Integrand size = 27, antiderivative size = 213 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {(a-3 b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac {(a-3 b) b^5}{16 (a-b) (a+b)^2 d \left (b^4-b^4 \sin (c+d x)\right )}-\frac {b^5 (a+3 b)}{16 (a-b)^2 (a+b) d \left (b^4+b^4 \sin (c+d x)\right )} \] Output:
-1/16*b*(a+3*b)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*(a-3*b)*b*ln(1+sin(d*x+c)) /(a-b)^3/d+a*b^4*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/4*sec(d*x+c)^4*(a-b*si n(d*x+c))/(a^2-b^2)/d+1/16*(a-3*b)*b^5/(a-b)/(a+b)^2/d/(b^4-b^4*sin(d*x+c) )-1/16*b^5*(a+3*b)/(a-b)^2/(a+b)/d/(b^4+b^4*sin(d*x+c))
Time = 6.17 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\frac {\frac {(a-b)^2 b (a+3 b) \log (1-\sin (c+d x))}{a+b}-\frac {(a-3 b) b (a+b)^2 \log (1+\sin (c+d x))}{a-b}-\frac {16 a b^4 \log (a+b \sin (c+d x))}{a^2-b^2}}{a^2-b^2}+\frac {2 \sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{a^2-b^2}}{16 \left (a^2-b^2\right ) d} \] Input:
Integrate[(Sec[c + d*x]^4*Tan[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(Sec[c + d*x]^4*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*d) - ((((a - b)^2*b*( a + 3*b)*Log[1 - Sin[c + d*x]])/(a + b) - ((a - 3*b)*b*(a + b)^2*Log[1 + S in[c + d*x]])/(a - b) - (16*a*b^4*Log[a + b*Sin[c + d*x]])/(a^2 - b^2))/(a ^2 - b^2) + (2*Sec[c + d*x]^2*(4*a*b^2 - b*(a^2 + 3*b^2)*Sin[c + d*x]))/(a ^2 - b^2))/(16*(a^2 - b^2)*d)
Time = 0.48 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3316, 27, 593, 25, 663, 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.
| \(\displaystyle \int \frac {\tan (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 593 |
| \(\displaystyle \frac {b^4 \left (\frac {a-b \sin (c+d x)}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {a-3 b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 \left (a^2-b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^4 \left (\frac {\int \frac {a-3 b \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 \left (a^2-b^2\right )}+\frac {a-b \sin (c+d x)}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^4 \left (\frac {\int \left (\frac {4 a}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {(a-b) (a+3 b)}{4 b^3 (a+b)^2 (b-b \sin (c+d x))}+\frac {(a-3 b) (a+b)}{4 (a-b)^2 b^3 (\sin (c+d x) b+b)}+\frac {a-3 b}{4 b^2 (a+b) (b-b \sin (c+d x))^2}+\frac {a+3 b}{4 (a-b) b^2 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{4 \left (a^2-b^2\right )}+\frac {a-b \sin (c+d x)}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^4 \left (\frac {a-b \sin (c+d x)}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {\frac {4 a \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {(a+b) (a-3 b) \log (b \sin (c+d x)+b)}{4 b^3 (a-b)^2}-\frac {(a-b) (a+3 b) \log (b-b \sin (c+d x))}{4 b^3 (a+b)^2}+\frac {a-3 b}{4 b^2 (a+b) (b-b \sin (c+d x))}-\frac {a+3 b}{4 b^2 (a-b) (b \sin (c+d x)+b)}}{4 \left (a^2-b^2\right )}\right )}{d}\) |
Input:
Int[(Sec[c + d*x]^4*Tan[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(b^4*((a - b*Sin[c + d*x])/(4*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2)^2) + (-1/4*((a - b)*(a + 3*b)*Log[b - b*Sin[c + d*x]])/(b^3*(a + b)^2) + (4*a*L og[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + ((a - 3*b)*(a + b)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^3) + (a - 3*b)/(4*b^2*(a + b)*(b - b*Sin[c + d*x] )) - (a + 3*b)/(4*(a - b)*b^2*(b + b*Sin[c + d*x])))/(4*(a^2 - b^2))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b* x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[1/c^p Int[ExpandI ntegrand[(d + e*x)^m*(f + g*x)^n*(-q + c*x)^p*(q + c*x)^p, x], x], x] /; ! FractionalPowerFactorQ[q]] /; FreeQ[{a, c, d, e, f, g}, x] && ILtQ[p, -1] & & IntegersQ[m, n] && NiceSqrtQ[(-a)*c]
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 = 3.52 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {a +3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\left (a +3 b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -3 b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {b^{4} a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(170\) |
| default | \(\frac {\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {a +3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\left (a +3 b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -3 b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {b^{4} a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(170\) |
| risch | \(-\frac {2 i a \,b^{4} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {2 i a \,b^{4} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {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} c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i b^{2} 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 b x}{8 a^{3}+24 a^{2} b +24 a \,b^{2}+8 b^{3}}+\frac {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 {-8 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-i a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 i b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+16 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-32 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+7 i a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-11 i b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-8 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+11 i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+3 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{4 \left (a^{2}-b^{2}\right )^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {a \,b^{4} \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 )}\) | \(761\) |
Input:
