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

Optimal. Leaf size=62 \[ -\frac {(c-d)^2 \cos (e+f x)}{a f (\sin (e+f x)+1)}+\frac {d x (2 c-d)}{a}-\frac {d^2 \cos (e+f x)}{a f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 62, 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 = {2746, 2735, 2648} \[ -\frac {(c-d)^2 \cos (e+f x)}{a f (\sin (e+f x)+1)}+\frac {d x (2 c-d)}{a}-\frac {d^2 \cos (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2746

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

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 122, normalized size = 1.97 \[ -\frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right ) \left (-2 c^2-2 c d (e+f x-2)+d^2 (e+f x-2)+d^2 \cos (e+f x)\right )+d \cos \left (\frac {1}{2} (e+f x)\right ) (d \cos (e+f x)-(2 c-d) (e+f x))\right )}{a f (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 142, normalized size = 2.29 \[ -\frac {d^{2} \cos \left (f x + e\right )^{2} - {\left (2 \, c d - d^{2}\right )} f x + c^{2} - 2 \, c d + d^{2} - {\left ({\left (2 \, c d - d^{2}\right )} f x - c^{2} + 2 \, c d - 2 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (2 \, c d - d^{2}\right )} f x - d^{2} \cos \left (f x + e\right ) + c^{2} - 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.15, size = 143, normalized size = 2.31 \[ \frac {\frac {{\left (2 \, c d - d^{2}\right )} {\left (f x + e\right )}}{a} - \frac {2 \, {\left (c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c^{2} - 2 \, c d + 2 \, d^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )} a}}{f} \]

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)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.24, size = 140, normalized size = 2.26 \[ -\frac {2 d^{2}}{a f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}+\frac {4 d \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c}{a f}-\frac {2 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2}}{a f}-\frac {2 c^{2}}{a f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {4 c d}{a f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d^{2}}{a f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]

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)),x)

[Out]

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

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maxima [B]  time = 0.45, size = 209, normalized size = 3.37 \[ -\frac {2 \, {\left (d^{2} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} - 2 \, c d {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac {c^{2}}{a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \]

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)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.38, size = 124, normalized size = 2.00 \[ -\frac {d^2\,f\,x-2\,c\,d\,f\,x}{a\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2-4\,c\,d+2\,d^2\right )-4\,c\,d+2\,c^2+4\,d^2+2\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )} \]

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)),x)

[Out]

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

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sympy [A]  time = 3.67, size = 940, normalized size = 15.16 \[ \text {result too large to display} \]

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)),x)

[Out]

Piecewise((-2*c**2*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/
2) + a*f) - 2*c**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*c*d*f*
x*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*c*d
*f*x*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*
c*d*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*
c*d*f*x/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 4*c*d*tan(e/2 + f*x
/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 4*c*d/(a*f*tan(e/2 +
f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - d**2*f*x*tan(e/2 + f*x/2)**3/(a*f*tan(e/2
+ f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - d**2*f*x*tan(e/2 + f*x/2)**2/(a*f*tan(e
/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - d**2*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/
2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - d**2*f*x/(a*f*tan(e/2 + f*x/2)**3 + a*
f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 2*d**2*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*
f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 2*d**2*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*t
an(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 4*d**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 +
a*f*tan(e/2 + f*x/2) + a*f), Ne(f, 0)), (x*(c + d*sin(e))**2/(a*sin(e) + a), True))

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