\(\int \frac {1}{(\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx\) [362]

Optimal result

Integrand size = 32, antiderivative size = 160 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{2 \sqrt {2} \left (b^2+c^2\right )^{3/4} e}-\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \sqrt {b^2+c^2} e \left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \] Output:

1/4*arctanh(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)^(3/ 
4)/e-1/2*(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))^(3/2)
 

Mathematica [F(-1)]

Timed out. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\text {\$Aborted} \] Input:

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

Output:

$Aborted
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 3595, 3042, 3594, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3595

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

ArcTanh[((b^2 + c^2)^(1/4)*Sin[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))
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(137)=274\).

Time = 0.73 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.19

method	result	size
default	\(-\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) \left (b^{2}+c^{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) b^{2}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right ) c^{2}+2 \sqrt {-\sqrt {b^{2}+c^{2}}\, \sin \left (e x +d -\arctan \left (-b , c\right )\right )+\sqrt {b^{2}+c^{2}}}\, \left (b^{2}+c^{2}\right )^{\frac {3}{4}}\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )-1\right )}}{4 \left (b^{2}+c^{2}\right )^{\frac {7}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+b^{2}+c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e}\)	\(350\)

Input:

int(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(3/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-1/4/(b^2+c^2)^(7/4)*(sin(e*x+d-arctan(-b,c))*2^(1/2)*arctanh(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)*arctanh(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)*arctanh( 
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)/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. 622 vs. \(2 (143) = 286\).

Time = 0.25 (sec) , antiderivative size = 622, normalized size of antiderivative = 3.89 \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

integrate(1/((b**2+c**2)**(1/2)+b*cos(e*x+d)+c*sin(e*x+d))**(3/2),x)
 

Output:

Integral((b*cos(d + e*x) + c*sin(d + e*x) + sqrt(b**2 + c**2))**(-3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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