Integrand size = 21, antiderivative size = 59 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{2 d} \]
Time = 0.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \]
((a^2 - b^2)^2*(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*S ec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c + d*x]*Tan[c + d*x] + (-6*a^3*b + 4*a* b^3)*Tan[c + d*x]^2)/(4*(-a^2 + b^2)*d)
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b^3 \int \frac {(a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (\frac {\left (a^2-b^2\right ) b^2}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {(a+b)^2 b^2}{4 (b-b \sin (c+d x))^2}+\frac {(a-b)^2 b^2}{4 (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} b \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {b^2 (a+b)^2}{4 (b-b \sin (c+d x))}-\frac {b^2 (a-b)^2}{4 (b \sin (c+d x)+b)}}{b d}\) |
((b*(a^2 - b^2)*ArcTanh[Sin[c + d*x]])/2 + (b^2*(a + b)^2)/(4*(b - b*Sin[c + d*x])) - ((a - b)^2*b^2)/(4*(b + b*Sin[c + d*x])))/(b*d)
3.4.91.3.1 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 = 1.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a b}{\cos \left (d x +c \right )^{2}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(99\) |
default | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a b}{\cos \left (d x +c \right )^{2}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(99\) |
parallelrisch | \(\frac {-\left (a -b \right ) \left (a +b \right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a -b \right ) \left (a +b \right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 a b \cos \left (2 d x +2 c \right )+\left (2 a^{2}+2 b^{2}\right ) \sin \left (d x +c \right )+2 a b}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(120\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-b^{2}+4 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}-\frac {a^{2} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}+\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b^{2}}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}\) | \(171\) |
norman | \(\frac {\frac {\left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (a^{2}+b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(228\) |
1/d*(a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a*b/cos (d*x+c)^2+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))))
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a b + 2 \, {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*((a^2 - b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (a^2 - b^2)*cos(d* x + c)^2*log(-sin(d*x + c) + 1) + 4*a*b + 2*(a^2 + b^2)*sin(d*x + c))/(d*c os(d*x + c)^2)
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.32 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
1/4*((a^2 - b^2)*log(sin(d*x + c) + 1) - (a^2 - b^2)*log(sin(d*x + c) - 1) - 2*(2*a*b + (a^2 + b^2)*sin(d*x + c))/(sin(d*x + c)^2 - 1))/d
Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.46 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right ) + b^{2} \sin \left (d x + c\right ) + 2 \, a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
1/4*((a^2 - b^2)*log(abs(sin(d*x + c) + 1)) - (a^2 - b^2)*log(abs(sin(d*x + c) - 1)) - 2*(a^2*sin(d*x + c) + b^2*sin(d*x + c) + 2*a*b)/(sin(d*x + c) ^2 - 1))/d
Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )}{d}-\frac {a\,b+\sin \left (c+d\,x\right )\,\left (\frac {a^2}{2}+\frac {b^2}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]