\(\int \frac {\sin ^3(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [141]

Optimal result

Integrand size = 25, antiderivative size = 168 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b (a+b) \sec (e+f x)}{3 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b (a+b) \sec (e+f x)}{3 (a-b)^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \]

[Out]

-(a+b)*cos(f*x+e)/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)^(3/2)+1/3*cos(f*x+e)^3/(a-b)/f/(a-b+b*sec(f*x+e)^2)^(3/2)-4/3
*b*(a+b)*sec(f*x+e)/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)^(3/2)-8/3*b*(a+b)*sec(f*x+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)
^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3745, 464, 277, 198, 197} \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {8 b (a+b) \sec (e+f x)}{3 f (a-b)^4 \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {4 b (a+b) \sec (e+f x)}{3 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}-\frac {(a+b) \cos (e+f x)}{f (a-b)^2 \left (a+b \sec ^2(e+f x)-b\right )^{3/2}} \]

[In]

Int[Sin[e + f*x]^3/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 277

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

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {-1+x^2}{x^4 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{x^2 \left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{(a-b) f} \\ & = -\frac {(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(4 b (a+b)) \text {Subst}\left (\int \frac {1}{\left (a-b+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{(a-b)^2 f} \\ & = -\frac {(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b (a+b) \sec (e+f x)}{3 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(8 b (a+b)) \text {Subst}\left (\int \frac {1}{\left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 (a-b)^3 f} \\ & = -\frac {(a+b) \cos (e+f x)}{(a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b (a+b) \sec (e+f x)}{3 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b (a+b) \sec (e+f x)}{3 (a-b)^4 f \sqrt {a-b+b \sec ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.91 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.22 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\cos (e+f x) \left (26 a^3+186 a^2 b+190 a b^2+110 b^3+3 \left (11 a^3+63 a^2 b-31 a b^2-43 b^3\right ) \cos (2 (e+f x))+6 (a-b)^2 (a+3 b) \cos (4 (e+f x))-a^3 \cos (6 (e+f x))+3 a^2 b \cos (6 (e+f x))-3 a b^2 \cos (6 (e+f x))+b^3 \cos (6 (e+f x))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{24 \sqrt {2} (a-b)^4 f (a+b+(a-b) \cos (2 (e+f x)))^2} \]

[In]

Integrate[Sin[e + f*x]^3/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-1/24*(Cos[e + f*x]*(26*a^3 + 186*a^2*b + 190*a*b^2 + 110*b^3 + 3*(11*a^3 + 63*a^2*b - 31*a*b^2 - 43*b^3)*Cos[
2*(e + f*x)] + 6*(a - b)^2*(a + 3*b)*Cos[4*(e + f*x)] - a^3*Cos[6*(e + f*x)] + 3*a^2*b*Cos[6*(e + f*x)] - 3*a*
b^2*Cos[6*(e + f*x)] + b^3*Cos[6*(e + f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt[2]
*(a - b)^4*f*(a + b + (a - b)*Cos[2*(e + f*x)])^2)

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.47

method	result	size
default	\(-\frac {a^{5} \left (\sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{4} b^{3}+a^{3} \cos \left (f x +e \right )^{6}-3 a^{2} b \cos \left (f x +e \right )^{6}+3 a \,b^{2} \cos \left (f x +e \right )^{6}+4 \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2} b^{3}-3 a^{3} \cos \left (f x +e \right )^{4}+3 a^{2} b \cos \left (f x +e \right )^{4}+3 a \,b^{2} \cos \left (f x +e \right )^{4}-8 \sin \left (f x +e \right )^{2} b^{3}-12 a^{2} b \cos \left (f x +e \right )^{2}-8 a \,b^{2}\right ) \left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (a -b \right ) \sec \left (f x +e \right )^{5}}{3 f \left (\sqrt {-b \left (a -b \right )}+a -b \right )^{5} \left (\sqrt {-b \left (a -b \right )}-a +b \right )^{5} \left (a +b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(247\)

[In]

int(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f*a^5/((-b*(a-b))^(1/2)+a-b)^5/((-b*(a-b))^(1/2)-a+b)^5*(sin(f*x+e)^2*cos(f*x+e)^4*b^3+a^3*cos(f*x+e)^6-3
*a^2*b*cos(f*x+e)^6+3*a*b^2*cos(f*x+e)^6+4*sin(f*x+e)^2*cos(f*x+e)^2*b^3-3*a^3*cos(f*x+e)^4+3*a^2*b*cos(f*x+e)
^4+3*a*b^2*cos(f*x+e)^4-8*sin(f*x+e)^2*b^3-12*a^2*b*cos(f*x+e)^2-8*a*b^2)*(a*cos(f*x+e)^2+b*sin(f*x+e)^2)*(a-b
)/(a+b*tan(f*x+e)^2)^(5/2)*sec(f*x+e)^5

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.61 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left ({\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 3 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} - 12 \, {\left (a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left ({\left (a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b - 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} - 10 \, a^{2} b^{4} + 5 \, a b^{5} - b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}\right )} f\right )}} \]

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/3*((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 3*(a^3 - a^2*b - a*b^2 + b^3)*cos(f*x + e)^5 - 12*(a^2*b
- b^3)*cos(f*x + e)^3 - 8*(a*b^2 + b^3)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^6
- 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*f*cos(f*x + e)^4 + 2*(a^5*b - 5*a^4*b^2 + 1
0*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*f*cos(f*x + e)^2 + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*f
)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**3/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.83 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {9 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}} + \frac {6 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}}{3 \, f} \]

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

[In]

int(sin(e + f*x)^3/(a + b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(sin(e + f*x)^3/(a + b*tan(e + f*x)^2)^(5/2), x)