Integrand size = 24, antiderivative size = 73 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {15 x}{8 a}+\frac {15 \tan (c+d x)}{8 a d}-\frac {5 \sin ^2(c+d x) \tan (c+d x)}{8 a d}-\frac {\sin ^4(c+d x) \tan (c+d x)}{4 a d} \]
-15/8*x/a+15/8*tan(d*x+c)/a/d-5/8*sin(d*x+c)^2*tan(d*x+c)/a/d-1/4*sin(d*x+ c)^4*tan(d*x+c)/a/d
Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {60 c+60 d x-16 \sin (2 (c+d x))+\sin (4 (c+d x))-32 \tan (c+d x)}{32 a d} \]
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3654, 3042, 3071, 252, 252, 262, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^6}{a-a \sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \frac {\int \sin ^4(c+d x) \tan ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin (c+d x)^4 \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3071 |
\(\displaystyle \frac {\int \frac {\tan ^6(c+d x)}{\left (\tan ^2(c+d x)+1\right )^3}d\tan (c+d x)}{a d}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {5}{4} \int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)-\frac {\tan ^5(c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}}{a d}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {3}{2} \int \frac {\tan ^2(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )-\frac {\tan ^5(c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}}{a d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {3}{2} \left (\tan (c+d x)-\int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)\right )-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )-\frac {\tan ^5(c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}}{a d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )-\frac {\tan ^5(c+d x)}{4 \left (\tan ^2(c+d x)+1\right )^2}}{a d}\) |
(-1/4*Tan[c + d*x]^5/(1 + Tan[c + d*x]^2)^2 + (5*((3*(-ArcTan[Tan[c + d*x] ] + Tan[c + d*x]))/2 - Tan[c + d*x]^3/(2*(1 + Tan[c + d*x]^2))))/4)/(a*d)
3.1.42.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(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]
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ 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]
Time = 0.63 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\tan \left (d x +c \right )-\frac {-\frac {9 \left (\tan ^{3}\left (d x +c \right )\right )}{8}-\frac {7 \tan \left (d x +c \right )}{8}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {15 \arctan \left (\tan \left (d x +c \right )\right )}{8}}{d a}\) | \(57\) |
default | \(\frac {\tan \left (d x +c \right )-\frac {-\frac {9 \left (\tan ^{3}\left (d x +c \right )\right )}{8}-\frac {7 \tan \left (d x +c \right )}{8}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {15 \arctan \left (\tan \left (d x +c \right )\right )}{8}}{d a}\) | \(57\) |
parallelrisch | \(\frac {-120 d x \cos \left (d x +c \right )+80 \sin \left (d x +c \right )-\sin \left (5 d x +5 c \right )+15 \sin \left (3 d x +3 c \right )}{64 a d \cos \left (d x +c \right )}\) | \(58\) |
risch | \(-\frac {15 x}{8 a}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 d a}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d a}+\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\sin \left (4 d x +4 c \right )}{32 a d}\) | \(83\) |
norman | \(\frac {\frac {15 x}{8 a}-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {35 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {113 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {29 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {113 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}-\frac {35 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {15 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {75 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {135 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}+\frac {75 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {75 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {135 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {75 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}-\frac {15 x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(289\) |
1/d/a*(tan(d*x+c)-(-9/8*tan(d*x+c)^3-7/8*tan(d*x+c))/(1+tan(d*x+c)^2)^2-15 /8*arctan(tan(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {15 \, d x \cos \left (d x + c\right ) + {\left (2 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right )}{8 \, a d \cos \left (d x + c\right )} \]
-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))
Leaf count of result is larger than twice the leaf count of optimal. 1161 vs. \(2 (61) = 122\).
Time = 6.57 (sec) , antiderivative size = 1161, normalized size of antiderivative = 15.90 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\text {Too large to display} \]
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 + 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*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/...
Time = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{a \tan \left (d x + c\right )^{4} + 2 \, a \tan \left (d x + c\right )^{2} + a} - \frac {15 \, {\left (d x + c\right )}}{a} + \frac {8 \, \tan \left (d x + c\right )}{a}}{8 \, d} \]
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
Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{a} - \frac {8 \, \tan \left (d x + c\right )}{a} - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} a}}{8 \, d} \]
-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
Time = 14.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^6(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{a\,d}-\frac {15\,x}{8\,a}+\frac {\frac {9\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8}+\frac {7\,\mathrm {tan}\left (c+d\,x\right )}{8}}{d\,\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )} \]