\(\int \frac {1}{(-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^{5/2}} \, dx\) [369]

Optimal result

Integrand size = 34, antiderivative size = 232 \[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=-\frac {3 \arctan \left (\frac {\sqrt [4]{b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {2} \sqrt {-\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{16 \sqrt {2} \left (b^2+c^2\right )^{5/4} e}+\frac {c \cos (d+e x)-b \sin (d+e x)}{4 \sqrt {b^2+c^2} e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 (c \cos (d+e x)-b \sin (d+e x))}{16 \left (b^2+c^2\right ) e \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \] Output:

-3/32*arctan(1/2*(b^2+c^2)^(1/4)*sin(d+e*x-arctan(c,b))*2^(1/2)/(-(b^2+c^2 
)^(1/2)+(b^2+c^2)^(1/2)*cos(d+e*x-arctan(c,b)))^(1/2))*2^(1/2)/(b^2+c^2)^( 
5/4)/e+1/4*(c*cos(e*x+d)-b*sin(e*x+d))/(b^2+c^2)^(1/2)/e/(-(b^2+c^2)^(1/2) 
+b*cos(e*x+d)+c*sin(e*x+d))^(5/2)-3/16*(c*cos(e*x+d)-b*sin(e*x+d))/(b^2+c^ 
2)/e/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2)
 

Mathematica [F(-1)]

Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {\$Aborted} \] Input:

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

Output:

$Aborted
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3595, 3042, 3595, 3042, 3594, 3042, 3128, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3595

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \int \frac {1}{\left (b \cos (d+e x)+c \sin (d+e x)-\sqrt {b^2+c^2}\right )^{3/2}}dx}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \int \frac {1}{\left (b \cos (d+e x)+c \sin (d+e x)-\sqrt {b^2+c^2}\right )^{3/2}}dx}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 3595

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \left (\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {b \cos (d+e x)+c \sin (d+e x)-\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}\right )}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \left (\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {b \cos (d+e x)+c \sin (d+e x)-\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}\right )}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 3594

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \left (\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )-\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}\right )}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \left (\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)+\frac {\pi }{2}\right )-\sqrt {b^2+c^2}}}dx}{4 \sqrt {b^2+c^2}}\right )}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \left (\frac {\int \frac {1}{-\frac {\left (b^2+c^2\right ) \sin ^2\left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )-\sqrt {b^2+c^2}}-2 \sqrt {b^2+c^2}}d\left (-\frac {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )-\sqrt {b^2+c^2}}}\right )}{2 e \sqrt {b^2+c^2}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\right )}{8 \sqrt {b^2+c^2}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {c \cos (d+e x)-b \sin (d+e x)}{4 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}}-\frac {3 \left (\frac {\arctan \left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )-\sqrt {b^2+c^2}}}\right )}{2 \sqrt {2} e \left (b^2+c^2\right )^{3/4}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{2 e \sqrt {b^2+c^2} \left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}}\right )}{8 \sqrt {b^2+c^2}}\)

Input:

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

Output:

(c*Cos[d + e*x] - b*Sin[d + e*x])/(4*Sqrt[b^2 + c^2]*e*(-Sqrt[b^2 + c^2] + 
 b*Cos[d + e*x] + c*Sin[d + e*x])^(5/2)) - (3*(ArcTan[((b^2 + c^2)^(1/4)*S 
in[d + e*x - ArcTan[b, c]])/(Sqrt[2]*Sqrt[-Sqrt[b^2 + c^2] + Sqrt[b^2 + c^ 
2]*Cos[d + e*x - ArcTan[b, c]]])]/(2*Sqrt[2]*(b^2 + c^2)^(3/4)*e) + (c*Cos 
[d + e*x] - b*Sin[d + e*x])/(2*Sqrt[b^2 + c^2]*e*(-Sqrt[b^2 + c^2] + b*Cos 
[d + e*x] + c*Sin[d + e*x])^(3/2))))/(8*Sqrt[b^2 + c^2])
 

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

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

rule 3594

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*( 
x_)]], x_Symbol] :> Int[1/Sqrt[a + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, 
c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
 

rule 3595

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_), x_Symbol] :> Simp[(c*Cos[d + e*x] - b*Sin[d + e*x])*((a + b*Cos[d + e 
*x] + c*Sin[d + e*x])^n/(a*e*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) 
 Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, 
c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.57

method	result	size
default	\(\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) \left (b^{2}+c^{2}\right )-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) b^{2}-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) c^{2}-2 \sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )-\sqrt {b^{2}+c^{2}}}\, \left (b^{2}+c^{2}\right )^{\frac {3}{4}}\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )+1\right )}}{4 \left (b^{2}+c^{2}\right )^{\frac {5}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )-b^{2}-c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e}\)	\(364\)

