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

Optimal. Leaf size=83 \[ \frac{2 a \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d) \sqrt{c^2-d^2}}-\frac{a \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.091673, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2754, 12, 2660, 618, 204} \[ \frac{2 a \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f (c+d) \sqrt{c^2-d^2}}-\frac{a \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.556515, size = 220, normalized size = 2.65 \[ \frac{a (\sin (e+f x)+1) \left (2 \sqrt{c^2-d^2} \csc (e) \sqrt{(\cos (e)-i \sin (e))^2} (c \cos (e)+d \sin (f x))+4 d (\cos (e)-i \sin (e)) (c+d \sin (e+f x)) \tan ^{-1}\left (\frac{(\cos (e)-i \sin (e)) \sec \left (\frac{f x}{2}\right ) \left (c \sin \left (\frac{f x}{2}\right )+d \cos \left (e+\frac{f x}{2}\right )\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )\right )}{2 d f (c+d) \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(1 + Sin[e + f*x])*(2*Sqrt[c^2 - d^2]*Csc[e]*Sqrt[(Cos[e] - I*Sin[e])^2]*(c*Cos[e] + d*Sin[f*x]) + 4*d*ArcT
an[(Sec[(f*x)/2]*(Cos[e] - I*Sin[e])*(d*Cos[e + (f*x)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2]*Sqrt[(Cos[e] - I*
Sin[e])^2])]*(Cos[e] - I*Sin[e])*(c + d*Sin[e + f*x])))/(2*d*(c + d)*Sqrt[c^2 - d^2]*f*Sqrt[(Cos[e] - I*Sin[e]
)^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*(c + d*Sin[e + f*x]))

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Maple [A]  time = 0.102, size = 147, normalized size = 1.8 \begin{align*} -2\,{\frac{da\tan \left ( 1/2\,fx+e/2 \right ) }{f \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{a}{f \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 ) }}+2\,{\frac{a}{f \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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((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 [A]  time = 1.42825, size = 790, normalized size = 9.52 \begin{align*} \left [-\frac{{\left (a d \sin \left (f x + e\right ) + a c\right )} \sqrt{-c^{2} + d^{2}} \log \left (\frac{{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \,{\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \,{\left (a c^{2} - a d^{2}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f \sin \left (f x + e\right ) +{\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f\right )}}, -\frac{{\left (a d \sin \left (f x + e\right ) + a c\right )} \sqrt{c^{2} - d^{2}} \arctan \left (-\frac{c \sin \left (f x + e\right ) + d}{\sqrt{c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) +{\left (a c^{2} - a d^{2}\right )} \cos \left (f x + e\right )}{{\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f \sin \left (f x + e\right ) +{\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.30054, size = 174, normalized size = 2.1 \begin{align*} \frac{2 \,{\left (\frac{{\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 )} a}{\sqrt{c^{2} - d^{2}}{\left (c + d\right )}} - \frac{a d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a c}{{\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 )}{\left (c^{2} + c d\right )}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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