3.2.58 \(\int \frac {\sin ^4(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [158]

3.2.58.1 Optimal result
3.2.58.2 Mathematica [A] (verified)
3.2.58.3 Rubi [A] (verified)
3.2.58.4 Maple [A] (verified)
3.2.58.5 Fricas [F]
3.2.58.6 Sympy [F(-1)]
3.2.58.7 Maxima [F]
3.2.58.8 Giac [F]
3.2.58.9 Mupad [F(-1)]

3.2.58.1 Optimal result

Integrand size = 25, antiderivative size = 202 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a \cos (e+f x) \sin (e+f x)}{b (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(2 a+b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{b^2 (a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {2 a \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

output
a*cos(f*x+e)*sin(f*x+e)/b/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)+(2*a+b)*Ellipti 
cE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e 
)^2)^(1/2)/b^2/(a+b)/f/(1+b*sin(f*x+e)^2/a)^(1/2)-2*a*EllipticF(sin(f*x+e) 
,(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/ 
b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
 
3.2.58.2 Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a \left (2 (2 a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-4 (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\sqrt {2} b \sin (2 (e+f x))\right )}{2 b^2 (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

input
Integrate[Sin[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(3/2),x]
 
output
(a*(2*(2*a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, 
-(b/a)] - 4*(a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f 
*x, -(b/a)] + Sqrt[2]*b*Sin[2*(e + f*x)]))/(2*b^2*(a + b)*f*Sqrt[2*a + b - 
 b*Cos[2*(e + f*x)]])
 
3.2.58.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3667, 372, 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.

\(\displaystyle \int \frac {\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^4}{\left (a+b \sin (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3667

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 399

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

\(\Big \downarrow \) 323

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

\(\Big \downarrow \) 321

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

\(\Big \downarrow \) 330

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

\(\Big \downarrow \) 327

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

input
Int[Sin[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(3/2),x]
 
output
(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*((a*Sin[e + f*x]*Sqrt[1 - Sin[e + f*x]^ 
2])/(b*(a + b)*Sqrt[a + b*Sin[e + f*x]^2]) - (-(((2*a + b)*EllipticE[ArcSi 
n[Sin[e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*Sqrt[1 + (b*Sin[e 
+ f*x]^2)/a])) + (2*a*(a + b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt 
[1 + (b*Sin[e + f*x]^2)/a])/(b*Sqrt[a + b*Sin[e + f*x]^2]))/(b*(a + b))))/ 
f
 

3.2.58.3.1 Defintions of rubi rules used

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 372
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

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 3667
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff^(m + 1 
)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x]))   Subst[Int[x^m*((a + b*ff^2*x^2) 
^p/Sqrt[1 - ff^2*x^2]), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, 
p}, x] && IntegerQ[m/2] &&  !IntegerQ[p]
 
3.2.58.4 Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.19

method result size
default \(-\frac {a \left (2 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b +\left (\sin ^{3}\left (f x +e \right )\right ) b -b \sin \left (f x +e \right )\right )}{b^{2} \left (a +b \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) \(241\)

input
int(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 
output
-a*(2*a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f* 
x+e),(-1/a*b)^(1/2))+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*E 
llipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-2*a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f 
*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))-(cos(f*x+e)^2)^(1/2 
)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b+sin( 
f*x+e)^3*b-b*sin(f*x+e))/b^2/(a+b)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 
3.2.58.5 Fricas [F]

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

input
integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 
output
integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b*cos(f*x + e)^2 + 
a + b)/(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + 
b^2), x)
 
3.2.58.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

input
integrate(sin(f*x+e)**4/(a+b*sin(f*x+e)**2)**(3/2),x)
 
output
Timed out
 
3.2.58.7 Maxima [F]

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

input
integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 
output
integrate(sin(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(3/2), x)
 
3.2.58.8 Giac [F]

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

input
integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 
output
sage0*x
 
3.2.58.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

input
int(sin(e + f*x)^4/(a + b*sin(e + f*x)^2)^(3/2),x)
 
output
int(sin(e + f*x)^4/(a + b*sin(e + f*x)^2)^(3/2), x)