Integrand size = 19, antiderivative size = 264 \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}-\frac {165 c^{7/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {165 c^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}} \]
-165/112/b^3/x^(7/2)+55/16*c/b^4/x^(3/2)+1/4/b/x^(7/2)/(c*x^2+b)^2+15/16/b ^2/x^(7/2)/(c*x^2+b)-165/64*c^(7/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/ 4))/b^(19/4)*2^(1/2)+165/64*c^(7/4)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/ 4))/b^(19/4)*2^(1/2)-165/128*c^(7/4)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)* 2^(1/2)*x^(1/2))/b^(19/4)*2^(1/2)+165/128*c^(7/4)*ln(b^(1/2)+x*c^(1/2)+b^( 1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(19/4)*2^(1/2)
Time = 0.41 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.61 \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {4 b^{3/4} \left (-32 b^3+160 b^2 c x^2+605 b c^2 x^4+385 c^3 x^6\right )}{x^{7/2} \left (b+c x^2\right )^2}-1155 \sqrt {2} c^{7/4} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+1155 \sqrt {2} c^{7/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{448 b^{19/4}} \]
((4*b^(3/4)*(-32*b^3 + 160*b^2*c*x^2 + 605*b*c^2*x^4 + 385*c^3*x^6))/(x^(7 /2)*(b + c*x^2)^2) - 1155*Sqrt[2]*c^(7/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sq rt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 1155*Sqrt[2]*c^(7/4)*ArcTanh[(Sqrt[2]*b^ (1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(448*b^(19/4))
Time = 0.48 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.17, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {9, 253, 253, 264, 264, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {1}{x^{9/2} \left (b+c x^2\right )^3}dx\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {15 \int \frac {1}{x^{9/2} \left (c x^2+b\right )^2}dx}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {15 \left (\frac {11 \int \frac {1}{x^{9/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \int \frac {1}{x^{5/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {c \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \int \frac {1}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {15 \left (\frac {11 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{4 b}+\frac {1}{2 b x^{7/2} \left (b+c x^2\right )}\right )}{8 b}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}\) |
1/(4*b*x^(7/2)*(b + c*x^2)^2) + (15*(1/(2*b*x^(7/2)*(b + c*x^2)) + (11*(-2 /(7*b*x^(7/2)) - (c*(-2/(3*b*x^(3/2)) - (2*c*((-(ArcTan[1 - (Sqrt[2]*c^(1/ 4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) + ArcTan[1 + (Sqrt[2]*c^(1 /4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[b]) + (-1/2*Log[S qrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]/(Sqrt[2]*b^(1/4)*c^( 1/4)) + Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]/(2*Sqrt [2]*b^(1/4)*c^(1/4)))/(2*Sqrt[b])))/b))/b))/(4*b)))/(8*b)
3.4.49.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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, 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.59
| method | result | size |
| risch | \(-\frac {2 \left (-7 c \,x^{2}+b \right )}{7 b^{4} x^{\frac {7}{2}}}+\frac {c^{2} \left (\frac {\frac {23 c \,x^{\frac {5}{2}}}{16}+\frac {27 b \sqrt {x}}{16}}{\left (c \,x^{2}+b \right )^{2}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}\right )}{b^{4}}\) | \(155\) |
| derivativedivides | \(-\frac {2}{7 b^{3} x^{\frac {7}{2}}}+\frac {2 c}{b^{4} x^{\frac {3}{2}}}+\frac {2 c^{2} \left (\frac {\frac {23 c \,x^{\frac {5}{2}}}{32}+\frac {27 b \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{4}}\) | \(156\) |
| default | \(-\frac {2}{7 b^{3} x^{\frac {7}{2}}}+\frac {2 c}{b^{4} x^{\frac {3}{2}}}+\frac {2 c^{2} \left (\frac {\frac {23 c \,x^{\frac {5}{2}}}{32}+\frac {27 b \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{4}}\) | \(156\) |
-2/7*(-7*c*x^2+b)/b^4/x^(7/2)+1/b^4*c^2*(2*(23/32*c*x^(5/2)+27/32*b*x^(1/2 ))/(c*x^2+b)^2+165/128*(b/c)^(1/4)/b*2^(1/2)*(ln((x+(b/c)^(1/4)*x^(1/2)*2^ (1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))+2*arctan(2 ^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)))
