\(\int \frac {\cos ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [288]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{3/2} f}+\frac {(a+b) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {b} (a+b) \sin (e+f x)-a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a}}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{a b^{3/2} f \sqrt {a+b \sin ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 73, 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, 298, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3669

\(\displaystyle \frac {\int \frac {1-\sin ^2(e+f x)}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-(ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]]/b^(3/2)) + ( 
(a + b)*Sin[e + f*x])/(a*b*Sqrt[a + b*Sin[e + f*x]^2]))/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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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 = 0.38 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (f x +e \right )}{b \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}+\frac {\sin \left (f x +e \right )}{a \sqrt {a +b \sin \left (f x +e \right )^{2}}}}{f}\)	\(85\)
default	\(\frac {\frac {\sin \left (f x +e \right )}{b \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}+\frac {\sin \left (f x +e \right )}{a \sqrt {a +b \sin \left (f x +e \right )^{2}}}}{f}\)	\(85\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (67) = 134\).

Time = 0.23 (sec) , antiderivative size = 559, normalized size of antiderivative = 7.45 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a 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} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left (a b^{3} f \cos \left (f x + e\right )^{2} - {\left (a^{2} b^{2} + a b^{3}\right )} f\right )}}, \frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a 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} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left (a b^{3} f \cos \left (f x + e\right )^{2} - {\left (a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \] Input:

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

Output:

[1/8*((a*b*cos(f*x + e)^2 - a^2 - a*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 + 24*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)*( 
a*b + b^2)*sin(f*x + e))/(a*b^3*f*cos(f*x + e)^2 - (a^2*b^2 + a*b^3)*f), 1 
/4*((a*b*cos(f*x + e)^2 - a^2 - a*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)*(a*b + b^2)*sin(f*x + e))/(a*b^3*f*cos(f*x + e)^2 - (a^2*b 
^2 + a*b^3)*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} b}}{f} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x)**3)/(sin(e + f*x)**4*b**2 + 
2*sin(e + f*x)**2*a*b + a**2),x)