\(\int \frac {\sin ^3(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [120]

Optimal result

Integrand size = 25, antiderivative size = 146 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b) \cos (e+f x)}{a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b (a+2 b) \sec (e+f x)}{3 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b (a+2 b) \sec (e+f x)}{3 a^4 f \sqrt {a+b \sec ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 146, 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 = {4219, 464, 277, 198, 197} \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {8 b (a+2 b) \sec (e+f x)}{3 a^4 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {4 b (a+2 b) \sec (e+f x)}{3 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {(a+2 b) \cos (e+f x)}{a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

-(((a + 2*b)*Cos[e + f*x])/(a^2*f*(a + b*Sec[e + f*x]^2)^(3/2))) + Cos[e + f*x]^3/(3*a*f*(a + b*Sec[e + f*x]^2
)^(3/2)) - (4*b*(a + 2*b)*Sec[e + f*x])/(3*a^3*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (8*b*(a + 2*b)*Sec[e + f*x])/
(3*a^4*f*Sqrt[a + 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 4219

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^
n)^p/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

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

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (26 a^3+264 a^2 b+640 a b^2+512 b^3+3 a \left (11 a^2+96 a b+128 b^2\right ) \cos (2 (e+f x))+6 a^2 (a+4 b) \cos (4 (e+f x))-a^3 \cos (6 (e+f x))\right ) \sec ^5(e+f x)}{192 a^4 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

[In]

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

[Out]

-1/192*((a + 2*b + a*Cos[2*(e + f*x)])*(26*a^3 + 264*a^2*b + 640*a*b^2 + 512*b^3 + 3*a*(11*a^2 + 96*a*b + 128*
b^2)*Cos[2*(e + f*x)] + 6*a^2*(a + 4*b)*Cos[4*(e + f*x)] - a^3*Cos[6*(e + f*x)])*Sec[e + f*x]^5)/(a^4*f*(a + b
*Sec[e + f*x]^2)^(5/2))

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {a \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (a +b \right )^{5} \left (a^{3} \cos \left (f x +e \right )^{6}-3 \cos \left (f x +e \right )^{4} a^{3}-6 \cos \left (f x +e \right )^{4} a^{2} b -12 \cos \left (f x +e \right )^{2} a^{2} b -24 \cos \left (f x +e \right )^{2} a \,b^{2}-8 a \,b^{2}-16 b^{3}\right ) \sec \left (f x +e \right )^{5}}{3 f \left (\sqrt {-a b}+a \right )^{5} \left (\sqrt {-a b}-a \right )^{5} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\)	\(147\)

[In]

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

[Out]

-1/3/f*a/((-a*b)^(1/2)+a)^5/((-a*b)^(1/2)-a)^5*(b+a*cos(f*x+e)^2)*(a+b)^5*(a^3*cos(f*x+e)^6-3*cos(f*x+e)^4*a^3
-6*cos(f*x+e)^4*a^2*b-12*cos(f*x+e)^2*a^2*b-24*cos(f*x+e)^2*a*b^2-8*a*b^2-16*b^3)/(a+b*sec(f*x+e)^2)^(5/2)*sec
(f*x+e)^5

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left (a^{3} \cos \left (f x + e\right )^{7} - 3 \, {\left (a^{3} + 2 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} - 12 \, {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{6} f \cos \left (f x + e\right )^{4} + 2 \, a^{5} b f \cos \left (f x + e\right )^{2} + a^{4} b^{2} f\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.34 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} - \frac {{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{4}} + \frac {6 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{3} \cos \left (f x + e\right )^{3}} + \frac {9 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{4} \cos \left (f x + e\right )^{3}}}{3 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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