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

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Rubi [A]  time = 0.07, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.43, size = 38, normalized size = 0.81 \[ -\frac {3 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} - 1}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \]

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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giac [B]  time = 0.16, size = 106, normalized size = 2.26 \[ \frac {2 \, {\left (\frac {3}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}} - \frac {\frac {12 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 5}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}\right )}}{3 \, d} \]

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

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maple [A]  time = 0.28, size = 37, normalized size = 0.79 \[ \frac {-\cos \left (d x +c \right )-\frac {2}{\cos \left (d x +c \right )}+\frac {1}{3 \cos \left (d x +c \right )^{3}}}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a-a*sin(d*x+c)^2)^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 = 0.34, size = 41, normalized size = 0.87 \[ -\frac {\frac {3 \, \cos \left (d x + c\right )}{a^{2}} + \frac {6 \, \cos \left (d x + c\right )^{2} - 1}{a^{2} \cos \left (d x + c\right )^{3}}}{3 \, d} \]

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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mupad [B]  time = 0.05, size = 36, normalized size = 0.77 \[ -\frac {{\cos \left (c+d\,x\right )}^4+2\,{\cos \left (c+d\,x\right )}^2-\frac {1}{3}}{a^2\,d\,{\cos \left (c+d\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 49.03, size = 156, normalized size = 3.32 \[ \begin {cases} - \frac {32 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} + \frac {16}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{5}{\relax (c )}}{\left (- a \sin ^{2}{\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

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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