Integrand size = 21, antiderivative size = 261 \[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=-\frac {\left (a-b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{2 \left (a^2-b^2 c\right ) x^2}-\frac {b d \sqrt {a+b \sqrt {c+d x}} \left (6 a b c-\left (a^2+5 b^2 c\right ) \sqrt {c+d x}\right )}{8 c \left (a^2-b^2 c\right )^2 x}+\frac {b \left (2 a-5 b \sqrt {c}\right ) d^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )}{16 \left (a-b \sqrt {c}\right )^{5/2} c^{3/2}}-\frac {b \left (2 a+5 b \sqrt {c}\right ) d^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right )}{16 \left (a+b \sqrt {c}\right )^{5/2} c^{3/2}} \]
1/16*b*d^2*arctanh((a+b*(d*x+c)^(1/2))^(1/2)/(a-b*c^(1/2))^(1/2))*(2*a-5*b *c^(1/2))/c^(3/2)/(a-b*c^(1/2))^(5/2)-1/16*b*d^2*arctanh((a+b*(d*x+c)^(1/2 ))^(1/2)/(a+b*c^(1/2))^(1/2))*(2*a+5*b*c^(1/2))/c^(3/2)/(a+b*c^(1/2))^(5/2 )-1/2*(a-b*(d*x+c)^(1/2))*(a+b*(d*x+c)^(1/2))^(1/2)/(-b^2*c+a^2)/x^2-1/8*b *d*(6*a*b*c-(5*b^2*c+a^2)*(d*x+c)^(1/2))*(a+b*(d*x+c)^(1/2))^(1/2)/c/(-b^2 *c+a^2)^2/x
Time = 1.31 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {a+b \sqrt {c+d x}} \left (4 a^3 c+b^3 c (4 c-5 d x) \sqrt {c+d x}-a^2 b \sqrt {c+d x} (4 c+d x)+2 a b^2 c (-2 c+3 d x)\right )}{\left (a^2-b^2 c\right )^2 x^2}+\frac {b \left (2 a+5 b \sqrt {c}\right ) d^2 \arctan \left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {-a-b \sqrt {c}}}\right )}{\left (-a-b \sqrt {c}\right )^{5/2}}+\frac {b \left (-2 a+5 b \sqrt {c}\right ) d^2 \arctan \left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {-a+b \sqrt {c}}}\right )}{\left (-a+b \sqrt {c}\right )^{5/2}}}{16 c^{3/2}} \]
((-2*Sqrt[c]*Sqrt[a + b*Sqrt[c + d*x]]*(4*a^3*c + b^3*c*(4*c - 5*d*x)*Sqrt [c + d*x] - a^2*b*Sqrt[c + d*x]*(4*c + d*x) + 2*a*b^2*c*(-2*c + 3*d*x)))/( (a^2 - b^2*c)^2*x^2) + (b*(2*a + 5*b*Sqrt[c])*d^2*ArcTan[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[-a - b*Sqrt[c]]])/(-a - b*Sqrt[c])^(5/2) + (b*(-2*a + 5*b*Sq rt[c])*d^2*ArcTan[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[-a + b*Sqrt[c]]])/(-a + b *Sqrt[c])^(5/2))/(16*c^(3/2))
Time = 0.61 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {896, 25, 1732, 561, 27, 1492, 27, 1492, 27, 1480, 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.
