\(\int \frac {\cos ^5(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [297]

Optimal result

Integrand size = 25, antiderivative size = 124 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{5/2} f}+\frac {(a+b)^2 \sin (e+f x)}{3 a b^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 (2 a-b) (a+b) \sin (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

arctanh(b^(1/2)*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2))/b^(5/2)/f+1/3*(a+b)^2 
*sin(f*x+e)/a/b^2/f/(a+b*sin(f*x+e)^2)^(3/2)-2/3*(2*a-b)*(a+b)*sin(f*x+e)/ 
a^2/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt {-b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{\sqrt {-b}}+\frac {2 \sqrt {2} (a+b) \left (-3 a^2+a b+b^2+(2 a-b) b \cos (2 (e+f x))\right ) \sin (e+f x)}{a^2 (2 a+b-b \cos (2 (e+f x)))^{3/2}}}{3 b^2 f} \] Input:

Integrate[Cos[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3669, 315, 25, 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.

\(\displaystyle \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (e+f x)^5}{\left (a+b \sin (e+f x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3669

\(\displaystyle \frac {\int \frac {\left (1-\sin ^2(e+f x)\right )^2}{\left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[Cos[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 315

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

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 = 0.56 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.68

method	result	size
derivativedivides	\(\frac {\sin \left (f x +e \right )}{3 a f \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} f \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {\sin \left (f x +e \right )^{3}}{3 f b \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\sin \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{f \,b^{\frac {5}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 f b \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 \sin \left (f x +e \right )}{3 f b a \sqrt {a +b \sin \left (f x +e \right )^{2}}}\)	\(208\)
default	\(\frac {\sin \left (f x +e \right )}{3 a f \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} f \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {\sin \left (f x +e \right )^{3}}{3 f b \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\sin \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \sin \left (f x +e \right )^{2}}\right )}{f \,b^{\frac {5}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 f b \left (a +b \sin \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {2 \sin \left (f x +e \right )}{3 f b a \sqrt {a +b \sin \left (f x +e \right )^{2}}}\)	\(208\)

Input:

int(cos(f*x+e)^5/(a+b*sin(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*sin(f*x+e)/a/f/(a+b*sin(f*x+e)^2)^(3/2)+2/3*sin(f*x+e)/a^2/f/(a+b*sin( 
f*x+e)^2)^(1/2)-1/3/f*sin(f*x+e)^3/b/(a+b*sin(f*x+e)^2)^(3/2)-1/f/b^2*sin( 
f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)+1/f/b^(5/2)*ln(b^(1/2)*sin(f*x+e)+(a+b*sin 
(f*x+e)^2)^(1/2))+2/3/f/b*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2)-2/3/f/b/a*si 
n(f*x+e)/(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. 345 vs. \(2 (110) = 220\).

Time = 0.86 (sec) , antiderivative size = 799, normalized size of antiderivative = 6.44 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(f*x+e)^5/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/24*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^ 
2*b^2)*cos(f*x + e)^2)*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)*s 
qrt(b)*sin(f*x + e)) - 8*(3*a^3*b + 4*a^2*b^2 - a*b^3 - 2*b^4 - 2*(2*a^2*b 
^2 + a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x 
+ e))/(a^2*b^5*f*cos(f*x + e)^4 - 2*(a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + 
 (a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f), -1/12*(3*(a^2*b^2*cos(f*x + e)^4 + a^ 
4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x + e)^2)*sqrt(-b)*arcta 
n(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*(3*a^3*b + 4*a^2*b^2 - a*b^3 - 2*b^4 - 2*(2*a^2*b^2 + a*b^3 - b^4)* 
cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(a^2*b^5*f*c 
os(f*x + e)^4 - 2*(a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^4*b^3 + 2*a^3* 
b^4 + a^2*b^5)*f)]
 

Sympy [F(-1)]

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

integrate(cos(f*x+e)**5/(a+b*sin(f*x+e)**2)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} \sin \left (f x + e\right ) - \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {\sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} + \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}} - \frac {2 \, \sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a b}}{3 \, f} \] Input:

integrate(cos(f*x+e)^5/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")
 

Output:

-1/3*((3*sin(f*x + e)^2/((b*sin(f*x + e)^2 + a)^(3/2)*b) + 2*a/((b*sin(f*x 
 + e)^2 + a)^(3/2)*b^2))*sin(f*x + e) - 3*arcsinh(b*sin(f*x + e)/sqrt(a*b) 
)/b^(5/2) - 2*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^2) - sin(f*x + e) 
/((b*sin(f*x + e)^2 + a)^(3/2)*a) + sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + 
a)*b^2) - 2*sin(f*x + e)/((b*sin(f*x + e)^2 + a)^(3/2)*b) + 2*sin(f*x + e) 
/(sqrt(b*sin(f*x + e)^2 + a)*a*b))/f
 

Giac [F]

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

integrate(cos(f*x+e)^5/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate(cos(f*x + e)^5/(b*sin(f*x + e)^2 + a)^(5/2), x)
 

Mupad [F(-1)]

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

int(cos(e + f*x)^5/(a + b*sin(e + f*x)^2)^(5/2),x)
 

Output:

int(cos(e + f*x)^5/(a + b*sin(e + f*x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \cos \left (f x +e \right )^{5}}{\sin \left (f x +e \right )^{6} b^{3}+3 \sin \left (f x +e \right )^{4} a \,b^{2}+3 \sin \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

int(cos(f*x+e)^5/(a+b*sin(f*x+e)^2)^(5/2),x)
 

Output:

int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x)**5)/(sin(e + f*x)**6*b**3 + 
3*sin(e + f*x)**4*a*b**2 + 3*sin(e + f*x)**2*a**2*b + a**3),x)