3.1.55 \(\int \frac {\sin ^6(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [55]

Optimal. Leaf size=69 \[ \frac {5 x}{2 a^2}-\frac {5 \tan (c+d x)}{2 a^2 d}+\frac {5 \tan ^3(c+d x)}{6 a^2 d}-\frac {\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d} \]

[Out]

5/2*x/a^2-5/2*tan(d*x+c)/a^2/d+5/6*tan(d*x+c)^3/a^2/d-1/2*sin(d*x+c)^2*tan(d*x+c)^3/a^2/d

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Rubi [A]
time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3254, 2671, 294, 308, 209} \begin {gather*} \frac {5 \tan ^3(c+d x)}{6 a^2 d}-\frac {5 \tan (c+d x)}{2 a^2 d}-\frac {\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {5 x}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

(5*x)/(2*a^2) - (5*Tan[c + d*x])/(2*a^2*d) + (5*Tan[c + d*x]^3)/(6*a^2*d) - (Sin[c + d*x]^2*Tan[c + d*x]^3)/(2
*a^2*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/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 ^6(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx &=\frac {\int \sin ^2(c+d x) \tan ^4(c+d x) \, dx}{a^2}\\ &=\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{a^2 d}\\ &=-\frac {\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {5 \text {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=-\frac {\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {5 \text {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=-\frac {5 \tan (c+d x)}{2 a^2 d}+\frac {5 \tan ^3(c+d x)}{6 a^2 d}-\frac {\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac {5 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=\frac {5 x}{2 a^2}-\frac {5 \tan (c+d x)}{2 a^2 d}+\frac {5 \tan ^3(c+d x)}{6 a^2 d}-\frac {\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 46, normalized size = 0.67 \begin {gather*} \frac {30 (c+d x)-3 \sin (2 (c+d x))+4 \left (-7+\sec ^2(c+d x)\right ) \tan (c+d x)}{12 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

(30*(c + d*x) - 3*Sin[2*(c + d*x)] + 4*(-7 + Sec[c + d*x]^2)*Tan[c + d*x])/(12*a^2*d)

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Maple [A]
time = 0.14, size = 56, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-2 \tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )}{2 \left (\tan ^{2}\left (d x +c \right )+1\right )}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{d \,a^{2}}\)	\(56\)
default	\(\frac {\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-2 \tan \left (d x +c \right )-\frac {\tan \left (d x +c \right )}{2 \left (\tan ^{2}\left (d x +c \right )+1\right )}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{d \,a^{2}}\)	\(56\)
risch	\(\frac {5 x}{2 a^{2}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {2 i \left (9 \,{\mathrm e}^{4 i \left (d x +c \right )}+12 \,{\mathrm e}^{2 i \left (d x +c \right )}+7\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a-a*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(1/3*tan(d*x+c)^3-2*tan(d*x+c)-1/2*tan(d*x+c)/(tan(d*x+c)^2+1)+5/2*arctan(tan(d*x+c)))

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Maxima [A]
time = 0.53, size = 64, normalized size = 0.93 \begin {gather*} -\frac {\frac {3 \, \tan \left (d x + c\right )}{a^{2} \tan \left (d x + c\right )^{2} + a^{2}} - \frac {2 \, {\left (\tan \left (d x + c\right )^{3} - 6 \, \tan \left (d x + c\right )\right )}}{a^{2}} - \frac {15 \, {\left (d x + c\right )}}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/6*(3*tan(d*x + c)/(a^2*tan(d*x + c)^2 + a^2) - 2*(tan(d*x + c)^3 - 6*tan(d*x + c))/a^2 - 15*(d*x + c)/a^2)/
d

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Fricas [A]
time = 0.39, size = 59, normalized size = 0.86 \begin {gather*} \frac {15 \, d x \cos \left (d x + c\right )^{3} - {\left (3 \, \cos \left (d x + c\right )^{4} + 14 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{6 \, 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)^6/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/6*(15*d*x*cos(d*x + c)^3 - (3*cos(d*x + c)^4 + 14*cos(d*x + c)^2 - 2)*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. 1275 vs. \(2 (63) = 126\).
time = 21.95, size = 1275, normalized size = 18.48 \begin {gather*} \begin {cases} \frac {15 d x \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} - \frac {15 d x \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} - \frac {30 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} + \frac {30 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} + \frac {15 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} - \frac {15 d x}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} + \frac {30 \tan ^{9}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} - \frac {40 \tan ^{7}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} - \frac {44 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} - \frac {40 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} + \frac {30 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{6}{\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)**6/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((15*d*x*tan(c/2 + d*x/2)**10/(6*a**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2
*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d) - 15*d*x*tan
(c/2 + d*x/2)**8/(6*a**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6
+ 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d) - 30*d*x*tan(c/2 + d*x/2)**6/(6*a*
*2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 +
d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d) + 30*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)*
*10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*
tan(c/2 + d*x/2)**2 - 6*a**2*d) + 15*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2
+ d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 -
6*a**2*d) - 15*d*x/(6*a**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)*
*6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d) + 30*tan(c/2 + d*x/2)**9/(6*a**2
*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 + d
*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d) - 40*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**10 -
6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/
2 + d*x/2)**2 - 6*a**2*d) - 44*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)*
*8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d)
- 40*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d
*x/2)**6 + 12*a**2*d*tan(c/2 + d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d) + 30*tan(c/2 + d*x/2)/(6*a
**2*d*tan(c/2 + d*x/2)**10 - 6*a**2*d*tan(c/2 + d*x/2)**8 - 12*a**2*d*tan(c/2 + d*x/2)**6 + 12*a**2*d*tan(c/2
+ d*x/2)**4 + 6*a**2*d*tan(c/2 + d*x/2)**2 - 6*a**2*d), Ne(d, 0)), (x*sin(c)**6/(-a*sin(c)**2 + a)**2, True))

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Giac [A]
time = 0.45, size = 68, normalized size = 0.99 \begin {gather*} \frac {\frac {15 \, {\left (d x + c\right )}}{a^{2}} - \frac {3 \, \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} a^{2}} + \frac {2 \, {\left (a^{4} \tan \left (d x + c\right )^{3} - 6 \, a^{4} \tan \left (d x + c\right )\right )}}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)/a^2 - 3*tan(d*x + c)/((tan(d*x + c)^2 + 1)*a^2) + 2*(a^4*tan(d*x + c)^3 - 6*a^4*tan(d*x + c)
)/a^6)/d

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Mupad [B]
time = 13.81, size = 66, normalized size = 0.96 \begin {gather*} \frac {5\,x}{2\,a^2}-\frac {\mathrm {tan}\left (c+d\,x\right )}{2\,d\,\left (a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2+a^2\right )}-\frac {2\,\mathrm {tan}\left (c+d\,x\right )}{a^2\,d}+\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)^6/(a - a*sin(c + d*x)^2)^2,x)

[Out]

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

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