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

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Rubi [A]  time = 0.14, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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 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]

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*}

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Mathematica [B]  time = 0.28, 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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fricas [B]  time = 0.45, size = 197, normalized size = 2.32 \[ -\frac {6 \, d^{2} f x - {\left (3 \, d^{2} f x + c^{2} + 4 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} + 2 \, c d - d^{2} + {\left (3 \, d^{2} f x - 2 \, c^{2} - 2 \, c d + 4 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left (6 \, d^{2} f x + c^{2} - 2 \, c d + d^{2} + {\left (3 \, d^{2} f x - c^{2} - 4 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

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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giac [A]  time = 0.17, size = 132, normalized size = 1.55 \[ \frac {\frac {3 \, {\left (f x + e\right )} d^{2}}{a^{2}} - \frac {2 \, {\left (3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, c^{2} + 2 \, c d - 4 \, d^{2}\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \]

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

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maple [B]  time = 0.25, size = 213, normalized size = 2.51 \[ \frac {2 d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{2} f}-\frac {2 c^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 d^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 c^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 c d}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {2 d^{2}}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 c^{2}}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8 c d}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {4 d^{2}}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}} \]

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

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maxima [B]  time = 0.43, size = 360, normalized size = 4.24 \[ \frac {2 \, {\left (d^{2} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 4}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} - \frac {c^{2} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac {2 \, c d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

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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mupad [B]  time = 7.43, size = 93, normalized size = 1.09 \[ \frac {d^2\,x}{a^2}-\frac {\frac {4\,c\,d}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2-2\,d^2\right )+\frac {4\,c^2}{3}-\frac {8\,d^2}{3}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c^2+4\,c\,d-6\,d^2\right )}{a^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.06, size = 915, normalized size = 10.76 \[ \text {result too large to display} \]

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((-6*c**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) - 6*c**2*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*tan(e/2 + f*x/2) + 3*a**2*f) - 4*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) - 12*c*d*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*tan(e/2 + f*x/2) + 3*a**2*f) - 4*c*d/(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*tan(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)**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) + 18*d**2*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*tan(e/2 + f*x/2) + 3*a**2*f) + 8*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*sin(e) + a
)**2, True))

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