Integrand size = 21, antiderivative size = 204 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\left (8 a^2+9 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {\left (8 a^2-9 a b+3 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^5 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \]
-1/16*(8*a^2+9*a*b+3*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(8*a^2-9*a*b+3*b ^2)*ln(1+sin(d*x+c))/(a-b)^3/d+a^5*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/4*se c(d*x+c)^4*(a-b*sin(d*x+c))/(a^2-b^2)/d-1/8*sec(d*x+c)^2*(4*a*(2*a^2-b^2)- b*(9*a^2-5*b^2)*sin(d*x+c))/(a^2-b^2)^2/d
Time = 0.75 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {\left (8 a^2+9 a b+3 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {\left (8 a^2-9 a b+3 b^2\right ) \log (1+\sin (c+d x))}{(a-b)^3}+\frac {16 a^5 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {7 a+5 b}{(a+b)^2 (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {-7 a+5 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \]
(-(((8*a^2 + 9*a*b + 3*b^2)*Log[1 - Sin[c + d*x]])/(a + b)^3) - ((8*a^2 - 9*a*b + 3*b^2)*Log[1 + Sin[c + d*x]])/(a - b)^3 + (16*a^5*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (7*a + 5*b)/((a + b)^2*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])^2) + (-7*a + 5*b)/((a - b)^2*(1 + Sin[c + d*x])))/(16*d)
Time = 0.62 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.29, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3200, 601, 25, 2178, 25, 27, 657, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^5 \sin ^5(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {\frac {a b^6}{a^2-b^2}-4 \sin ^3(c+d x) b^5-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) b^5}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {a b^6}{a^2-b^2}-4 \sin ^3(c+d x) b^5-\frac {\left (4 a^2-b^2\right ) \sin (c+d x) b^5}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2178 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {b^4 \left (a b^2 \left (7 a^2-3 b^2\right )-b \left (8 a^4-7 b^2 a^2+3 b^4\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {b^4 \left (a b^2 \left (7 a^2-3 b^2\right )-b \left (8 a^4-7 b^2 a^2+3 b^4\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {b^2 \int \frac {a b^2 \left (7 a^2-3 b^2\right )-b \left (8 a^4-7 b^2 a^2+3 b^4\right ) \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )^2}-\frac {b^4 \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {\frac {-\frac {b^2 \int \left (-\frac {8 a^5}{(a-b) (a+b) (a+b \sin (c+d x))}-\frac {(a-b)^2 \left (8 a^2+9 b a+3 b^2\right )}{2 (a+b) (b-b \sin (c+d x))}+\frac {(a+b)^2 \left (8 a^2-9 b a+3 b^2\right )}{2 (a-b) (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 \left (a^2-b^2\right )^2}-\frac {b^4 \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {-\frac {b^4 \left (4 a \left (2 a^2-b^2\right )-b \left (9 a^2-5 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {b^2 \left (\frac {(a-b)^2 \left (8 a^2+9 a b+3 b^2\right ) \log (b-b \sin (c+d x))}{2 (a+b)}+\frac {(a+b)^2 \left (8 a^2-9 a b+3 b^2\right ) \log (b \sin (c+d x)+b)}{2 (a-b)}-\frac {8 a^5 \log (a+b \sin (c+d x))}{a^2-b^2}\right )}{2 \left (a^2-b^2\right )^2}}{4 b^2}}{d}\) |
((b^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)^2) + (-1/2*(b^2*(((a - b)^2*(8*a^2 + 9*a*b + 3*b^2)*Log[b - b*Sin[c + d*x]])/( 2*(a + b)) - (8*a^5*Log[a + b*Sin[c + d*x]])/(a^2 - b^2) + ((a + b)^2*(8*a ^2 - 9*a*b + 3*b^2)*Log[b + b*Sin[c + d*x]])/(2*(a - b))))/(a^2 - b^2)^2 - (b^4*(4*a*(2*a^2 - b^2) - b*(9*a^2 - 5*b^2)*Sin[c + d*x]))/(2*(a^2 - b^2) ^2*(b^2 - b^2*Sin[c + d*x]^2)))/(4*b^2))/d
3.14.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[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[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x )^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
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 = 0.94 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.93
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 a -5 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-8 a^{2}+9 a b -3 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-7 a -5 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-8 a^{2}-9 a b -3 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(189\) |
| default | \(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 a -5 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-8 a^{2}+9 a b -3 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-7 a -5 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-8 a^{2}-9 a b -3 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(189\) |
| parallelrisch | \(\frac {4 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{5} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-4 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {9}{8} a b +\frac {3}{8} b^{2}\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-4 \left (a +b \right ) \left (\left (a^{2}-\frac {9}{8} a b +\frac {3}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {\left (a -b \right ) \left (\left (a^{3}-a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\frac {\left (-3 a^{3}+a \,b^{2}\right ) \cos \left (4 d x +4 c \right )}{4}+\frac {\left (-9 a^{2} b +5 b^{3}\right ) \sin \left (3 d x +3 c \right )}{4}+\frac {\left (-a^{2} b -3 b^{3}\right ) \sin \left (d x +c \right )}{4}-\frac {a^{3}}{4}+\frac {3 a \,b^{2}}{4}\right )}{4}\right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \left (a +b \right )^{3} \left (a -b \right )^{3}}\) | \(315\) |
| norman | \(\frac {-\frac {2 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {2 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {2 \left (4 a^{3}-2 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {b \left (7 a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {b \left (7 a^{2}-3 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (15 a^{2}-11 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (15 a^{2}-11 b^{2}\right ) b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{5} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {\left (8 a^{2}-9 a b +3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {\left (8 a^{2}+9 a b +3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}\) | \(484\) |
