\(\int \frac {\sqrt {x} (A+B x^2)}{(a+b x^2)^3} \, dx\) [386]

Optimal result

Integrand size = 22, antiderivative size = 298 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}-\frac {(5 A b+3 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(5 A b+3 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(5 A b+3 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}-\frac {(5 A b+3 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}} \]

[Out]

1/4*(A*b-B*a)*x^(3/2)/a/b/(b*x^2+a)^2+1/16*(5*A*b+3*B*a)*x^(3/2)/a^2/b/(b*x^2+a)-1/64*(5*A*b+3*B*a)*arctan(1-b
^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(9/4)/b^(7/4)*2^(1/2)+1/64*(5*A*b+3*B*a)*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^
(1/4))/a^(9/4)/b^(7/4)*2^(1/2)+1/128*(5*A*b+3*B*a)*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(9/
4)/b^(7/4)*2^(1/2)-1/128*(5*A*b+3*B*a)*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/a^(9/4)/b^(7/4)*2
^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {468, 296, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {(3 a B+5 A b) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(3 a B+5 A b) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(3 a B+5 A b) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}-\frac {(3 a B+5 A b) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}+\frac {x^{3/2} (3 a B+5 A b)}{16 a^2 b \left (a+b x^2\right )}+\frac {x^{3/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

[In]

Int[(Sqrt[x]*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

((A*b - a*B)*x^(3/2))/(4*a*b*(a + b*x^2)^2) + ((5*A*b + 3*a*B)*x^(3/2))/(16*a^2*b*(a + b*x^2)) - ((5*A*b + 3*a
*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*b^(7/4)) + ((5*A*b + 3*a*B)*ArcTan[1 +
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*b^(7/4)) + ((5*A*b + 3*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(9/4)*b^(7/4)) - ((5*A*b + 3*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(9/4)*b^(7/4))

Rule 210

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])

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {\left (\frac {5 A b}{2}+\frac {3 a B}{2}\right ) \int \frac {\sqrt {x}}{\left (a+b x^2\right )^2} \, dx}{4 a b} \\ & = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac {(5 A b+3 a B) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{32 a^2 b} \\ & = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a^2 b} \\ & = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}-\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a^2 b^{3/2}}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{32 a^2 b^{3/2}} \\ & = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{64 a^2 b^2}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{64 a^2 b^2}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} a^{9/4} b^{7/4}} \\ & = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac {(5 A b+3 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}-\frac {(5 A b+3 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}-\frac {(5 A b+3 a B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}} \\ & = \frac {(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac {(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}-\frac {(5 A b+3 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(5 A b+3 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{9/4} b^{7/4}}+\frac {(5 A b+3 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}}-\frac {(5 A b+3 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{9/4} b^{7/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {4 \sqrt [4]{a} b^{3/4} x^{3/2} \left (9 a A b-a^2 B+5 A b^2 x^2+3 a b B x^2\right )}{\left (a+b x^2\right )^2}-\sqrt {2} (5 A b+3 a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-\sqrt {2} (5 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{64 a^{9/4} b^{7/4}} \]

[In]

Integrate[(Sqrt[x]*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

((4*a^(1/4)*b^(3/4)*x^(3/2)*(9*a*A*b - a^2*B + 5*A*b^2*x^2 + 3*a*b*B*x^2))/(a + b*x^2)^2 - Sqrt[2]*(5*A*b + 3*
a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] - Sqrt[2]*(5*A*b + 3*a*B)*ArcTanh[(Sqrt[2
]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(64*a^(9/4)*b^(7/4))

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.56

method	result	size
derivativedivides	\(\frac {\frac {\left (5 A b +3 B a \right ) x^{\frac {7}{2}}}{16 a^{2}}+\frac {\left (9 A b -B a \right ) x^{\frac {3}{2}}}{16 a b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (5 A b +3 B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{128 a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(168\)
default	\(\frac {\frac {\left (5 A b +3 B a \right ) x^{\frac {7}{2}}}{16 a^{2}}+\frac {\left (9 A b -B a \right ) x^{\frac {3}{2}}}{16 a b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (5 A b +3 B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{128 a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(168\)

[In]

int((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

2*(1/32*(5*A*b+3*B*a)/a^2*x^(7/2)+1/32*(9*A*b-B*a)/a/b*x^(3/2))/(b*x^2+a)^2+1/128*(5*A*b+3*B*a)/a^2/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*a
rctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 878, normalized size of antiderivative = 2.95 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (a^{7} b^{5} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} a^{3} + 135 \, A B^{2} a^{2} b + 225 \, A^{2} B a b^{2} + 125 \, A^{3} b^{3}\right )} \sqrt {x}\right ) - {\left (i \, a^{2} b^{3} x^{4} + 2 i \, a^{3} b^{2} x^{2} + i \, a^{4} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (i \, a^{7} b^{5} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} a^{3} + 135 \, A B^{2} a^{2} b + 225 \, A^{2} B a b^{2} + 125 \, A^{3} b^{3}\right )} \sqrt {x}\right ) - {\left (-i \, a^{2} b^{3} x^{4} - 2 i \, a^{3} b^{2} x^{2} - i \, a^{4} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (-i \, a^{7} b^{5} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} a^{3} + 135 \, A B^{2} a^{2} b + 225 \, A^{2} B a b^{2} + 125 \, A^{3} b^{3}\right )} \sqrt {x}\right ) - {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {1}{4}} \log \left (-a^{7} b^{5} \left (-\frac {81 \, B^{4} a^{4} + 540 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 1500 \, A^{3} B a b^{3} + 625 \, A^{4} b^{4}}{a^{9} b^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} a^{3} + 135 \, A B^{2} a^{2} b + 225 \, A^{2} B a b^{2} + 125 \, A^{3} b^{3}\right )} \sqrt {x}\right ) + 4 \, {\left ({\left (3 \, B a b + 5 \, A b^{2}\right )} x^{3} - {\left (B a^{2} - 9 \, A a b\right )} x\right )} \sqrt {x}}{64 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} \]

