Integrand size = 21, antiderivative size = 59 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{2 d} \] Output:
1/2*(a^2-b^2)*arctanh(sin(d*x+c))/d+1/2*sec(d*x+c)^2*(b+a*sin(d*x+c))*(a+b *sin(d*x+c))/d
Time = 1.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \] Input:
Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
Output:
((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 = 0.31 (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 |
| \(\displaystyle \frac {\frac {1}{2} b \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))+\frac {b^2 (a+b)^2}{4 (b-b \sin (c+d x))}-\frac {b^2 (a-b)^2}{4 (b \sin (c+d x)+b)}}{b d}\) |
Input:
Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
Output:
((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)
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 = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a b}{\cos \left (d x +c \right )^{2}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(99\) |
| default | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {a b}{\cos \left (d x +c \right )^{2}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(99\) |
| parallelrisch | \(\frac {-\left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 a b \cos \left (2 d x +2 c \right )+\left (2 a^{2}+2 b^{2}\right ) \sin \left (d x +c \right )+2 a b}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(120\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-a^{2}-b^{2}+4 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}\) | \(171\) |
| norman | \(\frac {\frac {\left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {3 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {3 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {8 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(228\) |
Input:
int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
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 = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a b + 2 \, {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
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)
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**2,x)
Output:
Integral((a + b*sin(c + d*x))**2*sec(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.32 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
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 = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.54 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, d} - \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, d} - \frac {a^{2} \sin \left (d x + c\right ) + b^{2} \sin \left (d x + c\right ) + 2 \, a b}{2 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )} d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/4*(a^2 - b^2)*log(abs(sin(d*x + c) + 1))/d - 1/4*(a^2 - b^2)*log(abs(sin (d*x + c) - 1))/d - 1/2*(a^2*sin(d*x + c) + b^2*sin(d*x + c) + 2*a*b)/((si n(d*x + c)^2 - 1)*d)
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )}{d}-\frac {a\,b+\sin \left (c+d\,x\right )\,\left (\frac {a^2}{2}+\frac {b^2}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \] Input:
int((a + b*sin(c + d*x))^2/cos(c + d*x)^3,x)
Output:
(atanh(sin(c + d*x))*(a^2/2 - b^2/2))/d - (a*b + sin(c + d*x)*(a^2/2 + b^2 /2))/(d*(sin(c + d*x)^2 - 1))
Time = 0.16 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.66 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-2 \sin \left (d x +c \right )^{2} a b -\sin \left (d x +c \right ) a^{2}-\sin \left (d x +c \right ) b^{2}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2,x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + log(tan((c + d*x)/2) - 1)*a**2 - log(tan((c + d*x)/2) - 1)*b**2 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 - log(tan((c + d*x)/2) + 1)*a**2 + log(tan((c + d*x)/2) + 1)*b**2 - 2*sin(c + d*x)**2*a*b - sin(c + d*x)*a **2 - sin(c + d*x)*b**2)/(2*d*(sin(c + d*x)**2 - 1))