\(\int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [37]

Optimal result

Integrand size = 24, antiderivative size = 27 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\cos (c+d x)}{a d}+\frac {\sec (c+d x)}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3254, 2670, 14} \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\cos (c+d x)}{a d}+\frac {\sec (c+d x)}{a d} \]

[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 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]]

Rule 2670

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sin (c+d x) \tan ^2(c+d x) \, dx}{a} \\ & = -\frac {\text {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {\cos (c+d x)}{d}+\frac {\sec (c+d x)}{d}}{a} \]

[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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\cos \left (d x +c \right )+\frac {1}{\cos \left (d x +c \right )}}{d a}\)	\(23\)
default	\(\frac {\cos \left (d x +c \right )+\frac {1}{\cos \left (d x +c \right )}}{d a}\)	\(23\)
parallelrisch	\(\frac {3+\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )}{2 a d \cos \left (d x +c \right )}\)	\(36\)
risch	\(\frac {{\mathrm e}^{3 i \left (d x +c \right )}+7 \cos \left (d x +c \right )+5 i \sin \left (d x +c \right )}{2 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(49\)
norman	\(\frac {-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {4}{a d}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\cos \left (d x + c\right )^{2} + 1}{a d \cos \left (d x + c\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\begin {cases} - \frac {4}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\left (c \right )}}{- a \sin ^{2}{\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {\cos \left (d x + c\right )}{a} + \frac {1}{a \cos \left (d x + c\right )}}{d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\cos \left (d x + c\right )}{a d} + \frac {1}{a d \cos \left (d x + c\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {{\cos \left (c+d\,x\right )}^2+1}{a\,d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

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