Integrand size = 11, antiderivative size = 75 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=-\frac {1}{4} (a-2 b) (a+b)^2 \log (1-\sin (x))+\frac {1}{4} (a-b)^2 (a+2 b) \log (1+\sin (x))+\frac {1}{2} a b^2 \sin (x)+\frac {1}{2} \sec ^2(x) (b+a \sin (x)) (a+b \sin (x))^2 \]
-1/4*(a-2*b)*(a+b)^2*ln(1-sin(x))+1/4*(a-b)^2*(a+2*b)*ln(1+sin(x))+1/2*a*b ^2*sin(x)+1/2*sec(x)^2*(b+a*sin(x))*(a+b*sin(x))^2
Time = 0.45 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.64 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=\frac {\left (a^2-b^2\right ) \left ((a-2 b) (a+b)^2 \log (1-\sin (x))-(a-b)^2 (a+2 b) \log (1+\sin (x))\right )+2 a^4 b \sec ^2(x)-2 a \left (a^4+2 a^2 b^2-3 b^4\right ) \sec (x) \tan (x)+\left (-8 a^4 b+4 a^2 b^3+2 b^5\right ) \tan ^2(x)}{4 \left (-a^2+b^2\right )} \]
((a^2 - b^2)*((a - 2*b)*(a + b)^2*Log[1 - Sin[x]] - (a - b)^2*(a + 2*b)*Lo g[1 + Sin[x]]) + 2*a^4*b*Sec[x]^2 - 2*a*(a^4 + 2*a^2*b^2 - 3*b^4)*Sec[x]*T an[x] + (-8*a^4*b + 4*a^2*b^3 + 2*b^5)*Tan[x]^2)/(4*(-a^2 + b^2))
Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4891, 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.
\(\displaystyle \int (a \sec (x)+b \tan (x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sec (x)+b \tan (x))^3dx\) |
\(\Big \downarrow \) 4891 |
\(\displaystyle \int \sec ^3(x) (a+b \sin (x))^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (x))^3}{\cos (x)^3}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle b^3 \int \frac {(a+b \sin (x))^3}{\left (b^2-b^2 \sin ^2(x)\right )^2}d(b \sin (x))\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (\frac {b^2 (a-b)^3}{4 (\sin (x) b+b)^2}+\frac {b (a+2 b) (a-b)^2}{4 (\sin (x) b+b)}+\frac {(a-2 b) b (a+b)^2}{4 (b-b \sin (x))}+\frac {b^2 (a+b)^3}{4 (b-b \sin (x))^2}\right )d(b \sin (x))}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {b^2 (a-b)^3}{4 (b \sin (x)+b)}+\frac {b^2 (a+b)^3}{4 (b-b \sin (x))}+\frac {1}{4} b (a+2 b) (a-b)^2 \log (b \sin (x)+b)-\frac {1}{4} b (a-2 b) (a+b)^2 \log (b-b \sin (x))}{b}\) |
(-1/4*((a - 2*b)*b*(a + b)^2*Log[b - b*Sin[x]]) + ((a - b)^2*b*(a + 2*b)*L og[b + b*Sin[x]])/4 + (b^2*(a + b)^3)/(4*(b - b*Sin[x])) - ((a - b)^3*b^2) /(4*(b + b*Sin[x])))/b
3.3.65.3.1 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]
Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x _)]^(n_.))^(p_), x_Symbol] :> Int[ActivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a *Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]
Time = 4.86 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99
method | result | size |
default | \(a^{3} \left (\frac {\sec \left (x \right ) \tan \left (x \right )}{2}+\frac {\ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{2}\right )+\frac {3 a^{2} b}{2 \cos \left (x \right )^{2}}+3 a \,b^{2} \left (\frac {\sin \left (x \right )^{3}}{2 \cos \left (x \right )^{2}}+\frac {\sin \left (x \right )}{2}-\frac {\ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{2}\right )+b^{3} \left (\frac {\tan \left (x \right )^{2}}{2}+\ln \left (\cos \left (x \right )\right )\right )\) | \(74\) |
parts | \(a^{3} \left (\frac {\sec \left (x \right ) \tan \left (x \right )}{2}+\frac {\ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{2}\right )+b^{3} \left (\frac {\tan \left (x \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (x \right )^{2}\right )}{2}\right )+3 a \,b^{2} \left (\frac {\sin \left (x \right )^{3}}{2 \cos \left (x \right )^{2}}+\frac {\sin \left (x \right )}{2}-\frac {\ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{2}\right )+\frac {3 a^{2} b \sec \left (x \right )^{2}}{2}\) | \(80\) |
risch | \(-i x \,b^{3}+\frac {{\mathrm e}^{i x} \left (-i a^{3} {\mathrm e}^{2 i x}-3 i a \,b^{2} {\mathrm e}^{2 i x}+i a^{3}+3 i a \,b^{2}+6 a^{2} b \,{\mathrm e}^{i x}+2 b^{3} {\mathrm e}^{i x}\right )}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) a^{3}}{2}+\frac {3 \ln \left ({\mathrm e}^{i x}-i\right ) a \,b^{2}}{2}+\ln \left ({\mathrm e}^{i x}-i\right ) b^{3}+\frac {\ln \left (i+{\mathrm e}^{i x}\right ) a^{3}}{2}-\frac {3 \ln \left (i+{\mathrm e}^{i x}\right ) a \,b^{2}}{2}+\ln \left (i+{\mathrm e}^{i x}\right ) b^{3}\) | \(172\) |
