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

Optimal. Leaf size=150 \[ -\frac{2 d (2 c+d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}-\frac{d (c+2 d) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.178282, antiderivative size = 150, 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 = {2768, 2754, 12, 2660, 618, 204} \[ -\frac{2 d (2 c+d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}-\frac{d (c+2 d) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2768

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

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

Mathematica [A]  time = 0.638198, size = 162, normalized size = 1.08 \[ \frac{\cos (e+f x) \left (-\frac{d}{(\sin (e+f x)+1) (c+d \sin (e+f x))}+\frac{c+2 d}{(c-d) (\sin (e+f x)+1)}-\frac{2 d (2 c+d) \tan ^{-1}\left (\frac{\sqrt{d-c} \sqrt{1-\sin (e+f x)}}{\sqrt{-c-d} \sqrt{\sin (e+f x)+1}}\right )}{\sqrt{-c-d} (d-c)^{3/2} \sqrt{\cos ^2(e+f x)}}\right )}{a f (d-c) (c+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.104, size = 273, normalized size = 1.8 \begin{align*} -2\,{\frac{{d}^{3}\tan \left ( 1/2\,fx+e/2 \right ) }{af \left ( c-d \right ) ^{2} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) c}}-2\,{\frac{{d}^{2}}{af \left ( c-d \right ) ^{2} \left ( c \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}+2\,\tan \left ( 1/2\,fx+e/2 \right ) d+c \right ) \left ( c+d \right ) }}-4\,{\frac{cd}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \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{{d}^{2}}{af \left ( c-d \right ) ^{2} \left ( c+d \right ) \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{1}{af \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.92085, size = 2422, normalized size = 16.15 \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))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/2*(2*c^4 - 4*c^2*d^2 + 2*d^4 + 2*(c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e)^2 + (2*c^2*d + 3*c*d^2 +
d^3 - (2*c*d^2 + d^3)*cos(f*x + e)^2 + (2*c^2*d + c*d^2)*cos(f*x + e) + (2*c^2*d + 3*c*d^2 + d^3 + (2*c*d^2 +
d^3)*cos(f*x + e))*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*(c^4 + c^3*d - c*d^3 - d^4)*cos(f*x + e) - 2*(c^4 - 2*c^2*d^2 + d^4 - (c^3*d + 2*c^2
*d^2 - c*d^3 - 2*d^4)*cos(f*x + e))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5
- a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e)
- (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5
- a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e)), (c^4 - 2*c^2*d^2 + d^
4 + (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e)^2 - (2*c^2*d + 3*c*d^2 + d^3 - (2*c*d^2 + d^3)*cos(f*x +
e)^2 + (2*c^2*d + c*d^2)*cos(f*x + e) + (2*c^2*d + 3*c*d^2 + d^3 + (2*c*d^2 + d^3)*cos(f*x + e))*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))) + (c^4 + c^3*d - c*d^3 - d^4)*co
s(f*x + e) - (c^4 - 2*c^2*d^2 + d^4 - (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e))*sin(f*x + e))/((a*c^5*
d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2
+ 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e) - (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f - ((a*c^5
*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*
d^4 - a*d^6)*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))/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.62021, size = 726, normalized size = 4.84 \begin{align*} \frac{\frac{{\left (2 \, a c^{4} d - a c^{3} d^{2} - 3 \, a c^{2} d^{3} + a c d^{4} + a d^{5}\right )} \sqrt{-c^{2} + d^{2}} \log \left ({\left |{\left (d + \sqrt{-c^{2} + d^{2}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c \right |}\right )}{a^{2} c^{8} - 2 \, a^{2} c^{7} d - 2 \, a^{2} c^{6} d^{2} + 6 \, a^{2} c^{5} d^{3} - 6 \, a^{2} c^{3} d^{5} + 2 \, a^{2} c^{2} d^{6} + 2 \, a^{2} c d^{7} - a^{2} d^{8}} - \frac{{\left (2 \, a c^{4} d - a c^{3} d^{2} - 3 \, a c^{2} d^{3} + a c d^{4} + a d^{5}\right )} \sqrt{-c^{2} + d^{2}} \log \left ({\left | -{\left (d - \sqrt{-c^{2} + d^{2}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - c \right |}\right )}{a^{2} c^{8} - 2 \, a^{2} c^{7} d - 2 \, a^{2} c^{6} d^{2} + 6 \, a^{2} c^{5} d^{3} - 6 \, a^{2} c^{3} d^{5} + 2 \, a^{2} c^{2} d^{6} + 2 \, a^{2} c d^{7} - a^{2} d^{8}} - \frac{2 \,{\left (c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c^{3} + c^{2} d + c d^{2}\right )}}{{\left (a c^{4} - a c^{3} d - a c^{2} d^{2} + a c d^{3}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a*c^4*d - a*c^3*d^2 - 3*a*c^2*d^3 + a*c*d^4 + a*d^5)*sqrt(-c^2 + d^2)*log(abs((d + sqrt(-c^2 + d^2))*tan(1
/2*f*x + 1/2*e) + c))/(a^2*c^8 - 2*a^2*c^7*d - 2*a^2*c^6*d^2 + 6*a^2*c^5*d^3 - 6*a^2*c^3*d^5 + 2*a^2*c^2*d^6 +
2*a^2*c*d^7 - a^2*d^8) - (2*a*c^4*d - a*c^3*d^2 - 3*a*c^2*d^3 + a*c*d^4 + a*d^5)*sqrt(-c^2 + d^2)*log(abs(-(d
- sqrt(-c^2 + d^2))*tan(1/2*f*x + 1/2*e) - c))/(a^2*c^8 - 2*a^2*c^7*d - 2*a^2*c^6*d^2 + 6*a^2*c^5*d^3 - 6*a^2
*c^3*d^5 + 2*a^2*c^2*d^6 + 2*a^2*c*d^7 - a^2*d^8) - 2*(c^3*tan(1/2*f*x + 1/2*e)^2 + c^2*d*tan(1/2*f*x + 1/2*e)
^2 + d^3*tan(1/2*f*x + 1/2*e)^2 + 2*c^2*d*tan(1/2*f*x + 1/2*e) + 3*c*d^2*tan(1/2*f*x + 1/2*e) + d^3*tan(1/2*f*
x + 1/2*e) + c^3 + c^2*d + c*d^2)/((a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*(c*tan(1/2*f*x + 1/2*e)^3 + c*tan(1
/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e)^2 + c*tan(1/2*f*x + 1/2*e) + 2*d*tan(1/2*f*x + 1/2*e) + c)))/f