| \(\displaystyle \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx\) |
| \(\Big \downarrow \) 896 |
| \(\displaystyle d^2 \int \frac {1}{d^3 x^3 \sqrt {a+b \sqrt {c+d x}}}d(c+d x)\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -d^2 \int -\frac {1}{d^3 x^3 \sqrt {a+b \sqrt {c+d x}}}d(c+d x)\) |
| \(\Big \downarrow \) 1732 |
| \(\displaystyle -2 d^2 \int -\frac {\sqrt {c+d x}}{d^3 x^3 \sqrt {a+b \sqrt {c+d x}}}d\sqrt {c+d x}\) |
| \(\Big \downarrow \) 561 |
| \(\displaystyle -\frac {4 d^2 \int \frac {a-c-d x}{b \left (\frac {a^2}{b^2}-\frac {2 (c+d x) a}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^3}d\sqrt {a+b \sqrt {c+d x}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4 d^2 \int \frac {a-c-d x}{\left (\frac {a^2}{b^2}-\frac {2 (c+d x) a}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^3}d\sqrt {a+b \sqrt {c+d x}}}{b^2}\) |
| \(\Big \downarrow \) 1492 |
| \(\displaystyle -\frac {4 d^2 \left (\frac {b^2 (2 a-c-d x) \sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}-\frac {b^4 \int -\frac {2 c (6 a-5 (c+d x))}{b^2 \left (\frac {a^2}{b^2}-\frac {2 (c+d x) a}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}d\sqrt {a+b \sqrt {c+d x}}}{16 c \left (a^2-b^2 c\right )}\right )}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4 d^2 \left (\frac {b^2 \int \frac {6 a-5 (c+d x)}{\left (\frac {a^2}{b^2}-\frac {2 (c+d x) a}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}d\sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right )}+\frac {b^2 (2 a-c-d x) \sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}\right )}{b^2}\) |
| \(\Big \downarrow \) 1492 |
| \(\displaystyle -\frac {4 d^2 \left (\frac {b^2 \left (\frac {\sqrt {a+b \sqrt {c+d x}} \left (a \left (a^2+11 b^2 c\right )-\left (a^2+5 b^2 c\right ) (c+d x)\right )}{4 c \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )}-\frac {b^4 \int \frac {2 \left (a \left (a^2-13 b^2 c\right )+\left (a^2+5 b^2 c\right ) (c+d x)\right )}{b^4 \left (\frac {a^2}{b^2}-\frac {2 (c+d x) a}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )}d\sqrt {a+b \sqrt {c+d x}}}{8 c \left (a^2-b^2 c\right )}\right )}{8 \left (a^2-b^2 c\right )}+\frac {b^2 (2 a-c-d x) \sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}\right )}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4 d^2 \left (\frac {b^2 \left (\frac {\sqrt {a+b \sqrt {c+d x}} \left (a \left (a^2+11 b^2 c\right )-\left (a^2+5 b^2 c\right ) (c+d x)\right )}{4 c \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )}-\frac {\int \frac {a \left (a^2-13 b^2 c\right )+\left (a^2+5 b^2 c\right ) (c+d x)}{\frac {a^2}{b^2}-\frac {2 (c+d x) a}{b^2}+\frac {(c+d x)^2}{b^2}-c}d\sqrt {a+b \sqrt {c+d x}}}{4 c \left (a^2-b^2 c\right )}\right )}{8 \left (a^2-b^2 c\right )}+\frac {b^2 (2 a-c-d x) \sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}\right )}{b^2}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle -\frac {4 d^2 \left (\frac {b^2 \left (\frac {\sqrt {a+b \sqrt {c+d x}} \left (a \left (a^2+11 b^2 c\right )-\left (a^2+5 b^2 c\right ) (c+d x)\right )}{4 c \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )}-\frac {\frac {\left (a-b \sqrt {c}\right )^2 \left (2 a+5 b \sqrt {c}\right ) \int \frac {1}{\frac {c+d x}{b^2}-\frac {a+b \sqrt {c}}{b^2}}d\sqrt {a+b \sqrt {c+d x}}}{2 b \sqrt {c}}-\frac {\left (2 a-5 b \sqrt {c}\right ) \left (a+b \sqrt {c}\right )^2 \int \frac {1}{\frac {c+d x}{b^2}-\frac {a-b \sqrt {c}}{b^2}}d\sqrt {a+b \sqrt {c+d x}}}{2 b \sqrt {c}}}{4 c \left (a^2-b^2 c\right )}\right )}{8 \left (a^2-b^2 c\right )}+\frac {b^2 (2 a-c-d x) \sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}\right )}{b^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {4 d^2 \left (\frac {b^2 \left (\frac {\sqrt {a+b \sqrt {c+d x}} \left (a \left (a^2+11 b^2 c\right )-\left (a^2+5 b^2 c\right ) (c+d x)\right )}{4 c \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )}-\frac {\frac {b \left (2 a-5 b \sqrt {c}\right ) \left (a+b \sqrt {c}\right )^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )}{2 \sqrt {c} \sqrt {a-b \sqrt {c}}}-\frac {b \left (a-b \sqrt {c}\right )^2 \left (2 a+5 b \sqrt {c}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right )}{2 \sqrt {c} \sqrt {a+b \sqrt {c}}}}{4 c \left (a^2-b^2 c\right )}\right )}{8 \left (a^2-b^2 c\right )}+\frac {b^2 (2 a-c-d x) \sqrt {a+b \sqrt {c+d x}}}{8 \left (a^2-b^2 c\right ) \left (\frac {a^2}{b^2}-\frac {2 a (c+d x)}{b^2}+\frac {(c+d x)^2}{b^2}-c\right )^2}\right )}{b^2}\) |
(-4*d^2*((b^2*(2*a - c - d*x)*Sqrt[a + b*Sqrt[c + d*x]])/(8*(a^2 - b^2*c)* (a^2/b^2 - c - (2*a*(c + d*x))/b^2 + (c + d*x)^2/b^2)^2) + (b^2*((Sqrt[a + b*Sqrt[c + d*x]]*(a*(a^2 + 11*b^2*c) - (a^2 + 5*b^2*c)*(c + d*x)))/(4*c*( a^2 - b^2*c)*(a^2/b^2 - c - (2*a*(c + d*x))/b^2 + (c + d*x)^2/b^2)) - ((b* (2*a - 5*b*Sqrt[c])*(a + b*Sqrt[c])^2*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sq rt[a - b*Sqrt[c]]])/(2*Sqrt[a - b*Sqrt[c]]*Sqrt[c]) - (b*(a - b*Sqrt[c])^2 *(2*a + 5*b*Sqrt[c])*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]] ])/(2*Sqrt[a + b*Sqrt[c]]*Sqrt[c]))/(4*c*(a^2 - b^2*c))))/(8*(a^2 - b^2*c) )))/b^2
3.7.52.3.1 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_)^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_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k*(n + 1) - 1)*(-c /d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac tionQ[n] && IntegerQ[p] && IntegerQ[m]
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb ol] :> Simp[x*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb ol] :> With[{g = Denominator[n]}, Simp[g Subst[Int[x^(g - 1)*(d + e*x^(g* n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} , x] && EqQ[n2, 2*n] && FractionQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(426\) vs. \(2(209)=418\).
