3.1.56 \(\int \frac {\sin ^4(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [56]

Optimal. Leaf size=38 \[ \frac {x}{a^2}-\frac {\tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 38, 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, 3554, 8} \begin {gather*} \frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan (c+d x)}{a^2 d}+\frac {x}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 42, normalized size = 1.11 \begin {gather*} \frac {\frac {\tan ^{-1}(\tan (c+d x))}{d}-\frac {\tan (c+d x)}{d}+\frac {\tan ^3(c+d x)}{3 d}}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^4/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

(ArcTan[Tan[c + d*x]]/d - Tan[c + d*x]/d + Tan[c + d*x]^3/(3*d))/a^2

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Maple [A]
time = 0.20, size = 34, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )}{d \,a^{2}}\)	\(34\)
default	\(\frac {\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )}{d \,a^{2}}\)	\(34\)
risch	\(\frac {x}{a^{2}}-\frac {4 i \left (3 \,{\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}+2\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(53\)
norman	\(\frac {\frac {x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x}{a}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {38 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {24 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {38 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {4 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.52, size = 37, normalized size = 0.97 \begin {gather*} \frac {\frac {\tan \left (d x + c\right )^{3} - 3 \, \tan \left (d x + c\right )}{a^{2}} + \frac {3 \, {\left (d x + c\right )}}{a^{2}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 49, normalized size = 1.29 \begin {gather*} \frac {3 \, d x \cos \left (d x + c\right )^{3} - {\left (4 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{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)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/3*(3*d*x*cos(d*x + c)^3 - (4*cos(d*x + c)^2 - 1)*sin(d*x + c))/(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. 551 vs. \(2 (31) = 62\).
time = 8.06, size = 551, normalized size = 14.50 \begin {gather*} \begin {cases} \frac {3 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} - \frac {9 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} + \frac {9 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} - \frac {3 d x}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} + \frac {6 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} - \frac {20 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 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 ^{4}{\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)**4/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((3*d*x*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*t
an(c/2 + d*x/2)**2 - 3*a**2*d) - 9*d*x*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 +
d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) + 9*d*x*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**6
- 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) - 3*d*x/(3*a**2*d*tan(c/2 + d*x/2)*
*6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) + 6*tan(c/2 + d*x/2)**5/(3*a**2*d
*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) - 20*tan(c/2 +
d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2
*d) + 6*tan(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x
/2)**2 - 3*a**2*d), Ne(d, 0)), (x*sin(c)**4/(-a*sin(c)**2 + a)**2, True))

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Giac [A]
time = 0.44, size = 44, normalized size = 1.16 \begin {gather*} \frac {\frac {3 \, {\left (d x + c\right )}}{a^{2}} + \frac {a^{4} \tan \left (d x + c\right )^{3} - 3 \, a^{4} \tan \left (d x + c\right )}{a^{6}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 13.48, size = 31, normalized size = 0.82 \begin {gather*} \frac {x}{a^2}-\frac {\mathrm {tan}\left (c+d\,x\right )-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3}}{a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4/(a - a*sin(c + d*x)^2)^2,x)

[Out]

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

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