Integrand size = 28, antiderivative size = 125 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {\left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
-1/8*(4*a^2+b^2)*arctanh(sin(d*x+c))/d-2/3*a*b*tan(d*x+c)/d+1/8*(2*a^2-b^2 )*sec(d*x+c)*tan(d*x+c)/d+1/6*a*b*sec(d*x+c)^2*tan(d*x+c)/d+1/4*(b+a*cos(d *x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.62 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {-3 \left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2-b^2\right ) \sec (c+d x)+6 b^2 \sec ^3(c+d x)+16 a b \tan ^2(c+d x)\right )}{24 d} \]
(-3*(4*a^2 + b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(4*a^2 - b^2)*Se c[c + d*x] + 6*b^2*Sec[c + d*x]^3 + 16*a*b*Tan[c + d*x]^2))/(24*d)
Time = 1.14 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.09, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4897, 3042, 3368, 3042, 3527, 3042, 3510, 25, 3042, 3500, 25, 3042, 3227, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^3 (a \sin (c+d x)+b \tan (c+d x))^2dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \tan ^2(c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3368 |
| \(\displaystyle \int \left (1-\cos ^2(c+d x)\right ) \sec ^5(c+d x) (a \cos (c+d x)+b)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {1}{4} \int (b+a \cos (c+d x)) \left (-3 a \cos ^2(c+d x)-b \cos (c+d x)+2 a\right ) \sec ^4(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-3 a \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \sin \left (c+d x+\frac {\pi }{2}\right )+2 a\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (-9 a^2 \cos ^2(c+d x)-8 a b \cos (c+d x)+3 \left (2 a^2-b^2\right )\right ) \sec ^3(c+d x)\right )dx\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (-9 a^2 \cos ^2(c+d x)-8 a b \cos (c+d x)+3 \left (2 a^2-b^2\right )\right ) \sec ^3(c+d x)dx+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {-9 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2-8 a b \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (2 a^2-b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int -\left (\left (16 a b+3 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x)\right )dx+\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \left (16 a b+3 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \frac {16 a b+3 \left (4 a^2+b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-3 \left (4 a^2+b^2\right ) \int \sec (c+d x)dx-16 a b \int \sec ^2(c+d x)dx\right )+\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-3 \left (4 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-16 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {16 a b \int 1d(-\tan (c+d x))}{d}-3 \left (4 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )+\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-3 \left (4 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {16 a b \tan (c+d x)}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (-\frac {3 \left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {16 a b \tan (c+d x)}{d}\right )+\frac {3 \left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d}\) |
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*a*b*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*(2*a^2 - b^2)*Sec[c + d*x]*Tan[c + d*x ])/(2*d) + ((-3*(4*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/d - (16*a*b*Tan[c + d *x])/d)/2)/3)/4
3.3.42.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n }, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.97 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(138\) |
| default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(138\) |
| risch | \(-\frac {i \left (12 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+48 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+21 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+48 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-21 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+16 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{i \left (d x +c \right )}+16 a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2}}{2 d}-\frac {b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{2 d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(261\) |
1/d*(a^2*(1/2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+t an(d*x+c)))+2/3*a*b*sin(d*x+c)^3/cos(d*x+c)^3+b^2*(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))))
Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {3 \, {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, a b \cos \left (d x + c\right )^{3} - 16 \, a b \cos \left (d x + c\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
-1/48*(3*(4*a^2 + b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*a^2 + b ^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*a*b*cos(d*x + c)^3 - 16* a*b*cos(d*x + c) - 3*(4*a^2 - b^2)*cos(d*x + c)^2 - 6*b^2)*sin(d*x + c))/( d*cos(d*x + c)^4)
\[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {32 \, a b \tan \left (d x + c\right )^{3} + 3 \, b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]
1/48*(32*a*b*tan(d*x + c)^3 + 3*b^2*(2*(sin(d*x + c)^3 + sin(d*x + c))/(si n(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)))/d
Time = 1.00 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.81 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {3 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 64 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 64 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
-1/24*(3*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2 + b^2 )*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(12*a^2*tan(1/2*d*x + 1/2*c)^7 + 3*b^2*tan(1/2*d*x + 1/2*c)^7 - 12*a^2*tan(1/2*d*x + 1/2*c)^5 - 64*a*b*tan( 1/2*d*x + 1/2*c)^5 + 21*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*a^2*tan(1/2*d*x + 1/2*c)^3 + 64*a*b*tan(1/2*d*x + 1/2*c)^3 + 21*b^2*tan(1/2*d*x + 1/2*c)^3 + 12*a^2*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 25.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.42 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+\frac {b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-a^2-\frac {16\,a\,b}{3}+\frac {7\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-a^2+\frac {16\,a\,b}{3}+\frac {7\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^2+\frac {b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+\frac {b^2}{4}\right )}{d} \]
(tan(c/2 + (d*x)/2)^3*((16*a*b)/3 - a^2 + (7*b^2)/4) + tan(c/2 + (d*x)/2)* (a^2 + b^2/4) + tan(c/2 + (d*x)/2)^7*(a^2 + b^2/4) - tan(c/2 + (d*x)/2)^5* ((16*a*b)/3 + a^2 - (7*b^2)/4))/(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)) - (atanh(t an(c/2 + (d*x)/2))*(a^2 + b^2/4))/d