Integrand size = 28, antiderivative size = 335 \[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}+\frac {d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}+\frac {5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}-\frac {5 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{128 \sqrt {2} a^{9/4} b^{7/4}}+\frac {5 d^{5/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{9/4} b^{7/4}}-\frac {5 d^{5/2} \log \left (\sqrt {a} \sqrt {d}+\sqrt {b} \sqrt {d} x+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}\right )}{256 \sqrt {2} a^{9/4} b^{7/4}} \]
-1/6*d*(d*x)^(3/2)/b/(b*x^2+a)^3+1/16*d*(d*x)^(3/2)/a/b/(b*x^2+a)^2+5/64*d *(d*x)^(3/2)/a^2/b/(b*x^2+a)-5/256*d^(5/2)*arctan(1-b^(1/4)*2^(1/2)*(d*x)^ (1/2)/a^(1/4)/d^(1/2))/a^(9/4)/b^(7/4)*2^(1/2)+5/256*d^(5/2)*arctan(1+b^(1 /4)*2^(1/2)*(d*x)^(1/2)/a^(1/4)/d^(1/2))/a^(9/4)/b^(7/4)*2^(1/2)+5/512*d^( 5/2)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1 /2))/a^(9/4)/b^(7/4)*2^(1/2)-5/512*d^(5/2)*ln(a^(1/2)*d^(1/2)+x*b^(1/2)*d^ (1/2)+a^(1/4)*b^(1/4)*2^(1/2)*(d*x)^(1/2))/a^(9/4)/b^(7/4)*2^(1/2)
Time = 0.40 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.48 \[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {(d x)^{5/2} \left (\frac {4 \sqrt [4]{a} b^{3/4} x^{3/2} \left (-5 a^2+42 a b x^2+15 b^2 x^4\right )}{\left (a+b x^2\right )^3}-15 \sqrt {2} \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{768 a^{9/4} b^{7/4} x^{5/2}} \]
((d*x)^(5/2)*((4*a^(1/4)*b^(3/4)*x^(3/2)*(-5*a^2 + 42*a*b*x^2 + 15*b^2*x^4 ))/(a + b*x^2)^3 - 15*Sqrt[2]*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4 )*b^(1/4)*Sqrt[x])] - 15*Sqrt[2]*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x]) /(Sqrt[a] + Sqrt[b]*x)]))/(768*a^(9/4)*b^(7/4)*x^(5/2))
Time = 0.57 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {1380, 27, 252, 253, 253, 266, 27, 826, 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.
| \(\displaystyle \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle b^4 \int \frac {(d x)^{5/2}}{b^4 \left (b x^2+a\right )^4}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(d x)^{5/2}}{\left (a+b x^2\right )^4}dx\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {d^2 \int \frac {\sqrt {d x}}{\left (b x^2+a\right )^3}dx}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \int \frac {\sqrt {d x}}{\left (b x^2+a\right )^2}dx}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {\int \frac {\sqrt {d x}}{b x^2+a}dx}{4 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {\int \frac {d^3 x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 a d}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \int \frac {d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\int \frac {\sqrt {b} x d+\sqrt {a} d}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\int \frac {1}{x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt {b}}+\frac {\int \frac {1}{x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt {b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\int \frac {1}{-d x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\int \frac {1}{-d x-1}d\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d}-2 \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt [4]{a} \sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt {d x}\right )}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a} \sqrt {d}-2 \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt [4]{a} \sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt {d x}\right )}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a} \sqrt {d}-2 \sqrt [4]{b} \sqrt {d x}}{x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt {b} \sqrt {d}}+\frac {\int \frac {\sqrt [4]{a} \sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt [4]{a} \sqrt {b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {d^2 \left (\frac {5 \left (\frac {d \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}+\sqrt {a} d+\sqrt {b} d x\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}+\sqrt {a} d+\sqrt {b} d x\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 a}+\frac {(d x)^{3/2}}{2 a d \left (a+b x^2\right )}\right )}{8 a}+\frac {(d x)^{3/2}}{4 a d \left (a+b x^2\right )^2}\right )}{4 b}-\frac {d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3}\) |
