Integrand size = 29, antiderivative size = 93 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {\sec ^2(c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \]
-1/8*(a^2-3*b^2)*arctanh(sin(d*x+c))/d-1/8*sec(d*x+c)^2*(4*a*b+(a^2+3*b^2) *sin(d*x+c))/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)/d
Time = 0.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {1}{4} \sec ^4(c+d x) \left (16 a b \cos (2 (c+d x))+2 \left (-3 a^2+b^2+\left (a^2+5 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{8 d} \]
-1/8*((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]] + (Sec[c + d*x]^4*(16*a*b*Cos[2* (c + d*x)] + 2*(-3*a^2 + b^2 + (a^2 + 5*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x ]))/4)/d
Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.45, 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.
| \(\displaystyle \int \tan ^2(c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 (a+b \sin (c+d x))^2}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^2(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \int \frac {b^2 \sin ^2(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {b^3 \left (\frac {\int -\frac {b^2 (a+b \sin (c+d x)) (a+3 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {b^2 (a+b \sin (c+d x)) (a+3 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {1}{4} \int \frac {(a+b \sin (c+d x)) (a+3 b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))\right )}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^3 \left (\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {1}{4} \int \left (\frac {(a-3 b) (a-b)}{4 b^2 (\sin (c+d x) b+b)^2}+\frac {a^2-3 b^2}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {(a+b) (a+3 b)}{4 b^2 (b-b \sin (c+d x))^2}\right )d(b \sin (c+d x))\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^3 \left (\frac {1}{4} \left (-\frac {\left (a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 b^3}+\frac {(a-3 b) (a-b)}{4 b^2 (b \sin (c+d x)+b)}-\frac {(a+b) (a+3 b)}{4 b^2 (b-b \sin (c+d x))}\right )+\frac {b \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
(b^3*((b*Sin[c + d*x]*(a + b*Sin[c + d*x])^2)/(4*(b^2 - b^2*Sin[c + d*x]^2 )^2) + (-1/2*((a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/b^3 - ((a + b)*(a + 3*b ))/(4*b^2*(b - b*Sin[c + d*x])) + ((a - 3*b)*(a - b))/(4*b^2*(b + b*Sin[c + d*x])))/4))/d
3.15.96.3.1 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_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 = 0.57 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.78
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(166\) |
| default | \(\frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(166\) |
| parallelrisch | \(\frac {2 \left (a^{2}-3 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \left (a^{2}-3 b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 a b \cos \left (2 d x +2 c \right )+2 \cos \left (4 d x +4 c \right ) a b +\left (-a^{2}-5 b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (7 a^{2}+3 b^{2}\right ) \sin \left (d x +c \right )+6 a b}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(190\) |
| risch | \(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-7 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+16 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}+7 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-5 b^{2}+16 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\) | \(252\) |
| norman | \(\frac {\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (a^{2}-3 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (9 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (9 a^{2}+5 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (11 a^{2}+15 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (11 a^{2}+15 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {16 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(296\) |
1/d*(a^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)))+1/2*a*b*sin(d*x+c)^4/cos(d*x+c)^ 4+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))))
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, a b \cos \left (d x + c\right )^{2} - 8 \, a b + 2 \, {\left ({\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
-1/16*((a^2 - 3*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)* cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 16*a*b*cos(d*x + c)^2 - 8*a*b + 2* ((a^2 + 5*b^2)*cos(d*x + c)^2 - 2*a^2 - 2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.29 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (8 \, a b \sin \left (d x + c\right )^{2} + {\left (a^{2} + 5 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 4 \, a b + {\left (a^{2} - 3 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
-1/16*((a^2 - 3*b^2)*log(sin(d*x + c) + 1) - (a^2 - 3*b^2)*log(sin(d*x + c ) - 1) - 2*(8*a*b*sin(d*x + c)^2 + (a^2 + 5*b^2)*sin(d*x + c)^3 - 4*a*b + (a^2 - 3*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.50 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {{\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right )^{3} + 5 \, b^{2} \sin \left (d x + c\right )^{3} + 8 \, a b \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right ) - 3 \, b^{2} \sin \left (d x + c\right ) - 4 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*((a^2 - 3*b^2)*log(abs(sin(d*x + c) + 1)) - (a^2 - 3*b^2)*log(abs(si n(d*x + c) - 1)) - 2*(a^2*sin(d*x + c)^3 + 5*b^2*sin(d*x + c)^3 + 8*a*b*si n(d*x + c)^2 + a^2*sin(d*x + c) - 3*b^2*sin(d*x + c) - 4*a*b)/(sin(d*x + c )^2 - 1)^2)/d
Time = 17.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,a^2}{4}+\frac {11\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {7\,a^2}{4}+\frac {11\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^2}{4}-\frac {3\,b^2}{4}\right )}{d} \]
(tan(c/2 + (d*x)/2)^7*(a^2/4 - (3*b^2)/4) + tan(c/2 + (d*x)/2)^3*((7*a^2)/ 4 + (11*b^2)/4) + tan(c/2 + (d*x)/2)^5*((7*a^2)/4 + (11*b^2)/4) + tan(c/2 + (d*x)/2)*(a^2/4 - (3*b^2)/4) + 8*a*b*tan(c/2 + (d*x)/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(tan(c/2 + (d*x)/2))*(a^2/4 - (3*b^2)/4))/d