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

Optimal. Leaf size=45 \[ -\frac {\cos (c+d x)}{a d}-\frac {\cos (c+d x)}{a d (\sin (c+d x)+1)}-\frac {x}{a} \]

[Out]

-x/a-cos(d*x+c)/a/d-cos(d*x+c)/a/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2746, 12, 2735, 2648} \[ -\frac {\cos (c+d x)}{a d}-\frac {\cos (c+d x)}{a d (\sin (c+d x)+1)}-\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) - Cos[c + d*x]/(a*d) - Cos[c + d*x]/(a*d*(1 + Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\cos (c+d x)}{a d}-\frac {\int \frac {a \sin (c+d x)}{a+a \sin (c+d x)} \, dx}{a}\\ &=-\frac {\cos (c+d x)}{a d}-\int \frac {\sin (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac {x}{a}-\frac {\cos (c+d x)}{a d}+\int \frac {1}{a+a \sin (c+d x)} \, dx\\ &=-\frac {x}{a}-\frac {\cos (c+d x)}{a d}-\frac {\cos (c+d x)}{d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 85, normalized size = 1.89 \[ -\frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+c+d x)+\sin \left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+c+d x-2)\right )}{a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 69, normalized size = 1.53 \[ -\frac {d x + {\left (d x + 2\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + {\left (d x + \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1}{a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.63, size = 77, normalized size = 1.71 \[ -\frac {\frac {d x + c}{a} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 64, normalized size = 1.42 \[ -\frac {2}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.94, size = 129, normalized size = 2.87 \[ -\frac {2 \, {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.18, size = 69, normalized size = 1.53 \[ -\frac {x}{a}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4}{a\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.05, size = 422, normalized size = 9.38 \[ \begin {cases} - \frac {d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {2 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {4}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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