Time = 0.42 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.64
| method | result | size |
| derivativedivides | \(-4 d^{2} b^{4} \left (\frac {\frac {-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}+\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {-\sqrt {b^{2} c}-a}}\right )}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right ) \sqrt {-\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}-\frac {\frac {-\frac {\left (-5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c -2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (-7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{-4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}-\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}-2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {\sqrt {b^{2} c}-a}}\right )}{4 \left (-b^{2} c +2 a \sqrt {b^{2} c}-a^{2}\right ) \sqrt {\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}\right )\) | \(427\) |
| default | \(-4 d^{2} b^{4} \left (\frac {\frac {-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}+\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}+2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {-\sqrt {b^{2} c}-a}}\right )}{4 \left (b^{2} c +2 a \sqrt {b^{2} c}+a^{2}\right ) \sqrt {-\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}-\frac {\frac {-\frac {\left (-5 \sqrt {b^{2} c}+2 a \right ) \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{4 \left (b^{2} c -2 a \sqrt {b^{2} c}+a^{2}\right )}+\frac {\left (-7 \sqrt {b^{2} c}+2 a \right ) \sqrt {a +b \sqrt {d x +c}}}{-4 \sqrt {b^{2} c}+4 a}}{\left (-b \sqrt {d x +c}-\sqrt {b^{2} c}\right )^{2}}-\frac {\left (5 \sqrt {b^{2} c}-2 a \right ) \arctan \left (\frac {\sqrt {a +b \sqrt {d x +c}}}{\sqrt {\sqrt {b^{2} c}-a}}\right )}{4 \left (-b^{2} c +2 a \sqrt {b^{2} c}-a^{2}\right ) \sqrt {\sqrt {b^{2} c}-a}}}{16 b^{2} c \sqrt {b^{2} c}}\right )\) | \(427\) |
-4*d^2*b^4*(1/16/b^2/c/(b^2*c)^(1/2)*((-1/4*(5*(b^2*c)^(1/2)+2*a)/(b^2*c+2 *a*(b^2*c)^(1/2)+a^2)*(a+b*(d*x+c)^(1/2))^(3/2)+1/4*(7*(b^2*c)^(1/2)+2*a)/ ((b^2*c)^(1/2)+a)*(a+b*(d*x+c)^(1/2))^(1/2))/(-b*(d*x+c)^(1/2)+(b^2*c)^(1/ 2))^2-1/4*(5*(b^2*c)^(1/2)+2*a)/(b^2*c+2*a*(b^2*c)^(1/2)+a^2)/(-(b^2*c)^(1 /2)-a)^(1/2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2)))-1 /16/b^2/c/(b^2*c)^(1/2)*((-1/4*(-5*(b^2*c)^(1/2)+2*a)/(b^2*c-2*a*(b^2*c)^( 1/2)+a^2)*(a+b*(d*x+c)^(1/2))^(3/2)+1/4*(-7*(b^2*c)^(1/2)+2*a)/(-(b^2*c)^( 1/2)+a)*(a+b*(d*x+c)^(1/2))^(1/2))/(-b*(d*x+c)^(1/2)-(b^2*c)^(1/2))^2-1/4* (5*(b^2*c)^(1/2)-2*a)/(-b^2*c+2*a*(b^2*c)^(1/2)-a^2)/((b^2*c)^(1/2)-a)^(1/ 2)*arctan((a+b*(d*x+c)^(1/2))^(1/2)/((b^2*c)^(1/2)-a)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 4390 vs. \(2 (212) = 424\).
Time = 1.22 (sec) , antiderivative size = 4390, normalized size of antiderivative = 16.82 \[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=\text {Too large to display} \]
-1/32*((b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)*x^2*sqrt(-((105*a*b^8*c^3 + 70*a^ 3*b^6*c^2 - 35*a^5*b^4*c + 4*a^7*b^2)*d^4 + (b^10*c^8 - 5*a^2*b^8*c^7 + 10 *a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18*c ^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8* b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14* c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^8 + 210*a^12*b^8*c^7 - 120*a^14* b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^4 + a^20*c^3)))/(b^10*c^8 - 5*a^ 2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3))*l og((625*b^12*c^3 + 3750*a^2*b^10*c^2 - 1491*a^4*b^8*c + 140*a^6*b^6)*sqrt( sqrt(d*x + c)*b + a)*d^6 + ((325*a*b^12*c^5 + 1977*a^3*b^10*c^4 - 609*a^5* b^8*c^3 + 35*a^7*b^6*c^2)*d^4 - (5*b^14*c^10 - 16*a^2*b^12*c^9 + 3*a^4*b^1 0*c^8 + 50*a^6*b^8*c^7 - 85*a^8*b^6*c^6 + 60*a^10*b^4*c^5 - 19*a^12*b^2*c^ 4 + 2*a^14*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 21966*a^4*b^14*c^ 2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*a^2*b^18*c^12 + 45*a^4*b^16*c^11 - 120*a^6*b^14*c^10 + 210*a^8*b^12*c^9 - 252*a^10*b^10*c^ 8 + 210*a^12*b^8*c^7 - 120*a^14*b^6*c^6 + 45*a^16*b^4*c^5 - 10*a^18*b^2*c^ 4 + a^20*c^3)))*sqrt(-((105*a*b^8*c^3 + 70*a^3*b^6*c^2 - 35*a^5*b^4*c + 4* a^7*b^2)*d^4 + (b^10*c^8 - 5*a^2*b^8*c^7 + 10*a^4*b^6*c^6 - 10*a^6*b^4*c^5 + 5*a^8*b^2*c^4 - a^10*c^3)*sqrt((625*b^18*c^4 + 7700*a^2*b^16*c^3 + 2196 6*a^4*b^14*c^2 - 10780*a^6*b^12*c + 1225*a^8*b^10)*d^8/(b^20*c^13 - 10*...
