\(\int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx\) [156]

Optimal result

Integrand size = 21, antiderivative size = 131 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}+\frac {6 \left (a^2+b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{5 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 1.32 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.95 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {a \cos (c+d x)+b \sin (c+d x)} \left (6 a \left (a^2+b^2\right )-2 a b^2 \cos (2 (c+d x))+b \left (a^2-b^2\right ) \sin (2 (c+d x))\right )-\frac {3 \left (a^2+b^2\right )^2 \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right ) \left (b \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right ) \sin \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )+\sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )} \left (2 a \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )-b \sin \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right )\right )}{\left (a \sqrt {1+\frac {b^2}{a^2}} \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right )^{3/2} \sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )}}}{5 b d} \] Input:

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^(5/2),x]
 

Output:

(Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]]*(6*a*(a^2 + b^2) - 2*a*b^2*Cos[2*(c 
 + d*x)] + b*(a^2 - b^2)*Sin[2*(c + d*x)]) - (3*(a^2 + b^2)^2*Cos[c + d*x 
- ArcTan[b/a]]*(b*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[c + d*x - Arc 
Tan[b/a]]^2]*Sin[c + d*x - ArcTan[b/a]] + Sqrt[Sin[c + d*x - ArcTan[b/a]]^ 
2]*(2*a*Cos[c + d*x - ArcTan[b/a]] - b*Sin[c + d*x - ArcTan[b/a]])))/((a*S 
qrt[1 + b^2/a^2]*Cos[c + d*x - ArcTan[b/a]])^(3/2)*Sqrt[Sin[c + d*x - ArcT 
an[b/a]]^2]))/(5*b*d)
 

Rubi [A] (verified)

Time = 0.41 (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, 3119}

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.

\(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3552

\(\displaystyle \frac {3}{5} \left (a^2+b^2\right ) \int \sqrt {a \cos (c+d x)+b \sin (c+d x)}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{5} \left (a^2+b^2\right ) \int \sqrt {a \cos (c+d x)+b \sin (c+d x)}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3557

\(\displaystyle \frac {3 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)} \int \sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )}dx}{5 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)} \int \sqrt {\sin \left (c+d x-\tan ^{-1}(a,b)+\frac {\pi }{2}\right )}dx}{5 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {6 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{5 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}\)

Input:

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(5/2),x]
 

Output:

(-2*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(3 
/2))/(5*d) + (6*(a^2 + b^2)*EllipticE[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[ 
a*Cos[c + d*x] + b*Sin[c + d*x]])/(5*d*Sqrt[(a*Cos[c + d*x] + b*Sin[c + d* 
x])/Sqrt[a^2 + b^2]])
 

Defintions of rubi rules used

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(249\) vs. \(2(121)=242\).

Time = 0.37 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.91

method	result	size
default	\(-\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \left (6 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{4}+2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{2}\right )}{5 \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\)	\(250\)

Input:

int((cos(d*x+c)*a+b*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/5*(a^2+b^2)^(3/2)*(6*(-sin(d*x+c-arctan(-a,b))+1)^(1/2)*(2*sin(d*x+c-ar 
ctan(-a,b))+2)^(1/2)*sin(d*x+c-arctan(-a,b))^(1/2)*EllipticE((-sin(d*x+c-a 
rctan(-a,b))+1)^(1/2),1/2*2^(1/2))-3*(-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))^ 
4+2*sin(d*x+c-arctan(-a,b))^2)/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 = 244, normalized size of antiderivative = 1.86 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (3 \, {\left (-i \, a^{2} - i \, b^{2}\right )} \sqrt {\frac {1}{2} \, a - \frac {1}{2} i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, a^{2} + i \, b^{2}\right )} \sqrt {\frac {1}{2} \, a + \frac {1}{2} i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (2 \, a b \cos \left (d x + c\right )^{2} - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a b\right )} \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}\right )}}{5 \, d} \] Input:

