Integrand size = 21, antiderivative size = 111 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {(a-2 b) (a+b)^2 \log (1-\sin (c+d x))}{4 d}+\frac {(a-b)^2 (a+2 b) \log (1+\sin (c+d x))}{4 d}+\frac {a b^2 \sin (c+d x)}{2 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{2 d} \]
-1/4*(a-2*b)*(a+b)^2*ln(1-sin(d*x+c))/d+1/4*(a-b)^2*(a+2*b)*ln(1+sin(d*x+c ))/d+1/2*a*b^2*sin(d*x+c)/d+1/2*sec(d*x+c)^2*(b+a*sin(d*x+c))*(a+b*sin(d*x +c))^2/d
Time = 0.90 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.59 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\left (a^2-b^2\right ) \left ((a-2 b) (a+b)^2 \log (1-\sin (c+d x))-(a-b)^2 (a+2 b) \log (1+\sin (c+d x))\right )+2 a^4 b \sec ^2(c+d x)-a \left (2 a^4+4 a^2 b^2-7 b^4+b^4 \cos (2 (c+d x))\right ) \sec (c+d x) \tan (c+d x)+\left (-8 a^4 b+4 a^2 b^3+2 b^5-2 a b^4 \sin (c+d x)\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \]
((a^2 - b^2)*((a - 2*b)*(a + b)^2*Log[1 - Sin[c + d*x]] - (a - b)^2*(a + 2 *b)*Log[1 + Sin[c + d*x]]) + 2*a^4*b*Sec[c + d*x]^2 - a*(2*a^4 + 4*a^2*b^2 - 7*b^4 + b^4*Cos[2*(c + d*x)])*Sec[c + d*x]*Tan[c + d*x] + (-8*a^4*b + 4 *a^2*b^3 + 2*b^5 - 2*a*b^4*Sin[c + d*x])*Tan[c + d*x]^2)/(4*(-a^2 + b^2)*d )
Time = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 477, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\cos (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {b^3 \int \frac {(a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (\frac {b^2 (a-b)^3}{4 (\sin (c+d x) b+b)^2}+\frac {b (a+2 b) (a-b)^2}{4 (\sin (c+d x) b+b)}+\frac {(a-2 b) b (a+b)^2}{4 (b-b \sin (c+d x))}+\frac {b^2 (a+b)^3}{4 (b-b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b^2 (a-b)^3}{4 (b \sin (c+d x)+b)}+\frac {b^2 (a+b)^3}{4 (b-b \sin (c+d x))}+\frac {1}{4} b (a+2 b) (a-b)^2 \log (b \sin (c+d x)+b)-\frac {1}{4} b (a-2 b) (a+b)^2 \log (b-b \sin (c+d x))}{b d}\) |
(-1/4*((a - 2*b)*b*(a + b)^2*Log[b - b*Sin[c + d*x]]) + ((a - b)^2*b*(a + 2*b)*Log[b + b*Sin[c + d*x]])/4 + (b^2*(a + b)^3)/(4*(b - b*Sin[c + d*x])) - ((a - b)^3*b^2)/(4*(b + b*Sin[c + d*x])))/(b*d)
3.5.4.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(126\) |
| default | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(126\) |
| parallelrisch | \(\frac {-2 b^{3} \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (a -2 b \right ) \left (a +b \right )^{2} \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +2 b \right ) \left (a -b \right )^{2} \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-3 a^{2} b -b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (2 a^{3}+6 a \,b^{2}\right ) \sin \left (d x +c \right )+3 a^{2} b +b^{3}}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(169\) |
| risch | \(-i b^{3} x -\frac {2 i b^{3} c}{d}+\frac {{\mathrm e}^{i \left (d x +c \right )} \left (-i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+i a^{3}+3 i a \,b^{2}+6 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}-\frac {a^{3} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}+\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a \,b^{2}}{2 d}+\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b^{3}}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}\) | \(251\) |
| norman | \(\frac {\frac {a \left (a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{2} b +2 b^{3}}{d}-\frac {\left (6 a^{2} b +2 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (15 a^{2} b +5 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (15 a^{2} b +5 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {b^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (a^{3}-3 a \,b^{2}-2 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a^{3}-3 a \,b^{2}+2 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(347\) |
1/d*(a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+3/2*a^2 *b/cos(d*x+c)^2+3*a*b^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)+tan(d*x+c)))+b^3*(1/2*tan(d*x+c)^2+ln(cos(d*x+c))))
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*((a^3 - 3*a*b^2 + 2*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (a^3 - 3*a*b^2 - 2*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 6*a^2*b + 2*b^3 + 2*(a^3 + 3*a*b^2)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a^{2} b + b^{3} + {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
1/4*((a^3 - 3*a*b^2 + 2*b^3)*log(sin(d*x + c) + 1) - (a^3 - 3*a*b^2 - 2*b^ 3)*log(sin(d*x + c) - 1) - 2*(3*a^2*b + b^3 + (a^3 + 3*a*b^2)*sin(d*x + c) )/(sin(d*x + c)^2 - 1))/d
Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) + 3 \, a b^{2} \sin \left (d x + c\right ) + 3 \, a^{2} b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
1/4*((a^3 - 3*a*b^2 + 2*b^3)*log(abs(sin(d*x + c) + 1)) - (a^3 - 3*a*b^2 - 2*b^3)*log(abs(sin(d*x + c) - 1)) - 2*(b^3*sin(d*x + c)^2 + a^3*sin(d*x + c) + 3*a*b^2*sin(d*x + c) + 3*a^2*b)/(sin(d*x + c)^2 - 1))/d
Time = 4.72 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.89 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2\,\left (a+2\,b\right )}{4\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{4\,d}-\frac {\frac {3\,a^2\,b}{2}+\frac {b^3}{2}+\sin \left (c+d\,x\right )\,\left (\frac {a^3}{2}+\frac {3\,a\,b^2}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]