\(\int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [279]

Optimal result

Integrand size = 25, antiderivative size = 79 \[ \int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 b^{3/2} f}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b f} \] Output:

1/2*(a+2*b)*arctanh(b^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2))/b^(3/2)/f 
-1/2*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-\frac {(-a-2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 b^{3/2}}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b}}{f} \] Input:

Integrate[Cos[e + f*x]^3/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(-1/2*((-a - 2*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2 
]])/b^(3/2) - (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(2*b))/f
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3669, 299, 224, 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.

\(\displaystyle \int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (e+f x)^3}{\sqrt {a+b \sin (e+f x)^2}}dx\)

\(\Big \downarrow \) 3669

\(\displaystyle \frac {\int \frac {1-\sin ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {(a+2 b) \int \frac {1}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{2 b}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {(a+2 b) \int \frac {1}{1-\frac {b \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}}{2 b}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 b^{3/2}}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b}}{f}\)

Input:

Int[Cos[e + f*x]^3/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(((a + 2*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2 
*b^(3/2)) - (Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(2*b))/f
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 3042

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

rule 3669

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   S 
ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] 
/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {-\frac {\sin \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}}{2 b}+\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{\sqrt {b}}}{f}\)	\(93\)
default	\(\frac {-\frac {\sin \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}}{2 b}+\frac {a \ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{\sqrt {b}}}{f}\)	\(93\)

Input:

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

Output:

1/f*(-1/2*sin(f*x+e)/b*(a+b*sin(f*x+e)^2)^(1/2)+1/2*a/b^(3/2)*ln(b^(1/2)*s 
in(f*x+e)+(a+b*sin(f*x+e)^2)^(1/2))+ln(b^(1/2)*sin(f*x+e)+(a+b*sin(f*x+e)^ 
2)^(1/2))/b^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 176 vs. \(2 (67) = 134\).

Time = 0.16 (sec) , antiderivative size = 461, normalized size of antiderivative = 5.84 \[ \int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (a + 2 \, b\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) - 8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{16 \, b^{2} f}, -\frac {{\left (a + 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{8 \, b^{2} f}\right ] \] Input:

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

Output:

[1/16*((a + 2*b)*sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)* 
cos(f*x + e)^6 + 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 
 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 2 
4*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 
2*b^3)*cos(f*x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 
24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b) 
*sin(f*x + e)) - 8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*sin(f*x + e))/(b^2*f) 
, -1/8*((a + 2*b)*sqrt(-b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b 
^2)*cos(f*x + e)^2 + a^2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)* 
sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b 
^3)*cos(f*x + e)^2)*sin(f*x + e))) + 4*sqrt(-b*cos(f*x + e)^2 + a + b)*b*s 
in(f*x + e))/(b^2*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**3/(a+b*sin(f*x+e)**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\frac {a \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )}{b}}{2 \, f} \] Input:

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

Output:

1/2*(a*arcsinh(b*sin(f*x + e)/sqrt(a*b))/b^(3/2) + 2*arcsinh(b*sin(f*x + e 
)/sqrt(a*b))/sqrt(b) - sqrt(b*sin(f*x + e)^2 + a)*sin(f*x + e)/b)/f
 

Giac [F]

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

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

Output:

integrate(cos(f*x + e)^3/sqrt(b*sin(f*x + e)^2 + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x)**3)/(sin(e + f*x)**2*b + a), 
x)