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

Optimal. Leaf size=322 \[ \frac{11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}-\frac{7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac{7 (11 A b-3 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4} \sqrt [4]{b}}-\frac{7 (11 A b-3 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4} \sqrt [4]{b}}+\frac{7 (11 A b-3 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{15/4} \sqrt [4]{b}}-\frac{7 (11 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{15/4} \sqrt [4]{b}}+\frac{A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2} \]

[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))

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Rubi [A]  time = 0.236954, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {457, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{11 A b-3 a B}{16 a^2 b x^{3/2} \left (a+b x^2\right )}-\frac{7 (11 A b-3 a B)}{48 a^3 b x^{3/2}}+\frac{7 (11 A b-3 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4} \sqrt [4]{b}}-\frac{7 (11 A b-3 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4} \sqrt [4]{b}}+\frac{7 (11 A b-3 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{15/4} \sqrt [4]{b}}-\frac{7 (11 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{15/4} \sqrt [4]{b}}+\frac{A b-a B}{4 a b x^{3/2} \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[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 457

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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 329

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 211

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 1165

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]

Rule 628

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 1162

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 617

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^{5/2} \left (a+b x^2\right )^3} \, dx &=\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)) \operatorname{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)) \operatorname{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)) \operatorname{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)) \operatorname{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)) \operatorname{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)) \operatorname{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)) \operatorname{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)) \operatorname{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)) \operatorname{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]  time = 0.431408, size = 400, normalized size = 1.24 \[ \frac{-\frac{96 a^{7/4} A b \sqrt{x}}{\left (a+b x^2\right )^2}-\frac{360 a^{3/4} A b \sqrt{x}}{a+b x^2}-\frac{256 a^{3/4} A}{x^{3/2}}+\frac{96 a^{11/4} B \sqrt{x}}{\left (a+b x^2\right )^2}+\frac{168 a^{7/4} B \sqrt{x}}{a+b x^2}+231 \sqrt{2} A b^{3/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )-231 \sqrt{2} A b^{3/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )+\frac{42 \sqrt{2} (11 A b-3 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{\sqrt [4]{b}}-\frac{42 \sqrt{2} (11 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{\sqrt [4]{b}}-\frac{63 \sqrt{2} a B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{\sqrt [4]{b}}+\frac{63 \sqrt{2} a B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{\sqrt [4]{b}}}{384 a^{15/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-256*a^(3/4)*A)/x^(3/2) - (96*a^(7/4)*A*b*Sqrt[x])/(a + b*x^2)^2 + (96*a^(11/4)*B*Sqrt[x])/(a + b*x^2)^2 - (
360*a^(3/4)*A*b*Sqrt[x])/(a + b*x^2) + (168*a^(7/4)*B*Sqrt[x])/(a + b*x^2) + (42*Sqrt[2]*(11*A*b - 3*a*B)*ArcT
an[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/b^(1/4) - (42*Sqrt[2]*(11*A*b - 3*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*
Sqrt[x])/a^(1/4)])/b^(1/4) + 231*Sqrt[2]*A*b^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]
- (63*Sqrt[2]*a*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/b^(1/4) - 231*Sqrt[2]*A*b^(3/4)*
Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + (63*Sqrt[2]*a*B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(
1/4)*Sqrt[x] + Sqrt[b]*x])/b^(1/4))/(384*a^(15/4))

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Maple [A]  time = 0.017, size = 357, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{3\,{a}^{3}}{x}^{-{\frac{3}{2}}}}-{\frac{15\,A{b}^{2}}{16\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{7\,Bb}{16\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{19\,Ab}{16\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}\sqrt{x}}+{\frac{11\,B}{16\,a \left ( b{x}^{2}+a \right ) ^{2}}\sqrt{x}}-{\frac{77\,\sqrt{2}Ab}{64\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }-{\frac{77\,\sqrt{2}Ab}{64\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }-{\frac{77\,\sqrt{2}Ab}{128\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{21\,\sqrt{2}B}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{21\,\sqrt{2}B}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{21\,\sqrt{2}B}{128\,{a}^{3}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*A/a^3/x^(3/2)-15/16/a^3/(b*x^2+a)^2*x^(5/2)*A*b^2+7/16/a^2/(b*x^2+a)^2*x^(5/2)*B*b-19/16/a^2/(b*x^2+a)^2*
A*x^(1/2)*b+11/16/a/(b*x^2+a)^2*B*x^(1/2)-77/64/a^4*(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/
2)+1)*b-77/64/a^4*(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)*b-77/128/a^4*(1/b*a)^(1/4)*2
^(1/2)*A*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))*b
+21/64/a^3*(1/b*a)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+21/64/a^3*(1/b*a)^(1/4)*2^(1/2)*B*a
rctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)+21/128/a^3*(1/b*a)^(1/4)*2^(1/2)*B*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+
(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.11157, size = 1901, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(84*(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)*arctan((sqrt(a^8*sqrt(-(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)) + (9*B^2*a^2 - 66*A*B*a*b + 121*A^2*b^2)*x)*a^11
*b*(-(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))^(3/4
) + (3*B*a^12*b - 11*A*a^11*b^2)*sqrt(x)*(-(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))^(3/4))/(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)) + 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)) - 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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.18111, size = 410, normalized size = 1.27 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)