Integrand size = 37, antiderivative size = 76 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=a b^2 x+\frac {b \left (5 a^2+2 b^2\right ) \text {arctanh}(\sin (d+e x))}{2 e}+\frac {a \left (a^2+2 b^2\right ) \tan (d+e x)}{e}+\frac {a^2 b \sec (d+e x) \tan (d+e x)}{2 e} \] Output:
a*b^2*x+1/2*b*(5*a^2+2*b^2)*arctanh(sin(e*x+d))/e+a*(a^2+2*b^2)*tan(e*x+d) /e+1/2*a^2*b*sec(e*x+d)*tan(e*x+d)/e
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\frac {2 b \left (2 a^2+b^2\right ) \coth ^{-1}(\sin (d+e x))+a \left (2 b^2 e x+a b \text {arctanh}(\sin (d+e x))+\left (2 a^2+4 b^2+a b \sec (d+e x)\right ) \tan (d+e x)\right )}{2 e} \] Input:
Integrate[(a + b*Sec[d + e*x])*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x ]^2),x]
Output:
(2*b*(2*a^2 + b^2)*ArcCoth[Sin[d + e*x]] + a*(2*b^2*e*x + a*b*ArcTanh[Sin[ d + e*x]] + (2*a^2 + 4*b^2 + a*b*Sec[d + e*x])*Tan[d + e*x]))/(2*e)
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {3042, 4536, 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.
| \(\displaystyle \int (a+b \sec (d+e x)) \left (a^2 \sec ^2(d+e x)+2 a b \sec (d+e x)+b^2\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (d+e x+\frac {\pi }{2}\right )\right ) \left (a^2 \csc \left (d+e x+\frac {\pi }{2}\right )^2+2 a b \csc \left (d+e x+\frac {\pi }{2}\right )+b^2\right )dx\) |
| \(\Big \downarrow \) 4536 |
| \(\displaystyle \frac {1}{2} \int \left (2 a b^2+\left (5 a^2+2 b^2\right ) \sec (d+e x) b+2 a \left (a^2+2 b^2\right ) \sec ^2(d+e x)\right )dx+\frac {a^2 b \tan (d+e x) \sec (d+e x)}{2 e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \left (5 a^2+2 b^2\right ) \text {arctanh}(\sin (d+e x))}{e}+\frac {2 a \left (a^2+2 b^2\right ) \tan (d+e x)}{e}+2 a b^2 x\right )+\frac {a^2 b \tan (d+e x) \sec (d+e x)}{2 e}\) |
Input:
Int[(a + b*Sec[d + e*x])*(b^2 + 2*a*b*Sec[d + e*x] + a^2*Sec[d + e*x]^2),x ]
Output:
(a^2*b*Sec[d + e*x]*Tan[d + e*x])/(2*e) + (2*a*b^2*x + (b*(5*a^2 + 2*b^2)* ArcTanh[Sin[d + e*x]])/e + (2*a*(a^2 + 2*b^2)*Tan[d + e*x])/e)/2
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2 Int[Simp[2*A*a + (2*B*a + b*(2* A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B, C}, x]
Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.25
| method | result | size |
| parts | \(a \,b^{2} x +\frac {\left (2 a^{2} b +b^{3}\right ) \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{e}+\frac {\left (a^{3}+2 a \,b^{2}\right ) \tan \left (e x +d \right )}{e}+\frac {a^{2} b \left (\frac {\sec \left (e x +d \right ) \tan \left (e x +d \right )}{2}+\frac {\ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{2}\right )}{e}\) | \(95\) |
| derivativedivides | \(\frac {a \,b^{2} \left (e x +d \right )+2 a^{2} b \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )+\tan \left (e x +d \right ) a^{3}+b^{3} \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )+2 \tan \left (e x +d \right ) a \,b^{2}+a^{2} b \left (\frac {\sec \left (e x +d \right ) \tan \left (e x +d \right )}{2}+\frac {\ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{2}\right )}{e}\) | \(112\) |
| default | \(\frac {a \,b^{2} \left (e x +d \right )+2 a^{2} b \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )+\tan \left (e x +d \right ) a^{3}+b^{3} \ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )+2 \tan \left (e x +d \right ) a \,b^{2}+a^{2} b \left (\frac {\sec \left (e x +d \right ) \tan \left (e x +d \right )}{2}+\frac {\ln \left (\sec \left (e x +d \right )+\tan \left (e x +d \right )\right )}{2}\right )}{e}\) | \(112\) |
| parallelrisch | \(\frac {5 \left (a^{2}+\frac {2 b^{2}}{5}\right ) \left (1+\cos \left (2 e x +2 d \right )\right ) b \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )-5 \left (a^{2}+\frac {2 b^{2}}{5}\right ) \left (1+\cos \left (2 e x +2 d \right )\right ) b \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )+2 a \left (b^{2} e x \cos \left (2 e x +2 d \right )+b^{2} e x +\sin \left (2 e x +2 d \right ) a^{2}+a b \sin \left (e x +d \right )+2 \sin \left (2 e x +2 d \right ) b^{2}\right )}{2 e \left (1+\cos \left (2 e x +2 d \right )\right )}\) | \(151\) |
