Integrand size = 21, antiderivative size = 173 \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{8 e \left (c^2 d+e\right ) \left (d+e x^2\right )}+\frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {c^2 d+e}}\right )}{8 d e^{3/2} \left (c^2 d+e\right )^{3/2}} \]
1/4*x^4*(a+b*arcsech(c*x))/d/(e*x^2+d)^2-1/8*b*(c^2*d+2*e)*arctanh(e^(1/2) *(-c^2*x^2+1)^(1/2)/(c^2*d+e)^(1/2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/d/e^( 3/2)/(c^2*d+e)^(3/2)+1/8*b*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1 /2)/e/(c^2*d+e)/(e*x^2+d)
Result contains complex when optimal does not.
Time = 1.58 (sec) , antiderivative size = 486, normalized size of antiderivative = 2.81 \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {-\frac {4 a d}{\left (d+e x^2\right )^2}+\frac {8 a}{d+e x^2}-\frac {2 e \sqrt {\frac {1-c x}{1+c x}} (b+b c x)}{\left (c^2 d+e\right ) \left (d+e x^2\right )}+\frac {4 b \left (d+2 e x^2\right ) \text {sech}^{-1}(c x)}{\left (d+e x^2\right )^2}+\frac {4 b \log (x)}{d}-\frac {4 b \log \left (1+\sqrt {\frac {1-c x}{1+c x}}+c x \sqrt {\frac {1-c x}{1+c x}}\right )}{d}+\frac {b \sqrt {e} \left (c^2 d+2 e\right ) \log \left (\frac {16 d e^{3/2} \sqrt {c^2 d+e} \left (\sqrt {e}-i c^2 \sqrt {d} x+\sqrt {c^2 d+e} \sqrt {\frac {1-c x}{1+c x}}+c \sqrt {c^2 d+e} x \sqrt {\frac {1-c x}{1+c x}}\right )}{b \left (c^2 d+2 e\right ) \left (-i \sqrt {d}+\sqrt {e} x\right )}\right )}{d \left (c^2 d+e\right )^{3/2}}+\frac {b \sqrt {e} \left (c^2 d+2 e\right ) \log \left (\frac {16 d e^{3/2} \sqrt {c^2 d+e} \left (\sqrt {e}+i c^2 \sqrt {d} x+\sqrt {c^2 d+e} \sqrt {\frac {1-c x}{1+c x}}+c \sqrt {c^2 d+e} x \sqrt {\frac {1-c x}{1+c x}}\right )}{b \left (c^2 d+2 e\right ) \left (i \sqrt {d}+\sqrt {e} x\right )}\right )}{d \left (c^2 d+e\right )^{3/2}}}{16 e^2} \]
-1/16*((-4*a*d)/(d + e*x^2)^2 + (8*a)/(d + e*x^2) - (2*e*Sqrt[(1 - c*x)/(1 + c*x)]*(b + b*c*x))/((c^2*d + e)*(d + e*x^2)) + (4*b*(d + 2*e*x^2)*ArcSe ch[c*x])/(d + e*x^2)^2 + (4*b*Log[x])/d - (4*b*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/d + (b*Sqrt[e]*(c^2*d + 2*e)*Log[ (16*d*e^(3/2)*Sqrt[c^2*d + e]*(Sqrt[e] - I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e] *Sqrt[(1 - c*x)/(1 + c*x)] + c*Sqrt[c^2*d + e]*x*Sqrt[(1 - c*x)/(1 + c*x)] ))/(b*(c^2*d + 2*e)*((-I)*Sqrt[d] + Sqrt[e]*x))])/(d*(c^2*d + e)^(3/2)) + (b*Sqrt[e]*(c^2*d + 2*e)*Log[(16*d*e^(3/2)*Sqrt[c^2*d + e]*(Sqrt[e] + I*c^ 2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[(1 - c*x)/(1 + c*x)] + c*Sqrt[c^2*d + e ]*x*Sqrt[(1 - c*x)/(1 + c*x)]))/(b*(c^2*d + 2*e)*(I*Sqrt[d] + Sqrt[e]*x))] )/(d*(c^2*d + e)^(3/2)))/e^2
Time = 0.36 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6855, 27, 354, 87, 73, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 6855 |
| \(\displaystyle b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {x^3}{4 d \sqrt {1-c^2 x^2} \left (e x^2+d\right )^2}dx+\frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {x^3}{\sqrt {1-c^2 x^2} \left (e x^2+d\right )^2}dx}{4 d}+\frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \int \frac {x^2}{\sqrt {1-c^2 x^2} \left (e x^2+d\right )^2}dx^2}{8 d}+\frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {\left (c^2 d+2 e\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \left (e x^2+d\right )}dx^2}{2 e \left (c^2 d+e\right )}+\frac {d \sqrt {1-c^2 x^2}}{e \left (c^2 d+e\right ) \left (d+e x^2\right )}\right )}{8 d}+\frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {d \sqrt {1-c^2 x^2}}{e \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac {\left (c^2 d+2 e\right ) \int \frac {1}{-\frac {e x^4}{c^2}+d+\frac {e}{c^2}}d\sqrt {1-c^2 x^2}}{c^2 e \left (c^2 d+e\right )}\right )}{8 d}+\frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {x^4 \left (a+b \text {sech}^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (\frac {d \sqrt {1-c^2 x^2}}{e \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac {\left (c^2 d+2 e\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {c^2 d+e}}\right )}{e^{3/2} \left (c^2 d+e\right )^{3/2}}\right )}{8 d}\) |
(x^4*(a + b*ArcSech[c*x]))/(4*d*(d + e*x^2)^2) + (b*Sqrt[(1 + c*x)^(-1)]*S qrt[1 + c*x]*((d*Sqrt[1 - c^2*x^2])/(e*(c^2*d + e)*(d + e*x^2)) - ((c^2*d + 2*e)*ArcTanh[(Sqrt[e]*Sqrt[1 - c^2*x^2])/Sqrt[c^2*d + e]])/(e^(3/2)*(c^2 *d + e)^(3/2))))/(8*d)
3.2.24.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_.) + ArcSech[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*( x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Si mp[(a + b*ArcSech[c*x]) u, x] + Simp[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)] Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; Fre eQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2 *p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(1361\) vs. \(2(147)=294\).
Time = 5.09 (sec) , antiderivative size = 1362, normalized size of antiderivative = 7.87
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1362\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1395\) |
| default | \(\text {Expression too large to display}\) | \(1395\) |
a*(-1/2/e^2/(e*x^2+d)+1/4*d/e^2/(e*x^2+d)^2)+b/c^4*(-1/2*c^6*arcsech(c*x)/ e^2/(c^2*e*x^2+c^2*d)+1/4*c^8*arcsech(c*x)*d/e^2/(c^2*e*x^2+c^2*d)^2-1/16* c^5*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)*e*(4*((c^2*d+e)/e)^(1/2)*ar ctanh(1/(-c^2*x^2+1)^(1/2))*c^6*d^2*e*x^2+4*((c^2*d+e)/e)^(1/2)*arctanh(1/ (-c^2*x^2+1)^(1/2))*c^6*d^3-ln(-2*(((c^2*d+e)/e)^(1/2)*(-c^2*x^2+1)^(1/2)* e-(-c^2*d*e)^(1/2)*c*x+e)/(-c*e*x+(-c^2*d*e)^(1/2)))*x^2*c^6*d^2*e-ln(-2*( ((c^2*d+e)/e)^(1/2)*(-c^2*x^2+1)^(1/2)*e-(-c^2*d*e)^(1/2)*c*x+e)/(-c*e*x+( -c^2*d*e)^(1/2)))*c^6*d^3-ln(2*(((c^2*d+e)/e)^(1/2)*(-c^2*x^2+1)^(1/2)*e+( -c^2*d*e)^(1/2)*c*x+e)/(c*e*x+(-c^2*d*e)^(1/2)))*c^6*d^2*e*x^2-ln(2*(((c^2 *d+e)/e)^(1/2)*(-c^2*x^2+1)^(1/2)*e+(-c^2*d*e)^(1/2)*c*x+e)/(c*e*x+(-c^2*d *e)^(1/2)))*c^6*d^3+8*((c^2*d+e)/e)^(1/2)*arctanh(1/(-c^2*x^2+1)^(1/2))*c^ 4*d*e^2*x^2+8*((c^2*d+e)/e)^(1/2)*arctanh(1/(-c^2*x^2+1)^(1/2))*c^4*d^2*e+ 2*(-c^2*x^2+1)^(1/2)*((c^2*d+e)/e)^(1/2)*c^4*d^2*e-3*ln(-2*(((c^2*d+e)/e)^ (1/2)*(-c^2*x^2+1)^(1/2)*e-(-c^2*d*e)^(1/2)*c*x+e)/(-c*e*x+(-c^2*d*e)^(1/2 )))*x^2*c^4*d*e^2-3*ln(-2*(((c^2*d+e)/e)^(1/2)*(-c^2*x^2+1)^(1/2)*e-(-c^2* d*e)^(1/2)*c*x+e)/(-c*e*x+(-c^2*d*e)^(1/2)))*c^4*d^2*e-3*ln(2*(((c^2*d+e)/ e)^(1/2)*(-c^2*x^2+1)^(1/2)*e+(-c^2*d*e)^(1/2)*c*x+e)/(c*e*x+(-c^2*d*e)^(1 /2)))*c^4*d*e^2*x^2-3*ln(2*(((c^2*d+e)/e)^(1/2)*(-c^2*x^2+1)^(1/2)*e+(-c^2 *d*e)^(1/2)*c*x+e)/(c*e*x+(-c^2*d*e)^(1/2)))*c^4*d^2*e+4*((c^2*d+e)/e)^(1/ 2)*arctanh(1/(-c^2*x^2+1)^(1/2))*e^3*c^2*x^2+4*((c^2*d+e)/e)^(1/2)*arct...
