\(\int \frac {\sin ^4(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [113]

Optimal result

Integrand size = 25, antiderivative size = 175 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {3 (a+b) (a+5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 a^{7/2} f}-\frac {5 (a+b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f \sqrt {a+b+b \tan ^2(e+f x)}}-\frac {b (13 a+15 b) \tan (e+f x)}{8 a^3 f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:

3/8*(a+b)*(a+5*b)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^ 
(7/2)/f-5/8*(a+b)*cos(f*x+e)*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1 
/4*cos(f*x+e)^3*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^(1/2)-1/8*b*(13*a+15*b 
)*tan(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 2.49 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.31 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^3(e+f x) \left (24 \left (a^2+6 a b+5 b^2\right ) \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) (a+2 b+a \cos (2 (e+f x)))-2 \sqrt {2} \sqrt {a} \sqrt {a+b} \sqrt {\frac {a+2 b+a \cos (2 (e+f x))}{a+b}} \left (7 a^2+62 a b+60 b^2+2 a (3 a+5 b) \cos (2 (e+f x))-a^2 \cos (4 (e+f x))\right ) \sin (e+f x)\right )}{256 a^{7/2} \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right )^{3/2} \sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}} \] Input:

Integrate[Sin[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*(24*(a^2 + 6*a*b + 5*b^2)*A 
rcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*(a + 2*b + a*Cos[2*(e + f*x)]) - 
 2*Sqrt[2]*Sqrt[a]*Sqrt[a + b]*Sqrt[(a + 2*b + a*Cos[2*(e + f*x)])/(a + b) 
]*(7*a^2 + 62*a*b + 60*b^2 + 2*a*(3*a + 5*b)*Cos[2*(e + f*x)] - a^2*Cos[4* 
(e + f*x)])*Sin[e + f*x]))/(256*a^(7/2)*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^ 
2)^(3/2)*Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4620, 372, 402, 27, 402, 27, 291, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {5 (a+b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\int \frac {(a+b) \left (-10 b \tan ^2(e+f x)+3 a+5 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}}{4 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {5 (a+b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \int \frac {-10 b \tan ^2(e+f x)+3 a+5 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{2 a}}{4 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {5 (a+b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \left (\frac {\int \frac {3 (a+b) (a+5 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a}}{4 a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {5 (a+b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \left (\frac {3 (a+5 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a}}{4 a}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {5 (a+b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \left (\frac {3 (a+5 b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a}}{4 a}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (\tan ^2(e+f x)+1\right )^2 \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {\frac {5 (a+b) \tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \left (\frac {3 (a+5 b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b (13 a+15 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 a}}{4 a}}{f}\)

Input:

Int[Sin[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

(Tan[e + f*x]/(4*a*(1 + Tan[e + f*x]^2)^2*Sqrt[a + b + b*Tan[e + f*x]^2]) 
- ((5*(a + b)*Tan[e + f*x])/(2*a*(1 + Tan[e + f*x]^2)*Sqrt[a + b + b*Tan[e 
 + f*x]^2]) - ((a + b)*((3*(a + 5*b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a 
+ b + b*Tan[e + f*x]^2]])/a^(3/2) - (b*(13*a + 15*b)*Tan[e + f*x])/(a*(a + 
 b)*Sqrt[a + b + b*Tan[e + f*x]^2])))/(2*a))/(4*a))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 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 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

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

rule 4620

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, 
 x] && IntegerQ[m/2] && IntegerQ[n/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(681\) vs. \(2(155)=310\).

Time = 10.30 (sec) , antiderivative size = 682, normalized size of antiderivative = 3.90

method	result	size
default	\(\frac {\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (3+3 \sec \left (f x +e \right )\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a^{2} b \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (18+18 \sec \left (f x +e \right )+3 \sec \left (f x +e \right )^{2}+3 \sec \left (f x +e \right )^{3}\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a \,b^{2} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (15+15 \sec \left (f x +e \right )+18 \sec \left (f x +e \right )^{2}+18 \sec \left (f x +e \right )^{3}\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{3} \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) \left (15 \sec \left (f x +e \right )^{2}+15 \sec \left (f x +e \right )^{3}\right )+\sin \left (f x +e \right ) \cos \left (f x +e \right ) \left (2 \cos \left (f x +e \right )^{2}-5\right ) a^{3} \sqrt {-a}+\left (-3 \cos \left (f x +e \right )^{2}-18\right ) \sqrt {-a}\, a^{2} b \tan \left (f x +e \right )+\sqrt {-a}\, a \,b^{2} \left (-20 \tan \left (f x +e \right )-13 \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}\right )-15 \sqrt {-a}\, b^{3} \tan \left (f x +e \right ) \sec \left (f x +e \right )^{2}}{8 f \,a^{3} \sqrt {-a}\, \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(682\)

Input:

int(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/8/f/a^3/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(3/2)*(((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)*a^3*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2) 
^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2) 
-4*sin(f*x+e)*a)*(3+3*sec(f*x+e))+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1 
/2)*a^2*b*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos( 
f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+ 
e)*a)*(18+18*sec(f*x+e)+3*sec(f*x+e)^2+3*sec(f*x+e)^3)+((b+a*cos(f*x+e)^2) 
/(1+cos(f*x+e))^2)^(1/2)*a*b^2*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos( 
f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) 
)^2)^(1/2)-4*sin(f*x+e)*a)*(15+15*sec(f*x+e)+18*sec(f*x+e)^2+18*sec(f*x+e) 
^3)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^3*ln(4*(-a)^(1/2)*((b+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f* 
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(15*sec(f*x+e)^2+15*sec(f* 
x+e)^3)+sin(f*x+e)*cos(f*x+e)*(2*cos(f*x+e)^2-5)*a^3*(-a)^(1/2)+(-3*cos(f* 
x+e)^2-18)*(-a)^(1/2)*a^2*b*tan(f*x+e)+(-a)^(1/2)*a*b^2*(-20*tan(f*x+e)-13 
*tan(f*x+e)*sec(f*x+e)^2)-15*(-a)^(1/2)*b^3*tan(f*x+e)*sec(f*x+e)^2)
 

Fricas [A] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 703, normalized size of antiderivative = 4.02 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/64*(3*(a^2*b + 6*a*b^2 + 5*b^3 + (a^3 + 6*a^2*b + 5*a*b^2)*cos(f*x + e 
)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^ 
6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70 
*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x 
 + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*( 
5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^ 
3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin( 
f*x + e)) - 8*(2*a^3*cos(f*x + e)^5 - 5*(a^3 + a^2*b)*cos(f*x + e)^3 - (13 
*a^2*b + 15*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^ 
2)*sin(f*x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f), -1/32*(3*(a^2*b + 6*a*b 
^2 + 5*b^3 + (a^3 + 6*a^2*b + 5*a*b^2)*cos(f*x + e)^2)*sqrt(a)*arctan(1/4* 
(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2) 
*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3 
*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x 
+ e))) - 4*(2*a^3*cos(f*x + e)^5 - 5*(a^3 + a^2*b)*cos(f*x + e)^3 - (13*a^ 
2*b + 15*a*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)* 
sin(f*x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f)]
 

Sympy [F]

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

integrate(sin(f*x+e)**4/(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Integral(sin(e + f*x)**4/(a + b*sec(e + f*x)**2)**(3/2), x)
 

Maxima [F]

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

integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(sin(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Giac [F]

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

integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(sin(f*x + e)^4/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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