| risch | \(a \,b^{2} x -\frac {i a \left ({\mathrm e}^{3 i \left (e x +d \right )} a b -2 \,{\mathrm e}^{2 i \left (e x +d \right )} a^{2}-4 b^{2} {\mathrm e}^{2 i \left (e x +d \right )}-a b \,{\mathrm e}^{i \left (e x +d \right )}-2 a^{2}-4 b^{2}\right )}{e \left ({\mathrm e}^{2 i \left (e x +d \right )}+1\right )^{2}}+\frac {5 b \ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right ) a^{2}}{2 e}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (e x +d \right )}+i\right )}{e}-\frac {5 b \ln \left ({\mathrm e}^{i \left (e x +d \right )}-i\right ) a^{2}}{2 e}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (e x +d \right )}-i\right )}{e}\) | \(177\) |
| norman | \(\frac {a \,b^{2} x +a \,b^{2} x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\frac {a \left (2 a^{2}+a b +4 b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}-2 a \,b^{2} x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-\frac {a \left (2 a^{2}-a b +4 b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{e}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-1\right )^{2}}-\frac {b \left (5 a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right )}{2 e}+\frac {b \left (5 a^{2}+2 b^{2}\right ) \ln \left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 e}\) | \(178\) |
Input:
int((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2),x,method=_RET URNVERBOSE)
Output:
a*b^2*x+(2*a^2*b+b^3)/e*ln(sec(e*x+d)+tan(e*x+d))+(a^3+2*a*b^2)/e*tan(e*x+ d)+a^2*b/e*(1/2*sec(e*x+d)*tan(e*x+d)+1/2*ln(sec(e*x+d)+tan(e*x+d)))
Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.64 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\frac {4 \, a b^{2} e x \cos \left (e x + d\right )^{2} + {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (e x + d\right )^{2} \log \left (\sin \left (e x + d\right ) + 1\right ) - {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (e x + d\right )^{2} \log \left (-\sin \left (e x + d\right ) + 1\right ) + 2 \, {\left (a^{2} b + 2 \, {\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{4 \, e \cos \left (e x + d\right )^{2}} \] Input:
integrate((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2),x, algo rithm="fricas")
Output:
1/4*(4*a*b^2*e*x*cos(e*x + d)^2 + (5*a^2*b + 2*b^3)*cos(e*x + d)^2*log(sin (e*x + d) + 1) - (5*a^2*b + 2*b^3)*cos(e*x + d)^2*log(-sin(e*x + d) + 1) + 2*(a^2*b + 2*(a^3 + 2*a*b^2)*cos(e*x + d))*sin(e*x + d))/(e*cos(e*x + d)^ 2)
\[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\int \left (a + b \sec {\left (d + e x \right )}\right ) \left (a \sec {\left (d + e x \right )} + b\right )^{2}\, dx \] Input:
integrate((a+b*sec(e*x+d))*(b**2+2*a*b*sec(e*x+d)+a**2*sec(e*x+d)**2),x)
Output:
Integral((a + b*sec(d + e*x))*(a*sec(d + e*x) + b)**2, x)
Time = 0.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.66 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\frac {4 \, {\left (e x + d\right )} a b^{2} - a^{2} b {\left (\frac {2 \, \sin \left (e x + d\right )}{\sin \left (e x + d\right )^{2} - 1} - \log \left (\sin \left (e x + d\right ) + 1\right ) + \log \left (\sin \left (e x + d\right ) - 1\right )\right )} + 8 \, a^{2} b \log \left (\sec \left (e x + d\right ) + \tan \left (e x + d\right )\right ) + 4 \, b^{3} \log \left (\sec \left (e x + d\right ) + \tan \left (e x + d\right )\right ) + 4 \, a^{3} \tan \left (e x + d\right ) + 8 \, a b^{2} \tan \left (e x + d\right )}{4 \, e} \] Input:
integrate((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2),x, algo rithm="maxima")
Output:
1/4*(4*(e*x + d)*a*b^2 - a^2*b*(2*sin(e*x + d)/(sin(e*x + d)^2 - 1) - log( sin(e*x + d) + 1) + log(sin(e*x + d) - 1)) + 8*a^2*b*log(sec(e*x + d) + ta n(e*x + d)) + 4*b^3*log(sec(e*x + d) + tan(e*x + d)) + 4*a^3*tan(e*x + d) + 8*a*b^2*tan(e*x + d))/e
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (72) = 144\).
