3.56 \(\int \frac{\sin ^4(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{\tan ^3(c+d x)}{3 a^2 d}-\frac{\tan (c+d x)}{a^2 d}+\frac{x}{a^2} \]

[Out]

x/a^2 - Tan[c + d*x]/(a^2*d) + Tan[c + d*x]^3/(3*a^2*d)

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Rubi [A]  time = 0.0556989, antiderivative size = 38, 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, 3473, 8} \[ \frac{\tan ^3(c+d x)}{3 a^2 d}-\frac{\tan (c+d x)}{a^2 d}+\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

x/a^2 - Tan[c + d*x]/(a^2*d) + Tan[c + d*x]^3/(3*a^2*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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\sin ^4(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx &=\frac{\int \tan ^4(c+d x) \, dx}{a^2}\\ &=\frac{\tan ^3(c+d x)}{3 a^2 d}-\frac{\int \tan ^2(c+d x) \, dx}{a^2}\\ &=-\frac{\tan (c+d x)}{a^2 d}+\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{\int 1 \, dx}{a^2}\\ &=\frac{x}{a^2}-\frac{\tan (c+d x)}{a^2 d}+\frac{\tan ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.014339, size = 42, normalized size = 1.11 \[ \frac{\frac{\tan ^3(c+d x)}{3 d}+\frac{\tan ^{-1}(\tan (c+d x))}{d}-\frac{\tan (c+d x)}{d}}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

(ArcTan[Tan[c + d*x]]/d - Tan[c + d*x]/d + Tan[c + d*x]^3/(3*d))/a^2

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Maple [A]  time = 0.036, size = 46, normalized size = 1.2 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{2}d}}-{\frac{\tan \left ( dx+c \right ) }{{a}^{2}d}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a-sin(d*x+c)^2*a)^2,x)

[Out]

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

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Maxima [A]  time = 1.43872, size = 50, normalized size = 1.32 \begin{align*} \frac{\frac{\tan \left (d x + c\right )^{3} - 3 \, \tan \left (d x + c\right )}{a^{2}} + \frac{3 \,{\left (d x + c\right )}}{a^{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.5878, size = 120, normalized size = 3.16 \begin{align*} \frac{3 \, d x \cos \left (d x + c\right )^{3} -{\left (4 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/3*(3*d*x*cos(d*x + c)^3 - (4*cos(d*x + c)^2 - 1)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3)

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Sympy [A]  time = 61.3636, size = 551, normalized size = 14.5 \begin{align*} \begin{cases} \frac{3 d x \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} - \frac{9 d x \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} + \frac{9 d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} - \frac{3 d x}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} + \frac{6 \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} - \frac{20 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} + \frac{6 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{4}{\left (c \right )}}{\left (- a \sin ^{2}{\left (c \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**4/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*t
an(c/2 + d*x/2)**2 - 3*a**2*d) - 9*d*x*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 +
d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) + 9*d*x*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**6
- 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) - 3*d*x/(3*a**2*d*tan(c/2 + d*x/2)*
*6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) + 6*tan(c/2 + d*x/2)**5/(3*a**2*d
*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) - 20*tan(c/2 +
d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2
*d) + 6*tan(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x
/2)**2 - 3*a**2*d), Ne(d, 0)), (x*sin(c)**4/(-a*sin(c)**2 + a)**2, True))

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Giac [A]  time = 1.15668, size = 59, normalized size = 1.55 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}}{a^{2}} + \frac{a^{4} \tan \left (d x + c\right )^{3} - 3 \, a^{4} \tan \left (d x + c\right )}{a^{6}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)/a^2 + (a^4*tan(d*x + c)^3 - 3*a^4*tan(d*x + c))/a^6)/d