\(\int \frac {\sin ^3(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [107]

Optimal result

Integrand size = 25, antiderivative size = 114 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\cos ^3(e+f x)}{3 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt {a+b \sec ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4219, 464, 277, 197} \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\cos ^3(e+f x)}{3 a f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

-1/3*((3*a + 4*b)*Cos[e + f*x])/(a^2*f*Sqrt[a + b*Sec[e + f*x]^2]) + Cos[e + f*x]^3/(3*a*f*Sqrt[a + b*Sec[e +
f*x]^2]) - (2*b*(3*a + 4*b)*Sec[e + f*x])/(3*a^3*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 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 )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x)}{3 a f \sqrt {a+b \sec ^2(e+f x)}}+\frac {(3 a+4 b) \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a f} \\ & = -\frac {(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\cos ^3(e+f x)}{3 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {(2 b (3 a+4 b)) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a^2 f} \\ & = -\frac {(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\cos ^3(e+f x)}{3 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt {a+b \sec ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (9 a^2+64 a b+64 b^2+8 a (a+2 b) \cos (2 (e+f x))-a^2 \cos (4 (e+f x))\right ) \sec ^3(e+f x)}{48 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

-1/48*((a + 2*b + a*Cos[2*(e + f*x)])*(9*a^2 + 64*a*b + 64*b^2 + 8*a*(a + 2*b)*Cos[2*(e + f*x)] - a^2*Cos[4*(e
+ f*x)])*Sec[e + f*x]^3)/(a^3*f*(a + b*Sec[e + f*x]^2)^(3/2))

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {a \left (a +b \right )^{4} \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\cos \left (f x +e \right )^{4} a^{2}-3 \cos \left (f x +e \right )^{2} a^{2}-4 \cos \left (f x +e \right )^{2} a b -6 a b -8 b^{2}\right ) \sec \left (f x +e \right )^{3}}{3 f \left (\sqrt {-a b}-a \right )^{4} \left (\sqrt {-a b}+a \right )^{4} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(115\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {{\left (a^{2} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} - \frac {{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{3}} + \frac {3 \, b}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )} + \frac {3 \, b^{2}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{3} \cos \left (f x + e\right )}}{3 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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