-1/6*(d*(d*x)^(3/2))/(b*(a + b*x^2)^3) + (d^2*((d*x)^(3/2)/(4*a*d*(a + b*x ^2)^2) + (5*((d*x)^(3/2)/(2*a*d*(a + b*x^2)) + (d*((-(ArcTan[1 - (Sqrt[2]* b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4 )*b^(1/4)*Sqrt[d]))/(2*Sqrt[b]) - (-1/2*Log[Sqrt[a]*d + Sqrt[b]*d*x - Sqrt [2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x]]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]) + Log[Sqrt[a]*d + Sqrt[b]*d*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x]]/ (2*Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]))/(2*Sqrt[b])))/(2*a)))/(8*a)))/(4*b)
3.8.3.3.1 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, 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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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)/(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] :> 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) 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[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
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 = 2.24 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.61
| method | result | size |
| derivativedivides | \(2 d^{7} \left (\frac {\frac {5 b \left (d x \right )^{\frac {11}{2}}}{128 a^{2} d^{4}}+\frac {7 \left (d x \right )^{\frac {7}{2}}}{64 a \,d^{2}}-\frac {5 \left (d x \right )^{\frac {3}{2}}}{384 b}}{\left (b \,d^{2} x^{2}+a \,d^{2}\right )^{3}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{1024 a^{2} d^{4} b^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\) | \(206\) |
| default | \(2 d^{7} \left (\frac {\frac {5 b \left (d x \right )^{\frac {11}{2}}}{128 a^{2} d^{4}}+\frac {7 \left (d x \right )^{\frac {7}{2}}}{64 a \,d^{2}}-\frac {5 \left (d x \right )^{\frac {3}{2}}}{384 b}}{\left (b \,d^{2} x^{2}+a \,d^{2}\right )^{3}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{1024 a^{2} d^{4} b^{2} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\) | \(206\) |
| pseudoelliptic | \(-\frac {5 d^{2} \left (8 x b \sqrt {d x}\, \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (-3 b^{2} x^{4}-\frac {42}{5} a b \,x^{2}+a^{2}\right )-3 \sqrt {2}\, d \left (b \,x^{2}+a \right )^{3} \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )\right )\right )}{1536 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{2} a^{2} \left (b \,x^{2}+a \right )^{3}}\) | \(227\) |
2*d^7*((5/128/a^2/d^4*b*(d*x)^(11/2)+7/64/a/d^2*(d*x)^(7/2)-5/384/b*(d*x)^ (3/2))/(b*d^2*x^2+a*d^2)^3+5/1024/a^2/d^4/b^2/(a*d^2/b)^(1/4)*2^(1/2)*(ln( (d*x-(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2))/(d*x+(a*d^2/b)^( 1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))+2*arctan(2^(1/2)/(a*d^2/b)^(1/4 )*(d*x)^(1/2)+1)+2*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)-1)))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.28 \[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {15 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} b^{5} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d^{7}\right ) - 15 \, {\left (i \, a^{2} b^{4} x^{6} + 3 i \, a^{3} b^{3} x^{4} + 3 i \, a^{4} b^{2} x^{2} + i \, a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (125 i \, a^{7} b^{5} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d^{7}\right ) - 15 \, {\left (-i \, a^{2} b^{4} x^{6} - 3 i \, a^{3} b^{3} x^{4} - 3 i \, a^{4} b^{2} x^{2} - i \, a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (-125 i \, a^{7} b^{5} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d^{7}\right ) - 15 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} b^{5} \left (-\frac {d^{10}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + 125 \, \sqrt {d x} d^{7}\right ) + 4 \, {\left (15 \, b^{2} d^{2} x^{5} + 42 \, a b d^{2} x^{3} - 5 \, a^{2} d^{2} x\right )} \sqrt {d x}}{768 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} \]
