\(\int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [283]

Optimal result

Integrand size = 25, antiderivative size = 170 \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f}-\frac {2 (a+2 b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 b^2 f \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}+\frac {(a+b) (2 a+3 b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{3 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

-1/3*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f-2/3*(a+2*b)*Ellipt 
icE(sin(f*x+e),(-b/a)^(1/2))*(a+b*sin(f*x+e)^2)^(1/2)/b^2/f/((a+b*sin(f*x+ 
e)^2)/a)^(1/2)+1/3*(a+b)*(2*a+3*b)*InverseJacobiAM(f*x+e,(-b/a)^(1/2))*((a 
+b*sin(f*x+e)^2)/a)^(1/2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-4 \sqrt {2} a (a+2 b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+2 \sqrt {2} \left (2 a^2+5 a b+3 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+b (-2 a-b+b \cos (2 (e+f x))) \sin (2 (e+f x))}{6 \sqrt {2} b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:

Integrate[Cos[e + f*x]^4/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(-4*Sqrt[2]*a*(a + 2*b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e 
 + f*x, -(b/a)] + 2*Sqrt[2]*(2*a^2 + 5*a*b + 3*b^2)*Sqrt[(2*a + b - b*Cos[ 
2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + b*(-2*a - b + b*Cos[2*(e + f 
*x)])*Sin[2*(e + f*x)])/(6*Sqrt[2]*b^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)] 
])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3671, 318, 399, 323, 321, 330, 327}

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.

\(\displaystyle \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (e+f x)^4}{\sqrt {a+b \sin (e+f x)^2}}dx\)

\(\Big \downarrow \) 3671

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 399

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

\(\Big \downarrow \) 323

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

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {(a+b) (2 a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a+2 b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\)

\(\Big \downarrow \) 330

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {(a+b) (2 a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a+2 b) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {(a+b) (2 a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a+2 b) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\)

Input:

Int[Cos[e + f*x]^4/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*(-1/3*(Sin[e + f*x]*Sqrt[1 - Sin[e + f* 
x]^2]*Sqrt[a + b*Sin[e + f*x]^2])/b + ((-2*(a + 2*b)*EllipticE[ArcSin[Sin[ 
e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*Sqrt[1 + (b*Sin[e + f*x] 
^2)/a]) + ((a + b)*(2*a + 3*b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqr 
t[1 + (b*Sin[e + f*x]^2)/a])/(b*Sqrt[a + b*Sin[e + f*x]^2]))/(3*b)))/f
 

Defintions of rubi rules used

rule 318

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

rule 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 323

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2]   Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + ( 
d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 330

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]   Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^ 
2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ[a, 
0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 3042

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

rule 3671

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*(Sqrt[ 
Cos[e + f*x]^2]/(f*Cos[e + f*x]))   Subst[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/2] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.86

method	result	size
default	\(\frac {b^{2} \sin \left (f x +e \right )^{5}+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+5 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b +3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b +a b \sin \left (f x +e \right )^{3}-b^{2} \sin \left (f x +e \right )^{3}-a b \sin \left (f x +e \right )}{3 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(316\)

Input:

int(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*(b^2*sin(f*x+e)^5+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)* 
EllipticF(sin(f*x+e),(-b/a)^(1/2))*a^2+5*a*(cos(f*x+e)^2)^(1/2)*((a+b*sin( 
f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*b+3*(cos(f*x+e)^2)^( 
1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*b^2-2 
*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(- 
b/a)^(1/2))*a^2-4*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*Ellipt 
icE(sin(f*x+e),(-b/a)^(1/2))*a*b+a*b*sin(f*x+e)^3-b^2*sin(f*x+e)^3-a*b*sin 
(f*x+e))/b^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [F]

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

integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)^4/(b*cos(f*x + e)^2 
 - a - b), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**4/(a+b*sin(f*x+e)**2)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(cos(e + f*x)^4/(a + b*sin(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(cos(e + f*x)^4/(a + b*sin(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int(cos(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x)
 

Output:

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