Time = 0.36 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.39 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\frac {2 \, {\left (e x + d\right )} a b^{2} + {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 1 \right |}\right ) - {\left (5 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 4 \, a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 2 \, a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 4 \, a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 1\right )}^{2}}}{2 \, e} \] Input:
integrate((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2),x, algo rithm="giac")
Output:
1/2*(2*(e*x + d)*a*b^2 + (5*a^2*b + 2*b^3)*log(abs(tan(1/2*e*x + 1/2*d) + 1)) - (5*a^2*b + 2*b^3)*log(abs(tan(1/2*e*x + 1/2*d) - 1)) - 2*(2*a^3*tan( 1/2*e*x + 1/2*d)^3 - a^2*b*tan(1/2*e*x + 1/2*d)^3 + 4*a*b^2*tan(1/2*e*x + 1/2*d)^3 - 2*a^3*tan(1/2*e*x + 1/2*d) - a^2*b*tan(1/2*e*x + 1/2*d) - 4*a*b ^2*tan(1/2*e*x + 1/2*d))/(tan(1/2*e*x + 1/2*d)^2 - 1)^2)/e
Time = 16.01 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.11 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\frac {2\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )}{e}+\frac {a^3\,\sin \left (d+e\,x\right )}{e\,\cos \left (d+e\,x\right )}+\frac {2\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )}{e}+\frac {5\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {d}{2}+\frac {e\,x}{2}\right )}{\cos \left (\frac {d}{2}+\frac {e\,x}{2}\right )}\right )}{e}+\frac {2\,a\,b^2\,\sin \left (d+e\,x\right )}{e\,\cos \left (d+e\,x\right )}+\frac {a^2\,b\,\sin \left (d+e\,x\right )}{2\,e\,{\cos \left (d+e\,x\right )}^2} \] Input:
int((a + b/cos(d + e*x))*(b^2 + a^2/cos(d + e*x)^2 + (2*a*b)/cos(d + e*x)) ,x)
Output:
(2*b^3*atanh(sin(d/2 + (e*x)/2)/cos(d/2 + (e*x)/2)))/e + (a^3*sin(d + e*x) )/(e*cos(d + e*x)) + (2*a*b^2*atan(sin(d/2 + (e*x)/2)/cos(d/2 + (e*x)/2))) /e + (5*a^2*b*atanh(sin(d/2 + (e*x)/2)/cos(d/2 + (e*x)/2)))/e + (2*a*b^2*s in(d + e*x))/(e*cos(d + e*x)) + (a^2*b*sin(d + e*x))/(2*e*cos(d + e*x)^2)
Time = 0.16 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.43 \[ \int (a+b \sec (d+e x)) \left (b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)\right ) \, dx=\frac {-2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{3}-4 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,b^{2}-5 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) \sin \left (e x +d \right )^{2} a^{2} b -2 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) \sin \left (e x +d \right )^{2} b^{3}+5 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )-1\right ) b^{3}+5 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) \sin \left (e x +d \right )^{2} a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) \sin \left (e x +d \right )^{2} b^{3}-5 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) a^{2} b -2 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )+1\right ) b^{3}+2 \sin \left (e x +d \right )^{2} a \,b^{2} e x -\sin \left (e x +d \right ) a^{2} b -2 a \,b^{2} e x}{2 e \left (\sin \left (e x +d \right )^{2}-1\right )} \] Input:
int((a+b*sec(e*x+d))*(b^2+2*a*b*sec(e*x+d)+a^2*sec(e*x+d)^2),x)
Output:
( - 2*cos(d + e*x)*sin(d + e*x)*a**3 - 4*cos(d + e*x)*sin(d + e*x)*a*b**2 - 5*log(tan((d + e*x)/2) - 1)*sin(d + e*x)**2*a**2*b - 2*log(tan((d + e*x) /2) - 1)*sin(d + e*x)**2*b**3 + 5*log(tan((d + e*x)/2) - 1)*a**2*b + 2*log (tan((d + e*x)/2) - 1)*b**3 + 5*log(tan((d + e*x)/2) + 1)*sin(d + e*x)**2* a**2*b + 2*log(tan((d + e*x)/2) + 1)*sin(d + e*x)**2*b**3 - 5*log(tan((d + e*x)/2) + 1)*a**2*b - 2*log(tan((d + e*x)/2) + 1)*b**3 + 2*sin(d + e*x)** 2*a*b**2*e*x - sin(d + e*x)*a**2*b - 2*a*b**2*e*x)/(2*e*(sin(d + e*x)**2 - 1))