3.467 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx\)

Optimal. Leaf size=131 \[ \frac{2 d^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^2 \sqrt{c^2-d^2}}-\frac{(c-4 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2} \]

[Out]

(2*d^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^2*Sqrt[c^2 - d^2]*f) - ((c - 4*d)*Cos[e
+ f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])) - Cos[e + f*x]/(3*(c - d)*f*(a + a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.282265, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2766, 2978, 12, 2660, 618, 204} \[ \frac{2 d^2 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^2 \sqrt{c^2-d^2}}-\frac{(c-4 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])),x]

[Out]

(2*d^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^2*Sqrt[c^2 - d^2]*f) - ((c - 4*d)*Cos[e
+ f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])) - Cos[e + f*x]/(3*(c - d)*f*(a + a*Sin[e + f*x])^2)

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

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

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

Mathematica [A]  time = 0.363462, size = 204, normalized size = 1.56 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\frac{6 d^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}+2 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )+2 (c-4 d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-(c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)*Sin[(e + f*x)/2] - (c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)
/2]) + 2*(c - 4*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (6*d^2*ArcTan[(d + c*Tan[(e + f*
x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/Sqrt[c^2 - d^2]))/(3*a^2*(c - d)^2*f*(1 + Sin
[e + f*x])^2)

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Maple [A]  time = 0.093, size = 175, normalized size = 1.3 \begin{align*} 2\,{\frac{{d}^{2}}{{a}^{2}f \left ( c-d \right ) ^{2}\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{c}{{a}^{2}f \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+4\,{\frac{d}{{a}^{2}f \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-{\frac{4}{3\,{a}^{2}f \left ( c-d \right ) } \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{{a}^{2}f \left ( c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x)

[Out]

2/f/a^2*d^2/(c-d)^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))-2/f/a^2/(c-d)^2/(
tan(1/2*f*x+1/2*e)+1)*c+4/f/a^2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)*d-4/3/f/a^2/(c-d)/(tan(1/2*f*x+1/2*e)+1)^3+2/f/
a^2/(c-d)/(tan(1/2*f*x+1/2*e)+1)^2

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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+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82903, size = 2129, normalized size = 16.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/6*(2*c^3 - 2*c^2*d - 2*c*d^2 + 2*d^3 + 2*(c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e)^2 - 3*(d^2*cos(f*x +
e)^2 - d^2*cos(f*x + e) - 2*d^2 - (d^2*cos(f*x + e) + 2*d^2)*sin(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)
*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2
+ d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*(2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x
+ e) - 2*(c^3 - c^2*d - c*d^2 + d^3 - (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 -
2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 - (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos
(f*x + e) - 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*
d^4)*f*cos(f*x + e) + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f)*sin(f*x + e)), 1/3*(c^3 - c^2*d - c
*d^2 + d^3 + (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e)^2 - 3*(d^2*cos(f*x + e)^2 - d^2*cos(f*x + e) - 2*d^2
- (d^2*cos(f*x + e) + 2*d^2)*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(
f*x + e))) + (2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^3 - (c^3 - 4*c^2*d -
c*d^2 + 4*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 -
(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) - 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2
*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c
*d^3 - a^2*d^4)*f)*sin(f*x + e))]

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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+a*sin(f*x+e))**2/(c+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [A]  time = 1.34709, size = 263, normalized size = 2.01 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )} d^{2}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt{c^{2} - d^{2}}} - \frac{3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 6 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 9 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, c - 5 \, d}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

2/3*(3*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))*d^2/((
a^2*c^2 - 2*a^2*c*d + a^2*d^2)*sqrt(c^2 - d^2)) - (3*c*tan(1/2*f*x + 1/2*e)^2 - 6*d*tan(1/2*f*x + 1/2*e)^2 + 3
*c*tan(1/2*f*x + 1/2*e) - 9*d*tan(1/2*f*x + 1/2*e) + 2*c - 5*d)/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*(tan(1/2*f*x
+ 1/2*e) + 1)^3))/f