Integrand size = 46, antiderivative size = 97 \[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=a^2 (b B-a C) x+\frac {b \left (4 a b B-2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b^2 (2 b B+a C) \tan (c+d x)}{2 d}+\frac {b^2 C (a+b \sec (c+d x)) \tan (c+d x)}{2 d} \]
a^2*(B*b-C*a)*x+1/2*b*(4*B*a*b-2*C*a^2+C*b^2)*arctanh(sin(d*x+c))/d+1/2*b^ 2*(2*B*b+C*a)*tan(d*x+c)/d+1/2*b^2*C*(a+b*sec(d*x+c))*tan(d*x+c)/d
Time = 0.68 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 (b B-a C) d x+b \left (4 a b B-2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))+b^2 (2 b B+2 a C+b C \sec (c+d x)) \tan (c+d x)}{2 d} \]
Integrate[(a + b*Sec[c + d*x])*(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C *Sec[c + d*x]^2),x]
(2*a^2*(b*B - a*C)*d*x + b*(4*a*b*B - 2*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x ]] + b^2*(2*b*B + 2*a*C + b*C*Sec[c + d*x])*Tan[c + d*x])/(2*d)
Time = 0.42 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {3042, 4529, 3042, 4406, 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.
| \(\displaystyle \int (a+b \sec (c+d x)) \left (a^2 (-C)+a b B+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a^2 (-C)+a b B+b^2 B \csc \left (c+d x+\frac {\pi }{2}\right )+b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4529 |
| \(\displaystyle \frac {\int (a+b \sec (c+d x))^2 \left (C \sec (c+d x) b^3+(b B-a C) b^2\right )dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (C \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2\right )dx}{b^2}\) |
| \(\Big \downarrow \) 4406 |
| \(\displaystyle \frac {\frac {1}{2} \int \left ((2 b B+a C) \sec ^2(c+d x) b^4+\left (-2 C a^2+4 b B a+b^2 C\right ) \sec (c+d x) b^3+2 a^2 (b B-a C) b^2\right )dx+\frac {b^4 C \tan (c+d x) (a+b \sec (c+d x))}{2 d}}{b^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {b^3 \left (-2 a^2 C+4 a b B+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{d}+2 a^2 b^2 x (b B-a C)+\frac {b^4 (a C+2 b B) \tan (c+d x)}{d}\right )+\frac {b^4 C \tan (c+d x) (a+b \sec (c+d x))}{2 d}}{b^2}\) |
((b^4*C*(a + b*Sec[c + d*x])*Tan[c + d*x])/(2*d) + (2*a^2*b^2*(b*B - a*C)* x + (b^3*(4*a*b*B - 2*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/d + (b^4*(2*b* B + a*C)*Tan[c + d*x])/d)/2)/b^2
3.9.99.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[1/m Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b* c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[b*B - a*C + b*C*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.68 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12
| method | result | size |
| parts | \(a^{2} \left (B b -C a \right ) x +\frac {\left (B \,b^{3}+C a \,b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 B a \,b^{2}-a^{2} b C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(109\) |
| derivativedivides | \(\frac {B \,a^{2} b \left (d x +c \right )-a^{3} C \left (d x +c \right )+2 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \,b^{2} \tan \left (d x +c \right )-a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{3}+C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(129\) |
| default | \(\frac {B \,a^{2} b \left (d x +c \right )-a^{3} C \left (d x +c \right )+2 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C a \,b^{2} \tan \left (d x +c \right )-a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{3}+C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(129\) |
| parallelrisch | \(\frac {-2 b \left (B a b -\frac {1}{2} C \,a^{2}+\frac {1}{4} C \,b^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 b \left (B a b -\frac {1}{2} C \,a^{2}+\frac {1}{4} C \,b^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+a^{2} d x \left (B b -C a \right ) \cos \left (2 d x +2 c \right )+\left (B \,b^{3}+C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+C \sin \left (d x +c \right ) b^{3}+a^{2} d x \left (B b -C a \right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(175\) |
| norman | \(\frac {\left (B \,a^{2} b -a^{3} C \right ) x +\left (-2 B \,a^{2} b +2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (B \,a^{2} b -a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {b^{2} \left (2 B b +2 C a +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b^{2} \left (2 B b +2 C a -C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {b \left (4 B a b -2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (4 B a b -2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(217\) |
