Integrand size = 27, antiderivative size = 190 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {a (3 a-b) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^4 b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \]
-1/16*a*(3*a+b)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*a*(3*a-b)*ln(1+sin(d*x+c)) /(a-b)^3/d-a^4*b*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d-1/4*sec(d*x+c)^4*(b-a*si n(d*x+c))/(a^2-b^2)/d+1/8*sec(d*x+c)^2*(4*b*(2*a^2-b^2)-a*(5*a^2-b^2)*sin( d*x+c))/(a^2-b^2)^2/d
Time = 0.98 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {a (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {a (3 a-b) \log (1+\sin (c+d x))}{(a-b)^3}-\frac {16 a^4 b \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {5 a+3 b}{(a+b)^2 (-1+\sin (c+d x))}-\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {5 a-3 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \]
(-((a*(3*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^3) + (a*(3*a - b)*Log[1 + S in[c + d*x]])/(a - b)^3 - (16*a^4*b*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (5*a + 3*b)/((a + b)^2*(-1 + Sin[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x])^2) + (5*a - 3*b)/((a - b) ^2*(1 + Sin[c + d*x])))/(16*d)
Time = 0.64 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.43, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3316, 27, 601, 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.
| \(\displaystyle \int \frac {\tan ^4(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^4(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {b^4 \sin ^4(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {b \left (-\frac {\int \frac {-\frac {3 a \sin (c+d x) b^5}{a^2-b^2}+4 \sin ^2(c+d x) b^4+\frac {a^2 b^4}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2178 |
| \(\displaystyle \frac {b \left (-\frac {\frac {\int -\frac {a b^4 \left (a \left (3 a^2+b^2\right )-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (-\frac {-\frac {\int \frac {a b^4 \left (a \left (3 a^2+b^2\right )-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \left (-\frac {-\frac {a b^2 \int \frac {a \left (3 a^2+b^2\right )-b \left (5 a^2-b^2\right ) \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b \left (-\frac {-\frac {a b^2 \int \left (-\frac {8 a^3}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {(a-b)^2 (3 a+b)}{2 b (a+b) (b-b \sin (c+d x))}+\frac {(3 a-b) (a+b)^2}{2 (a-b) b (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 \left (a^2-b^2\right )^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {a b^2 \left (-\frac {8 a^3 \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {(a-b)^2 (3 a+b) \log (b-b \sin (c+d x))}{2 b (a+b)}+\frac {(3 a-b) (a+b)^2 \log (b \sin (c+d x)+b)}{2 b (a-b)}\right )}{2 \left (a^2-b^2\right )^2}}{4 b^2}\right )}{d}\) |
(b*(-1/4*(b^2*(b^2/(a^2 - b^2) - (a*b*Sin[c + d*x])/(a^2 - b^2)))/(b^2 - b ^2*Sin[c + d*x]^2)^2 - (-1/2*(a*b^2*(-1/2*((a - b)^2*(3*a + b)*Log[b - b*S in[c + d*x]])/(b*(a + b)) - (8*a^3*Log[a + b*Sin[c + d*x]])/(a^2 - b^2) + ((3*a - b)*(a + b)^2*Log[b + b*Sin[c + d*x]])/(2*(a - b)*b)))/(a^2 - b^2)^ 2 - (b^2*(4*b^2*(2*a^2 - b^2) - a*b*(5*a^2 - 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
3.14.63.3.1 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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.93 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-5 a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {a \left (3 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {a^{4} b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-5 a -3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {a \left (3 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(175\) |
| default | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-5 a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {a \left (3 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {a^{4} b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-5 a -3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {a \left (3 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(175\) |
| parallelrisch | \(\frac {-8 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{4} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-3 \left (a +\frac {b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{3} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {b}{3}\right ) \left (a +b \right )^{2} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {\left (\frac {4 \left (a^{2} b -b^{3}\right ) \cos \left (2 d x +2 c \right )}{3}+\left (-a^{2} b +\frac {1}{3} b^{3}\right ) \cos \left (4 d x +4 c \right )+\frac {\left (-5 a^{3}+a \,b^{2}\right ) \sin \left (3 d x +3 c \right )}{3}+\left (a^{3}-\frac {7}{3} a \,b^{2}\right ) \sin \left (d x +c \right )-\frac {a^{2} b}{3}+b^{3}\right ) \left (a -b \right )}{2}\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(299\) |
| norman | \(\frac {\frac {2 a^{2} b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {2 a^{2} b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {2 \left (-4 a^{2} b +2 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {a \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {a \left (3 a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {\left (11 a^{2}-7 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {\left (11 a^{2}-7 b^{2}\right ) a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a \left (3 a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {a \left (3 a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {a^{4} b \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(468\) |
