3.42 \(\int \frac{\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx\)
Optimal. Leaf size=73 \[ \frac{15 \tan (c+d x)}{8 a d}-\frac{\sin ^4(c+d x) \tan (c+d x)}{4 a d}-\frac{5 \sin ^2(c+d x) \tan (c+d x)}{8 a d}-\frac{15 x}{8 a} \]
[Out]
(-15*x)/(8*a) + (15*Tan[c + d*x])/(8*a*d) - (5*Sin[c + d*x]^2*Tan[c + d*x])/(8*a*d) - (Sin[c + d*x]^4*Tan[c +
d*x])/(4*a*d)
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Rubi [A] time = 0.0902474, antiderivative size = 73, 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, 321, 203} \[ \frac{15 \tan (c+d x)}{8 a d}-\frac{\sin ^4(c+d x) \tan (c+d x)}{4 a d}-\frac{5 \sin ^2(c+d x) \tan (c+d x)}{8 a d}-\frac{15 x}{8 a} \]
Antiderivative was successfully verified.
[In]
Int[Sin[c + d*x]^6/(a - a*Sin[c + d*x]^2),x]
[Out]
(-15*x)/(8*a) + (15*Tan[c + d*x])/(8*a*d) - (5*Sin[c + d*x]^2*Tan[c + d*x])/(8*a*d) - (Sin[c + d*x]^4*Tan[c +
d*x])/(4*a*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 321
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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
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)}{a-a \sin ^2(c+d x)} \, dx &=\frac{\int \sin ^4(c+d x) \tan ^2(c+d x) \, dx}{a}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac{\sin ^4(c+d x) \tan (c+d x)}{4 a d}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 a d}\\ &=-\frac{5 \sin ^2(c+d x) \tan (c+d x)}{8 a d}-\frac{\sin ^4(c+d x) \tan (c+d x)}{4 a d}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 a d}\\ &=\frac{15 \tan (c+d x)}{8 a d}-\frac{5 \sin ^2(c+d x) \tan (c+d x)}{8 a d}-\frac{\sin ^4(c+d x) \tan (c+d x)}{4 a d}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 a d}\\ &=-\frac{15 x}{8 a}+\frac{15 \tan (c+d x)}{8 a d}-\frac{5 \sin ^2(c+d x) \tan (c+d x)}{8 a d}-\frac{\sin ^4(c+d x) \tan (c+d x)}{4 a d}\\ \end{align*}
Mathematica [A] time = 0.189153, size = 44, normalized size = 0.6 \[ -\frac{-16 \sin (2 (c+d x))+\sin (4 (c+d x))-32 \tan (c+d x)+60 c+60 d x}{32 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[c + d*x]^6/(a - a*Sin[c + d*x]^2),x]
[Out]
-(60*c + 60*d*x - 16*Sin[2*(c + d*x)] + Sin[4*(c + d*x)] - 32*Tan[c + d*x])/(32*a*d)
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Maple [A] time = 0.046, size = 84, normalized size = 1.2 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{da}}+{\frac{9\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{8\,da \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{7\,\tan \left ( dx+c \right ) }{8\,da \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{15\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{8\,da}} \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),x)
[Out]
tan(d*x+c)/d/a+9/8/d/a/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3+7/8/d/a/(tan(d*x+c)^2+1)^2*tan(d*x+c)-15/8/d/a*arctan(t
an(d*x+c))
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Maxima [A] time = 1.45398, size = 97, normalized size = 1.33 \begin{align*} \frac{\frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{a \tan \left (d x + c\right )^{4} + 2 \, a \tan \left (d x + c\right )^{2} + a} - \frac{15 \,{\left (d x + c\right )}}{a} + \frac{8 \, \tan \left (d x + c\right )}{a}}{8 \, 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),x, algorithm="maxima")
[Out]
1/8*((9*tan(d*x + c)^3 + 7*tan(d*x + c))/(a*tan(d*x + c)^4 + 2*a*tan(d*x + c)^2 + a) - 15*(d*x + c)/a + 8*tan(
d*x + c)/a)/d
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Fricas [A] time = 1.64148, size = 140, normalized size = 1.92 \begin{align*} -\frac{15 \, d x \cos \left (d x + c\right ) +{\left (2 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right )}{8 \, a d \cos \left (d x + c\right )} \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),x, algorithm="fricas")
[Out]
-1/8*(15*d*x*cos(d*x + c) + (2*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - 8)*sin(d*x + c))/(a*d*cos(d*x + c))
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Sympy [A] time = 107.4, size = 1161, normalized size = 15.9 \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),x)
[Out]
Piecewise((-15*d*x*tan(c/2 + d*x/2)**10/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(
c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) - 45*d*x*tan(c/2 + d*x/2)**
8/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x
/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) - 30*d*x*tan(c/2 + d*x/2)**6/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d
*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 -
8*a*d) + 30*d*x*tan(c/2 + d*x/2)**4/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2
+ d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) + 45*d*x*tan(c/2 + d*x/2)**2/(8
*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)*
*4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) + 15*d*x/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 1
6*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) - 30*tan(c/2 + d*
x/2)**9/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2
+ d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) - 80*tan(c/2 + d*x/2)**7/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a
*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2
- 8*a*d) - 36*tan(c/2 + d*x/2)**5/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 +
d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) - 80*tan(c/2 + d*x/2)**3/(8*a*d*t
an(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 2
4*a*d*tan(c/2 + d*x/2)**2 - 8*a*d) - 30*tan(c/2 + d*x/2)/(8*a*d*tan(c/2 + d*x/2)**10 + 24*a*d*tan(c/2 + d*x/2)
**8 + 16*a*d*tan(c/2 + d*x/2)**6 - 16*a*d*tan(c/2 + d*x/2)**4 - 24*a*d*tan(c/2 + d*x/2)**2 - 8*a*d), Ne(d, 0))
, (x*sin(c)**6/(-a*sin(c)**2 + a), True))
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Giac [A] time = 1.14889, size = 85, normalized size = 1.16 \begin{align*} -\frac{\frac{15 \,{\left (d x + c\right )}}{a} - \frac{8 \, \tan \left (d x + c\right )}{a} - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} a}}{8 \, 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),x, algorithm="giac")
[Out]
-1/8*(15*(d*x + c)/a - 8*tan(d*x + c)/a - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/((tan(d*x + c)^2 + 1)^2*a))/d