\(\int \frac {\sin ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [117]

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {(3 a-b) \cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3745, 464, 270} \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cos ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{3 f (a-b)}-\frac {(3 a-b) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{3 f (a-b)^2} \]

[In]

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

[Out]

-1/3*((3*a - b)*Cos[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/((a - b)^2*f) + (Cos[e + f*x]^3*Sqrt[a - b + b*Se
c[e + f*x]^2])/(3*(a - b)*f)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(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 \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b) f}+\frac {(3 a-b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{3 (a-b) f} \\ & = -\frac {(3 a-b) \cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{3 (a-b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cos (e+f x) (-5 a+b+(a-b) \cos (2 (e+f x))) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} (a-b)^2 f} \]

[In]

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

[Out]

(Cos[e + f*x]*(-5*a + b + (a - b)*Cos[2*(e + f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(
6*Sqrt[2]*(a - b)^2*f)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\left (a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}\right ) \left (b \sin \left (f x +e \right )^{2}+a \cos \left (f x +e \right )^{2}-3 a \right ) \sec \left (f x +e \right )}{3 f \left (a -b \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(78\)

[In]

int(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.85 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f} \]

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate(sin(f*x+e)**3/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\frac {3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a - b} - \frac {{\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}}}{3 \, f} \]

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(sin(e + f*x)^3/(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^3/(a + b*tan(e + f*x)^2)^(1/2), x)