Integrand size = 14, antiderivative size = 58 \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{\left (c+d x^2\right )^2}}\right )}{2 d} \] Output:
1/2*a*x^2+1/2*b*(d*x^2+c)*arcsec(d*x^2+c)/d-1/2*b*arctanh((1-1/(d*x^2+c)^2 )^(1/2))/d
Result contains complex when optimal does not.
Time = 2.01 (sec) , antiderivative size = 516, normalized size of antiderivative = 8.90 \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \sec ^{-1}\left (c+d x^2\right )+\frac {b \left (c+d x^2\right ) \sqrt {\frac {-1+c^2+2 c d x^2+d^2 x^4}{\left (c+d x^2\right )^2}} \left (\sqrt [4]{-1} \left (-i+\sqrt {-1+c^2}\right ) \sqrt {2 i-i c^2+2 \sqrt {-1+c^2}} \arctan \left (\frac {(-1)^{3/4} \sqrt {2 i-i c^2+2 \sqrt {-1+c^2}} d x^2}{c \sqrt {-1+c^2}-c \sqrt {-1+c^2+2 c d x^2+d^2 x^4}}\right )+(-1)^{3/4} \left (i+\sqrt {-1+c^2}\right ) \sqrt {-2 i+i c^2+2 \sqrt {-1+c^2}} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-2 i+i c^2+2 \sqrt {-1+c^2}} d x^2}{c \sqrt {-1+c^2}-c \sqrt {-1+c^2+2 c d x^2+d^2 x^4}}\right )+c \left (c \arctan \left (\frac {\sqrt {-1+c^2} d^2 x^4}{c^4+c^3 d x^2+d^2 x^4-c^2 \left (1+\sqrt {-1+c^2} \sqrt {-1+c^2+2 c d x^2+d^2 x^4}\right )}\right )-\log \left (\sqrt {-1+c^2}-d x^2-\sqrt {-1+c^2+2 c d x^2+d^2 x^4}\right )+\log \left (d^2 \left (\sqrt {-1+c^2}+d x^2-\sqrt {-1+c^2+2 c d x^2+d^2 x^4}\right )\right )\right )\right )}{2 c d \sqrt {-1+c^2+2 c d x^2+d^2 x^4}} \] Input:
Integrate[x*(a + b*ArcSec[c + d*x^2]),x]
Output:
(a*x^2)/2 + (b*x^2*ArcSec[c + d*x^2])/2 + (b*(c + d*x^2)*Sqrt[(-1 + c^2 + 2*c*d*x^2 + d^2*x^4)/(c + d*x^2)^2]*((-1)^(1/4)*(-I + Sqrt[-1 + c^2])*Sqrt [2*I - I*c^2 + 2*Sqrt[-1 + c^2]]*ArcTan[((-1)^(3/4)*Sqrt[2*I - I*c^2 + 2*S qrt[-1 + c^2]]*d*x^2)/(c*Sqrt[-1 + c^2] - c*Sqrt[-1 + c^2 + 2*c*d*x^2 + d^ 2*x^4])] + (-1)^(3/4)*(I + Sqrt[-1 + c^2])*Sqrt[-2*I + I*c^2 + 2*Sqrt[-1 + c^2]]*ArcTan[((-1)^(1/4)*Sqrt[-2*I + I*c^2 + 2*Sqrt[-1 + c^2]]*d*x^2)/(c* Sqrt[-1 + c^2] - c*Sqrt[-1 + c^2 + 2*c*d*x^2 + d^2*x^4])] + c*(c*ArcTan[(S qrt[-1 + c^2]*d^2*x^4)/(c^4 + c^3*d*x^2 + d^2*x^4 - c^2*(1 + Sqrt[-1 + c^2 ]*Sqrt[-1 + c^2 + 2*c*d*x^2 + d^2*x^4]))] - Log[Sqrt[-1 + c^2] - d*x^2 - S qrt[-1 + c^2 + 2*c*d*x^2 + d^2*x^4]] + Log[d^2*(Sqrt[-1 + c^2] + d*x^2 - S qrt[-1 + c^2 + 2*c*d*x^2 + d^2*x^4])])))/(2*c*d*Sqrt[-1 + c^2 + 2*c*d*x^2 + d^2*x^4])
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {7266, 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.
