\(\int \frac {\sin (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [154]

Optimal result

Integrand size = 23, antiderivative size = 34 \[ \int \frac {\sin (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\cos (e+f x)}{(a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3265, 197} \[ \int \frac {\sin (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\cos (e+f x)}{f (a+b) \sqrt {a-b \cos ^2(e+f x)+b}} \]

[In]

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

[Out]

-(Cos[e + f*x]/((a + b)*f*Sqrt[a + b - b*Cos[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 3265

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[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}{\left (a+b-b x^2\right )^{3/2}} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x)}{(a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {\sin (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {2} \cos (e+f x)}{(a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {\cos \left (f x +e \right )}{\left (a +b \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.68 \[ \int \frac {\sin (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{{\left (a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {\sin (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 14.90 (sec) , antiderivative size = 119, normalized size of antiderivative = 3.50 \[ \int \frac {\sin (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {2}\,\sqrt {2\,a+b-b\,\cos \left (2\,e+2\,f\,x\right )}\,\left (4\,a\,\cos \left (e+f\,x\right )+b\,\cos \left (e+f\,x\right )-b\,\cos \left (3\,e+3\,f\,x\right )\right )}{f\,\left (a+b\right )\,\left (8\,a\,b+8\,a^2+3\,b^2-4\,b^2\,\cos \left (2\,e+2\,f\,x\right )+b^2\,\cos \left (4\,e+4\,f\,x\right )-8\,a\,b\,\cos \left (2\,e+2\,f\,x\right )\right )} \]

[In]

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

[Out]

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