3.55 \(\int \frac{\sin ^6(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\)
Optimal. Leaf size=69 \[ \frac{5 \tan ^3(c+d x)}{6 a^2 d}-\frac{5 \tan (c+d x)}{2 a^2 d}-\frac{\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac{5 x}{2 a^2} \]
[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)
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Rubi [A] time = 0.0818862, antiderivative size = 69, normalized size of antiderivative = 1.,
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 =
{3175, 2591, 288, 302, 203} \[ \frac{5 \tan ^3(c+d x)}{6 a^2 d}-\frac{5 \tan (c+d x)}{2 a^2 d}-\frac{\sin ^2(c+d x) \tan ^3(c+d x)}{2 a^2 d}+\frac{5 x}{2 a^2} \]
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 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 2591
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])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Rule 288
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 302
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
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{\operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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*}
Mathematica [A] time = 0.210398, size = 46, normalized size = 0.67 \[ \frac{30 (c+d x)-3 \sin (2 (c+d x))+4 \tan (c+d x) \left (\sec ^2(c+d x)-7\right )}{12 a^2 d} \]
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 = 0.046, size = 73, normalized size = 1.1 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{2}d}}-2\,{\frac{\tan \left ( dx+c \right ) }{{a}^{2}d}}-{\frac{\tan \left ( dx+c \right ) }{2\,{a}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{5\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{2\,{a}^{2}d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(d*x+c)^6/(a-sin(d*x+c)^2*a)^2,x)
[Out]
1/3*tan(d*x+c)^3/a^2/d-2*tan(d*x+c)/a^2/d-1/2/d/a^2*tan(d*x+c)/(tan(d*x+c)^2+1)+5/2/d/a^2*arctan(tan(d*x+c))
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Maxima [A] time = 1.46678, size = 86, normalized size = 1.25 \begin{align*} -\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{align*}
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 = 1.70056, size = 149, normalized size = 2.16 \begin{align*} \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{align*}
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 [A] time = 144.461, size = 1275, normalized size = 18.48 \begin{align*} \text{result too large to display} \end{align*}
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 = 1.16731, size = 92, normalized size = 1.33 \begin{align*} \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{align*}
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