\(\int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx\) [160]

Optimal result

Integrand size = 21, antiderivative size = 138 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{\left (a^2+b^2\right ) d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \] Output:

(-2*b*cos(d*x+c)+2*a*sin(d*x+c))/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^( 
1/2)-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)/(a^2+b^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 = 2.33 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\frac {-\frac {2 b \cos (c+d x)}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}+\frac {2 a \sin (c+d x)}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\sqrt {a \sqrt {1+\frac {b^2}{a^2}} \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )} \tan \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )+\frac {\sqrt {a \sqrt {1+\frac {b^2}{a^2}} \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right ) \tan \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )}{\sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )}}}{\left (a^2+b^2\right ) d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 138, 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, 3555, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3555

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3557

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

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]))/((a^2 + b^2)*d*Sqrt[a*Cos[c + d*x] 
+ b*Sin[c + d*x]]) - (2*EllipticE[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[a*Co 
s[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)*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 3555

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 + 1)*(a^2 + b^2))), x] + Simp[(n + 2)/((n + 1)*(a^ 
2 + b^2))   Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1] && NeQ[n, -2]
 

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 = 0.34 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.68

method	result	size
default	\(\frac {2 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c -\arctan \left (-a , b\right )\right )^{2}}{\sqrt {a^{2}+b^{2}}\, \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\)	\(232\)

Input:

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

Output:

(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)*EllipticE((-sin(d*x+c-arctan(-a,b))+1)^(1/2) 
,1/2*2^(1/2))-(-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*cos(d*x+c-arctan(-a,b))^2)/(a^2+b^2)^(1/2)/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 = 270, normalized size of antiderivative = 1.96 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left ({\left (i \, a \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {1}{2} \, a - \frac {1}{2} i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (-i \, a \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {\frac {1}{2} \, a + \frac {1}{2} i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}\right )}}{{\left (a^{3} + a b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{2} b + b^{3}\right )} d \sin \left (d x + c\right )} \] Input:

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-2*((I*a*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(1/2*a - 1/2*I*b)*weierstras 
sZeta(-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))) + (-I*a* 
cos(d*x + c) - I*b*sin(d*x + c))*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))) + sqrt(a*cos(d*x 
+ c) + b*sin(d*x + c))*(b*cos(d*x + c) - a*sin(d*x + c)))/((a^3 + a*b^2)*d 
*cos(d*x + c) + (a^2*b + b^3)*d*sin(d*x + c))
 

Sympy [F]

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

integrate(1/(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]

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

integrate(1/(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]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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