\(\int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx\) [157]

Optimal result

Integrand size = 21, antiderivative size = 131 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}+\frac {2 \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}{3 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\frac {2 \left ((-b \cos (c+d x)+a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}+\frac {\left (a^2+b^2\right ) \sqrt {\cos ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )\right ) \tan \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}{\sqrt {\sqrt {1+\frac {a^2}{b^2}} b \sin \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}}\right )}{3 d} \] Input:

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

Output:

(2*((-(b*Cos[c + d*x]) + a*Sin[c + d*x])*Sqrt[a*Cos[c + d*x] + b*Sin[c + d 
*x]] + ((a^2 + b^2)*Sqrt[Cos[c + d*x + ArcTan[a/b]]^2]*HypergeometricPFQ[{ 
1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/b]]^2]*Tan[c + d*x + ArcTan[a/b]] 
)/Sqrt[Sqrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]]))/(3*d)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3552, 3042, 3557, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3552

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3557

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, 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 3552

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x 
_Symbol] :> Simp[(-(b*Cos[c + d*x] - a*Sin[c + d*x]))*((a*Cos[c + d*x] + b* 
Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[(n - 1)*((a^2 + b^2)/n)   Int[(a*Co 
s[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && N 
eQ[a^2 + b^2, 0] &&  !IntegerQ[(n - 1)/2] && GtQ[n, 1]
 

rule 3557

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x 
_Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S 
in[c + d*x])/Sqrt[a^2 + b^2])^n   Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] 
 /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 
+ b^2, 0] || EqQ[a^2 + b^2, 0])
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\left (a^{2}+b^{2}\right ) \left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{3}+2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )\right )}{3 \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\)	\(165\)

Input:

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

Output:

-1/3*(a^2+b^2)*((-sin(d*x+c-arctan(-a,b))+1)^(1/2)*(2*sin(d*x+c-arctan(-a, 
b))+2)^(1/2)*sin(d*x+c-arctan(-a,b))^(1/2)*EllipticF((-sin(d*x+c-arctan(-a 
,b))+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c-arctan(-a,b))^3+2*sin(d*x+c-arctan( 
-a,b)))/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-arctan(-a,b))*(a^2+b^2)^(1/2))^ 
(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.22 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2} \, a - \frac {1}{2} i \, b} {\left (i \, a - b\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {\frac {1}{2} \, a + \frac {1}{2} i \, b} {\left (-i \, a - b\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}\right )}}{3 \, d} \] Input:

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

Output:

-2/3*(sqrt(1/2*a - 1/2*I*b)*(I*a - b)*weierstrassPInverse(-4*(a^2 + 2*I*a* 
b - b^2)/(a^2 + b^2), 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(1/2*a + 1/2 
*I*b)*(-I*a - b)*weierstrassPInverse(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 
 0, cos(d*x + c) - I*sin(d*x + c)) + sqrt(a*cos(d*x + c) + b*sin(d*x + c)) 
*(b*cos(d*x + c) - a*sin(d*x + c)))/d
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a \cos (c+d x)+b \sin (c+d x))^{3/2} \, dx=\frac {-2 \sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \cos \left (d x +c \right ) a^{3}-4 \sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \cos \left (d x +c \right ) a \,b^{2}+4 \sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \sin \left (d x +c \right ) a^{2} b +2 \sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \sin \left (d x +c \right ) b^{3}-3 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )}{\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}d x \right ) a^{4} d -6 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )}{\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}d x \right ) a^{2} b^{2} d -3 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )}{\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}d x \right ) b^{4} d}{3 a b d} \] Input:

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

Output:

( - 2*sqrt(cos(c + d*x)*a + sin(c + d*x)*b)*cos(c + d*x)*a**3 - 4*sqrt(cos 
(c + d*x)*a + sin(c + d*x)*b)*cos(c + d*x)*a*b**2 + 4*sqrt(cos(c + d*x)*a 
+ sin(c + d*x)*b)*sin(c + d*x)*a**2*b + 2*sqrt(cos(c + d*x)*a + sin(c + d* 
x)*b)*sin(c + d*x)*b**3 - 3*int((sqrt(cos(c + d*x)*a + sin(c + d*x)*b)*cos 
(c + d*x)*sin(c + d*x))/(cos(c + d*x)*a + sin(c + d*x)*b),x)*a**4*d - 6*in 
t((sqrt(cos(c + d*x)*a + sin(c + d*x)*b)*cos(c + d*x)*sin(c + d*x))/(cos(c 
 + d*x)*a + sin(c + d*x)*b),x)*a**2*b**2*d - 3*int((sqrt(cos(c + d*x)*a + 
sin(c + d*x)*b)*cos(c + d*x)*sin(c + d*x))/(cos(c + d*x)*a + sin(c + d*x)* 
b),x)*b**4*d)/(3*a*b*d)