Timed out. \[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=\int { \frac {1}{\sqrt {\sqrt {d x + c} b + a} x^{3}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1303 vs. \(2 (212) = 424\).
Time = 0.44 (sec) , antiderivative size = 1303, normalized size of antiderivative = 4.99 \[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=\text {Too large to display} \]
1/16*(((b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)^2*(5*b^6*c + a^2*b^4)*d^3 - (13 *a*b^10*c^(7/2) - 27*a^3*b^8*c^(5/2) + 15*a^5*b^6*c^(3/2) - a^7*b^4*sqrt(c ))*d^3*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c) + 2*(4*a^2*b^14*c^6 - 17*a^4 *b^12*c^5 + 28*a^6*b^10*c^4 - 22*a^8*b^8*c^3 + 8*a^10*b^6*c^2 - a^12*b^4*c )*d^3)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-(a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c + sqrt((a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c)^2 + (b^6*c^4 - 3*a^2*b^4 *c^3 + 3*a^4*b^2*c^2 - a^6*c)*(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/((b^9*c^6 - a*b^8*c^(11/2) - 4*a^2*b^7*c^5 + 4 *a^3*b^6*c^(9/2) + 6*a^4*b^5*c^4 - 6*a^5*b^4*c^(7/2) - 4*a^6*b^3*c^3 + 4*a ^7*b^2*c^(5/2) + a^8*b*c^2 - a^9*c^(3/2))*sqrt(-b*sqrt(c) - a)*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)) + ((b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)^2*(5*b ^6*c + a^2*b^4)*d^3 + (13*a*b^10*c^(7/2) - 27*a^3*b^8*c^(5/2) + 15*a^5*b^6 *c^(3/2) - a^7*b^4*sqrt(c))*d^3*abs(b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c) + 2 *(4*a^2*b^14*c^6 - 17*a^4*b^12*c^5 + 28*a^6*b^10*c^4 - 22*a^8*b^8*c^3 + 8* a^10*b^6*c^2 - a^12*b^4*c)*d^3)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-(a* b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c - sqrt((a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c) ^2 + (b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*(b^4*c^3 - 2*a^2*b^ 2*c^2 + a^4*c)))/(b^4*c^3 - 2*a^2*b^2*c^2 + a^4*c)))/((b^9*c^6 + a*b^8*c^( 11/2) - 4*a^2*b^7*c^5 - 4*a^3*b^6*c^(9/2) + 6*a^4*b^5*c^4 + 6*a^5*b^4*c^(7 /2) - 4*a^6*b^3*c^3 - 4*a^7*b^2*c^(5/2) + a^8*b*c^2 + a^9*c^(3/2))*sqrt...
Timed out. \[ \int \frac {1}{x^3 \sqrt {a+b \sqrt {c+d x}}} \, dx=\int \frac {1}{x^3\,\sqrt {a+b\,\sqrt {c+d\,x}}} \,d x \]