Integrand size = 19, antiderivative size = 116 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {(8 a+15 b) \log (1-\sin (c+d x))}{16 d}-\frac {(8 a-15 b) \log (1+\sin (c+d x))}{16 d}-\frac {15 b \sin (c+d x)}{8 d}-\frac {(4 a+5 b \sin (c+d x)) \tan ^2(c+d x)}{8 d}+\frac {(a+b \sin (c+d x)) \tan ^4(c+d x)}{4 d} \]
-1/16*(8*a+15*b)*ln(1-sin(d*x+c))/d-1/16*(8*a-15*b)*ln(1+sin(d*x+c))/d-15/ 8*b*sin(d*x+c)/d-1/8*(4*a+5*b*sin(d*x+c))*tan(d*x+c)^2/d+1/4*(a+b*sin(d*x+ c))*tan(d*x+c)^4/d
Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.17 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {15 b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 b \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {b \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {a \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]
(15*b*ArcTanh[Sin[c + d*x]])/(8*d) + (15*b*Sec[c + d*x]*Tan[c + d*x])/(8*d ) - (15*b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*b*Sec[c + d*x]*Tan[c + d *x]^3)/d - (b*Sin[c + d*x]*Tan[c + d*x]^4)/d - (a*(4*Log[Cos[c + d*x]] + 2 *Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d)
Time = 0.43 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3200, 530, 2345, 2341, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^5 (a+b \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^5 \sin ^5(c+d x) (a+b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 530 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {4 \sin ^4(c+d x) b^6+4 \sin ^2(c+d x) b^6+b^6+4 a \sin ^3(c+d x) b^5+4 a \sin (c+d x) b^5}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (8 a+9 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {8 \sin ^2(c+d x) b^6+7 b^6+8 a \sin (c+d x) b^5}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (8 a+9 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (\frac {15 b^6+8 a \sin (c+d x) b^5}{b^2-b^2 \sin ^2(c+d x)}-8 b^4\right )d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 (a+b \sin (c+d x))}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (8 a+9 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {-4 a b^4 \log \left (b^2-b^2 \sin ^2(c+d x)\right )+15 b^5 \text {arctanh}(\sin (c+d x))-8 b^5 \sin (c+d x)}{2 b^2}}{4 b^2}}{d}\) |
((b^4*(a + b*Sin[c + d*x]))/(4*(b^2 - b^2*Sin[c + d*x]^2)^2) - (-1/2*(15*b ^5*ArcTanh[Sin[c + d*x]] - 4*a*b^4*Log[b^2 - b^2*Sin[c + d*x]^2] - 8*b^5*S in[c + d*x])/b^2 + (b^4*(8*a + 9*b*Sin[c + d*x]))/(2*(b^2 - b^2*Sin[c + d* x]^2)))/(4*b^2))/d
3.15.83.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x )*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.74 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(121\) |
| default | \(\frac {a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(121\) |
| parallelrisch | \(\frac {16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {15 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (a -\frac {15 b}{8}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 a \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right ) a -2 b \sin \left (5 d x +5 c \right )-5 b \sin \left (d x +c \right )-15 b \sin \left (3 d x +3 c \right )+a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(208\) |
| risch | \(i a x +\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a c}{d}+\frac {i \left (16 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+9 b \,{\mathrm e}^{7 i \left (d x +c \right )}+16 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{5 i \left (d x +c \right )}+16 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{3 i \left (d x +c \right )}-9 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}\) | \(229\) |
| norman | \(\frac {-\frac {15 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {10 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {9 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {10 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {15 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {6 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a -15 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a +15 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(252\) |
1/d*(a*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+b*(1/4*sin(d*x+c )^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d* x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a - 15 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 15 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b \cos \left (d x + c\right )^{4} + 9 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]
-1/16*((8*a - 15*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (8*a + 15*b)*co s(d*x + c)^4*log(-sin(d*x + c) + 1) + 16*a*cos(d*x + c)^2 + 2*(8*b*cos(d*x + c)^4 + 9*b*cos(d*x + c)^2 - 2*b)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a - 15 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 15 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, b \sin \left (d x + c\right )^{3} + 8 \, a \sin \left (d x + c\right )^{2} - 7 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
-1/16*((8*a - 15*b)*log(sin(d*x + c) + 1) + (8*a + 15*b)*log(sin(d*x + c) - 1) + 16*b*sin(d*x + c) - 2*(9*b*sin(d*x + c)^3 + 8*a*sin(d*x + c)^2 - 7* b*sin(d*x + c) - 6*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (8 \, a - 15 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (8 \, a + 15 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \sin \left (d x + c\right ) - \frac {2 \, {\left (6 \, a \sin \left (d x + c\right )^{4} + 9 \, b \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} - 7 \, b \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*((8*a - 15*b)*log(abs(sin(d*x + c) + 1)) + (8*a + 15*b)*log(abs(sin( d*x + c) - 1)) + 16*b*sin(d*x + c) - 2*(6*a*sin(d*x + c)^4 + 9*b*sin(d*x + c)^3 - 4*a*sin(d*x + c)^2 - 7*b*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d
Time = 12.39 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.25 \[ \int (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-\frac {15\,b}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+\frac {15\,b}{8}\right )}{d}-\frac {\frac {15\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {9\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {15\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (log(tan(c/2 + (d*x)/2) + 1)*(a - (1 5*b)/8))/d - (log(tan(c/2 + (d*x)/2) - 1)*(a + (15*b)/8))/d - ((15*b*tan(c /2 + (d*x)/2))/4 + 2*a*tan(c/2 + (d*x)/2)^2 - 6*a*tan(c/2 + (d*x)/2)^4 - 6 *a*tan(c/2 + (d*x)/2)^6 + 2*a*tan(c/2 + (d*x)/2)^8 - 10*b*tan(c/2 + (d*x)/ 2)^3 + (9*b*tan(c/2 + (d*x)/2)^5)/2 - 10*b*tan(c/2 + (d*x)/2)^7 + (15*b*ta n(c/2 + (d*x)/2)^9)/4)/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1))