| risch | \(B \,a^{2} b x -a^{3} x C -\frac {i b^{2} \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -2 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b C}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a \,b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b C}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(233\) |
int((a+b*sec(d*x+c))*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2),x,m ethod=_RETURNVERBOSE)
a^2*(B*b-C*a)*x+(B*b^3+C*a*b^2)/d*tan(d*x+c)+(2*B*a*b^2-C*a^2*b)/d*ln(sec( d*x+c)+tan(d*x+c))+C*b^3/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+ta n(d*x+c)))
Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.59 \[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (C a^{3} - B a^{2} b\right )} d x \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C b^{3} + 2 \, {\left (C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
integrate((a+b*sec(d*x+c))*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^ 2),x, algorithm="fricas")
-1/4*(4*(C*a^3 - B*a^2*b)*d*x*cos(d*x + c)^2 + (2*C*a^2*b - 4*B*a*b^2 - C* b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*C*a^2*b - 4*B*a*b^2 - C*b^3 )*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(C*b^3 + 2*(C*a*b^2 + B*b^3)*c os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=- \int C a^{3}\, dx - \int \left (- B a^{2} b\right )\, dx - \int \left (- B b^{3} \sec ^{2}{\left (c + d x \right )}\right )\, dx - \int \left (- C b^{3} \sec ^{3}{\left (c + d x \right )}\right )\, dx - \int \left (- 2 B a b^{2} \sec {\left (c + d x \right )}\right )\, dx - \int \left (- C a b^{2} \sec ^{2}{\left (c + d x \right )}\right )\, dx - \int C a^{2} b \sec {\left (c + d x \right )}\, dx \]
-Integral(C*a**3, x) - Integral(-B*a**2*b, x) - Integral(-B*b**3*sec(c + d *x)**2, x) - Integral(-C*b**3*sec(c + d*x)**3, x) - Integral(-2*B*a*b**2*s ec(c + d*x), x) - Integral(-C*a*b**2*sec(c + d*x)**2, x) - Integral(C*a**2 *b*sec(c + d*x), x)
Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.46 \[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (d x + c\right )} C a^{3} - 4 \, {\left (d x + c\right )} B a^{2} b + C b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 8 \, B a b^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, C a b^{2} \tan \left (d x + c\right ) - 4 \, B b^{3} \tan \left (d x + c\right )}{4 \, d} \]
integrate((a+b*sec(d*x+c))*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^ 2),x, algorithm="maxima")
-1/4*(4*(d*x + c)*C*a^3 - 4*(d*x + c)*B*a^2*b + C*b^3*(2*sin(d*x + c)/(sin (d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 4*C*a^ 2*b*log(sec(d*x + c) + tan(d*x + c)) - 8*B*a*b^2*log(sec(d*x + c) + tan(d* x + c)) - 4*C*a*b^2*tan(d*x + c) - 4*B*b^3*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (93) = 186\).
Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.20 \[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (C a^{3} - B a^{2} b\right )} {\left (d x + c\right )} + {\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, C a^{2} b - 4 \, B a b^{2} - C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (2 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
integrate((a+b*sec(d*x+c))*(B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^ 2),x, algorithm="giac")
-1/2*(2*(C*a^3 - B*a^2*b)*(d*x + c) + (2*C*a^2*b - 4*B*a*b^2 - C*b^3)*log( abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*C*a^2*b - 4*B*a*b^2 - C*b^3)*log(abs(t an(1/2*d*x + 1/2*c) - 1)) + 2*(2*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^3* tan(1/2*d*x + 1/2*c)^3 - C*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*b^2*tan(1/2* d*x + 1/2*c) - 2*B*b^3*tan(1/2*d*x + 1/2*c) - C*b^3*tan(1/2*d*x + 1/2*c))/ (tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 18.78 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.28 \[ \int (a+b \sec (c+d x)) \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{2}+\frac {C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{2}-B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-C\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}\right )}{d} \]
((B*b^3*sin(2*c + 2*d*x))/2 + (C*b^3*sin(c + d*x))/2 + (C*a*b^2*sin(2*c + 2*d*x))/2)/(d*(cos(2*c + 2*d*x)/2 + 1/2)) - (2*(C*a^3*atan(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2)) + (C*b^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d *x)/2))*1i)/2 - B*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + B*a* b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i - C*a^2*b*atan((si n(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i))/d