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

Optimal. Leaf size=112 \[ -\frac{2 a^2 (c-d)^2 (c+2 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f \left (c^2-d^2\right )^{3/2}}+\frac{a^2 (c-d) \cos (e+f x)}{d f (c+d) (c+d \sin (e+f x))}+\frac{a^2 x}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.182345, antiderivative size = 115, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2762, 2735, 2660, 618, 204} \[ -\frac{2 a^2 (c-d) (c+2 d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^2 f (c+d) \sqrt{c^2-d^2}}+\frac{a^2 (c-d) \cos (e+f x)}{d f (c+d) (c+d \sin (e+f x))}+\frac{a^2 x}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], 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] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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{(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx &=\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f (c+d \sin (e+f x))}-\frac{a \int \frac{-2 a d-a (c+d) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=\frac{a^2 x}{d^2}+\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f (c+d \sin (e+f x))}-\frac{\left (a^2 (c-d) (c+2 d)\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{d^2 (c+d)}\\ &=\frac{a^2 x}{d^2}+\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f (c+d \sin (e+f x))}-\frac{\left (2 a^2 (c-d) (c+2 d)\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 )}{d^2 (c+d) f}\\ &=\frac{a^2 x}{d^2}+\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f (c+d \sin (e+f x))}+\frac{\left (4 a^2 (c-d) (c+2 d)\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 )}{d^2 (c+d) f}\\ &=\frac{a^2 x}{d^2}-\frac{2 a^2 (c-d) (c+2 d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{d^2 (c+d) \sqrt{c^2-d^2} f}+\frac{a^2 (c-d) \cos (e+f x)}{d (c+d) f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.458566, size = 139, normalized size = 1.24 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (-\frac{2 \left (c^2+c d-2 d^2\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}+\frac{d (c-d) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}+e+f x\right )}{d^2 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.112, size = 389, normalized size = 3.5 \begin{align*} 2\,{\frac{{a}^{2}\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 ) }}-2\,{\frac{{a}^{2}d\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}^{2}c}{df \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}^{2}}{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}^{2}{c}^{2}}{f{d}^{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{{a}^{2}c}{df \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 ) }+4\,{\frac{{a}^{2}}{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 ) }+2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.837, size = 1054, normalized size = 9.41 \begin{align*} \left [\frac{2 \,{\left (a^{2} c d + a^{2} d^{2}\right )} f x \sin \left (f x + e\right ) + 2 \,{\left (a^{2} c^{2} + a^{2} c d\right )} f x +{\left (a^{2} c^{2} + 2 \, a^{2} c d +{\left (a^{2} c d + 2 \, a^{2} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-\frac{c - d}{c + d}} \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 ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{c - d}{c + d}}}{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^{2} c d - a^{2} d^{2}\right )} \cos \left (f x + e\right )}{2 \,{\left ({\left (c d^{3} + d^{4}\right )} f \sin \left (f x + e\right ) +{\left (c^{2} d^{2} + c d^{3}\right )} f\right )}}, \frac{{\left (a^{2} c d + a^{2} d^{2}\right )} f x \sin \left (f x + e\right ) +{\left (a^{2} c^{2} + a^{2} c d\right )} f x +{\left (a^{2} c^{2} + 2 \, a^{2} c d +{\left (a^{2} c d + 2 \, a^{2} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{\frac{c - d}{c + d}} \arctan \left (-\frac{{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt{\frac{c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) +{\left (a^{2} c d - a^{2} d^{2}\right )} \cos \left (f x + e\right )}{{\left (c d^{3} + d^{4}\right )} f \sin \left (f x + e\right ) +{\left (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))^2/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.38669, size = 277, normalized size = 2.47 \begin{align*} \frac{\frac{{\left (f x + e\right )} a^{2}}{d^{2}} - \frac{2 \,{\left (a^{2} c^{2} + a^{2} c d - 2 \, a^{2} d^{2}\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 (c d^{2} + d^{3}\right )} \sqrt{c^{2} - d^{2}}} + \frac{2 \,{\left (a^{2} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{2} c^{2} - a^{2} c d\right )}}{{\left (c^{2} d + c d^{2}\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 )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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