(a*d) + (2 
*Sqrt[a + b]*(2*a + 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 - Se 
c[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d - (6*a*Sqrt[ 
a + b]*Cot[c + d*x]*EllipticPi[(a + b)/b, 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)])/d)/2 + (b*Sqrt[a + 
 b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])
 

3.7.12.3.1 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 3288
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c 
*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti 
cPi[(c + d)/d, 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}, x] && NeQ[c^2 - 
 d^2, 0] && PosQ[(c + d)/b]
 

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 3300
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] 
)^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n))   I 
nt[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a^2*c*d*( 
m + n) + b*d*(b*c*(m - 1) + a*d*n) + (a*d*(2*b*c + a*d)*(m + n) - b*d*(a*c 
- b*d*(m + n - 1)))*Sin[e + f*x] + b*d*(b*c*n + a*d*(2*m + n - 1))*Sin[e + 
f*x]^2, x], x], 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] && LtQ[0, m, 2] && LtQ[-1, n, 2] && 
NeQ[m + n, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

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 3532
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2   Int[Sqrt[a + b*Sin[e + f*x]] 
/Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2   Int[(A*b^2 - a^2*C + b*(b* 
B - 2*a*C)*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, C}, x] && NeQ[b*c - a*d, 0] & 
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 
3.7.12.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1362\) vs. \(2(347)=694\).

Time = 9.36 (sec) , antiderivative size = 1363, normalized size of antiderivative = 3.63

method result size
default \(\text {Expression too large to display}\) \(1363\)

input
int((a+cos(d*x+c)*b)^(3/2)/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 
output
1/d*(-EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c) 
*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b*cos( 
d*x+c)^2-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c) 
)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2*c 
os(d*x+c)^2-6*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d 
*x+c))/(a+b))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/ 
2))*a*b*cos(d*x+c)^2-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b) 
/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b) 
)^(1/2))*a^2*cos(d*x+c)^2+4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b)) 
^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+cos(d 
*x+c)))^(1/2)*a*b*cos(d*x+c)^2-2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/( 
a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*(cos(d*x+c)/(1+ 
cos(d*x+c)))^(1/2)*a*b*cos(d*x+c)-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+ 
cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c), 
(-(a-b)/(a+b))^(1/2))*b^2*cos(d*x+c)-12*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* 
((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi(cot(d*x+c)-csc(d* 
x+c),-1,(-(a-b)/(a+b))^(1/2))*a*b*cos(d*x+c)-4*(cos(d*x+c)/(1+cos(d*x+c))) 
^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF(cot(d*x+c)- 
csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*cos(d*x+c)+8*EllipticF(cot(d*x+c)-csc 
(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(...
 
3.7.12.5 Fricas [F]

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

input
integrate((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")
 
output
integral((b*cos(d*x + c) + a)^(3/2)/sqrt(cos(d*x + c)), x)
 
3.7.12.6 Sympy [F]

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

input
integrate((a+b*cos(d*x+c))**(3/2)/cos(d*x+c)**(1/2),x)
 
output
Integral((a + b*cos(c + d*x))**(3/2)/sqrt(cos(c + d*x)), x)
 
3.7.12.7 Maxima [F]

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

input
integrate((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")
 
output
integrate((b*cos(d*x + c) + a)^(3/2)/sqrt(cos(d*x + c)), x)
 
3.7.12.8 Giac [F]

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

input
integrate((a+b*cos(d*x+c))^(3/2)/cos(d*x+c)^(1/2),x, algorithm="giac")
 
output
integrate((b*cos(d*x + c) + a)^(3/2)/sqrt(cos(d*x + c)), x)
 
3.7.12.9 Mupad [F(-1)]

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

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