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

Optimal. Leaf size=185 \[ \frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2} (b c-a d)^2}+\frac{2 d \left (a c d-b \left (2 c^2-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2} (b c-a d)^2}-\frac{d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.468232, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2802, 3001, 2660, 618, 204} \[ \frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2} (b c-a d)^2}+\frac{2 d \left (a c d-b \left (2 c^2-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2} (b c-a d)^2}-\frac{d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

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

Mathematica [A]  time = 0.883794, size = 165, normalized size = 0.89 \[ \frac{\frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{2 d \left (a c d+b \left (d^2-2 c^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac{d^2 (a d-b c) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{f (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.121, size = 514, normalized size = 2.8 \begin{align*} 2\,{\frac{{d}^{4}\tan \left ( 1/2\,fx+e/2 \right ) a}{f \left ( da-cb \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 ) c \left ({c}^{2}-{d}^{2} \right ) }}-2\,{\frac{{d}^{3}\tan \left ( 1/2\,fx+e/2 \right ) b}{f \left ( da-cb \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}^{2}-{d}^{2} \right ) }}+2\,{\frac{a{d}^{3}}{f \left ( da-cb \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}^{2}-{d}^{2} \right ) }}-2\,{\frac{c{d}^{2}b}{f \left ( da-cb \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}^{2}-{d}^{2} \right ) }}+2\,{\frac{ac{d}^{2}}{f \left ( da-cb \right ) ^{2} \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-4\,{\frac{b{c}^{2}d}{f \left ( da-cb \right ) ^{2} \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\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}^{3}b}{f \left ( da-cb \right ) ^{2} \left ({c}^{2}-{d}^{2} \right ) ^{3/2}}\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{{b}^{2}}{f \left ({a}^{2}{d}^{2}-2\,abcd+{c}^{2}{b}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [A]  time = 1.28399, size = 416, normalized size = 2.25 \begin{align*} \frac{2 \,{\left (\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 )} b^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a^{2} - b^{2}}} - \frac{{\left (2 \, b c^{2} d - a c d^{2} - b d^{3}\right )}{\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 )}}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} - b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - a^{2} d^{4}\right )} \sqrt{c^{2} - d^{2}}} - \frac{d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c d^{2}}{{\left (b c^{4} - a c^{3} d - b c^{2} d^{2} + a c d^{3}\right )}{\left (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 ) + c\right )}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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