Integrand size = 29, antiderivative size = 156 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))}{16 d}-\frac {(a+b)^2 (a+4 b)}{16 d (1-\sin (c+d x))}+\frac {(a-4 b) (a-b)^2}{16 d (1+\sin (c+d x))}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \] Output:
1/16*(a^3-9*a*b^2-8*b^3)*ln(1-sin(d*x+c))/d-1/16*(a^3-9*a*b^2+8*b^3)*ln(1+ sin(d*x+c))/d-1/16*(a+b)^2*(a+4*b)/d/(1-sin(d*x+c))+1/16*(a-4*b)*(a-b)^2/d /(1+sin(d*x+c))+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)/d
Time = 0.45 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.90 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))-\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {(a+b)^2 (a+7 b)}{-1+\sin (c+d x)}-\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(a-7 b) (a-b)^2}{1+\sin (c+d x)}}{16 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*Tan[c + d*x]^2,x]
Output:
((a^3 - 9*a*b^2 - 8*b^3)*Log[1 - Sin[c + d*x]] - (a^3 - 9*a*b^2 + 8*b^3)*L og[1 + Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(a + 7 *b))/(-1 + Sin[c + d*x]) - (a - b)^3/(1 + Sin[c + d*x])^2 + ((a - 7*b)*(a - b)^2)/(1 + Sin[c + d*x]))/(16*d)
Time = 0.47 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3316, 27, 531, 25, 27, 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.
| \(\displaystyle \int \tan ^2(c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 (a+b \sin (c+d x))^3}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^2(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \int \frac {b^2 \sin ^2(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {b^3 \left (\frac {\int -\frac {b^2 (a+b \sin (c+d x))^2 (a+4 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {b^2 (a+b \sin (c+d x))^2 (a+4 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {(a+b \sin (c+d x))^2 (a+4 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))\right )}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {1}{4} \int \left (\frac {(a-4 b) (a-b)^2}{4 b^2 (\sin (c+d x) b+b)^2}+\frac {a^3-9 b^2 a-8 b^3}{4 b^3 (b-b \sin (c+d x))}+\frac {a^3-9 b^2 a+8 b^3}{4 b^3 (\sin (c+d x) b+b)}+\frac {(a+b)^2 (a+4 b)}{4 b^2 (b-b \sin (c+d x))^2}\right )d(b \sin (c+d x))\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^3 \left (\frac {1}{4} \left (\frac {\left (a^3-9 a b^2-8 b^3\right ) \log (b-b \sin (c+d x))}{4 b^3}-\frac {\left (a^3-9 a b^2+8 b^3\right ) \log (b \sin (c+d x)+b)}{4 b^3}+\frac {(a-4 b) (a-b)^2}{4 b^2 (b \sin (c+d x)+b)}-\frac {(a+b)^2 (a+4 b)}{4 b^2 (b-b \sin (c+d x))}\right )+\frac {b \sin (c+d x) (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^3*Tan[c + d*x]^2,x]
Output:
(b^3*((b*Sin[c + d*x]*(a + b*Sin[c + d*x])^3)/(4*(b^2 - b^2*Sin[c + d*x]^2 )^2) + (((a^3 - 9*a*b^2 - 8*b^3)*Log[b - b*Sin[c + d*x]])/(4*b^3) - ((a^3 - 9*a*b^2 + 8*b^3)*Log[b + b*Sin[c + d*x]])/(4*b^3) - ((a + b)^2*(a + 4*b) )/(4*b^2*(b - b*Sin[c + d*x])) + ((a - 4*b)*(a - b)^2)/(4*b^2*(b + b*Sin[c + d*x])))/4))/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_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b *x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
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 = 3.59 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.31
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(204\) |
| default | \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{3} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(204\) |
| risch | \(i b^{3} x +\frac {2 i b^{3} c}{d}-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (-i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-15 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+7 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+9 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-7 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-9 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{3}+15 i a \,b^{2}+24 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{8 d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}\) | \(373\) |
Input:
int(sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8* sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c)))+3/4*a^2*b*sin(d*x+c)^4/cos(d*x+c )^4+3*a*b^2*(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)))+b^3*(1/4*tan (d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))
Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 12 \, a^{2} b - 4 \, b^{3} + 8 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{3} + 6 \, a b^{2} - {\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="frica s")
Output:
-1/16*((a^3 - 9*a*b^2 + 8*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^3 - 9*a*b^2 - 8*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 12*a^2*b - 4*b ^3 + 8*(3*a^2*b + 2*b^3)*cos(d*x + c)^2 - 2*(2*a^3 + 6*a*b^2 - (a^3 + 15*a *b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**3*tan(d*x+c)**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (a^{3} + 15 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{2} b - 6 \, b^{3} + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="maxim a")
Output:
-1/16*((a^3 - 9*a*b^2 + 8*b^3)*log(sin(d*x + c) + 1) - (a^3 - 9*a*b^2 - 8* b^3)*log(sin(d*x + c) - 1) - 2*((a^3 + 15*a*b^2)*sin(d*x + c)^3 - 6*a^2*b - 6*b^3 + 4*(3*a^2*b + 2*b^3)*sin(d*x + c)^2 + (a^3 - 9*a*b^2)*sin(d*x + c ))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.46 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} + \frac {{\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {{\left (a^{3} + 15 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{2} b - 6 \, b^{3} + 4 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="giac" )
Output:
-1/16*(a^3 - 9*a*b^2 + 8*b^3)*log(abs(sin(d*x + c) + 1))/d + 1/16*(a^3 - 9 *a*b^2 - 8*b^3)*log(abs(sin(d*x + c) - 1))/d + 1/8*((a^3 + 15*a*b^2)*sin(d *x + c)^3 - 6*a^2*b - 6*b^3 + 4*(3*a^2*b + 2*b^3)*sin(d*x + c)^2 + (a^3 - 9*a*b^2)*sin(d*x + c))/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 19.66 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {b^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (-\frac {a^3}{8}+\frac {9\,a\,b^2}{8}+b^3\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {a^3}{8}-\frac {9\,a\,b^2}{8}+b^3\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {9\,a\,b^2}{4}-\frac {a^3}{4}\right )+2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {9\,a\,b^2}{4}-\frac {a^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {7\,a^3}{4}+\frac {33\,a\,b^2}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {7\,a^3}{4}+\frac {33\,a\,b^2}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^2\,b+8\,b^3\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((tan(c + d*x)^2*(a + b*sin(c + d*x))^3)/cos(c + d*x)^3,x)
Output:
(b^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (log(tan(c/2 + (d*x)/2) - 1)*((9*a *b^2)/8 - a^3/8 + b^3))/d - (log(tan(c/2 + (d*x)/2) + 1)*(a^3/8 - (9*a*b^2 )/8 + b^3))/d - (tan(c/2 + (d*x)/2)*((9*a*b^2)/4 - a^3/4) + 2*b^3*tan(c/2 + (d*x)/2)^2 + 2*b^3*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^7*((9*a*b^2 )/4 - a^3/4) - tan(c/2 + (d*x)/2)^3*((33*a*b^2)/4 + (7*a^3)/4) - tan(c/2 + (d*x)/2)^5*((33*a*b^2)/4 + (7*a^3)/4) - tan(c/2 + (d*x)/2)^4*(12*a^2*b + 8*b^3))/(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))
Time = 0.19 (sec) , antiderivative size = 595, normalized size of antiderivative = 3.81 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3*(a+b*sin(d*x+c))^3*tan(d*x+c)^2,x)
Output:
(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*b**3 - 16*log(tan((c + d*x )/2)**2 + 1)*sin(c + d*x)**2*b**3 + 8*log(tan((c + d*x)/2)**2 + 1)*b**3 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**2 - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b* *3 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 18*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**2*a*b**2 + 16*log(tan((c + d*x)/2) - 1)*sin(c + d *x)**2*b**3 + log(tan((c + d*x)/2) - 1)*a**3 - 9*log(tan((c + d*x)/2) - 1) *a*b**2 - 8*log(tan((c + d*x)/2) - 1)*b**3 - log(tan((c + d*x)/2) + 1)*sin (c + d*x)**4*a**3 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**2 - 8 *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**3 + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a* b**2 + 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - log(tan((c + d* x)/2) + 1)*a**3 + 9*log(tan((c + d*x)/2) + 1)*a*b**2 - 8*log(tan((c + d*x) /2) + 1)*b**3 + 6*sin(c + d*x)**4*a**2*b + 6*sin(c + d*x)**4*b**3 + sin(c + d*x)**3*a**3 + 15*sin(c + d*x)**3*a*b**2 - 4*sin(c + d*x)**2*b**3 + sin( c + d*x)*a**3 - 9*sin(c + d*x)*a*b**2)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d *x)**2 + 1))