3.37 \(\int \frac{\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=27 \[ \frac{\cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d} \]

[Out]

Cos[c + d*x]/(a*d) + Sec[c + d*x]/(a*d)

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Rubi [A]  time = 0.0654911, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3175, 2590, 14} \[ \frac{\cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a - a*Sin[c + d*x]^2),x]

[Out]

Cos[c + d*x]/(a*d) + Sec[c + d*x]/(a*d)

Rule 3175

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 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx &=\frac{\int \sin (c+d x) \tan ^2(c+d x) \, dx}{a}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac{\cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d}\\ \end{align*}

Mathematica [A]  time = 0.0331787, size = 25, normalized size = 0.93 \[ \frac{\frac{\cos (c+d x)}{d}+\frac{\sec (c+d x)}{d}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a - a*Sin[c + d*x]^2),x]

[Out]

(Cos[c + d*x]/d + Sec[c + d*x]/d)/a

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Maple [A]  time = 0.04, size = 23, normalized size = 0.9 \begin{align*}{\frac{\cos \left ( dx+c \right ) + \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a-sin(d*x+c)^2*a),x)

[Out]

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

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Maxima [A]  time = 0.949182, size = 36, normalized size = 1.33 \begin{align*} \frac{\frac{\cos \left (d x + c\right )}{a} + \frac{1}{a \cos \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-a*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.91117, size = 55, normalized size = 2.04 \begin{align*} \frac{\cos \left (d x + c\right )^{2} + 1}{a d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-a*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 15.0018, size = 36, normalized size = 1.33 \begin{align*} \begin{cases} - \frac{4}{a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - a d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{3}{\left (c \right )}}{- a \sin ^{2}{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**3/(a-a*sin(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.10316, size = 39, normalized size = 1.44 \begin{align*} \frac{\cos \left (d x + c\right )}{a d} + \frac{1}{a d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-a*sin(d*x+c)^2),x, algorithm="giac")

[Out]

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