3.2.88 \(\int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [188]

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

[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 = 0.06, 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 = {2825, 12, 2814, 2727} \begin {gather*} -\frac {\cos (c+d x)}{a d}-\frac {\cos (c+d x)}{a d (\sin (c+d x)+1)}-\frac {x}{a} \end {gather*}

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 2727

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 2814

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 2825

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 = 0.10, size = 85, normalized size = 1.89 \begin {gather*} -\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) (c+d x+\cos (c+d x))+(-2+c+d x+\cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{a d (1+\sin (c+d x))} \end {gather*}

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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Maple [A]
time = 0.07, size = 54, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(54\)
default	\(\frac {-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(54\)
risch	\(-\frac {x}{a}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}-\frac {2}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\)	\(64\)
norman	\(\frac {-\frac {2}{a d}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {x}{a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {2 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (45) = 90\).
time = 0.50, size = 129, normalized size = 2.87 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.35, size = 69, normalized size = 1.53 \begin {gather*} -\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} \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (32) = 64\).
time = 1.25, size = 422, normalized size = 9.38 \begin {gather*} \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}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \end {gather*}

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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Giac [A]
time = 5.19, size = 77, normalized size = 1.71 \begin {gather*} -\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} \end {gather*}

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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Mupad [B]
time = 1.18, size = 69, normalized size = 1.53 \begin {gather*} -\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 )} \end {gather*}

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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