[In]

integrate((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*((a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*
B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*log(a^7*b^5*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1
500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(3/4) + (27*B^3*a^3 + 135*A*B^2*a^2*b + 225*A^2*B*a*b^2 + 125*A^3*b^
3)*sqrt(x)) - (I*a^2*b^3*x^4 + 2*I*a^3*b^2*x^2 + I*a^4*b)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b
^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*log(I*a^7*b^5*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2
*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(3/4) + (27*B^3*a^3 + 135*A*B^2*a^2*b + 225*A^2*B*a*
b^2 + 125*A^3*b^3)*sqrt(x)) - (-I*a^2*b^3*x^4 - 2*I*a^3*b^2*x^2 - I*a^4*b)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1
350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*log(-I*a^7*b^5*(-(81*B^4*a^4 + 540*A*B^
3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(3/4) + (27*B^3*a^3 + 135*A*B^2*a^
2*b + 225*A^2*B*a*b^2 + 125*A^3*b^3)*sqrt(x)) - (a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(81*B^4*a^4 + 540*A*B^
3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*log(-a^7*b^5*(-(81*B^4*a^4 +
540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(3/4) + (27*B^3*a^3 + 135
*A*B^2*a^2*b + 225*A^2*B*a*b^2 + 125*A^3*b^3)*sqrt(x)) + 4*((3*B*a*b + 5*A*b^2)*x^3 - (B*a^2 - 9*A*a*b)*x)*sqr
t(x))/(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((B*x**2+A)*x**(1/2)/(b*x**2+a)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, B a b + 5 \, A b^{2}\right )} x^{\frac {7}{2}} - {\left (B a^{2} - 9 \, A a b\right )} x^{\frac {3}{2}}}{16 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} + \frac {{\left (3 \, B a + 5 \, A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{128 \, a^{2} b} \]

[In]

integrate((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/16*((3*B*a*b + 5*A*b^2)*x^(7/2) - (B*a^2 - 9*A*a*b)*x^(3/2))/(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b) + 1/128*(
3*B*a + 5*A*b)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b
)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(
x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqr
t(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(
1/4)*b^(3/4)))/(a^2*b)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {3 \, B a b x^{\frac {7}{2}} + 5 \, A b^{2} x^{\frac {7}{2}} - B a^{2} x^{\frac {3}{2}} + 9 \, A a b x^{\frac {3}{2}}}{16 \, {\left (b x^{2} + a\right )}^{2} a^{2} b} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{64 \, a^{3} b^{4}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{64 \, a^{3} b^{4}} - \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{3} b^{4}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{3} b^{4}} \]

[In]

integrate((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/16*(3*B*a*b*x^(7/2) + 5*A*b^2*x^(7/2) - B*a^2*x^(3/2) + 9*A*a*b*x^(3/2))/((b*x^2 + a)^2*a^2*b) + 1/64*sqrt(2
)*(3*(a*b^3)^(3/4)*B*a + 5*(a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4)
)/(a^3*b^4) + 1/64*sqrt(2)*(3*(a*b^3)^(3/4)*B*a + 5*(a*b^3)^(3/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4
) - 2*sqrt(x))/(a/b)^(1/4))/(a^3*b^4) - 1/128*sqrt(2)*(3*(a*b^3)^(3/4)*B*a + 5*(a*b^3)^(3/4)*A*b)*log(sqrt(2)*
sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^4) + 1/128*sqrt(2)*(3*(a*b^3)^(3/4)*B*a + 5*(a*b^3)^(3/4)*A*b)*log
(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^4)

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {x^{7/2}\,\left (5\,A\,b+3\,B\,a\right )}{16\,a^2}+\frac {x^{3/2}\,\left (9\,A\,b-B\,a\right )}{16\,a\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (5\,A\,b+3\,B\,a\right )}{32\,{\left (-a\right )}^{9/4}\,b^{7/4}}-\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (5\,A\,b+3\,B\,a\right )}{32\,{\left (-a\right )}^{9/4}\,b^{7/4}} \]

[In]

int((x^(1/2)*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

((x^(7/2)*(5*A*b + 3*B*a))/(16*a^2) + (x^(3/2)*(9*A*b - B*a))/(16*a*b))/(a^2 + b^2*x^4 + 2*a*b*x^2) + (atan((b
^(1/4)*x^(1/2))/(-a)^(1/4))*(5*A*b + 3*B*a))/(32*(-a)^(9/4)*b^(7/4)) - (atanh((b^(1/4)*x^(1/2))/(-a)^(1/4))*(5
*A*b + 3*B*a))/(32*(-a)^(9/4)*b^(7/4))