1/768*(15*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)*(-d^10/(a^ 9*b^7))^(1/4)*log(125*a^7*b^5*(-d^10/(a^9*b^7))^(3/4) + 125*sqrt(d*x)*d^7) - 15*(I*a^2*b^4*x^6 + 3*I*a^3*b^3*x^4 + 3*I*a^4*b^2*x^2 + I*a^5*b)*(-d^10 /(a^9*b^7))^(1/4)*log(125*I*a^7*b^5*(-d^10/(a^9*b^7))^(3/4) + 125*sqrt(d*x )*d^7) - 15*(-I*a^2*b^4*x^6 - 3*I*a^3*b^3*x^4 - 3*I*a^4*b^2*x^2 - I*a^5*b) *(-d^10/(a^9*b^7))^(1/4)*log(-125*I*a^7*b^5*(-d^10/(a^9*b^7))^(3/4) + 125* sqrt(d*x)*d^7) - 15*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)* (-d^10/(a^9*b^7))^(1/4)*log(-125*a^7*b^5*(-d^10/(a^9*b^7))^(3/4) + 125*sqr t(d*x)*d^7) + 4*(15*b^2*d^2*x^5 + 42*a*b*d^2*x^3 - 5*a^2*d^2*x)*sqrt(d*x)) /(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)
\[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\int \frac {\left (d x\right )^{\frac {5}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.96 \[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {15 \, d^{4} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{a^{2} b} + \frac {8 \, {\left (15 \, \left (d x\right )^{\frac {11}{2}} b^{2} d^{4} + 42 \, \left (d x\right )^{\frac {7}{2}} a b d^{6} - 5 \, \left (d x\right )^{\frac {3}{2}} a^{2} d^{8}\right )}}{a^{2} b^{4} d^{6} x^{6} + 3 \, a^{3} b^{3} d^{6} x^{4} + 3 \, a^{4} b^{2} d^{6} x^{2} + a^{5} b d^{6}}}{1536 \, d} \]
1/1536*(15*d^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4 ) + 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d) *sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^(1/4) - 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b)*d)*sq rt(b)) - sqrt(2)*log(sqrt(b)*d*x + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*d*x - sqrt(2)* (a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)))/(a^2 *b) + 8*(15*(d*x)^(11/2)*b^2*d^4 + 42*(d*x)^(7/2)*a*b*d^6 - 5*(d*x)^(3/2)* a^2*d^8)/(a^2*b^4*d^6*x^6 + 3*a^3*b^3*d^6*x^4 + 3*a^4*b^2*d^6*x^2 + a^5*b* d^6))/d
Time = 0.31 (sec) , antiderivative size = 317, normalized size of antiderivative = 0.95 \[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {1}{1536} \, d^{2} {\left (\frac {8 \, {\left (15 \, \sqrt {d x} b^{2} d^{6} x^{5} + 42 \, \sqrt {d x} a b d^{6} x^{3} - 5 \, \sqrt {d x} a^{2} d^{6} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a^{2} b} + \frac {30 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} b^{4} d} + \frac {30 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a^{3} b^{4} d} - \frac {15 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{3} b^{4} d} + \frac {15 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a^{3} b^{4} d}\right )} \]
1/1536*d^2*(8*(15*sqrt(d*x)*b^2*d^6*x^5 + 42*sqrt(d*x)*a*b*d^6*x^3 - 5*sqr t(d*x)*a^2*d^6*x)/((b*d^2*x^2 + a*d^2)^3*a^2*b) + 30*sqrt(2)*(a*b^3*d^2)^( 3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^ (1/4))/(a^3*b^4*d) + 30*sqrt(2)*(a*b^3*d^2)^(3/4)*arctan(-1/2*sqrt(2)*(sqr t(2)*(a*d^2/b)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(a^3*b^4*d) - 15*sqrt (2)*(a*b^3*d^2)^(3/4)*log(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a *d^2/b))/(a^3*b^4*d) + 15*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x - sqrt(2)*(a*d ^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(a^3*b^4*d))
Time = 13.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.44 \[ \int \frac {(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {7\,d^5\,{\left (d\,x\right )}^{7/2}}{32\,a}-\frac {5\,d^7\,{\left (d\,x\right )}^{3/2}}{192\,b}+\frac {5\,b\,d^3\,{\left (d\,x\right )}^{11/2}}{64\,a^2}}{a^3\,d^6+3\,a^2\,b\,d^6\,x^2+3\,a\,b^2\,d^6\,x^4+b^3\,d^6\,x^6}+\frac {5\,d^{5/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{9/4}\,b^{7/4}}-\frac {5\,d^{5/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{128\,{\left (-a\right )}^{9/4}\,b^{7/4}} \]
((7*d^5*(d*x)^(7/2))/(32*a) - (5*d^7*(d*x)^(3/2))/(192*b) + (5*b*d^3*(d*x) ^(11/2))/(64*a^2))/(a^3*d^6 + b^3*d^6*x^6 + 3*a^2*b*d^6*x^2 + 3*a*b^2*d^6* x^4) + (5*d^(5/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(128*( -a)^(9/4)*b^(7/4)) - (5*d^(5/2)*atanh((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^ (1/2))))/(128*(-a)^(9/4)*b^(7/4))