Integrand size = 17, antiderivative size = 60 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {2 \sin (a+b x)}{b}+\frac {\sin ^3(a+b x)}{3 b}+\frac {\sec (a+b x) \tan (a+b x)}{2 b} \] Output:
-5/2*arctanh(sin(b*x+a))/b+2*sin(b*x+a)/b+1/3*sin(b*x+a)^3/b+1/2*sec(b*x+a )*tan(b*x+a)/b
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {5 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {5 \sin (a+b x) \tan ^2(a+b x)}{3 b}-\frac {\sin ^3(a+b x) \tan ^2(a+b x)}{3 b} \] Input:
Integrate[Sin[a + b*x]^3*Tan[a + b*x]^3,x]
Output:
(-5*ArcTanh[Sin[a + b*x]])/(2*b) + (5*Sec[a + b*x]*Tan[a + b*x])/(2*b) - ( 5*Sin[a + b*x]*Tan[a + b*x]^2)/(3*b) - (Sin[a + b*x]^3*Tan[a + b*x]^2)/(3* b)
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 3072, 252, 254, 2009}
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.
\(\displaystyle \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^3 \tan (a+b x)^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {\int \frac {\sin ^6(a+b x)}{\left (1-\sin ^2(a+b x)\right )^2}d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sin ^5(a+b x)}{2 \left (1-\sin ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\sin ^4(a+b x)}{1-\sin ^2(a+b x)}d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\frac {\sin ^5(a+b x)}{2 \left (1-\sin ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\sin ^2(a+b x)+\frac {1}{1-\sin ^2(a+b x)}-1\right )d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sin ^5(a+b x)}{2 \left (1-\sin ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sin (a+b x))-\frac {1}{3} \sin ^3(a+b x)-\sin (a+b x)\right )}{b}\) |
Input:
Int[Sin[a + b*x]^3*Tan[a + b*x]^3,x]
Output:
(Sin[a + b*x]^5/(2*(1 - Sin[a + b*x]^2)) - (5*(ArcTanh[Sin[a + b*x]] - Sin [a + b*x] - Sin[a + b*x]^3/3))/2)/b
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[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Time = 1.75 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{7}}{2 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )^{5}}{2}+\frac {5 \sin \left (b x +a \right )^{3}}{6}+\frac {5 \sin \left (b x +a \right )}{2}-\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(68\) |
default | \(\frac {\frac {\sin \left (b x +a \right )^{7}}{2 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )^{5}}{2}+\frac {5 \sin \left (b x +a \right )^{3}}{6}+\frac {5 \sin \left (b x +a \right )}{2}-\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(68\) |
risch | \(\frac {i {\mathrm e}^{3 i \left (b x +a \right )}}{24 b}-\frac {9 i {\mathrm e}^{i \left (b x +a \right )}}{8 b}+\frac {9 i {\mathrm e}^{-i \left (b x +a \right )}}{8 b}-\frac {i {\mathrm e}^{-3 i \left (b x +a \right )}}{24 b}-\frac {i \left ({\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}\) | \(138\) |
Input:
int(sin(b*x+a)^3*tan(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(1/2*sin(b*x+a)^7/cos(b*x+a)^2+1/2*sin(b*x+a)^5+5/6*sin(b*x+a)^3+5/2*s in(b*x+a)-5/2*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.40 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {15 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 15 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (b x + a\right )^{4} - 14 \, \cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right )}{12 \, b \cos \left (b x + a\right )^{2}} \] Input:
integrate(sin(b*x+a)^3*tan(b*x+a)^3,x, algorithm="fricas")
Output:
-1/12*(15*cos(b*x + a)^2*log(sin(b*x + a) + 1) - 15*cos(b*x + a)^2*log(-si n(b*x + a) + 1) + 2*(2*cos(b*x + a)^4 - 14*cos(b*x + a)^2 - 3)*sin(b*x + a ))/(b*cos(b*x + a)^2)
Timed out. \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**3*tan(b*x+a)**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {4 \, \sin \left (b x + a\right )^{3} - \frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right ) + 24 \, \sin \left (b x + a\right )}{12 \, b} \] Input:
integrate(sin(b*x+a)^3*tan(b*x+a)^3,x, algorithm="maxima")
Output:
1/12*(4*sin(b*x + a)^3 - 6*sin(b*x + a)/(sin(b*x + a)^2 - 1) - 15*log(sin( b*x + a) + 1) + 15*log(sin(b*x + a) - 1) + 24*sin(b*x + a))/b
Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=-\frac {5 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} + \frac {5 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} - \frac {\sin \left (b x + a\right )}{2 \, {\left (\sin \left (b x + a\right )^{2} - 1\right )} b} + \frac {b^{2} \sin \left (b x + a\right )^{3} + 6 \, b^{2} \sin \left (b x + a\right )}{3 \, b^{3}} \] Input:
integrate(sin(b*x+a)^3*tan(b*x+a)^3,x, algorithm="giac")
Output:
-5/4*log(abs(sin(b*x + a) + 1))/b + 5/4*log(abs(sin(b*x + a) - 1))/b - 1/2 *sin(b*x + a)/((sin(b*x + a)^2 - 1)*b) + 1/3*(b^2*sin(b*x + a)^3 + 6*b^2*s in(b*x + a))/b^3
Time = 28.92 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.45 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {5\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b} \] Input:
int(sin(a + b*x)^3*tan(a + b*x)^3,x)
Output:
(5*tan(a/2 + (b*x)/2) + (20*tan(a/2 + (b*x)/2)^3)/3 - (22*tan(a/2 + (b*x)/ 2)^5)/3 + (20*tan(a/2 + (b*x)/2)^7)/3 + 5*tan(a/2 + (b*x)/2)^9)/(b*(tan(a/ 2 + (b*x)/2)^2 - 2*tan(a/2 + (b*x)/2)^4 - 2*tan(a/2 + (b*x)/2)^6 + tan(a/2 + (b*x)/2)^8 + tan(a/2 + (b*x)/2)^10 + 1)) - (5*atanh(tan(a/2 + (b*x)/2)) )/b
Time = 0.17 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.97 \[ \int \sin ^3(a+b x) \tan ^3(a+b x) \, dx=\frac {15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}+15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+2 \sin \left (b x +a \right )^{5}+10 \sin \left (b x +a \right )^{3}-15 \sin \left (b x +a \right )}{6 b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:
int(sin(b*x+a)^3*tan(b*x+a)^3,x)
Output:
(15*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**2 - 15*log(tan((a + b*x)/2) - 1) - 15*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2 + 15*log(tan((a + b*x)/2 ) + 1) + 2*sin(a + b*x)**5 + 10*sin(a + b*x)**3 - 15*sin(a + b*x))/(6*b*(s in(a + b*x)**2 - 1))