| risch | \(\frac {3 i b^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i a^{5} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {i a^{2} c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i b^{2} c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {9 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i a^{2} c}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {2 i a^{5} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {3 i b^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {9 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {9 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i b^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a^{2} x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i \left (16 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-8 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-9 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+5 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-8 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-5 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} d}+\frac {i a^{2} x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {9 i a b c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(982\) |
1/d*(1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(7*a-5*b)/(a-b)^2/(1+sin(d*x+c))+ 1/16/(a-b)^3*(-8*a^2+9*a*b-3*b^2)*ln(1+sin(d*x+c))+a^5/(a+b)^3/(a-b)^3*ln( a+b*sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(-7*a-5*b)/(a+b)^2/(si n(d*x+c)-1)+1/16/(a+b)^3*(-8*a^2-9*a*b-3*b^2)*ln(sin(d*x+c)-1))
Time = 0.47 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a^{5} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (8 \, a^{5} + 15 \, a^{4} b - 10 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{5} - 15 \, a^{4} b + 10 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (9 \, a^{4} b - 14 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]
1/16*(16*a^5*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (8*a^5 - 15*a^4 *b + 10*a^2*b^3 - 3*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^5 - 8 *a^3*b^2 + 4*a*b^4 - 8*(2*a^5 - 3*a^3*b^2 + a*b^4)*cos(d*x + c)^2 - 2*(2*a ^4*b - 4*a^2*b^3 + 2*b^5 - (9*a^4*b - 14*a^2*b^3 + 5*b^5)*cos(d*x + c)^2)* sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)
Timed out. \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{5} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (8 \, a^{2} - 9 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 9 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (9 \, a^{2} b - 5 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{3} - 2 \, a b^{2} - 4 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (7 \, a^{2} b - 3 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]
1/16*(16*a^5*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (8*a^2 - 9*a*b + 3*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 9*a*b + 3*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a* b^2 + b^3) - 2*((9*a^2*b - 5*b^3)*sin(d*x + c)^3 + 6*a^3 - 2*a*b^2 - 4*(2* a^3 - a*b^2)*sin(d*x + c)^2 - (7*a^2*b - 3*b^3)*sin(d*x + c))/((a^4 - 2*a^ 2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2))/d
Time = 0.40 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.68 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{5} b \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (8 \, a^{2} - 9 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 9 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{5} \sin \left (d x + c\right )^{4} - 9 \, a^{4} b \sin \left (d x + c\right )^{3} + 14 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - 5 \, b^{5} \sin \left (d x + c\right )^{3} - 4 \, a^{5} \sin \left (d x + c\right )^{2} - 12 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b^{4} \sin \left (d x + c\right )^{2} + 7 \, a^{4} b \sin \left (d x + c\right ) - 10 \, a^{2} b^{3} \sin \left (d x + c\right ) + 3 \, b^{5} \sin \left (d x + c\right ) + 8 \, a^{3} b^{2} - 2 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*(16*a^5*b*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (8*a^2 - 9*a*b + 3*b^2)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2* b + 3*a*b^2 - b^3) - (8*a^2 + 9*a*b + 3*b^2)*log(abs(sin(d*x + c) - 1))/(a ^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(6*a^5*sin(d*x + c)^4 - 9*a^4*b*sin(d*x + c)^3 + 14*a^2*b^3*sin(d*x + c)^3 - 5*b^5*sin(d*x + c)^3 - 4*a^5*sin(d*x + c)^2 - 12*a^3*b^2*sin(d*x + c)^2 + 4*a*b^4*sin(d*x + c)^2 + 7*a^4*b*sin( d*x + c) - 10*a^2*b^3*sin(d*x + c) + 3*b^5*sin(d*x + c) + 8*a^3*b^2 - 2*a* b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d
Time = 13.07 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.44 \[ \int \frac {\tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^5\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {1}{a+b}-\frac {7\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {7\,b}{8\,{\left (a-b\right )}^2}+\frac {1}{a-b}\right )}{d}-\frac {\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a\,b^2-2\,a^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (15\,a^2\,b-11\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (15\,a^2\,b-11\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,a^2-3\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(a^5*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))/(d*(a^6 - b ^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (log(tan(c/2 + (d*x)/2) - 1)*(1/(a + b) - ( 7*b)/(8*(a + b)^2) + b^2/(4*(a + b)^3)))/d - (log(tan(c/2 + (d*x)/2) + 1)* (b^2/(4*(a - b)^3) + (7*b)/(8*(a - b)^2) + 1/(a - b)))/d - ((2*a^3*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) + (2*a^3*tan(c/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2) + (4*tan(c/2 + (d*x)/2)^4*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^7*(7*a^2*b - 3*b^3))/(4*(a^4 + b^4 - 2* a^2*b^2)) + (tan(c/2 + (d*x)/2)^3*(15*a^2*b - 11*b^3))/(4*(a^4 + b^4 - 2*a ^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(15*a^2*b - 11*b^3))/(4*(a^4 + b^4 - 2*a^ 2*b^2)) - (b*tan(c/2 + (d*x)/2)*(7*a^2 - 3*b^2))/(4*(a^4 + b^4 - 2*a^2*b^2 )))/(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))