3.192 \(\int \frac{A+B x^2}{x^{5/2} (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=257 \[ \frac{c^{3/4} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}+\frac{c^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{11/4}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}-\frac{2 A}{7 b x^{7/2}} \]

[Out]

(-2*A)/(7*b*x^(7/2)) - (2*(b*B - A*c))/(3*b^2*x^(3/2)) + (c^(3/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt
[x])/b^(1/4)])/(Sqrt[2]*b^(11/4)) - (c^(3/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[
2]*b^(11/4)) + (c^(3/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(
11/4)) - (c^(3/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(11/4))

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Rubi [A]  time = 0.213796, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {1584, 453, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{c^{3/4} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}+\frac{c^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{11/4}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}-\frac{2 A}{7 b x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^(5/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*A)/(7*b*x^(7/2)) - (2*(b*B - A*c))/(3*b^2*x^(3/2)) + (c^(3/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt
[x])/b^(1/4)])/(Sqrt[2]*b^(11/4)) - (c^(3/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[
2]*b^(11/4)) + (c^(3/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(
11/4)) - (c^(3/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(11/4))

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 453

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

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 (b x^2+c x^4\right )} \, dx &=\int \frac{A+B x^2}{x^{9/2} \left (b+c x^2\right )} \, dx\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{\left (2 \left (-\frac{7 b B}{2}+\frac{7 A c}{2}\right )\right ) \int \frac{1}{x^{5/2} \left (b+c x^2\right )} \, dx}{7 b}\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}-\frac{(c (b B-A c)) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{b^2}\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}-\frac{(2 c (b B-A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{b^2}\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}-\frac{(c (b B-A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{b^{5/2}}-\frac{(c (b B-A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{b^{5/2}}\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}-\frac{\left (\sqrt{c} (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{2 b^{5/2}}-\frac{\left (\sqrt{c} (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{2 b^{5/2}}+\frac{\left (c^{3/4} (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} b^{11/4}}+\frac{\left (c^{3/4} (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{2 \sqrt{2} b^{11/4}}\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}+\frac{c^{3/4} (b B-A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}-\frac{\left (c^{3/4} (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{11/4}}+\frac{\left (c^{3/4} (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{11/4}}\\ &=-\frac{2 A}{7 b x^{7/2}}-\frac{2 (b B-A c)}{3 b^2 x^{3/2}}+\frac{c^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{11/4}}+\frac{c^{3/4} (b B-A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}-\frac{c^{3/4} (b B-A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{2 \sqrt{2} b^{11/4}}\\ \end{align*}

Mathematica [C]  time = 0.0178511, size = 47, normalized size = 0.18 \[ \frac{14 x^2 (A c-b B) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\frac{c x^2}{b}\right )-6 A b}{21 b^2 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^(5/2)*(b*x^2 + c*x^4)),x]

[Out]

(-6*A*b + 14*(-(b*B) + A*c)*x^2*Hypergeometric2F1[-3/4, 1, 1/4, -((c*x^2)/b)])/(21*b^2*x^(7/2))

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Maple [A]  time = 0.011, size = 308, normalized size = 1.2 \begin{align*} -{\frac{2\,A}{7\,b}{x}^{-{\frac{7}{2}}}}+{\frac{2\,Ac}{3\,{b}^{2}}{x}^{-{\frac{3}{2}}}}-{\frac{2\,B}{3\,b}{x}^{-{\frac{3}{2}}}}+{\frac{{c}^{2}\sqrt{2}A}{4\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }+{\frac{{c}^{2}\sqrt{2}A}{2\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{{c}^{2}\sqrt{2}A}{2\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{c\sqrt{2}B}{4\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }-{\frac{c\sqrt{2}B}{2\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{c\sqrt{2}B}{2\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^(5/2)/(c*x^4+b*x^2),x)

[Out]

-2/7*A/b/x^(7/2)+2/3/b^2/x^(3/2)*A*c-2/3/b/x^(3/2)*B+1/4*c^2/b^3*(b/c)^(1/4)*2^(1/2)*A*ln((x+(b/c)^(1/4)*x^(1/
2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))+1/2*c^2/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(
2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/2*c^2/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-1/4*c/b
^2*(b/c)^(1/4)*2^(1/2)*B*ln((x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(
1/2)))-1/2*c/b^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-1/2*c/b^2*(b/c)^(1/4)*2^(1/2)*B*a
rctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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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)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31124, size = 1453, normalized size = 5.65 \begin{align*} \frac{84 \, b^{2} x^{4} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{b^{6} \sqrt{-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}} +{\left (B^{2} b^{2} c^{2} - 2 \, A B b c^{3} + A^{2} c^{4}\right )} x} b^{8} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{3}{4}} +{\left (B b^{9} c - A b^{8} c^{2}\right )} \sqrt{x} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{3}{4}}}{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}\right ) + 21 \, b^{2} x^{4} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{1}{4}} \log \left (b^{3} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{1}{4}} -{\left (B b c - A c^{2}\right )} \sqrt{x}\right ) - 21 \, b^{2} x^{4} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{1}{4}} \log \left (-b^{3} \left (-\frac{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac{1}{4}} -{\left (B b c - A c^{2}\right )} \sqrt{x}\right ) - 4 \,{\left (7 \,{\left (B b - A c\right )} x^{2} + 3 \, A b\right )} \sqrt{x}}{42 \, b^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(5/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/42*(84*b^2*x^4*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(1/4)*a
rctan((sqrt(b^6*sqrt(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11) + (B
^2*b^2*c^2 - 2*A*B*b*c^3 + A^2*c^4)*x)*b^8*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^
6 + A^4*c^7)/b^11)^(3/4) + (B*b^9*c - A*b^8*c^2)*sqrt(x)*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5
- 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(3/4))/(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A
^4*c^7)) + 21*b^2*x^4*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(1
/4)*log(b^3*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(1/4) - (B*b
*c - A*c^2)*sqrt(x)) - 21*b^2*x^4*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c
^7)/b^11)^(1/4)*log(-b^3*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)
^(1/4) - (B*b*c - A*c^2)*sqrt(x)) - 4*(7*(B*b - A*c)*x^2 + 3*A*b)*sqrt(x))/(b^2*x^4)

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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)/(c*x**4+b*x**2),x)

[Out]

Timed out

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Giac [A]  time = 1.18617, size = 347, normalized size = 1.35 \begin{align*} -\frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{2 \, b^{3}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{2 \, b^{3}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{4 \, b^{3}} + \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{4 \, b^{3}} - \frac{2 \,{\left (7 \, B b x^{2} - 7 \, A c x^{2} + 3 \, A b\right )}}{21 \, b^{2} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(5/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c
)^(1/4))/b^3 - 1/2*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) -
2*sqrt(x))/(b/c)^(1/4))/b^3 - 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1
/4) + x + sqrt(b/c))/b^3 + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4
) + x + sqrt(b/c))/b^3 - 2/21*(7*B*b*x^2 - 7*A*c*x^2 + 3*A*b)/(b^2*x^(7/2))