\(\int \frac {A+B x^2}{x^{5/2} (a+b x^2)^3} \, dx\) [389]

Optimal result

Integrand size = 22, antiderivative size = 322 \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=-\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}+\frac {7 (11 A b-3 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4} \sqrt [4]{b}}-\frac {7 (11 A b-3 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4} \sqrt [4]{b}}+\frac {7 (11 A b-3 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4} \sqrt [4]{b}}-\frac {7 (11 A b-3 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4} \sqrt [4]{b}} \]

[Out]

-7/48*(11*A*b-3*B*a)/a^3/b/x^(3/2)+1/4*(A*b-B*a)/a/b/x^(3/2)/(b*x^2+a)^2+1/16*(11*A*b-3*B*a)/a^2/b/x^(3/2)/(b*
x^2+a)+7/64*(11*A*b-3*B*a)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(15/4)/b^(1/4)*2^(1/2)-7/64*(11*A*b-3*B
*a)*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/a^(15/4)/b^(1/4)*2^(1/2)+7/128*(11*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^(15/4)/b^(1/4)*2^(1/2)-7/128*(11*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^(15/4)/b^(1/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {468, 296, 331, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=\frac {7 (11 A b-3 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{15/4} \sqrt [4]{b}}-\frac {7 (11 A b-3 a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{15/4} \sqrt [4]{b}}+\frac {7 (11 A b-3 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4} \sqrt [4]{b}}-\frac {7 (11 A b-3 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{15/4} \sqrt [4]{b}}-\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2} \]

[In]

Int[(A + B*x^2)/(x^(5/2)*(a + b*x^2)^3),x]

[Out]

(-7*(11*A*b - 3*a*B))/(48*a^3*b*x^(3/2)) + (A*b - a*B)/(4*a*b*x^(3/2)*(a + b*x^2)^2) + (11*A*b - 3*a*B)/(16*a^
2*b*x^(3/2)*(a + b*x^2)) + (7*(11*A*b - 3*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(1
5/4)*b^(1/4)) - (7*(11*A*b - 3*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(15/4)*b^(1/4
)) + (7*(11*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^(15/4)*b^(1
/4)) - (7*(11*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^(15/4)*b^
(1/4))

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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

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 331

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

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}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {\left (\frac {11 A b}{2}-\frac {3 a B}{2}\right ) \int \frac {1}{x^{5/2} \left (a+b x^2\right )^2} \, dx}{4 a b} \\ & = \frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}+\frac {(7 (11 A b-3 a B)) \int \frac {1}{x^{5/2} \left (a+b x^2\right )} \, dx}{32 a^2 b} \\ & = -\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}-\frac {(7 (11 A b-3 a B)) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{32 a^3} \\ & = -\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}-\frac {(7 (11 A b-3 a B)) \text {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a^3} \\ & = -\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}-\frac {(7 (11 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^{7/2}}-\frac {(7 (11 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^{7/2}} \\ & = -\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}-\frac {(7 (11 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^{7/2} \sqrt {b}}-\frac {(7 (11 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^{7/2} \sqrt {b}}+\frac {(7 (11 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^{15/4} \sqrt [4]{b}}+\frac {(7 (11 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^{15/4} \sqrt [4]{b}} \\ & = -\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}+\frac {7 (11 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^{15/4} \sqrt [4]{b}}-\frac {7 (11 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^{15/4} \sqrt [4]{b}}-\frac {(7 (11 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^{15/4} \sqrt [4]{b}}+\frac {(7 (11 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^{15/4} \sqrt [4]{b}} \\ & = -\frac {7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac {A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2}+\frac {11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}+\frac {7 (11 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^{15/4} \sqrt [4]{b}}-\frac {7 (11 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^{15/4} \sqrt [4]{b}}+\frac {7 (11 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^{15/4} \sqrt [4]{b}}-\frac {7 (11 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^{15/4} \sqrt [4]{b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.58 \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=\frac {-\frac {4 a^{3/4} \left (77 A b^2 x^4+a^2 \left (32 A-33 B x^2\right )+a b x^2 \left (121 A-21 B x^2\right )\right )}{x^{3/2} \left (a+b x^2\right )^2}+\frac {21 \sqrt {2} (11 A b-3 a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt [4]{b}}+\frac {21 \sqrt {2} (-11 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt [4]{b}}}{192 a^{15/4}} \]