Leaf count of result is larger than twice the leaf count of optimal. 638 vs. \(2 (115) = 230\).
Time = 0.38 (sec) , antiderivative size = 1346, normalized size of antiderivative = 7.78 \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Too large to display} \]
[-1/16*(4*a*c^4*d^4 + 2*(4*a - b)*c^2*d^3*e + 2*(2*a - b)*d^2*e^2 - 2*(b*c ^2*d*e^3 + b*e^4)*x^4 + 4*(2*a*c^4*d^3*e + (4*a - b)*c^2*d^2*e^2 + (2*a - b)*d*e^3)*x^2 - (b*c^2*d^3 + (b*c^2*d*e^2 + 2*b*e^3)*x^4 + 2*b*d^2*e + 2*( b*c^2*d^2*e + 2*b*d*e^2)*x^2)*sqrt(c^2*d*e + e^2)*log((c^4*d^2 + 4*c^2*d*e - (c^4*d*e + 2*c^2*e^2)*x^2 + 4*(c^3*d*e + c*e^2)*x*sqrt(-(c^2*x^2 - 1)/( c^2*x^2)) + 4*e^2 + 2*(c^2*e*x^2 - c^2*d - (c^3*d + 2*c*e)*x*sqrt(-(c^2*x^ 2 - 1)/(c^2*x^2)) - 2*e)*sqrt(c^2*d*e + e^2))/(e*x^2 + d)) + 4*(b*c^4*d^4 + 2*b*c^2*d^3*e + b*d^2*e^2 + (b*c^4*d^2*e^2 + 2*b*c^2*d*e^3 + b*e^4)*x^4 + 2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) + 4*(b*c^4*d^4 + 2*b*c^2*d^3*e + b*d^2*e^2 + 2*(b *c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*log((c*x*sqrt(-(c^2*x^2 - 1)/ (c^2*x^2)) + 1)/(c*x)) - 2*((b*c^3*d^2*e^2 + b*c*d*e^3)*x^3 + (b*c^3*d^3*e + b*c*d^2*e^2)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/(c^4*d^5*e^2 + 2*c^2*d^ 4*e^3 + d^3*e^4 + (c^4*d^3*e^4 + 2*c^2*d^2*e^5 + d*e^6)*x^4 + 2*(c^4*d^4*e ^3 + 2*c^2*d^3*e^4 + d^2*e^5)*x^2), -1/8*(2*a*c^4*d^4 + (4*a - b)*c^2*d^3* e + (2*a - b)*d^2*e^2 - (b*c^2*d*e^3 + b*e^4)*x^4 + 2*(2*a*c^4*d^3*e + (4* a - b)*c^2*d^2*e^2 + (2*a - b)*d*e^3)*x^2 + (b*c^2*d^3 + (b*c^2*d*e^2 + 2* b*e^3)*x^4 + 2*b*d^2*e + 2*(b*c^2*d^2*e + 2*b*d*e^2)*x^2)*sqrt(-c^2*d*e - e^2)*arctan((sqrt(-c^2*d*e - e^2)*c*d*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - s qrt(-c^2*d*e - e^2)*(e*x^2 + d))/((c^2*d*e + e^2)*x^2)) + 2*(b*c^4*d^4 ...
Timed out. \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]