\(\int (a+b \cos (c+d x))^{3/2} \, dx\) [496]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 14, antiderivative size = 157 \[ \int (a+b \cos (c+d x))^{3/2} \, dx=\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.85 \[ \int (a+b \cos (c+d x))^{3/2} \, dx=\frac {8 a (a+b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+2 b (a+b \cos (c+d x)) \sin (c+d x)}{3 d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(8*a*(a + b)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2* 
b)/(a + b)] - 2*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[( 
c + d*x)/2, (2*b)/(a + b)] + 2*b*(a + b*Cos[c + d*x])*Sin[c + d*x])/(3*d*S 
qrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 3135, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3135

\(\displaystyle \frac {2}{3} \int \frac {3 a^2+4 b \cos (c+d x) a+b^2}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 a^2+4 b \cos (c+d x) a+b^2}{\sqrt {a+b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {3 a^2+4 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+b^2}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{3} \left (4 a \int \sqrt {a+b \cos (c+d x)}dx-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (4 a \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{3} \left (\frac {4 a \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {4 a \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}\right )+\frac {2 b \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\)

Input:

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

Output:

((8*a*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*S 
qrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*Sqrt[(a + b*Cos[c + d* 
x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + 
d*x]]))/3 + (2*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)
 

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 3132
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3135
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[1/n   Int[(a + b* 
Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*x] 
, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && 
 IntegerQ[2*n]
 

rule 3140
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(150)=300\).

Time = 4.79 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.87

method result size
default \(-\frac {2 \sqrt {\left (2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b^{2}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a b -6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}-a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a b -2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}\right )}{3 \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\) \(450\)

Input:

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

Output:

-2/3*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/ 
2*d*x+1/2*c)^5*b^2+2*cos(1/2*d*x+1/2*c)^3*a*b-6*cos(1/2*d*x+1/2*c)^3*b^2-a 
^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/ 
2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+b^2*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2* 
d*x+1/2*c),(-2*b/(a-b))^(1/2))+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)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b)) 
^(1/2))*a^2-4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b) 
/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b-2*cos(1 
/2*d*x+1/2*c)*a*b+2*cos(1/2*d*x+1/2*c)*b^2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+ 
b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2 
*b+a+b)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.54 \[ \int (a+b \cos (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (-12 i \, \sqrt {\frac {1}{2}} a b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 12 i \, \sqrt {\frac {1}{2}} a b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 \, \sqrt {b \cos \left (d x + c\right ) + a} b^{2} \sin \left (d x + c\right ) + \sqrt {\frac {1}{2}} {\left (i \, a^{2} + 3 i \, b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + \sqrt {\frac {1}{2}} {\left (-i \, a^{2} - 3 i \, b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right )}}{9 \, b d} \] Input:

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

Output:

-2/9*(-12*I*sqrt(1/2)*a*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, - 
8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, - 
8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2 
*a)/b)) + 12*I*sqrt(1/2)*a*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2 
, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2 
, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) 
+ 2*a)/b)) - 3*sqrt(b*cos(d*x + c) + a)*b^2*sin(d*x + c) + sqrt(1/2)*(I*a^ 
2 + 3*I*b^2)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8 
*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) 
+ sqrt(1/2)*(-I*a^2 - 3*I*b^2)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3* 
b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d 
*x + c) + 2*a)/b))/(b*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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