| risch | \(\frac {3 i a^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a b c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a b x}{8 a^{3}+24 a^{2} b +24 a \,b^{2}+8 b^{3}}+\frac {2 i a^{4} b x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {3 i a^{2} c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {i a b x}{8 a^{3}-24 a^{2} b +24 a \,b^{2}-8 b^{3}}-\frac {3 i a^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {2 i a^{4} b c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {-5 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-7 i b^{2} a \,{\mathrm e}^{5 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+7 i b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+5 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{2 i \left (d x +c \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 {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{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 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 ) a^{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 b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {a^{4} b \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 )}\) | \(782\) |
1/d*(-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(-5*a+3*b)/(a-b)^2/(1+sin(d*x+c) )+1/16*a*(3*a-b)/(a-b)^3*ln(1+sin(d*x+c))-a^4*b/(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*(-5*a-3*b)/(a+b)^2/(sin(d*x+c )-1)-1/16*a*(3*a+b)/(a+b)^3*ln(sin(d*x+c)-1))
Time = 0.52 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.37 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, a^{4} b \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]
-1/16*(16*a^4*b*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^5 + 8*a^4*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (3*a^5 - 8*a^4 *b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^4*b - 8*a^2*b^3 + 4*b^5 - 8*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(d*x + c)^2 - 2*(2*a^ 5 - 4*a^3*b^2 + 2*a*b^4 - (5*a^5 - 6*a^3*b^2 + a*b^4)*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. \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.45 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, a^{4} b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a^{2} - a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (5 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b - 2 \, b^{3} - 4 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (3 \, a^{3} + a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]
-1/16*(16*a^4*b*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 ) - (3*a^2 - a*b)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a^2 + a*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(( 5*a^3 - a*b^2)*sin(d*x + c)^3 + 6*a^2*b - 2*b^3 - 4*(2*a^2*b - b^3)*sin(d* x + c)^2 - (3*a^3 + a*b^2)*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 = 0.52 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.75 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, a^{4} b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (3 \, a^{2} - a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{4} b \sin \left (d x + c\right )^{4} - 5 \, a^{5} \sin \left (d x + c\right )^{3} + 6 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} - a b^{4} \sin \left (d x + c\right )^{3} - 4 \, a^{4} b \sin \left (d x + c\right )^{2} - 12 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} + 4 \, b^{5} \sin \left (d x + c\right )^{2} + 3 \, a^{5} \sin \left (d x + c\right ) - 2 \, a^{3} b^{2} \sin \left (d x + c\right ) - a b^{4} \sin \left (d x + c\right ) + 8 \, a^{2} b^{3} - 2 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*(16*a^4*b^2*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2* b^5 - b^7) - (3*a^2 - a*b)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a *b^2 - b^3) + (3*a^2 + a*b)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3* a*b^2 + b^3) + 2*(6*a^4*b*sin(d*x + c)^4 - 5*a^5*sin(d*x + c)^3 + 6*a^3*b^ 2*sin(d*x + c)^3 - a*b^4*sin(d*x + c)^3 - 4*a^4*b*sin(d*x + c)^2 - 12*a^2* b^3*sin(d*x + c)^2 + 4*b^5*sin(d*x + c)^2 + 3*a^5*sin(d*x + c) - 2*a^3*b^2 *sin(d*x + c) - a*b^4*sin(d*x + c) + 8*a^2*b^3 - 2*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d
Time = 12.47 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.67 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {5\,b}{8\,{\left (a-b\right )}^2}+\frac {3}{8\,\left (a-b\right )}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^3+a\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a\,b^2-11\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (7\,a\,b^2-11\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3}{8\,\left (a+b\right )}-\frac {5\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a^4\,b\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]
(log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + (5*b)/(8*(a - b)^2) + 3/ (8*(a - b))))/d - ((tan(c/2 + (d*x)/2)^7*(a*b^2 + 3*a^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^3*(7*a*b^2 - 11*a^3))/(4*(a^4 + b^4 - 2* a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(7*a*b^2 - 11*a^3))/(4*(a^4 + b^4 - 2*a^ 2*b^2)) + (4*tan(c/2 + (d*x)/2)^4*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2*b^2) - (2*a^2*b*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) - (2*a^2*b*tan(c /2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2) + (a*tan(c/2 + (d*x)/2)*(3*a^2 + 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)) - (log( tan(c/2 + (d*x)/2) - 1)*(3/(8*(a + b)) - (5*b)/(8*(a + b)^2) + b^2/(4*(a + b)^3)))/d - (a^4*b*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))