\(\int \frac {\sin ^3(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [129]

Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {a b \sec (e+f x)}{(a-b)^3 f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {(3 a+2 b) \cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b)^3 f}+\frac {\cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b)^2 f} \] Output:

-a*b*sec(f*x+e)/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)^(1/2)-1/3*(3*a+2*b)*cos(f*x 
+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/(a-b)^3/f+1/3*cos(f*x+e)^3*(a-b+b*sec(f*x+e 
)^2)^(1/2)/(a-b)^2/f
 

Mathematica [A] (verified)

Time = 3.42 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (9 a^2+46 a b+9 b^2+8 \left (a^2-b^2\right ) \cos (2 (e+f x))-(a-b)^2 \cos (4 (e+f x))\right ) \sec (e+f x)}{12 \sqrt {2} (a-b)^3 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \] Input:

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

Output:

-1/12*((9*a^2 + 46*a*b + 9*b^2 + 8*(a^2 - b^2)*Cos[2*(e + f*x)] - (a - b)^ 
2*Cos[4*(e + f*x)])*Sec[e + f*x])/(Sqrt[2]*(a - b)^3*f*Sqrt[(a + b + (a - 
b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4147, 25, 359, 245, 208}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 245

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(Cos[e + f*x]^3/(3*(a - b)*Sqrt[a - b + b*Sec[e + f*x]^2]) + ((3*a + b)*(- 
(Cos[e + f*x]/((a - b)*Sqrt[a - b + b*Sec[e + f*x]^2])) - (2*b*Sec[e + f*x 
])/((a - b)^2*Sqrt[a - b + b*Sec[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 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 245

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + 
 b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) 
   Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si 
mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
 

rule 359

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

rule 3042

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

rule 4147

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

Maple [A] (verified)

Time = 6.77 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {\left (a -b \right ) a^{4} \left (a^{3} \left (\cos \left (f x +e \right )^{3}-3 \cos \left (f x +e \right )\right )+a^{2} b \left (-3 \cos \left (f x +e \right )^{3}+6 \cos \left (f x +e \right )-9 \sec \left (f x +e \right )\right )+\left (-3 \cos \left (f x +e \right )^{4}-6\right ) a \,b^{2} \tan \left (f x +e \right )^{2} \sec \left (f x +e \right )+\left (-\cos \left (f x +e \right )^{2}-2\right ) b^{3} \sin \left (f x +e \right ) \tan \left (f x +e \right )^{3}\right )}{3 f \left (\sqrt {-b \left (a -b \right )}-a +b \right )^{4} \left (\sqrt {-b \left (a -b \right )}+a -b \right )^{4} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(177\)

Input:

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

Output:

1/3/f*(a-b)/((-b*(a-b))^(1/2)-a+b)^4/((-b*(a-b))^(1/2)+a-b)^4*a^4/(a+b*tan 
(f*x+e)^2)^(3/2)*(a^3*(cos(f*x+e)^3-3*cos(f*x+e))+a^2*b*(-3*cos(f*x+e)^3+6 
*cos(f*x+e)-9*sec(f*x+e))+(-3*cos(f*x+e)^4-6)*a*b^2*tan(f*x+e)^2*sec(f*x+e 
)+(-cos(f*x+e)^2-2)*b^3*sin(f*x+e)*tan(f*x+e)^3)
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {{\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{2} - 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}} \] Input:

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

Output:

1/3*((a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (3*a^2 - 2*a*b - b^2)*cos(f*x + 
e)^3 - 2*(3*a*b + b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos 
(f*x + e)^2)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 
 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.70 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {3 \, b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )} + \frac {3 \, b}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}}{3 \, f} \] Input:

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

Output:

-1/3*(3*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a^2 - 2*a*b + b^2) - 
((a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 6*sqrt(a - b + b/cos(f* 
x + e)^2)*b*cos(f*x + e))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 3*b^2/((a^3 - 
3*a^2*b + 3*a*b^2 - b^3)*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)) + 3* 
b/((a^2 - 2*a*b + b^2)*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1414 vs. \(2 (117) = 234\).

Time = 2.86 (sec) , antiderivative size = 1414, normalized size of antiderivative = 11.13 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/3*(3*((a^3*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 2*a^2*b^3*sgn(tan(1/2*f 
*x + 1/2*e)^2 - 1) + a*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))*tan(1/2*f*x + 
1/2*e)^2/(a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6) + ( 
a^3*b^2*sgn(tan(1/2*f*x + 1/2*e)^2 - 1) - 2*a^2*b^3*sgn(tan(1/2*f*x + 1/2* 
e)^2 - 1) + a*b^4*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))/(a^5*b - 5*a^4*b^2 + 10 
*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6))/sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2* 
a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) + 4*(3*(sqrt(a) 
*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x 
+ 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*b + 12*(sqrt(a)*tan(1/2*f* 
x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 
+ 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^4*a^(3/2) - 3*(sqrt(a)*tan(1/2*f*x + 1/ 
2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b* 
tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(a)*b - 16*(sqrt(a)*tan(1/2*f*x + 1/2*e 
)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan 
(1/2*f*x + 1/2*e)^2 + a))^3*a^2 + 14*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqr 
t(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x 
+ 1/2*e)^2 + a))^3*a*b + 8*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/ 
2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 
 + a))^3*b^2 - 24*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1 
/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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