3.1.44 \(\int \frac {\sin ^2(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [44]

Optimal. Leaf size=20 \[ -\frac {x}{a}+\frac {\tan (c+d x)}{a d} \]

[Out]

-x/a+tan(d*x+c)/a/d

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Rubi [A]
time = 0.04, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3250, 3254, 3852, 8} \begin {gather*} \frac {\tan (c+d x)}{a d}-\frac {x}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) + Tan[c + d*x]/(a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3250

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[B*(x
/b), x] + Dist[(A*b - a*B)/b, Int[1/(a + b*Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3254

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^2(c+d x)}{a-a \sin ^2(c+d x)} \, dx &=-\frac {x}{a}+\int \frac {1}{a-a \sin ^2(c+d x)} \, dx\\ &=-\frac {x}{a}+\frac {\int \sec ^2(c+d x) \, dx}{a}\\ &=-\frac {x}{a}-\frac {\text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d}\\ &=-\frac {x}{a}+\frac {\tan (c+d x)}{a d}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 27, normalized size = 1.35 \begin {gather*} \frac {-\frac {\tan ^{-1}(\tan (c+d x))}{d}+\frac {\tan (c+d x)}{d}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(ArcTan[Tan[c + d*x]]/d) + Tan[c + d*x]/d)/a

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Maple [A]
time = 0.15, size = 24, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )}{d a}\)	\(24\)
default	\(\frac {\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )}{d a}\)	\(24\)
risch	\(-\frac {x}{a}+\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(30\)
norman	\(\frac {\frac {x}{a}+\frac {x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 26, normalized size = 1.30 \begin {gather*} -\frac {\frac {d x + c}{a} - \frac {\tan \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)^2),x, algorithm="maxima")

[Out]

-((d*x + c)/a - tan(d*x + c)/a)/d

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Fricas [A]
time = 0.39, size = 34, normalized size = 1.70 \begin {gather*} -\frac {d x \cos \left (d x + c\right ) - \sin \left (d x + c\right )}{a d \cos \left (d x + c\right )} \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)^2),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (12) = 24\).
time = 1.01, size = 100, normalized size = 5.00 \begin {gather*} \begin {cases} - \frac {d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - a d} + \frac {d x}{a d \tan ^{2}{\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 ^{2}{\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 ^{2}{\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)**2),x)

[Out]

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

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Giac [A]
time = 0.50, size = 26, normalized size = 1.30 \begin {gather*} -\frac {\frac {d x + c}{a} - \frac {\tan \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)^2),x, algorithm="giac")

[Out]

-((d*x + c)/a - tan(d*x + c)/a)/d

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Mupad [B]
time = 13.74, size = 20, normalized size = 1.00 \begin {gather*} \frac {\mathrm {tan}\left (c+d\,x\right )}{a\,d}-\frac {x}{a} \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)^2),x)

[Out]

tan(c + d*x)/(a*d) - x/a

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