Integrand size = 21, antiderivative size = 195 \[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}+\frac {b c d^{3/2} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{3 e \sqrt {c^2 x^2}}-\frac {b \left (3 c^2 d+e\right ) x \text {arctanh}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{6 c^2 \sqrt {e} \sqrt {c^2 x^2}} \]
1/3*(e*x^2+d)^(3/2)*(a+b*arcsec(c*x))/e+1/3*b*c*d^(3/2)*x*arctan((e*x^2+d) ^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/e/(c^2*x^2)^(1/2)-1/6*b*(3*c^2*d+e)*x*ar ctanh(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/c^2/e^(1/2)/(c^2*x^2)^( 1/2)-1/6*b*x*(c^2*x^2-1)^(1/2)*(e*x^2+d)^(1/2)/c/(c^2*x^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.59 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.09 \[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {\frac {2 b d^2 \sqrt {1+\frac {d}{e x^2}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )}{c e x}+\frac {\frac {b \left (3 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x^3 \sqrt {1+\frac {e x^2}{d}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,c^2 x^2,-\frac {e x^2}{d}\right )}{\sqrt {1-c^2 x^2}}+\frac {2 \left (d+e x^2\right ) \left (-b e \sqrt {1-\frac {1}{c^2 x^2}} x+2 a c \left (d+e x^2\right )+2 b c \left (d+e x^2\right ) \sec ^{-1}(c x)\right )}{e}}{c}}{12 \sqrt {d+e x^2}} \]
((2*b*d^2*Sqrt[1 + d/(e*x^2)]*AppellF1[1, 1/2, 1/2, 2, 1/(c^2*x^2), -(d/(e *x^2))])/(c*e*x) + ((b*(3*c^2*d + e)*Sqrt[1 - 1/(c^2*x^2)]*x^3*Sqrt[1 + (e *x^2)/d]*AppellF1[1, 1/2, 1/2, 2, c^2*x^2, -((e*x^2)/d)])/Sqrt[1 - c^2*x^2 ] + (2*(d + e*x^2)*(-(b*e*Sqrt[1 - 1/(c^2*x^2)]*x) + 2*a*c*(d + e*x^2) + 2 *b*c*(d + e*x^2)*ArcSec[c*x]))/e)/c)/(12*Sqrt[d + e*x^2])
Time = 0.40 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5759, 354, 113, 27, 175, 66, 104, 217, 221}
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.
| \(\displaystyle \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx\) |
| \(\Big \downarrow \) 5759 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \int \frac {\left (e x^2+d\right )^{3/2}}{x \sqrt {c^2 x^2-1}}dx}{3 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \int \frac {\left (e x^2+d\right )^{3/2}}{x^2 \sqrt {c^2 x^2-1}}dx^2}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 113 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {\int \frac {2 c^2 d^2+e \left (3 d c^2+e\right ) x^2}{2 x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {\int \frac {2 c^2 d^2+e \left (3 d c^2+e\right ) x^2}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {2 c^2 d^2 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+e \left (3 c^2 d+e\right ) \int \frac {1}{\sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {2 c^2 d^2 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+2 e \left (3 c^2 d+e\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {4 c^2 d^2 \int \frac {1}{-x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {c^2 x^2-1}}+2 e \left (3 c^2 d+e\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {2 e \left (3 c^2 d+e\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}-4 c^2 d^{3/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac {b c x \left (\frac {\frac {2 \sqrt {e} \left (3 c^2 d+e\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{c}-4 c^2 d^{3/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e \sqrt {c^2 x^2}}\) |
((d + e*x^2)^(3/2)*(a + b*ArcSec[c*x]))/(3*e) - (b*c*x*((e*Sqrt[-1 + c^2*x ^2]*Sqrt[d + e*x^2])/c^2 + (-4*c^2*d^(3/2)*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d] *Sqrt[-1 + c^2*x^2])] + (2*Sqrt[e]*(3*c^2*d + e)*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[d + e*x^2])])/c)/(2*c^2)))/(6*e*Sqrt[c^2*x^2])
3.2.13.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSec[c*x])/(2*e*(p + 1))), x ] - Simp[b*c*(x/(2*e*(p + 1)*Sqrt[c^2*x^2])) Int[(d + e*x^2)^(p + 1)/(x*S qrt[c^2*x^2 - 1]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
\[\int x \left (a +b \,\operatorname {arcsec}\left (c x \right )\right ) \sqrt {e \,x^{2}+d}d x\]
Time = 0.52 (sec) , antiderivative size = 1100, normalized size of antiderivative = 5.64 \[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\text {Too large to display} \]
[1/24*(2*b*c^3*sqrt(-d)*d*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^ 2 - d*e)*x^2 - 4*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d) *sqrt(-d) + 8*d^2)/x^4) + (3*b*c^2*d + b*e)*sqrt(e)*log(8*c^4*e^2*x^4 + c^ 4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d - c *e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) + 4*(2*a*c^3*e*x^2 + 2*a*c^3*d - sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b*c^3*d)*arcsec(c*x ))*sqrt(e*x^2 + d))/(c^3*e), 1/24*(4*b*c^3*d^(3/2)*arctan(-1/2*sqrt(c^2*x^ 2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2 *d^2 - d*e)*x^2 - d^2)) + (3*b*c^2*d + b*e)*sqrt(e)*log(8*c^4*e^2*x^4 + c^ 4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d - c *e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) + 4*(2*a*c^3*e*x^2 + 2*a*c^3*d - sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b*c^3*d)*arcsec(c*x ))*sqrt(e*x^2 + d))/(c^3*e), 1/12*(b*c^3*sqrt(-d)*d*log(((c^4*d^2 - 6*c^2* d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 - 4*sqrt(c^2*x^2 - 1)*((c^2*d - e)* x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) + (3*b*c^2*d + b*e)*sqrt (-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d )*sqrt(-e)/(c^3*e^2*x^4 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 2*(2*a*c^3*e*x ^2 + 2*a*c^3*d - sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b*c^3*d)*arcse c(c*x))*sqrt(e*x^2 + d))/(c^3*e), 1/12*(2*b*c^3*d^(3/2)*arctan(-1/2*sqrt(c ^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^...
\[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \sqrt {d + e x^{2}}\, dx \]
\[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x \,d x } \]
1/3*(e*x^2 + d)^(3/2)*a/e + 1/3*((e*x^2 + d)^(3/2)*arctan(sqrt(c*x + 1)*sq rt(c*x - 1)) - 3*e*integrate((3*(c^2*e*x^3 - e*x + (c^2*e*x^3 - e*x)*e^(lo g(c*x + 1) + log(c*x - 1)))*sqrt(e*x^2 + d)*log(x) + (3*c^2*e*x^3*log(c) - 3*e*x*log(c) + ((3*c^2*log(c) + c^2)*e*x^3 + (c^2*d - 3*e*log(c))*x)*e^(l og(c*x + 1) + log(c*x - 1)))*sqrt(e*x^2 + d))/(c^2*e*x^2 + (c^2*e*x^2 - e) *e^(log(c*x + 1) + log(c*x - 1)) - e), x))*b/e
\[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x \,d x } \]
Timed out. \[ \int x \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x\,\sqrt {e\,x^2+d}\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]