\(\int \frac {\sin ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [163]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {2 \cos (e+f x)}{3 (a+b)^2 f \sqrt {a+b-b \cos ^2(e+f x)}}-\frac {\cos (e+f x) \sin ^2(e+f x)}{3 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, 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 = {3265, 386, 197} \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {2 \cos (e+f x)}{3 f (a+b)^2 \sqrt {a-b \cos ^2(e+f x)+b}}-\frac {\sin ^2(e+f x) \cos (e+f x)}{3 f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

[In]

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

[Out]

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

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 386

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*(
(c + d*x^n)^q/(a*n*(p + 1))), x] - Dist[c*(q/(a*(p + 1))), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sqrt {2} \cos (e+f x) (-5 a-3 b+(a+3 b) \cos (2 (e+f x)))}{3 (a+b)^2 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \]

[In]

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

[Out]

(Sqrt[2]*Cos[e + f*x]*(-5*a - 3*b + (a + 3*b)*Cos[2*(e + f*x)]))/(3*(a + b)^2*f*(2*a + b - b*Cos[2*(e + f*x)])
^(3/2))

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {\left (a \left (\sin ^{2}\left (f x +e \right )\right )+3 b \left (\sin ^{2}\left (f x +e \right )\right )+2 a \right ) \cos \left (f x +e \right )}{3 {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} \left (a^{2}+2 a b +b^{2}\right ) f}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.69 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left ({\left (a + 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.49 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {2 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} + \frac {\cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} - \frac {\cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {\cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 20.11 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.17 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+b\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,\left (a+3\,b-10\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}-6\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+3\,b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}{3\,f\,{\left (a+b\right )}^2\,{\left (b-4\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-2\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}^2} \]

[In]

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

[Out]

(2*exp(e*1i + f*x*1i)*(exp(e*2i + f*x*2i) + 1)*(a + b*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2
)^2)^(1/2)*(a + 3*b - 10*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x*4i) - 6*b*exp(e*2i + f*x*2i) + 3*b*exp(e*4i +
f*x*4i)))/(3*f*(a + b)^2*(b - 4*a*exp(e*2i + f*x*2i) - 2*b*exp(e*2i + f*x*2i) + b*exp(e*4i + f*x*4i))^2)