a^3*(1/2*sec(x)*tan(x)+1/2*ln(sec(x)+tan(x)))+3/2*a^2*b/cos(x)^2+3*a*b^2*( 1/2*sin(x)^3/cos(x)^2+1/2*sin(x)-1/2*ln(sec(x)+tan(x)))+b^3*(1/2*tan(x)^2+ ln(cos(x)))
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=\frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 6 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (x\right )}{4 \, \cos \left (x\right )^{2}} \]
1/4*((a^3 - 3*a*b^2 + 2*b^3)*cos(x)^2*log(sin(x) + 1) - (a^3 - 3*a*b^2 - 2 *b^3)*cos(x)^2*log(-sin(x) + 1) + 6*a^2*b + 2*b^3 + 2*(a^3 + 3*a*b^2)*sin( x))/cos(x)^2
Time = 1.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.63 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=- \frac {a^{3} \log {\left (\sin {\left (x \right )} - 1 \right )}}{4} + \frac {a^{3} \log {\left (\sin {\left (x \right )} + 1 \right )}}{4} - \frac {a^{3} \sin {\left (x \right )}}{2 \sin ^{2}{\left (x \right )} - 2} + \frac {3 a^{2} b \sec ^{2}{\left (x \right )}}{2} + \frac {3 a b^{2} \log {\left (\sin {\left (x \right )} - 1 \right )}}{4} - \frac {3 a b^{2} \log {\left (\sin {\left (x \right )} + 1 \right )}}{4} - \frac {3 a b^{2} \sin {\left (x \right )}}{2 \sin ^{2}{\left (x \right )} - 2} - \frac {b^{3} \log {\left (\sec ^{2}{\left (x \right )} \right )}}{2} + \frac {b^{3} \sec ^{2}{\left (x \right )}}{2} \]
-a**3*log(sin(x) - 1)/4 + a**3*log(sin(x) + 1)/4 - a**3*sin(x)/(2*sin(x)** 2 - 2) + 3*a**2*b*sec(x)**2/2 + 3*a*b**2*log(sin(x) - 1)/4 - 3*a*b**2*log( sin(x) + 1)/4 - 3*a*b**2*sin(x)/(2*sin(x)**2 - 2) - b**3*log(sec(x)**2)/2 + b**3*sec(x)**2/2
Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.27 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=\frac {3}{2} \, a^{2} b \tan \left (x\right )^{2} - \frac {3}{4} \, a b^{2} {\left (\frac {2 \, \sin \left (x\right )}{\sin \left (x\right )^{2} - 1} + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (\sin \left (x\right ) - 1\right )\right )} - \frac {1}{4} \, a^{3} {\left (\frac {2 \, \sin \left (x\right )}{\sin \left (x\right )^{2} - 1} - \log \left (\sin \left (x\right ) + 1\right ) + \log \left (\sin \left (x\right ) - 1\right )\right )} - \frac {1}{2} \, b^{3} {\left (\frac {1}{\sin \left (x\right )^{2} - 1} - \log \left (\sin \left (x\right )^{2} - 1\right )\right )} \]
3/2*a^2*b*tan(x)^2 - 3/4*a*b^2*(2*sin(x)/(sin(x)^2 - 1) + log(sin(x) + 1) - log(sin(x) - 1)) - 1/4*a^3*(2*sin(x)/(sin(x)^2 - 1) - log(sin(x) + 1) + log(sin(x) - 1)) - 1/2*b^3*(1/(sin(x)^2 - 1) - log(sin(x)^2 - 1))
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.15 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=\frac {1}{4} \, {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \log \left (\sin \left (x\right ) + 1\right ) - \frac {1}{4} \, {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \log \left (-\sin \left (x\right ) + 1\right ) - \frac {b^{3} \sin \left (x\right )^{2} + a^{3} \sin \left (x\right ) + 3 \, a b^{2} \sin \left (x\right ) + 3 \, a^{2} b}{2 \, {\left (\sin \left (x\right )^{2} - 1\right )}} \]
1/4*(a^3 - 3*a*b^2 + 2*b^3)*log(sin(x) + 1) - 1/4*(a^3 - 3*a*b^2 - 2*b^3)* log(-sin(x) + 1) - 1/2*(b^3*sin(x)^2 + a^3*sin(x) + 3*a*b^2*sin(x) + 3*a^2 *b)/(sin(x)^2 - 1)
Time = 28.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.68 \[ \int (a \sec (x)+b \tan (x))^3 \, dx=\frac {\left (a^3+3\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (6\,a^2\,b+2\,b^3\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\left (a^3+3\,a\,b^2\right )\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-b^3\,\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{2}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )\,{\left (a-b\right )}^2\,\left (a+2\,b\right )}{2} \]