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

Optimal. Leaf size=46 \[ -\frac{\cos ^3(c+d x)}{3 a d}+\frac{2 \cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0802002, antiderivative size = 46, 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} \[ -\frac{\cos ^3(c+d x)}{3 a d}+\frac{2 \cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0436003, size = 43, normalized size = 0.93 \[ \frac{\frac{7 \cos (c+d x)}{4 d}-\frac{\cos (3 (c+d x))}{12 d}+\frac{\sec (c+d x)}{d}}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*Cos[c + d*x])/(4*d) - Cos[3*(c + d*x)]/(12*d) + Sec[c + d*x]/d)/a

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Maple [A]  time = 0.04, size = 35, normalized size = 0.8 \begin{align*}{\frac{1}{da} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3}}+2\,\cos \left ( dx+c \right ) + \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \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),x)

[Out]

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

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Maxima [A]  time = 0.941479, size = 54, normalized size = 1.17 \begin{align*} -\frac{\frac{\cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )}{a} - \frac{3}{a \cos \left (d x + c\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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.86087, size = 88, normalized size = 1.91 \begin{align*} -\frac{\cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{2} - 3}{3 \, 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)^5/(a-a*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 42.9334, size = 143, normalized size = 3.11 \begin{align*} \begin{cases} - \frac{32 \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 6 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a d} - \frac{16}{3 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 6 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 6 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{5}{\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)**5/(a-a*sin(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 1.1816, size = 142, normalized size = 3.09 \begin{align*} \frac{2 \,{\left (\frac{3}{a{\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{\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),x, algorithm="giac")

[Out]

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