\(\displaystyle \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{2} \int \left (a+b \sec ^{-1}\left (d x^2+c\right )\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (a x^2-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{\left (c+d x^2\right )^2}}\right )}{d}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{d}\right )\) |
Input:
Int[x*(a + b*ArcSec[c + d*x^2]),x]
Output:
(a*x^2 + (b*(c + d*x^2)*ArcSec[c + d*x^2])/d - (b*ArcTanh[Sqrt[1 - (c + d* x^2)^(-2)]])/d)/2
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10
method | result | size |
parts | \(\frac {a \,x^{2}}{2}+\frac {b \left (\left (d \,x^{2}+c \right ) \operatorname {arcsec}\left (d \,x^{2}+c \right )-\ln \left (d \,x^{2}+c +\left (d \,x^{2}+c \right ) \sqrt {1-\frac {1}{\left (d \,x^{2}+c \right )^{2}}}\right )\right )}{2 d}\) | \(64\) |
derivativedivides | \(\frac {\left (d \,x^{2}+c \right ) a +b \left (\left (d \,x^{2}+c \right ) \operatorname {arcsec}\left (d \,x^{2}+c \right )-\ln \left (d \,x^{2}+c +\left (d \,x^{2}+c \right ) \sqrt {1-\frac {1}{\left (d \,x^{2}+c \right )^{2}}}\right )\right )}{2 d}\) | \(68\) |
default | \(\frac {\left (d \,x^{2}+c \right ) a +b \left (\left (d \,x^{2}+c \right ) \operatorname {arcsec}\left (d \,x^{2}+c \right )-\ln \left (d \,x^{2}+c +\left (d \,x^{2}+c \right ) \sqrt {1-\frac {1}{\left (d \,x^{2}+c \right )^{2}}}\right )\right )}{2 d}\) | \(68\) |
Input:
int(x*(a+b*arcsec(d*x^2+c)),x,method=_RETURNVERBOSE)
Output:
1/2*a*x^2+1/2*b/d*((d*x^2+c)*arcsec(d*x^2+c)-ln(d*x^2+c+(d*x^2+c)*(1-1/(d* x^2+c)^2)^(1/2)))
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.66 \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\frac {b d x^{2} \operatorname {arcsec}\left (d x^{2} + c\right ) + a d x^{2} + 2 \, b c \arctan \left (-d x^{2} - c + \sqrt {d^{2} x^{4} + 2 \, c d x^{2} + c^{2} - 1}\right ) + b \log \left (-d x^{2} - c + \sqrt {d^{2} x^{4} + 2 \, c d x^{2} + c^{2} - 1}\right )}{2 \, d} \] Input:
integrate(x*(a+b*arcsec(d*x^2+c)),x, algorithm="fricas")
Output:
1/2*(b*d*x^2*arcsec(d*x^2 + c) + a*d*x^2 + 2*b*c*arctan(-d*x^2 - c + sqrt( d^2*x^4 + 2*c*d*x^2 + c^2 - 1)) + b*log(-d*x^2 - c + sqrt(d^2*x^4 + 2*c*d* x^2 + c^2 - 1)))/d
Timed out. \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x*(a+b*asec(d*x**2+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.22 \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {{\left (2 \, {\left (d x^{2} + c\right )} \operatorname {arcsec}\left (d x^{2} + c\right ) - \log \left (\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right ) + \log \left (-\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right )\right )} b}{4 \, d} \] Input:
integrate(x*(a+b*arcsec(d*x^2+c)),x, algorithm="maxima")
Output:
1/2*a*x^2 + 1/4*(2*(d*x^2 + c)*arcsec(d*x^2 + c) - log(sqrt(-1/(d*x^2 + c) ^2 + 1) + 1) + log(-sqrt(-1/(d*x^2 + c)^2 + 1) + 1))*b/d
Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.72 \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\frac {1}{2} \, a x^{2} + \frac {1}{4} \, b d {\left (\frac {2 \, {\left (d x^{2} + c\right )} \arccos \left (-\frac {1}{{\left (d x^{2} + c\right )} {\left (\frac {c}{d x^{2} + c} - 1\right )} - c}\right )}{d^{2}} - \frac {\log \left (\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right )}{d^{2}}\right )} \] Input:
integrate(x*(a+b*arcsec(d*x^2+c)),x, algorithm="giac")
Output:
1/2*a*x^2 + 1/4*b*d*(2*(d*x^2 + c)*arccos(-1/((d*x^2 + c)*(c/(d*x^2 + c) - 1) - c))/d^2 - (log(sqrt(-1/(d*x^2 + c)^2 + 1) + 1) - log(-sqrt(-1/(d*x^2 + c)^2 + 1) + 1))/d^2)
Time = 1.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\frac {a\,x^2}{2}-\frac {b\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{{\left (d\,x^2+c\right )}^2}}}\right )}{2\,d}+\frac {b\,\mathrm {acos}\left (\frac {1}{d\,x^2+c}\right )\,\left (d\,x^2+c\right )}{2\,d} \] Input:
int(x*(a + b*acos(1/(c + d*x^2))),x)
Output:
(a*x^2)/2 - (b*atanh(1/(1 - 1/(c + d*x^2)^2)^(1/2)))/(2*d) + (b*acos(1/(c + d*x^2))*(c + d*x^2))/(2*d)
\[ \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx=\left (\int \mathit {asec} \left (d \,x^{2}+c \right ) x d x \right ) b +\frac {a \,x^{2}}{2} \] Input:
int(x*(a+b*asec(d*x^2+c)),x)
Output:
(2*int(asec(c + d*x**2)*x,x)*b + a*x**2)/2