Integrand size = 16, antiderivative size = 44 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^2}{2}+\frac {a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )}{d}+\frac {b^2 \tan \left (c+d x^2\right )}{2 d} \] Output:
1/2*a^2*x^2+a*b*arctanh(sin(d*x^2+c))/d+1/2*b^2*tan(d*x^2+c)/d
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 d x^2+2 a b \coth ^{-1}\left (\sin \left (c+d x^2\right )\right )+b^2 \tan \left (c+d x^2\right )}{2 d} \] Input:
Integrate[x*(a + b*Sec[c + d*x^2])^2,x]
Output:
(a^2*d*x^2 + 2*a*b*ArcCoth[Sin[c + d*x^2]] + b^2*Tan[c + d*x^2])/(2*d)
Time = 0.35 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4692, 3042, 4260, 3042, 4254, 24, 4257}
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.
\(\displaystyle \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 4692 |
\(\displaystyle \frac {1}{2} \int \left (a+b \sec \left (d x^2+c\right )\right )^2dx^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \left (a+b \csc \left (d x^2+c+\frac {\pi }{2}\right )\right )^2dx^2\) |
\(\Big \downarrow \) 4260 |
\(\displaystyle \frac {1}{2} \left (2 a b \int \sec \left (d x^2+c\right )dx^2+b^2 \int \sec ^2\left (d x^2+c\right )dx^2+a^2 x^2\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (2 a b \int \csc \left (d x^2+c+\frac {\pi }{2}\right )dx^2+b^2 \int \csc \left (d x^2+c+\frac {\pi }{2}\right )^2dx^2+a^2 x^2\right )\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{2} \left (2 a b \int \csc \left (d x^2+c+\frac {\pi }{2}\right )dx^2-\frac {b^2 \int 1d\left (-\tan \left (d x^2+c\right )\right )}{d}+a^2 x^2\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{2} \left (2 a b \int \csc \left (d x^2+c+\frac {\pi }{2}\right )dx^2+a^2 x^2+\frac {b^2 \tan \left (c+d x^2\right )}{d}\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{2} \left (a^2 x^2+\frac {2 a b \text {arctanh}\left (\sin \left (c+d x^2\right )\right )}{d}+\frac {b^2 \tan \left (c+d x^2\right )}{d}\right )\) |
Input:
Int[x*(a + b*Sec[c + d*x^2])^2,x]
Output:
(a^2*x^2 + (2*a*b*ArcTanh[Sin[c + d*x^2]])/d + (b^2*Tan[c + d*x^2])/d)/2
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Simp[2*a*b Int[Csc[c + d*x], x], x] + Simp[b^2 Int[Csc[c + d*x]^2, x] , x]) /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14
method | result | size |
parts | \(\frac {a^{2} x^{2}}{2}+\frac {b^{2} \tan \left (d \,x^{2}+c \right )}{2 d}+\frac {a b \ln \left (\sec \left (d \,x^{2}+c \right )+\tan \left (d \,x^{2}+c \right )\right )}{d}\) | \(50\) |
derivativedivides | \(\frac {a^{2} \left (d \,x^{2}+c \right )+2 a b \ln \left (\sec \left (d \,x^{2}+c \right )+\tan \left (d \,x^{2}+c \right )\right )+b^{2} \tan \left (d \,x^{2}+c \right )}{2 d}\) | \(52\) |
default | \(\frac {a^{2} \left (d \,x^{2}+c \right )+2 a b \ln \left (\sec \left (d \,x^{2}+c \right )+\tan \left (d \,x^{2}+c \right )\right )+b^{2} \tan \left (d \,x^{2}+c \right )}{2 d}\) | \(52\) |
risch | \(\frac {a^{2} x^{2}}{2}+\frac {i b^{2}}{d \left (1+{\mathrm e}^{2 i \left (d \,x^{2}+c \right )}\right )}+\frac {a b \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+i\right )}{d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}-i\right )}{d}\) | \(77\) |
parallelrisch | \(\frac {a^{2} d \,x^{2} \cos \left (d \,x^{2}+c \right )-2 \cos \left (d \,x^{2}+c \right ) a b \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )+2 \cos \left (d \,x^{2}+c \right ) a b \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right )+b^{2} \sin \left (d \,x^{2}+c \right )}{2 \cos \left (d \,x^{2}+c \right ) d}\) | \(97\) |
norman | \(\frac {-\frac {a^{2} x^{2}}{2}+\frac {a^{2} x^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d}}{\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}-1}+\frac {a b \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a b \ln \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(110\) |
Input:
int(x*(a+b*sec(d*x^2+c))^2,x,method=_RETURNVERBOSE)
Output:
1/2*a^2*x^2+1/2*b^2*tan(d*x^2+c)/d+a*b/d*ln(sec(d*x^2+c)+tan(d*x^2+c))
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (40) = 80\).
Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.07 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b \cos \left (d x^{2} + c\right ) \log \left (\sin \left (d x^{2} + c\right ) + 1\right ) - a b \cos \left (d x^{2} + c\right ) \log \left (-\sin \left (d x^{2} + c\right ) + 1\right ) + b^{2} \sin \left (d x^{2} + c\right )}{2 \, d \cos \left (d x^{2} + c\right )} \] Input:
integrate(x*(a+b*sec(d*x^2+c))^2,x, algorithm="fricas")
Output:
1/2*(a^2*d*x^2*cos(d*x^2 + c) + a*b*cos(d*x^2 + c)*log(sin(d*x^2 + c) + 1) - a*b*cos(d*x^2 + c)*log(-sin(d*x^2 + c) + 1) + b^2*sin(d*x^2 + c))/(d*co s(d*x^2 + c))
\[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\int x \left (a + b \sec {\left (c + d x^{2} \right )}\right )^{2}\, dx \] Input:
integrate(x*(a+b*sec(d*x**2+c))**2,x)
Output:
Integral(x*(a + b*sec(c + d*x**2))**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (40) = 80\).
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.18 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {1}{2} \, a^{2} x^{2} + \frac {a b \log \left (\sec \left (d x^{2} + c\right ) + \tan \left (d x^{2} + c\right )\right )}{d} + \frac {b^{2} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d} \] Input:
integrate(x*(a+b*sec(d*x^2+c))^2,x, algorithm="maxima")
Output:
1/2*a^2*x^2 + a*b*log(sec(d*x^2 + c) + tan(d*x^2 + c))/d + b^2*sin(2*d*x^2 + 2*c)/(d*cos(2*d*x^2 + 2*c)^2 + d*sin(2*d*x^2 + 2*c)^2 + 2*d*cos(2*d*x^2 + 2*c) + d)
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (40) = 80\).
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.00 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {{\left (d x^{2} + c\right )} a^{2} + 2 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )^{2} - 1}}{2 \, d} \] Input:
integrate(x*(a+b*sec(d*x^2+c))^2,x, algorithm="giac")
Output:
1/2*((d*x^2 + c)*a^2 + 2*a*b*log(abs(tan(1/2*d*x^2 + 1/2*c) + 1)) - 2*a*b* log(abs(tan(1/2*d*x^2 + 1/2*c) - 1)) - 2*b^2*tan(1/2*d*x^2 + 1/2*c)/(tan(1 /2*d*x^2 + 1/2*c)^2 - 1))/d
Time = 19.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.27 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2\,x^2}{2}+\frac {b^2\,1{}\mathrm {i}}{d\,\left ({\mathrm {e}}^{2{}\mathrm {i}\,d\,x^2+c\,2{}\mathrm {i}}+1\right )}+\frac {a\,b\,\ln \left (-a\,b\,x\,4{}\mathrm {i}-4\,a\,b\,x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{d}-\frac {a\,b\,\ln \left (a\,b\,x\,4{}\mathrm {i}-4\,a\,b\,x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{d} \] Input:
int(x*(a + b/cos(c + d*x^2))^2,x)
Output:
(a^2*x^2)/2 + (b^2*1i)/(d*(exp(c*2i + d*x^2*2i) + 1)) + (a*b*log(- a*b*x*4 i - 4*a*b*x*exp(d*x^2*1i)*exp(c*1i)))/d - (a*b*log(a*b*x*4i - 4*a*b*x*exp( d*x^2*1i)*exp(c*1i)))/d
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.18 \[ \int x \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {-2 \cos \left (d \,x^{2}+c \right ) \mathrm {log}\left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right ) a b +2 \cos \left (d \,x^{2}+c \right ) \mathrm {log}\left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+1\right ) a b +\cos \left (d \,x^{2}+c \right ) a^{2} d \,x^{2}+\sin \left (d \,x^{2}+c \right ) b^{2}}{2 \cos \left (d \,x^{2}+c \right ) d} \] Input:
int(x*(a+b*sec(d*x^2+c))^2,x)
Output:
( - 2*cos(c + d*x**2)*log(tan((c + d*x**2)/2) - 1)*a*b + 2*cos(c + d*x**2) *log(tan((c + d*x**2)/2) + 1)*a*b + cos(c + d*x**2)*a**2*d*x**2 + sin(c + d*x**2)*b**2)/(2*cos(c + d*x**2)*d)