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

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Rubi [A]  time = 0.08, antiderivative size = 46, 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} \[ -\frac {\cos ^3(c+d x)}{3 a d}+\frac {2 \cos (c+d x)}{a d}+\frac {\sec (c+d x)}{a d} \]

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 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)}{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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.44, size = 36, normalized size = 0.78 \[ -\frac {\cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{2} - 3}{3 \, a d \cos \left (d x + c\right )} \]

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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giac [B]  time = 0.14, size = 105, normalized size = 2.28 \[ \frac {2 \, {\left (\frac {3}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}} + \frac {\frac {12 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 5}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{3}}\right )}}{3 \, d} \]

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

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maple [A]  time = 0.31, size = 35, normalized size = 0.76 \[ \frac {-\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+2 \cos \left (d x +c \right )+\frac {1}{\cos \left (d x +c \right )}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a-a*sin(d*x+c)^2),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 = 0.36, size = 40, normalized size = 0.87 \[ -\frac {\frac {\cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )}{a} - \frac {3}{a \cos \left (d x + c\right )}}{3 \, d} \]

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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mupad [B]  time = 0.06, size = 38, normalized size = 0.83 \[ \frac {-{\cos \left (c+d\,x\right )}^4+6\,{\cos \left (c+d\,x\right )}^2+3}{3\,a\,d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 14.35, size = 143, normalized size = 3.11 \[ \begin {cases} - \frac {32 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a d} - \frac {16}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{5}{\relax (c )}}{- a \sin ^{2}{\relax (c )} + a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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