[In]

Integrate[(A + B*x^2)/(x^(5/2)*(a + b*x^2)^3),x]

[Out]

((-4*a^(3/4)*(77*A*b^2*x^4 + a^2*(32*A - 33*B*x^2) + a*b*x^2*(121*A - 21*B*x^2)))/(x^(3/2)*(a + b*x^2)^2) + (2
1*Sqrt[2]*(11*A*b - 3*a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/b^(1/4) + (21*Sqrt
[2]*(-11*A*b + 3*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/b^(1/4))/(192*a^(15/4)
)

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.54

method	result	size
derivativedivides	\(-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 \left (\frac {\left (\frac {15}{32} b^{2} A -\frac {7}{32} a b B \right ) x^{\frac {5}{2}}+\frac {a \left (19 A b -11 B a \right ) \sqrt {x}}{32}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (11 A b -3 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a}\right )}{a^{3}}\)	\(173\)
default	\(-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 \left (\frac {\left (\frac {15}{32} b^{2} A -\frac {7}{32} a b B \right ) x^{\frac {5}{2}}+\frac {a \left (19 A b -11 B a \right ) \sqrt {x}}{32}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (11 A b -3 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{256 a}\right )}{a^{3}}\)	\(173\)
risch	\(-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {\frac {2 \left (\frac {15}{32} b^{2} A -\frac {7}{32} a b B \right ) x^{\frac {5}{2}}+\frac {a \left (19 A b -11 B a \right ) \sqrt {x}}{16}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (11 A b -3 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{128 a}}{a^{3}}\)	\(174\)

[In]

int((B*x^2+A)/x^(5/2)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

-2/3*A/a^3/x^(3/2)-2/a^3*(((15/32*b^2*A-7/32*a*b*B)*x^(5/2)+1/32*a*(19*A*b-11*B*a)*x^(1/2))/(b*x^2+a)^2+7/256*
(11*A*b-3*B*a)*(a/b)^(1/4)/a*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)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 770, normalized size of antiderivative = 2.39 \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=-\frac {21 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} \log \left (7 \, a^{4} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} - 7 \, {\left (3 \, B a - 11 \, A b\right )} \sqrt {x}\right ) + 21 \, {\left (i \, a^{3} b^{2} x^{6} + 2 i \, a^{4} b x^{4} + i \, a^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} \log \left (7 i \, a^{4} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} - 7 \, {\left (3 \, B a - 11 \, A b\right )} \sqrt {x}\right ) + 21 \, {\left (-i \, a^{3} b^{2} x^{6} - 2 i \, a^{4} b x^{4} - i \, a^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} \log \left (-7 i \, a^{4} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} - 7 \, {\left (3 \, B a - 11 \, A b\right )} \sqrt {x}\right ) - 21 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} \log \left (-7 \, a^{4} \left (-\frac {81 \, B^{4} a^{4} - 1188 \, A B^{3} a^{3} b + 6534 \, A^{2} B^{2} a^{2} b^{2} - 15972 \, A^{3} B a b^{3} + 14641 \, A^{4} b^{4}}{a^{15} b}\right )^{\frac {1}{4}} - 7 \, {\left (3 \, B a - 11 \, A b\right )} \sqrt {x}\right ) - 4 \, {\left (7 \, {\left (3 \, B a b - 11 \, A b^{2}\right )} x^{4} - 32 \, A a^{2} + 11 \, {\left (3 \, B a^{2} - 11 \, A a b\right )} x^{2}\right )} \sqrt {x}}{192 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} \]