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.25 \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {1155 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (165 \, b^{5} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} + 165 \, c^{2} \sqrt {x}\right ) - 1155 \, {\left (-i \, b^{4} c^{2} x^{8} - 2 i \, b^{5} c x^{6} - i \, b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (165 i \, b^{5} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} + 165 \, c^{2} \sqrt {x}\right ) - 1155 \, {\left (i \, b^{4} c^{2} x^{8} + 2 i \, b^{5} c x^{6} + i \, b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (-165 i \, b^{5} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} + 165 \, c^{2} \sqrt {x}\right ) - 1155 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (-165 \, b^{5} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} + 165 \, c^{2} \sqrt {x}\right ) + 4 \, {\left (385 \, c^{3} x^{6} + 605 \, b c^{2} x^{4} + 160 \, b^{2} c x^{2} - 32 \, b^{3}\right )} \sqrt {x}}{448 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} \]
1/448*(1155*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-c^7/b^19)^(1/4)*log(16 5*b^5*(-c^7/b^19)^(1/4) + 165*c^2*sqrt(x)) - 1155*(-I*b^4*c^2*x^8 - 2*I*b^ 5*c*x^6 - I*b^6*x^4)*(-c^7/b^19)^(1/4)*log(165*I*b^5*(-c^7/b^19)^(1/4) + 1 65*c^2*sqrt(x)) - 1155*(I*b^4*c^2*x^8 + 2*I*b^5*c*x^6 + I*b^6*x^4)*(-c^7/b ^19)^(1/4)*log(-165*I*b^5*(-c^7/b^19)^(1/4) + 165*c^2*sqrt(x)) - 1155*(b^4 *c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-c^7/b^19)^(1/4)*log(-165*b^5*(-c^7/b^1 9)^(1/4) + 165*c^2*sqrt(x)) + 4*(385*c^3*x^6 + 605*b*c^2*x^4 + 160*b^2*c*x ^2 - 32*b^3)*sqrt(x))/(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)
Timed out. \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {385 \, c^{3} x^{6} + 605 \, b c^{2} x^{4} + 160 \, b^{2} c x^{2} - 32 \, b^{3}}{112 \, {\left (b^{4} c^{2} x^{\frac {15}{2}} + 2 \, b^{5} c x^{\frac {11}{2}} + b^{6} x^{\frac {7}{2}}\right )}} + \frac {165 \, {\left (\frac {2 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} c^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b^{4}} \]
1/112*(385*c^3*x^6 + 605*b*c^2*x^4 + 160*b^2*c*x^2 - 32*b^3)/(b^4*c^2*x^(1 5/2) + 2*b^5*c*x^(11/2) + b^6*x^(7/2)) + 165/128*(2*sqrt(2)*c^2*arctan(1/2 *sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c )))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*c^2*arctan(-1/2*sqrt(2)*(s qrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b )*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*c^(7/4)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqr t(x) + sqrt(c)*x + sqrt(b))/b^(3/4) - sqrt(2)*c^(7/4)*log(-sqrt(2)*b^(1/4) *c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4))/b^4
Time = 0.30 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.85 \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{5}} + \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{5}} + \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{5}} - \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{5}} + \frac {23 \, c^{3} x^{\frac {5}{2}} + 27 \, b c^{2} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{4}} + \frac {2 \, {\left (7 \, c x^{2} - b\right )}}{7 \, b^{4} x^{\frac {7}{2}}} \]
165/64*sqrt(2)*(b*c^3)^(1/4)*c*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2 *sqrt(x))/(b/c)^(1/4))/b^5 + 165/64*sqrt(2)*(b*c^3)^(1/4)*c*arctan(-1/2*sq rt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^5 + 165/128*sqrt(2) *(b*c^3)^(1/4)*c*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 - 16 5/128*sqrt(2)*(b*c^3)^(1/4)*c*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt( b/c))/b^5 + 1/16*(23*c^3*x^(5/2) + 27*b*c^2*sqrt(x))/((c*x^2 + b)^2*b^4) + 2/7*(7*c*x^2 - b)/(b^4*x^(7/2))
Time = 13.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.41 \[ \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {10\,c\,x^2}{7\,b^2}-\frac {2}{7\,b}+\frac {605\,c^2\,x^4}{112\,b^3}+\frac {55\,c^3\,x^6}{16\,b^4}}{b^2\,x^{7/2}+c^2\,x^{15/2}+2\,b\,c\,x^{11/2}}+\frac {165\,{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{19/4}}+\frac {165\,{\left (-c\right )}^{7/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{19/4}} \]