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

Optimal. Leaf size=85 \[ -\frac{(c-d) (c+4 d) \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}+\frac{d^2 x}{a^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.141571, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2760, 2735, 2648} \[ -\frac{(c-d) (c+4 d) \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}+\frac{d^2 x}{a^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2760

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]))/(a*f*(2*m + 1)), x] + Dist[1/(a*b*(2*m +
1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*c*d*(m - 1) + b*(d^2 + c^2*(m + 1)) + d*(a*d*(m - 1) + b*c*(m + 2
))*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] && LtQ[m
, -1]

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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [B]  time = 0.283661, size = 172, normalized size = 2.02 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 \left (c^2+4 c d-5 d^2\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+2 (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-(c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+3 d^2 (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{3 a^2 f (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)^2*Sin[(e + f*x)/2] - (c - d)^2*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) + 2*(c^2 + 4*c*d - 5*d^2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 3*d^2*(e + f*x)*
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3))/(3*a^2*f*(1 + Sin[e + f*x])^2)

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Maple [B]  time = 0.059, size = 213, normalized size = 2.5 \begin{align*} 2\,{\frac{{d}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{a}^{2}f}}-2\,{\frac{{c}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{{d}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{{c}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-4\,{\frac{cd}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+2\,{\frac{{d}^{2}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-{\frac{4\,{c}^{2}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+{\frac{8\,cd}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}-{\frac{4\,{d}^{2}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^2*d^2*arctan(tan(1/2*f*x+1/2*e))-2/f/a^2/(tan(1/2*f*x+1/2*e)+1)*c^2+2/f/a^2/(tan(1/2*f*x+1/2*e)+1)*d^2+2
/f/a^2/(tan(1/2*f*x+1/2*e)+1)^2*c^2-4/f/a^2/(tan(1/2*f*x+1/2*e)+1)^2*c*d+2/f/a^2/(tan(1/2*f*x+1/2*e)+1)^2*d^2-
4/3/f/a^2/(tan(1/2*f*x+1/2*e)+1)^3*c^2+8/3/f/a^2/(tan(1/2*f*x+1/2*e)+1)^3*c*d-4/3/f/a^2/(tan(1/2*f*x+1/2*e)+1)
^3*d^2

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Maxima [B]  time = 1.70652, size = 486, normalized size = 5.72 \begin{align*} \frac{2 \,{\left (d^{2}{\left (\frac{\frac{9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 4}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{3 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} - \frac{c^{2}{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac{2 \, c d{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(d^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x
+ e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - c^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2
/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) - 2*c*d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*
sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3))/f

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Fricas [B]  time = 1.62509, size = 454, normalized size = 5.34 \begin{align*} -\frac{6 \, d^{2} f x -{\left (3 \, d^{2} f x + c^{2} + 4 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} + 2 \, c d - d^{2} +{\left (3 \, d^{2} f x - 2 \, c^{2} - 2 \, c d + 4 \, d^{2}\right )} \cos \left (f x + e\right ) +{\left (6 \, d^{2} f x + c^{2} - 2 \, c d + d^{2} +{\left (3 \, d^{2} f x - c^{2} - 4 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.74888, size = 853, normalized size = 10.04 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*c**2*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*
tan(e/2 + f*x/2) + 3*a**2*f) - 2*c**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*
tan(e/2 + f*x/2) + 3*a**2*f) + 4*c*d*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*
x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 12*c*d*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9
*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 3*d**2*f*x*tan(e/2 + f*x/2)**3/(3*a**2*f
*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 9*d**2*f*x*tan(e
/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a*
*2*f) + 9*d**2*f*x*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*ta
n(e/2 + f*x/2) + 3*a**2*f) + 3*d**2*f*x/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*
f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*d**2*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 +
f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 12*d**2*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3
+ 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 2*d**2/(3*a**2*f*tan(e/2 + f*x/2)**3
+ 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(c + d*sin(e))**2/(a*si
n(e) + a)**2, True))

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Giac [A]  time = 1.2666, size = 178, normalized size = 2.09 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} d^{2}}{a^{2}} - \frac{2 \,{\left (3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 9 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, c^{2} + 2 \, c d - 4 \, d^{2}\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(f*x + e)*d^2/a^2 - 2*(3*c^2*tan(1/2*f*x + 1/2*e)^2 - 3*d^2*tan(1/2*f*x + 1/2*e)^2 + 3*c^2*tan(1/2*f*x
+ 1/2*e) + 6*c*d*tan(1/2*f*x + 1/2*e) - 9*d^2*tan(1/2*f*x + 1/2*e) + 2*c^2 + 2*c*d - 4*d^2)/(a^2*(tan(1/2*f*x
+ 1/2*e) + 1)^3))/f