3.1.49 \(\int \frac {\sin ^7(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [49]

Optimal. Leaf size=65 \[ -\frac {3 \cos (c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {3 \sec (c+d x)}{a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 65, 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 = {3254, 2670, 276} \begin {gather*} \frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {3 \cos (c+d x)}{a^2 d}+\frac {\sec ^3(c+d x)}{3 a^2 d}-\frac {3 \sec (c+d x)}{a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^7/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

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

Rule 276

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 2670

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 3254

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 ^7(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx &=\frac {\int \sin ^3(c+d x) \tan ^4(c+d x) \, dx}{a^2}\\ &=-\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\text {Subst}\left (\int \left (3+\frac {1}{x^4}-\frac {3}{x^2}-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {3 \cos (c+d x)}{a^2 d}+\frac {\cos ^3(c+d x)}{3 a^2 d}-\frac {3 \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 = 0.04, size = 59, normalized size = 0.91 \begin {gather*} \frac {-\frac {11 \cos (c+d x)}{4 d}+\frac {\cos (3 (c+d x))}{12 d}-\frac {3 \sec (c+d x)}{d}+\frac {\sec ^3(c+d x)}{3 d}}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^7/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

((-11*Cos[c + d*x])/(4*d) + Cos[3*(c + d*x)]/(12*d) - (3*Sec[c + d*x])/d + Sec[c + d*x]^3/(3*d))/a^2

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Maple [A]
time = 0.16, size = 47, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}-3 \cos \left (d x +c \right )-\frac {3}{\cos \left (d x +c \right )}+\frac {1}{3 \cos \left (d x +c \right )^{3}}}{d \,a^{2}}\)	\(47\)
default	\(\frac {\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}-3 \cos \left (d x +c \right )-\frac {3}{\cos \left (d x +c \right )}+\frac {1}{3 \cos \left (d x +c \right )^{3}}}{d \,a^{2}}\)	\(47\)
risch	\(\frac {{\mathrm e}^{3 i \left (d x +c \right )}}{24 d \,a^{2}}-\frac {11 \,{\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {11 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{2}}+\frac {{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d \,a^{2}}-\frac {2 \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}+14 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^7/(a-a*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 52, normalized size = 0.80 \begin {gather*} \frac {\frac {\cos \left (d x + c\right )^{3} - 9 \, \cos \left (d x + c\right )}{a^{2}} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{a^{2} \cos \left (d x + c\right )^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/3*((cos(d*x + c)^3 - 9*cos(d*x + c))/a^2 - (9*cos(d*x + c)^2 - 1)/(a^2*cos(d*x + c)^3))/d

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Fricas [A]
time = 0.40, size = 46, normalized size = 0.71 \begin {gather*} \frac {\cos \left (d x + c\right )^{6} - 9 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} + 1}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/3*(cos(d*x + c)^6 - 9*cos(d*x + c)^4 - 9*cos(d*x + c)^2 + 1)/(a^2*d*cos(d*x + c)^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (56) = 112\).
time = 29.63, size = 156, normalized size = 2.40 \begin {gather*} \begin {cases} - \frac {96 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} + \frac {32}{3 a^{2} d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{7}{\left (c \right )}}{\left (- a \sin ^{2}{\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**7/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((-96*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**12 - 9*a**2*d*tan(c/2 + d*x/2)**8 + 9*a**2*d*ta
n(c/2 + d*x/2)**4 - 3*a**2*d) + 32/(3*a**2*d*tan(c/2 + d*x/2)**12 - 9*a**2*d*tan(c/2 + d*x/2)**8 + 9*a**2*d*ta
n(c/2 + d*x/2)**4 - 3*a**2*d), Ne(d, 0)), (x*sin(c)**7/(-a*sin(c)**2 + a)**2, True))

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Giac [A]
time = 0.45, size = 57, normalized size = 0.88 \begin {gather*} -\frac {32 \, {\left (\frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}}{3 \, a^{2} d {\left (\frac {{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-32/3*(3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1)/(a^2*d*((cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1)
^3)

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Mupad [B]
time = 13.64, size = 48, normalized size = 0.74 \begin {gather*} -\frac {-{\cos \left (c+d\,x\right )}^6+9\,{\cos \left (c+d\,x\right )}^4+9\,{\cos \left (c+d\,x\right )}^2-1}{3\,a^2\,d\,{\cos \left (c+d\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^7/(a - a*sin(c + d*x)^2)^2,x)

[Out]

-(9*cos(c + d*x)^2 + 9*cos(c + d*x)^4 - cos(c + d*x)^6 - 1)/(3*a^2*d*cos(c + d*x)^3)

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