Integrand size = 22, antiderivative size = 255 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 (A b-a B) \sqrt {x}}{b^2}+\frac {2 B x^{5/2}}{5 b}+\frac {\sqrt [4]{a} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{9/4}}-\frac {\sqrt [4]{a} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{9/4}}+\frac {\sqrt [4]{a} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {\sqrt [4]{a} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}} \]
2/5*B*x^(5/2)/b+1/2*a^(1/4)*(A*b-B*a)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^( 1/4))/b^(9/4)*2^(1/2)-1/2*a^(1/4)*(A*b-B*a)*arctan(1+b^(1/4)*2^(1/2)*x^(1/ 2)/a^(1/4))/b^(9/4)*2^(1/2)+1/4*a^(1/4)*(A*b-B*a)*ln(a^(1/2)+x*b^(1/2)-a^( 1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(9/4)*2^(1/2)-1/4*a^(1/4)*(A*b-B*a)*ln(a^( 1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(9/4)*2^(1/2)+2*(A*b-B*a )*x^(1/2)/b^2
Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.59 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 \sqrt {x} \left (5 A b-5 a B+b B x^2\right )}{5 b^2}-\frac {\sqrt [4]{a} (-A b+a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt {2} b^{9/4}}+\frac {\sqrt [4]{a} (-A b+a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} b^{9/4}} \]
(2*Sqrt[x]*(5*A*b - 5*a*B + b*B*x^2))/(5*b^2) - (a^(1/4)*(-(A*b) + a*B)*Ar cTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(Sqrt[2]*b^ (9/4)) + (a^(1/4)*(-(A*b) + a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x]) /(Sqrt[a] + Sqrt[b]*x)])/(Sqrt[2]*b^(9/4))
Time = 0.44 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {363, 262, 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.
| \(\displaystyle \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(A b-a B) \int \frac {x^{3/2}}{b x^2+a}dx}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \left (b x^2+a\right )}dx}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{b x^2+a}d\sqrt {x}}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{b x^2+a}d\sqrt {x}}{2 \sqrt {a}}+\frac {\int \frac {\sqrt {b} x+\sqrt {a}}{b x^2+a}d\sqrt {x}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{b x^2+a}d\sqrt {x}}{2 \sqrt {a}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}d\sqrt {x}}{2 \sqrt {b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{b x^2+a}d\sqrt {x}}{2 \sqrt {a}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{b x^2+a}d\sqrt {x}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{b} \left (x-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} \sqrt {x}+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{b} \left (x-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} \sqrt {x}+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt {b}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}+\sqrt [4]{a}}{x+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}d\sqrt {x}}{2 \sqrt [4]{a} \sqrt {b}}}{2 \sqrt {a}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {a}}\right )}{b}\right )}{b}+\frac {2 B x^{5/2}}{5 b}\) |
(2*B*x^(5/2))/(5*b) + ((A*b - a*B)*((2*Sqrt[x])/b - (2*a*((-(ArcTan[1 - (S qrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(Sqrt[2]*a^(1/4)*b^(1/4))) + ArcTan[1 + ( Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(Sqrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[a]) + (-1/2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]/(Sqrt[2] *a^(1/4)*b^(1/4)) + Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b ]*x]/(2*Sqrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[a])))/b))/b
3.4.69.3.1 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)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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 = 2.77 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.54
| method | result | size |
| risch | \(\frac {2 \left (b B \,x^{2}+5 A b -5 B a \right ) \sqrt {x}}{5 b^{2}}-\frac {\left (A b -B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2}}\) | \(138\) |
| derivativedivides | \(\frac {\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 A b \sqrt {x}-2 B a \sqrt {x}}{b^{2}}-\frac {\left (A b -B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2}}\) | \(141\) |
| default | \(\frac {\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 A b \sqrt {x}-2 B a \sqrt {x}}{b^{2}}-\frac {\left (A b -B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2}}\) | \(141\) |
2/5*(B*b*x^2+5*A*b-5*B*a)*x^(1/2)/b^2-1/4*(A*b-B*a)/b^2*(a/b)^(1/4)*2^(1/2 )*(ln((x+(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/(x-(a/b)^(1/4)*x^(1/2)*2 ^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+2*arctan(2^(1 /2)/(a/b)^(1/4)*x^(1/2)-1))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 597, normalized size of antiderivative = 2.34 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=-\frac {5 \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) + 5 i \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (i \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - 5 i \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-i \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - 5 \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - 4 \, {\left (B b x^{2} - 5 \, B a + 5 \, A b\right )} \sqrt {x}}{10 \, b^{2}} \]
-1/10*(5*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2* b^3 + A^4*a*b^4)/b^9)^(1/4)*log(b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2 *a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4) - (B*a - A*b)*sqrt(x)) + 5*I*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4)*log(I*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2* a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4) - (B*a - A*b)*sqrt(x)) - 5*I*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4)*log(-I*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2* a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4) - (B*a - A*b)*sqrt(x)) - 5*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4)*log(-b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3* b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4) - (B*a - A*b)*sqrt(x)) - 4*( B*b*x^2 - 5*B*a + 5*A*b)*sqrt(x))/b^2