integrate((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-2/5*(3*(-I*a^2 - I*b^2)*sqrt(1/2*a - 1/2*I*b)*weierstrassZeta(-4*(a^2 + 2 
*I*a*b - b^2)/(a^2 + b^2), 0, weierstrassPInverse(-4*(a^2 + 2*I*a*b - b^2) 
/(a^2 + b^2), 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(I*a^2 + I*b^2)*sqrt( 
1/2*a + 1/2*I*b)*weierstrassZeta(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 0, 
weierstrassPInverse(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) 
- I*sin(d*x + c))) + (2*a*b*cos(d*x + c)^2 - (a^2 - b^2)*cos(d*x + c)*sin( 
d*x + c) - a*b)*sqrt(a*cos(d*x + c) + b*sin(d*x + c)))/d
 

Sympy [F(-1)]

Timed out. \[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((a*cos(d*x+c)+b*sin(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=\int { {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + b*sin(d*x + c))^(5/2), x)
 

Giac [F]

\[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=\int { {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + b*sin(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx=\int {\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int((a*cos(c + d*x) + b*sin(c + d*x))^(5/2),x)
 

Output:

int((a*cos(c + d*x) + b*sin(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx =\text {Too large to display} \] Input:

int((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(8*sqrt(cos(c + d*x)*a + sin(c + d*x)*b)*cos(c + d*x)**2*a**5 + 6*sqrt(cos 
(c + d*x)*a + sin(c + d*x)*b)*cos(c + d*x)**2*a**3*b**2 + 6*sqrt(cos(c + d 
*x)*a + sin(c + d*x)*b)*cos(c + d*x)*sin(c + d*x)*a**4*b + 2*sqrt(cos(c + 
d*x)*a + sin(c + d*x)*b)*cos(c + d*x)*sin(c + d*x)*a**2*b**3 + 10*sqrt(cos 
(c + d*x)*a + sin(c + d*x)*b)*sin(c + d*x)**2*a**5 + 18*sqrt(cos(c + d*x)* 
a + sin(c + d*x)*b)*sin(c + d*x)**2*a**3*b**2 + 6*sqrt(cos(c + d*x)*a + si 
n(c + d*x)*b)*sin(c + d*x)**2*a*b**4 + 10*int((sqrt(cos(c + d*x)*a + sin(c 
 + d*x)*b)*sin(c + d*x)**3)/(2*cos(c + d*x)*a**3 + cos(c + d*x)*a*b**2 + 2 
*sin(c + d*x)*a**2*b + sin(c + d*x)*b**3),x)*a**8*d + 35*int((sqrt(cos(c + 
 d*x)*a + sin(c + d*x)*b)*sin(c + d*x)**3)/(2*cos(c + d*x)*a**3 + cos(c + 
d*x)*a*b**2 + 2*sin(c + d*x)*a**2*b + sin(c + d*x)*b**3),x)*a**6*b**2*d + 
45*int((sqrt(cos(c + d*x)*a + sin(c + d*x)*b)*sin(c + d*x)**3)/(2*cos(c + 
d*x)*a**3 + cos(c + d*x)*a*b**2 + 2*sin(c + d*x)*a**2*b + sin(c + d*x)*b** 
3),x)*a**4*b**4*d + 25*int((sqrt(cos(c + d*x)*a + sin(c + d*x)*b)*sin(c + 
d*x)**3)/(2*cos(c + d*x)*a**3 + cos(c + d*x)*a*b**2 + 2*sin(c + d*x)*a**2* 
b + sin(c + d*x)*b**3),x)*a**2*b**6*d + 5*int((sqrt(cos(c + d*x)*a + sin(c 
 + d*x)*b)*sin(c + d*x)**3)/(2*cos(c + d*x)*a**3 + cos(c + d*x)*a*b**2 + 2 
*sin(c + d*x)*a**2*b + sin(c + d*x)*b**3),x)*b**8*d)/(5*b*d*(2*a**2 + b**2 
))