3.719 \(\int \frac{1}{(a+b \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=131 \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac{3 a b \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac{b \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.113618, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2664, 2754, 12, 2660, 618, 204} \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac{3 a b \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac{b \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^(-3),x]

[Out]

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

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(a+b \sin (e+f x))^3} \, dx &=\frac{b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}-\frac{\int \frac{-2 a+b \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac{b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\int \frac{2 a^2+b^2}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\left (2 a^2+b^2\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\left (2 a^2+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right )^2 f}\\ &=\frac{b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac{\left (2 \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right )^2 f}\\ &=\frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} f}+\frac{b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.382713, size = 114, normalized size = 0.87 \[ \frac{\frac{2 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{b \cos (e+f x) \left (4 a^2+3 a b \sin (e+f x)-b^2\right )}{(a-b)^2 (a+b)^2 (a+b \sin (e+f x))^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b*Cos[e + f*x]*(4*a^2
- b^2 + 3*a*b*Sin[e + f*x]))/((a - b)^2*(a + b)^2*(a + b*Sin[e + f*x])^2))/(2*f)

________________________________________________________________________________________

Maple [B]  time = 0.075, size = 705, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2*e)^3-2/f/(ta
n(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^4/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^3+4/f/(tan(1/2*f
*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b/(a^4-2*a^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^2+7/f/(tan(1/2*f*x+1/2*
e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^2-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*
tan(1/2*f*x+1/2*e)*b+a)^2*b^5/(a^4-2*a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2+11/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(
1/2*f*x+1/2*e)*b+a)^2*b^2*a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1
/2*e)*b+a)^2*b^4/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)+4/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a
)^2*b/(a^4-2*a^2*b^2+b^4)*a^2-1/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^3/(a^4-2*a^2*b^2+b^4)+
2/f/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2+1/f/(a^4-
2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*b^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.65302, size = 1346, normalized size = 10.27 \begin{align*} \left [-\frac{6 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} -{\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (f x + e\right )}{4 \,{\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) -{\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}, -\frac{3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} -{\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) +{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) -{\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*(a^3*b^2 - a*b^4)*cos(f*x + e)*sin(f*x + e) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2 + b^4)*cos(f*x +
e)^2 + 2*(2*a^3*b + a*b^3)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x +
e) - a^2 - b^2 + 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 2*a*
b*sin(f*x + e) - a^2 - b^2)) + 2*(4*a^4*b - 5*a^2*b^3 + b^5)*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 -
b^8)*f*cos(f*x + e)^2 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e) - (a^8 - 2*a^6*b^2 + 2*a^2*b
^6 - b^8)*f), -1/2*(3*(a^3*b^2 - a*b^4)*cos(f*x + e)*sin(f*x + e) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2 + b^
4)*cos(f*x + e)^2 + 2*(2*a^3*b + a*b^3)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 -
b^2)*cos(f*x + e))) + (4*a^4*b - 5*a^2*b^3 + b^5)*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*co
s(f*x + e)^2 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*
f)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.2564, size = 383, normalized size = 2.92 \begin{align*} \frac{\frac{{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a b^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 4 \, a^{4} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 7 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, b^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a b^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, a^{4} b - a^{2} b^{3}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a\right )}^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))*(2*a^2 + b^2
)/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*tan(1/2*f*x + 1/2*e)^3 - 2*a*b^4*tan(1/2*f*x + 1/2*e)
^3 + 4*a^4*b*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^3*tan(1/2*f*x + 1/2*e)^2 - 2*b^5*tan(1/2*f*x + 1/2*e)^2 + 11*a^3
*b^2*tan(1/2*f*x + 1/2*e) - 2*a*b^4*tan(1/2*f*x + 1/2*e) + 4*a^4*b - a^2*b^3)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*
tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)^2))/f