[In]

integrate((B*x^2+A)/x^(5/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/192*(21*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(-(81*B^4*a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 159
72*A^3*B*a*b^3 + 14641*A^4*b^4)/(a^15*b))^(1/4)*log(7*a^4*(-(81*B^4*a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*
b^2 - 15972*A^3*B*a*b^3 + 14641*A^4*b^4)/(a^15*b))^(1/4) - 7*(3*B*a - 11*A*b)*sqrt(x)) + 21*(I*a^3*b^2*x^6 + 2
*I*a^4*b*x^4 + I*a^5*x^2)*(-(81*B^4*a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 15972*A^3*B*a*b^3 + 14641*
A^4*b^4)/(a^15*b))^(1/4)*log(7*I*a^4*(-(81*B^4*a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 15972*A^3*B*a*b
^3 + 14641*A^4*b^4)/(a^15*b))^(1/4) - 7*(3*B*a - 11*A*b)*sqrt(x)) + 21*(-I*a^3*b^2*x^6 - 2*I*a^4*b*x^4 - I*a^5
*x^2)*(-(81*B^4*a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 15972*A^3*B*a*b^3 + 14641*A^4*b^4)/(a^15*b))^(
1/4)*log(-7*I*a^4*(-(81*B^4*a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 15972*A^3*B*a*b^3 + 14641*A^4*b^4)
/(a^15*b))^(1/4) - 7*(3*B*a - 11*A*b)*sqrt(x)) - 21*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(-(81*B^4*a^4 - 1188
*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 15972*A^3*B*a*b^3 + 14641*A^4*b^4)/(a^15*b))^(1/4)*log(-7*a^4*(-(81*B^4*
a^4 - 1188*A*B^3*a^3*b + 6534*A^2*B^2*a^2*b^2 - 15972*A^3*B*a*b^3 + 14641*A^4*b^4)/(a^15*b))^(1/4) - 7*(3*B*a
- 11*A*b)*sqrt(x)) - 4*(7*(3*B*a*b - 11*A*b^2)*x^4 - 32*A*a^2 + 11*(3*B*a^2 - 11*A*a*b)*x^2)*sqrt(x))/(a^3*b^2
*x^6 + 2*a^4*b*x^4 + a^5*x^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((B*x**2+A)/x**(5/2)/(b*x**2+a)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=\frac {7 \, {\left (3 \, B a b - 11 \, A b^{2}\right )} x^{4} - 32 \, A a^{2} + 11 \, {\left (3 \, B a^{2} - 11 \, A a b\right )} x^{2}}{48 \, {\left (a^{3} b^{2} x^{\frac {11}{2}} + 2 \, a^{4} b x^{\frac {7}{2}} + a^{5} x^{\frac {3}{2}}\right )}} + \frac {7 \, {\left (\frac {2 \, \sqrt {2} {\left (3 \, B a - 11 \, A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (3 \, B a - 11 \, A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (3 \, B a - 11 \, A b\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, B a - 11 \, A b\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )}}{128 \, a^{3}} \]

[In]

integrate((B*x^2+A)/x^(5/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/48*(7*(3*B*a*b - 11*A*b^2)*x^4 - 32*A*a^2 + 11*(3*B*a^2 - 11*A*a*b)*x^2)/(a^3*b^2*x^(11/2) + 2*a^4*b*x^(7/2)
+ a^5*x^(3/2)) + 7/128*(2*sqrt(2)*(3*B*a - 11*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sq
rt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(3*B*a - 11*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))) + sqrt(2
)*(3*B*a - 11*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)*(3*B
*a - 11*A*b)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/a^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=\frac {7 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{64 \, a^{4} b} + \frac {7 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{64 \, a^{4} b} + \frac {7 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{4} b} - \frac {7 \, \sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{128 \, a^{4} b} - \frac {2 \, A}{3 \, a^{3} x^{\frac {3}{2}}} + \frac {7 \, B a b x^{\frac {5}{2}} - 15 \, A b^{2} x^{\frac {5}{2}} + 11 \, B a^{2} \sqrt {x} - 19 \, A a b \sqrt {x}}{16 \, {\left (b x^{2} + a\right )}^{2} a^{3}} \]