int(sec(d*x+c)^4*tan(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(a+3*b)/(a+b)^2/(sin(d*x+c)-1)-1/ 16*(a+3*b)/(a+b)^3*b*ln(sin(d*x+c)-1)+1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16* (-a+3*b)/(a-b)^2/(1+sin(d*x+c))+1/16*(a-3*b)/(a-b)^3*b*ln(1+sin(d*x+c))+b^ 4*a/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c)))
Time = 0.17 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a b^{4} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{4} b - 6 \, a^{2} b^{3} - 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} b - 6 \, a^{2} b^{3} + 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^4*tan(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
1/16*(16*a*b^4*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + (a^4*b - 6*a^2*b^3 - 8*a*b^4 - 3*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^4*b - 6*a^2* b^3 + 8*a*b^4 - 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*(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 - (a^4*b + 2*a^2*b^3 - 3*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)
\[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**4*tan(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)*sec(c + d*x)**4/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a b - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a b^{2} \sin \left (d x + c\right )^{2} - {\left (a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, a b^{2} - {\left (a^{2} b - 5 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \] Input:
integrate(sec(d*x+c)^4*tan(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
1/16*(16*a*b^4*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (a*b - 3*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - ( a*b + 3*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(4* a*b^2*sin(d*x + c)^2 - (a^2*b + 3*b^3)*sin(d*x + c)^3 + 2*a^3 - 6*a*b^2 - (a^2*b - 5*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 = 0.19 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {{\left (a b + 3 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {{\left (a b - 3 \, b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} + \frac {2 \, a^{5} - 8 \, a^{3} b^{2} + 6 \, a b^{4} - {\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \sin \left (d x + c\right )^{3} + 4 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2} - {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^4*tan(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
a*b^5*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3*a^2*b^5*d - b^7*d) - 1/16*(a*b + 3*b^2)*log(abs(-sin(d*x + c) + 1))/(a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + 1/16*(a*b - 3*b^2)*log(abs(-sin(d*x + c) - 1))/(a^ 3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) + 1/8*(2*a^5 - 8*a^3*b^2 + 6*a*b^4 - (a^4*b + 2*a^2*b^3 - 3*b^5)*sin(d*x + c)^3 + 4*(a^3*b^2 - a*b^4)*sin(d*x + c)^2 - (a^4*b - 6*a^2*b^3 + 5*b^5)*sin(d*x + c))/((a + b)^3*(a - b)^3*d*( sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 21.52 (sec) , antiderivative size = 483, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a\,b^4\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2\,b-5\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-5\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-3\,b\right )}{8\,d\,{\left (a-b\right )}^3} \] Input:
int(tan(c + d*x)/(cos(c + d*x)^4*(a + b*sin(c + d*x))),x)
Output:
(a*b^4*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)) - ((2*tan(c/2 + (d*x)/2)^2*(2*a*b^2 - a^3)) /(a^4 + b^4 - 2*a^2*b^2) + (2*tan(c/2 + (d*x)/2)^6*(2*a*b^2 - a^3))/(a^4 + b^4 - 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^7*(a^2*b - 5*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^3*(7*a^2*b - 3*b^3))/(4*(a^4 + b^4 - 2* a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(7*a^2*b - 3*b^3))/(4*(a^4 + b^4 - 2*a^2 *b^2)) - (4*a*b^2*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^2) + (b*tan(c /2 + (d*x)/2)*(a^2 - 5*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)*(b/(8*(a + b)^2) + b^2/(4*( a + b)^3)))/d + (b*log(tan(c/2 + (d*x)/2) + 1)*(a - 3*b))/(8*d*(a - b)^3)
Time = 0.20 (sec) , antiderivative size = 1015, normalized size of antiderivative = 4.77 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4*tan(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**4*b + 6*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**4*a**2*b**3 - 8*log(tan((c + d*x)/2) - 1)*sin(c + d *x)**4*a*b**4 + 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**5 + 2*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b - 12*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2*a**2*b**3 + 16*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a*b**4 - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**5 - log(tan((c + d *x)/2) - 1)*a**4*b + 6*log(tan((c + d*x)/2) - 1)*a**2*b**3 - 8*log(tan((c + d*x)/2) - 1)*a*b**4 + 3*log(tan((c + d*x)/2) - 1)*b**5 + log(tan((c + d* x)/2) + 1)*sin(c + d*x)**4*a**4*b - 6*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**4*a**2*b**3 - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**4 - 3*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**5 - 2*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)**2*a**4*b + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a* *2*b**3 + 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**4 + 6*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)**2*b**5 + log(tan((c + d*x)/2) + 1)*a**4*b - 6*log(tan((c + d*x)/2) + 1)*a**2*b**3 - 8*log(tan((c + d*x)/2) + 1)*a*b* *4 - 3*log(tan((c + d*x)/2) + 1)*b**5 + 8*log(tan((c + d*x)/2)**2*a + 2*ta n((c + d*x)/2)*b + a)*sin(c + d*x)**4*a*b**4 - 16*log(tan((c + d*x)/2)**2* a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a*b**4 + 8*log(tan((c + d*x) /2)**2*a + 2*tan((c + d*x)/2)*b + a)*a*b**4 + 2*sin(c + d*x)**4*a**5 - 4*s in(c + d*x)**4*a**3*b**2 + 2*sin(c + d*x)**4*a*b**4 - sin(c + d*x)**3*a...