Integrand size = 24, antiderivative size = 49 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {3 x}{2 a}+\frac {3 \tan (c+d x)}{2 a d}-\frac {\sin ^2(c+d x) \tan (c+d x)}{2 a d} \]
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {-6 (c+d x)+\sin (2 (c+d x))+4 \tan (c+d x)}{4 a d} \]
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3654, 3042, 3071, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4}{a-a \sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \frac {\int \sin ^2(c+d x) \tan ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin (c+d x)^2 \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3071 |
\(\displaystyle \frac {\int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)}{a d}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {3}{2} \int \frac {\tan ^2(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}}{a d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {3}{2} \left (\tan (c+d x)-\int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)\right )-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}}{a d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}}{a d}\) |
((3*(-ArcTan[Tan[c + d*x]] + Tan[c + d*x]))/2 - Tan[c + d*x]^3/(2*(1 + Tan [c + d*x]^2)))/(a*d)
3.1.43.3.1 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 = 0.46 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )}{2+2 \left (\tan ^{2}\left (d x +c \right )\right )}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{d a}\) | \(44\) |
default | \(\frac {\tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )}{2+2 \left (\tan ^{2}\left (d x +c \right )\right )}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{d a}\) | \(44\) |
parallelrisch | \(\frac {-12 d x \cos \left (d x +c \right )+9 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )}{8 a d \cos \left (d x +c \right )}\) | \(45\) |
risch | \(-\frac {3 x}{2 a}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d a}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d a}+\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\) | \(66\) |
norman | \(\frac {\frac {3 x}{2 a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {9 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {3 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {9 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(217\) |
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {3 \, d x \cos \left (d x + c\right ) - {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{2 \, a d \cos \left (d x + c\right )} \]
Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (39) = 78\).
Time = 2.56 (sec) , antiderivative size = 502, normalized size of antiderivative = 10.24 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\begin {cases} - \frac {3 d x \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} - \frac {3 d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} + \frac {3 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} + \frac {3 d x}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} - \frac {6 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} - \frac {4 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} - \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )}}{- a \sin ^{2}{\left (c \right )} + a} & \text {otherwise} \end {cases} \]
Piecewise((-3*d*x*tan(c/2 + d*x/2)**6/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*t an(c/2 + d*x/2)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) - 3*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*t an(c/2 + d*x/2)**2 - 2*a*d) + 3*d*x*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d *x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) + 3*d*x/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*tan (c/2 + d*x/2)**2 - 2*a*d) - 6*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)* *6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) - 4*ta n(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) - 6*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) , Ne(d, 0)), (x*sin(c)**4/(-a*sin(c)**2 + a), True))
Time = 0.38 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )}}{a} - \frac {\tan \left (d x + c\right )}{a \tan \left (d x + c\right )^{2} + a} - \frac {2 \, \tan \left (d x + c\right )}{a}}{2 \, d} \]
Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )}}{a} - \frac {2 \, \tan \left (d x + c\right )}{a} - \frac {\tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} a}}{2 \, d} \]
Time = 13.72 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{2\,d\,\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}-\frac {3\,x}{2\,a}+\frac {\mathrm {tan}\left (c+d\,x\right )}{a\,d} \]