Input:

int(1/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/4*(sin(e*x+d-arctan(-b,c))*2^(1/2)*arctan(1/2*(-(b^2+c^2)^(1/2)*sin(e*x+ 
d-arctan(-b,c))-(b^2+c^2)^(1/2))^(1/2)*2^(1/2)/(b^2+c^2)^(1/4))*(b^2+c^2)- 
2^(1/2)*arctan(1/2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-(b^2+c^2)^(1/ 
2))^(1/2)*2^(1/2)/(b^2+c^2)^(1/4))*b^2-2^(1/2)*arctan(1/2*(-(b^2+c^2)^(1/2 
)*sin(e*x+d-arctan(-b,c))-(b^2+c^2)^(1/2))^(1/2)*2^(1/2)/(b^2+c^2)^(1/4))* 
c^2-2*(-(b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))-(b^2+c^2)^(1/2))^(1/2)*(b^ 
2+c^2)^(3/4))*(-(b^2+c^2)^(1/2)*(sin(e*x+d-arctan(-b,c))+1))^(1/2)/(b^2+c^ 
2)^(5/4)/cos(e*x+d-arctan(-b,c))/((b^2*sin(e*x+d-arctan(-b,c))+c^2*sin(e*x 
+d-arctan(-b,c))-b^2-c^2)/(b^2+c^2)^(1/2))^(1/2)/e
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 655 vs. \(2 (209) = 418\).

Time = 0.19 (sec) , antiderivative size = 655, normalized size of antiderivative = 2.82 \[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(5/2),x, algorith 
m="fricas")
 

Output:

1/16*(3*sqrt(1/2)*(5*b^4*c*cos(e*x + d) + (5*b^4*c - 10*b^2*c^3 + c^5)*cos 
(e*x + d)^5 - 10*(b^4*c - b^2*c^3)*cos(e*x + d)^3 - (b^5 + (b^5 - 10*b^3*c 
^2 + 5*b*c^4)*cos(e*x + d)^4 - 2*(b^5 - 5*b^3*c^2)*cos(e*x + d)^2)*sin(e*x 
 + d))*arctan(-sqrt(1/2)*(b*cos(e*x + d) + c*sin(e*x + d) + sqrt(b^2 + c^2 
))*sqrt(b*cos(e*x + d) + c*sin(e*x + d) - sqrt(b^2 + c^2))/((b^2 + c^2)^(1 
/4)*(c*cos(e*x + d) - b*sin(e*x + d))))/(b^2 + c^2)^(1/4) + (3*(b^4 - 6*b^ 
2*c^2 + c^4)*cos(e*x + d)^4 - 7*b^4 - 26*b^2*c^2 - 16*c^4 - 6*(2*b^4 - 3*b 
^2*c^2 - c^4)*cos(e*x + d)^2 + 12*((b^3*c - b*c^3)*cos(e*x + d)^3 - (2*b^3 
*c + b*c^3)*cos(e*x + d))*sin(e*x + d) + 2*((b^3 - 3*b*c^2)*cos(e*x + d)^3 
 - 3*(3*b^3 + 2*b*c^2)*cos(e*x + d) - (9*b^2*c + 8*c^3 - (3*b^2*c - c^3)*c 
os(e*x + d)^2)*sin(e*x + d))*sqrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin( 
e*x + d) - sqrt(b^2 + c^2)))/((5*b^6*c - 5*b^4*c^3 - 9*b^2*c^5 + c^7)*e*co 
s(e*x + d)^5 - 10*(b^6*c - b^2*c^5)*e*cos(e*x + d)^3 + 5*(b^6*c + b^4*c^3) 
*e*cos(e*x + d) - ((b^7 - 9*b^5*c^2 - 5*b^3*c^4 + 5*b*c^6)*e*cos(e*x + d)^ 
4 - 2*(b^7 - 4*b^5*c^2 - 5*b^3*c^4)*e*cos(e*x + d)^2 + (b^7 + b^5*c^2)*e)* 
sin(e*x + d))
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \cos {\left (d + e x \right )} + c \sin {\left (d + e x \right )} - \sqrt {b^{2} + c^{2}}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(-(b**2+c**2)**(1/2)+b*cos(e*x+d)+c*sin(e*x+d))**(5/2),x)
 

Output:

Integral((b*cos(d + e*x) + c*sin(d + e*x) - sqrt(b**2 + c**2))**(-5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(5/2),x, algorith 
m="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(5/2),x, algorith 
m="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,\cos \left (d+e\,x\right )+c\,\sin \left (d+e\,x\right )-\sqrt {b^2+c^2}\right )}^{5/2}} \,d x \] Input:

int(1/(b*cos(d + e*x) + c*sin(d + e*x) - (b^2 + c^2)^(1/2))^(5/2),x)
 

Output:

int(1/(b*cos(d + e*x) + c*sin(d + e*x) - (b^2 + c^2)^(1/2))^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (-\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{5/2}} \, dx=\int \frac {1}{\left (-\sqrt {b^{2}+c^{2}}+b \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right )^{\frac {5}{2}}}d x \] Input:

int(1/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(5/2),x)
 

Output:

int(1/(-(b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(5/2),x)