Time = 4.74 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.08 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {9}{2}}}{9}}{a} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: a = 0 \\\frac {2 A \sqrt {x}}{b} + \frac {A \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b} - \frac {A \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b} - \frac {A \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b} - \frac {2 B a \sqrt {x}}{b^{2}} - \frac {B a \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {B a \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {B a \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 b} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(5/2)/5), Eq(a, 0) & Eq(b, 0)), ((2*A *x**(5/2)/5 + 2*B*x**(9/2)/9)/a, Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(5/2)/5 )/b, Eq(a, 0)), (2*A*sqrt(x)/b + A*(-a/b)**(1/4)*log(sqrt(x) - (-a/b)**(1/ 4))/(2*b) - A*(-a/b)**(1/4)*log(sqrt(x) + (-a/b)**(1/4))/(2*b) - A*(-a/b)* *(1/4)*atan(sqrt(x)/(-a/b)**(1/4))/b - 2*B*a*sqrt(x)/b**2 - B*a*(-a/b)**(1 /4)*log(sqrt(x) - (-a/b)**(1/4))/(2*b**2) + B*a*(-a/b)**(1/4)*log(sqrt(x) + (-a/b)**(1/4))/(2*b**2) + B*a*(-a/b)**(1/4)*atan(sqrt(x)/(-a/b)**(1/4))/ b**2 + 2*B*x**(5/2)/(5*b), True))
Time = 0.28 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {{\left (\frac {2 \, \sqrt {2} {\left (B a - A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (B a - A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (B a - A b\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B a - A b\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )} a}{4 \, b^{2}} + \frac {2 \, {\left (B b x^{\frac {5}{2}} - 5 \, {\left (B a - A b\right )} \sqrt {x}\right )}}{5 \, b^{2}} \]
1/4*(2*sqrt(2)*(B*a - A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2 *sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(B*a - A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sq rt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + sq rt(2)*(B*a - A*b)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a ))/(a^(3/4)*b^(1/4)) - sqrt(2)*(B*a - A*b)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sq rt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)))*a/b^2 + 2/5*(B*b*x^(5/2) - 5*(B*a - A*b)*sqrt(x))/b^2
Time = 0.28 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.03 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{3}} + \frac {2 \, {\left (B b^{4} x^{\frac {5}{2}} - 5 \, B a b^{3} \sqrt {x} + 5 \, A b^{4} \sqrt {x}\right )}}{5 \, b^{5}} \]
1/2*sqrt(2)*((a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(sq rt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/b^3 + 1/2*sqrt(2)*((a*b^3)^(1/ 4)*B*a - (a*b^3)^(1/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*s qrt(x))/(a/b)^(1/4))/b^3 + 1/4*sqrt(2)*((a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)* A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^3 - 1/4*sqrt(2)*(( a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^3 + 2/5*(B*b^4*x^(5/2) - 5*B*a*b^3*sqrt(x) + 5*A*b^4*sqrt( x))/b^5
Time = 5.16 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.09 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )+\frac {2\,B\,x^{5/2}}{5\,b}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}+\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}}{\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )}{2\,b^{9/4}}-\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )}{2\,b^{9/4}}}\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{b^{9/4}}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )}{2\,b^{9/4}}+\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )}{2\,b^{9/4}}}{\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}-\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}}\right )\,\left (A\,b-B\,a\right )}{b^{9/4}} \]
x^(1/2)*((2*A)/b - (2*B*a)/b^2) + (2*B*x^(5/2))/(5*b) - ((-a)^(1/4)*atan(( ((-a)^(1/4)*(A*b - B*a)*((16*x^(1/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b) )/b - ((-a)^(1/4)*(32*A*a^2*b^2 - 32*B*a^3*b)*(A*b - B*a))/(2*b^(9/4)))*1i )/(2*b^(9/4)) + ((-a)^(1/4)*(A*b - B*a)*((16*x^(1/2)*(B^2*a^4 + A^2*a^2*b^ 2 - 2*A*B*a^3*b))/b + ((-a)^(1/4)*(32*A*a^2*b^2 - 32*B*a^3*b)*(A*b - B*a)) /(2*b^(9/4)))*1i)/(2*b^(9/4)))/(((-a)^(1/4)*(A*b - B*a)*((16*x^(1/2)*(B^2* a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b))/b - ((-a)^(1/4)*(32*A*a^2*b^2 - 32*B*a^3 *b)*(A*b - B*a))/(2*b^(9/4))))/(2*b^(9/4)) - ((-a)^(1/4)*(A*b - B*a)*((16* x^(1/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b))/b + ((-a)^(1/4)*(32*A*a^2*b ^2 - 32*B*a^3*b)*(A*b - B*a))/(2*b^(9/4))))/(2*b^(9/4))))*(A*b - B*a)*1i)/ b^(9/4) - ((-a)^(1/4)*atan((((-a)^(1/4)*(A*b - B*a)*((16*x^(1/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b))/b - ((-a)^(1/4)*(32*A*a^2*b^2 - 32*B*a^3*b)* (A*b - B*a)*1i)/(2*b^(9/4))))/(2*b^(9/4)) + ((-a)^(1/4)*(A*b - B*a)*((16*x ^(1/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b))/b + ((-a)^(1/4)*(32*A*a^2*b^ 2 - 32*B*a^3*b)*(A*b - B*a)*1i)/(2*b^(9/4))))/(2*b^(9/4)))/(((-a)^(1/4)*(A *b - B*a)*((16*x^(1/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a^3*b))/b - ((-a)^(1 /4)*(32*A*a^2*b^2 - 32*B*a^3*b)*(A*b - B*a)*1i)/(2*b^(9/4)))*1i)/(2*b^(9/4 )) - ((-a)^(1/4)*(A*b - B*a)*((16*x^(1/2)*(B^2*a^4 + A^2*a^2*b^2 - 2*A*B*a ^3*b))/b + ((-a)^(1/4)*(32*A*a^2*b^2 - 32*B*a^3*b)*(A*b - B*a)*1i)/(2*b^(9 /4)))*1i)/(2*b^(9/4))))*(A*b - B*a))/b^(9/4)