Integrand size = 15, antiderivative size = 44 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {\sin (a+b x)}{b}+\frac {\sec (a+b x) \tan (a+b x)}{2 b} \] Output:
-3/2*arctanh(sin(b*x+a))/b+sin(b*x+a)/b+1/2*sec(b*x+a)*tan(b*x+a)/b
Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {3 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {\sin (a+b x) \tan ^2(a+b x)}{b} \] Input:
Integrate[Sin[a + b*x]*Tan[a + b*x]^3,x]
Output:
(-3*ArcTanh[Sin[a + b*x]])/(2*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(2*b) - ( Sin[a + b*x]*Tan[a + b*x]^2)/b
Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3072, 252, 262, 219}
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.
\(\displaystyle \int \sin (a+b x) \tan ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x) \tan (a+b x)^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {\int \frac {\sin ^4(a+b x)}{\left (1-\sin ^2(a+b x)\right )^2}d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sin ^3(a+b x)}{2 \left (1-\sin ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\sin ^2(a+b x)}{1-\sin ^2(a+b x)}d\sin (a+b x)}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\sin ^3(a+b x)}{2 \left (1-\sin ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sin ^2(a+b x)}d\sin (a+b x)-\sin (a+b x)\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sin ^3(a+b x)}{2 \left (1-\sin ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\sin (a+b x))-\sin (a+b x))}{b}\) |
Input:
Int[Sin[a + b*x]*Tan[a + b*x]^3,x]
Output:
((-3*(ArcTanh[Sin[a + b*x]] - Sin[a + b*x]))/2 + Sin[a + b*x]^3/(2*(1 - Si n[a + b*x]^2)))/b
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[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[((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 = 1.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{5}}{2 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )^{3}}{2}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(58\) |
default | \(\frac {\frac {\sin \left (b x +a \right )^{5}}{2 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )^{3}}{2}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(58\) |
risch | \(\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {i \left ({\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}\) | \(108\) |
Input:
int(sin(b*x+a)*tan(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(1/2*sin(b*x+a)^5/cos(b*x+a)^2+1/2*sin(b*x+a)^3+3/2*sin(b*x+a)-3/2*ln( sec(b*x+a)+tan(b*x+a)))
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )^{2}} \] Input:
integrate(sin(b*x+a)*tan(b*x+a)^3,x, algorithm="fricas")
Output:
-1/4*(3*cos(b*x + a)^2*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^2*log(-sin(b *x + a) + 1) - 2*(2*cos(b*x + a)^2 + 1)*sin(b*x + a))/(b*cos(b*x + a)^2)
\[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=\int \sin {\left (a + b x \right )} \tan ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(sin(b*x+a)*tan(b*x+a)**3,x)
Output:
Integral(sin(a + b*x)*tan(a + b*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.27 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\sin \left (b x + a\right ) - 1\right ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \] Input:
integrate(sin(b*x+a)*tan(b*x+a)^3,x, algorithm="maxima")
Output:
-1/4*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 3*log(sin(b*x + a) + 1) - 3*lo g(sin(b*x + a) - 1) - 4*sin(b*x + a))/b
Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.45 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} + \frac {3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} + \frac {\sin \left (b x + a\right )}{b} - \frac {\sin \left (b x + a\right )}{2 \, {\left (\sin \left (b x + a\right )^{2} - 1\right )} b} \] Input:
integrate(sin(b*x+a)*tan(b*x+a)^3,x, algorithm="giac")
Output:
-3/4*log(abs(sin(b*x + a) + 1))/b + 3/4*log(abs(sin(b*x + a) - 1))/b + sin (b*x + a)/b - 1/2*sin(b*x + a)/((sin(b*x + a)^2 - 1)*b)
Time = 27.47 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.23 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b}-\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+3\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{b\,\left (-{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )} \] Input:
int(sin(a + b*x)*tan(a + b*x)^3,x)
Output:
- (3*atanh(tan(a/2 + (b*x)/2)))/b - (3*tan(a/2 + (b*x)/2) - 2*tan(a/2 + (b *x)/2)^3 + 3*tan(a/2 + (b*x)/2)^5)/(b*(tan(a/2 + (b*x)/2)^2 + tan(a/2 + (b *x)/2)^4 - tan(a/2 + (b*x)/2)^6 - 1))
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.61 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=\frac {-\cos \left (b x +a \right ) \tan \left (b x +a \right )+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\sin \left (b x +a \right ) \tan \left (b x +a \right )^{2}+4 \sin \left (b x +a \right )}{2 b} \] Input:
int(sin(b*x+a)*tan(b*x+a)^3,x)
Output:
( - cos(a + b*x)*tan(a + b*x) + 3*log(tan((a + b*x)/2) - 1) - 3*log(tan((a + b*x)/2) + 1) + sin(a + b*x)*tan(a + b*x)**2 + 4*sin(a + b*x))/(2*b)