Integrand size = 30, antiderivative size = 106 \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=a^4 x+\frac {a b \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 d}-\frac {a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d} \]
[Out]
Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4127, 4003, 4133, 3855, 3852, 8} \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=a^4 x+\frac {a b \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 d}-\frac {a b^3 \tan (c+d x) \sec (c+d x)}{3 d}-\frac {b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]
[In]
[Out]
Rule 8
Rule 3852
Rule 3855
Rule 4003
Rule 4127
Rule 4133
Rubi steps \begin{align*} \text {integral}& = -\int (-a+b \sec (c+d x)) (a+b \sec (c+d x))^3 \, dx \\ & = -\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {1}{3} \int (a+b \sec (c+d x)) \left (-3 a^3-b \left (3 a^2-2 b^2\right ) \sec (c+d x)+2 a b^2 \sec ^2(c+d x)\right ) \, dx \\ & = -\frac {a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-6 a^4-6 a b \left (2 a^2-b^2\right ) \sec (c+d x)-2 b^2 \left (a^2-2 b^2\right ) \sec ^2(c+d x)\right ) \, dx \\ & = a^4 x-\frac {a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {1}{3} \left (b^2 \left (a^2-2 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\left (a b \left (2 a^2-b^2\right )\right ) \int \sec (c+d x) \, dx \\ & = a^4 x+\frac {a b \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac {\left (b^2 \left (a^2-2 b^2\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = a^4 x+\frac {a b \left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 d}-\frac {a b^3 \sec (c+d x) \tan (c+d x)}{3 d}-\frac {b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=a^4 x+\frac {2 a^3 b \text {arctanh}(\sin (c+d x))}{d}-\frac {a b^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {a b^3 \sec (c+d x) \tan (c+d x)}{d}-\frac {b^4 \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \]
[In]
[Out]
Time = 0.62 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {a^{4} \left (d x +c \right )+2 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(94\) |
default | \(\frac {a^{4} \left (d x +c \right )+2 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(94\) |
parts | \(a^{4} x +\frac {b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {a \,b^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}-\frac {a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(99\) |
risch | \(a^{4} x +\frac {2 i b^{3} \left (3 a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 a \,{\mathrm e}^{i \left (d x +c \right )}-2 b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(156\) |
parallelrisch | \(\frac {-18 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) b a \left (a^{2}-\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) b a \left (a^{2}-\frac {b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+9 a^{4} x d \cos \left (d x +c \right )+3 a^{4} x d \cos \left (3 d x +3 c \right )-6 a \,b^{3} \sin \left (2 d x +2 c \right )-2 b^{4} \sin \left (3 d x +3 c \right )-6 \sin \left (d x +c \right ) b^{4}}{3 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(182\) |
norman | \(\frac {a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-a^{4} x +3 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 a^{4} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {4 b^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 b^{3} \left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 b^{3} \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}+\frac {a b \left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a b \left (2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(197\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.25 \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\frac {6 \, a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{2} + 3 \, a b^{3} \cos \left (d x + c\right ) + b^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]
[In]
[Out]
\[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\int \left (a - b \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3}\, dx \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (d x + c\right )} a^{4} - 2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{4} + 3 \, a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{6 \, d} \]
[In]
[Out]
none
Time = 0.31 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.58 \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (d x + c\right )} a^{4} + 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]
[In]
[Out]
Time = 17.00 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.51 \[ \int (a+b \sec (c+d x))^2 \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=\frac {2\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}-\frac {b^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}-\frac {2\,a\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a^3\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,b^3\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2} \]
[In]
[Out]