[In]

integrate((B*x^2+A)/x^(5/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

7/64*sqrt(2)*(3*(a*b^3)^(1/4)*B*a - 11*(a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))
/(a/b)^(1/4))/(a^4*b) + 7/64*sqrt(2)*(3*(a*b^3)^(1/4)*B*a - 11*(a*b^3)^(1/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)
*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^4*b) + 7/128*sqrt(2)*(3*(a*b^3)^(1/4)*B*a - 11*(a*b^3)^(1/4)*A*b)*lo
g(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b) - 7/128*sqrt(2)*(3*(a*b^3)^(1/4)*B*a - 11*(a*b^3)^(1/4)
*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b) - 2/3*A/(a^3*x^(3/2)) + 1/16*(7*B*a*b*x^(5/2)
- 15*A*b^2*x^(5/2) + 11*B*a^2*sqrt(x) - 19*A*a*b*sqrt(x))/((b*x^2 + a)^2*a^3)

Mupad [B] (verification not implemented)

Time = 5.52 (sec) , antiderivative size = 888, normalized size of antiderivative = 2.76 \[ \int \frac {A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

int((A + B*x^2)/(x^(5/2)*(a + b*x^2)^3),x)

[Out]

- ((2*A)/(3*a) + (11*x^2*(11*A*b - 3*B*a))/(48*a^2) + (7*b*x^4*(11*A*b - 3*B*a))/(48*a^3))/(a^2*x^(3/2) + b^2*
x^(11/2) + 2*a*b*x^(7/2)) - (atan((((11*A*b - 3*B*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3 - 5
2985856*A*B*a^10*b^4) - (7*(11*A*b - 3*B*a)*(80740352*A*a^13*b^4 - 22020096*B*a^14*b^3))/(64*(-a)^(15/4)*b^(1/
4)))*7i)/(64*(-a)^(15/4)*b^(1/4)) + ((11*A*b - 3*B*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3 -
52985856*A*B*a^10*b^4) + (7*(11*A*b - 3*B*a)*(80740352*A*a^13*b^4 - 22020096*B*a^14*b^3))/(64*(-a)^(15/4)*b^(1
/4)))*7i)/(64*(-a)^(15/4)*b^(1/4)))/((7*(11*A*b - 3*B*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3
- 52985856*A*B*a^10*b^4) - (7*(11*A*b - 3*B*a)*(80740352*A*a^13*b^4 - 22020096*B*a^14*b^3))/(64*(-a)^(15/4)*b
^(1/4))))/(64*(-a)^(15/4)*b^(1/4)) - (7*(11*A*b - 3*B*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3
- 52985856*A*B*a^10*b^4) + (7*(11*A*b - 3*B*a)*(80740352*A*a^13*b^4 - 22020096*B*a^14*b^3))/(64*(-a)^(15/4)*b
^(1/4))))/(64*(-a)^(15/4)*b^(1/4))))*(11*A*b - 3*B*a)*7i)/(32*(-a)^(15/4)*b^(1/4)) - (7*atan(((7*(11*A*b - 3*B
*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3 - 52985856*A*B*a^10*b^4) - ((11*A*b - 3*B*a)*(807403
52*A*a^13*b^4 - 22020096*B*a^14*b^3)*7i)/(64*(-a)^(15/4)*b^(1/4))))/(64*(-a)^(15/4)*b^(1/4)) + (7*(11*A*b - 3*
B*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3 - 52985856*A*B*a^10*b^4) + ((11*A*b - 3*B*a)*(80740
352*A*a^13*b^4 - 22020096*B*a^14*b^3)*7i)/(64*(-a)^(15/4)*b^(1/4))))/(64*(-a)^(15/4)*b^(1/4)))/(((11*A*b - 3*B
*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3 - 52985856*A*B*a^10*b^4) - ((11*A*b - 3*B*a)*(807403
52*A*a^13*b^4 - 22020096*B*a^14*b^3)*7i)/(64*(-a)^(15/4)*b^(1/4)))*7i)/(64*(-a)^(15/4)*b^(1/4)) - ((11*A*b - 3
*B*a)*(x^(1/2)*(97140736*A^2*a^9*b^5 + 7225344*B^2*a^11*b^3 - 52985856*A*B*a^10*b^4) + ((11*A*b - 3*B*a)*(8074
0352*A*a^13*b^4 - 22020096*B*a^14*b^3)*7i)/(64*(-a)^(15/4)*b^(1/4)))*7i)/(64*(-a)^(15/4)*b^(1/4))))*(11*A*b -
3*B*a))/(32*(-a)^(15/4)*b^(1/4))