3.50 \(\int \frac{\sin ^5(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=47 \[ -\frac{\cos (c+d x)}{a^2 d}+\frac{\sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0694904, antiderivative size = 47, 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, 270} \[ -\frac{\cos (c+d x)}{a^2 d}+\frac{\sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

-(Cos[c + d*x]/(a^2*d)) - (2*Sec[c + d*x])/(a^2*d) + Sec[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 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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^5(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx &=\frac{\int \sin (c+d x) \tan ^4(c+d x) \, dx}{a^2}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^4} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}-\frac{2}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac{\cos (c+d x)}{a^2 d}-\frac{2 \sec (c+d x)}{a^2 d}+\frac{\sec ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0367936, size = 42, normalized size = 0.89 \[ \frac{-\frac{\cos (c+d x)}{d}+\frac{\sec ^3(c+d x)}{3 d}-\frac{2 \sec (c+d x)}{d}}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.042, size = 37, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{2}d} \left ( -\cos \left ( dx+c \right ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{-1}+{\frac{1}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a-sin(d*x+c)^2*a)^2,x)

[Out]

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

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Maxima [A]  time = 0.950131, size = 55, normalized size = 1.17 \begin{align*} -\frac{\frac{3 \, \cos \left (d x + c\right )}{a^{2}} + \frac{6 \, \cos \left (d x + c\right )^{2} - 1}{a^{2} \cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.6276, size = 96, normalized size = 2.04 \begin{align*} -\frac{3 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} - 1}{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)^5/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 93.1717, size = 156, normalized size = 3.32 \begin{align*} \begin{cases} - \frac{32 \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 6 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} + \frac{16}{3 a^{2} d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 6 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 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 ^{5}{\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)**5/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((-32*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**8 - 6*a**2*d*tan(c/2 + d*x/2)**6 + 6*a**2*d*tan
(c/2 + d*x/2)**2 - 3*a**2*d) + 16/(3*a**2*d*tan(c/2 + d*x/2)**8 - 6*a**2*d*tan(c/2 + d*x/2)**6 + 6*a**2*d*tan(
c/2 + d*x/2)**2 - 3*a**2*d), Ne(d, 0)), (x*sin(c)**5/(-a*sin(c)**2 + a)**2, True))

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Giac [B]  time = 1.17277, size = 143, normalized size = 3.04 \begin{align*} \frac{2 \,{\left (\frac{3}{a^{2}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}} - \frac{\frac{12 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 5}{a^{2}{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

2/3*(3/(a^2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) - (12*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*(cos(
d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 5)/(a^2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3))/d