Integrand size = 27, antiderivative size = 74 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {a \tan ^4(c+d x)}{4 d} \]
3/8*b*arctanh(sin(d*x+c))/d-3/8*b*sec(d*x+c)*tan(d*x+c)/d+1/4*b*sec(d*x+c) *tan(d*x+c)^3/d+1/4*a*tan(d*x+c)^4/d
Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b \sec (c+d x) \tan ^3(c+d x)}{d}+\frac {a \tan ^4(c+d x)}{4 d} \]
(3*b*ArcTanh[Sin[c + d*x]])/(8*d) + (3*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) - (3*b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (b*Sec[c + d*x]*Tan[c + d*x]^3 )/d + (a*Tan[c + d*x]^4)/(4*d)
Time = 0.55 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3313, 3042, 3087, 15, 3091, 3042, 3091, 3042, 4257}
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.
| \(\displaystyle \int \tan ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \sec ^2(c+d x) \tan ^3(c+d x)dx+b \int \sec (c+d x) \tan ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sec (c+d x)^2 \tan (c+d x)^3dx+b \int \sec (c+d x) \tan (c+d x)^4dx\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {a \int \tan ^3(c+d x)d\tan (c+d x)}{d}+b \int \sec (c+d x) \tan (c+d x)^4dx\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle b \int \sec (c+d x) \tan (c+d x)^4dx+\frac {a \tan ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle b \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx\right )+\frac {a \tan ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx\right )+\frac {a \tan ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle b \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )\right )+\frac {a \tan ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+\frac {a \tan ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a \tan ^4(c+d x)}{4 d}+b \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )\right )\) |
(a*Tan[c + d*x]^4)/(4*d) + b*((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1 /2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
3.15.85.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.57 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32
| method | result | size |
| derivativedivides | \(\frac {\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(98\) |
| default | \(\frac {\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(98\) |
| risch | \(\frac {i \left (8 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+5 b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 b \,{\mathrm e}^{5 i \left (d x +c \right )}+8 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}\) | \(134\) |
| parallelrisch | \(\frac {-6 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 a \cos \left (2 d x +2 c \right )+\cos \left (4 d x +4 c \right ) a +3 b \sin \left (d x +c \right )-5 b \sin \left (3 d x +3 c \right )+3 a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(152\) |
| norman | \(\frac {-\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(187\) |
1/d*(1/4*a*sin(d*x+c)^4/cos(d*x+c)^4+b*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8* sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c )+tan(d*x+c))))
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 \, b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, a \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*(3*b*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*b*cos(d*x + c)^4*log(-s in(d*x + c) + 1) - 8*a*cos(d*x + c)^2 - 2*(5*b*cos(d*x + c)^2 - 2*b)*sin(d *x + c) + 4*a)/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
1/16*(3*b*log(sin(d*x + c) + 1) - 3*b*log(sin(d*x + c) - 1) + 2*(5*b*sin(d *x + c)^3 + 4*a*sin(d*x + c)^2 - 3*b*sin(d*x + c) - 2*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.37 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {3 \, b \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, b \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 3 \, b \sin \left (d x + c\right ) - 2 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*(3*b*log(abs(sin(d*x + c) + 1)) - 3*b*log(abs(sin(d*x + c) - 1)) + 2* (5*b*sin(d*x + c)^3 + 4*a*sin(d*x + c)^2 - 3*b*sin(d*x + c) - 2*a)/(sin(d* x + c)^2 - 1)^2)/d
Time = 17.75 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.95 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]
(4*a*tan(c/2 + (d*x)/2)^4 - (3*b*tan(c/2 + (d*x)/2))/4 + (11*b*tan(c/2 + ( d*x)/2)^3)/4 + (11*b*tan(c/2 + (d*x)/2)^5)/4 - (3*b*tan(c/2 + (d*x)/2)^7)/ 4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x) /2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (3*b*atanh(tan(c/2 + (d*x)/2)))/(4*d)