3.191 \(\int \frac{A+B x^2}{x^{3/2} (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=255 \[ -\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{\sqrt [4]{c} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}+\frac{\sqrt [4]{c} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}+\frac{\sqrt [4]{c} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{9/4}}-\frac{\sqrt [4]{c} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{9/4}}-\frac{2 A}{5 b x^{5/2}} \]

[Out]

(-2*A)/(5*b*x^(5/2)) - (2*(b*B - A*c))/(b^2*Sqrt[x]) + (c^(1/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x
])/b^(1/4)])/(Sqrt[2]*b^(9/4)) - (c^(1/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*
b^(9/4)) - (c^(1/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(9/4)
) + (c^(1/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(9/4))

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Rubi [A]  time = 0.214583, antiderivative size = 255, 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, 297, 1162, 617, 204, 1165, 628} \[ -\frac{2 (b B-A c)}{b^2 \sqrt{x}}-\frac{\sqrt [4]{c} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}+\frac{\sqrt [4]{c} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{9/4}}+\frac{\sqrt [4]{c} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{9/4}}-\frac{\sqrt [4]{c} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{9/4}}-\frac{2 A}{5 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^(3/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*A)/(5*b*x^(5/2)) - (2*(b*B - A*c))/(b^2*Sqrt[x]) + (c^(1/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x
])/b^(1/4)])/(Sqrt[2]*b^(9/4)) - (c^(1/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*
b^(9/4)) - (c^(1/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(9/4)
) + (c^(1/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(9/4))

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 297

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 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])

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]

Rubi steps

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

Mathematica [C]  time = 0.0200648, size = 46, normalized size = 0.18 \[ -\frac{2 \left (5 x^2 (b B-A c) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\frac{c x^2}{b}\right )+A b\right )}{5 b^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^(3/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*(A*b + 5*(b*B - A*c)*x^2*Hypergeometric2F1[-1/4, 1, 3/4, -((c*x^2)/b)]))/(5*b^2*x^(5/2))

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Maple [A]  time = 0.011, size = 299, normalized size = 1.2 \begin{align*} -{\frac{2\,A}{5\,b}{x}^{-{\frac{5}{2}}}}+2\,{\frac{Ac}{{b}^{2}\sqrt{x}}}-2\,{\frac{B}{b\sqrt{x}}}+{\frac{\sqrt{2}Ac}{2\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{\sqrt{2}Ac}{4\,{b}^{2}}\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{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{\sqrt{2}Ac}{2\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{\sqrt{2}B}{2\,b}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{\sqrt{2}B}{4\,b}\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{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{\sqrt{2}B}{2\,b}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^(3/2)/(c*x^4+b*x^2),x)

[Out]

-2/5*A/b/x^(5/2)+2/b^2/x^(1/2)*A*c-2/b/x^(1/2)*B+1/2/b^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1
/2)-1)*c+1/4/b^2/(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)))*c+1/2/b^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)*c-1/2/b/(b/c)^(1/4)
*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-1/4/b/(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/b/(b/c)^(1/4)*2^(1/2)*B*arctan(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^(3/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.2636, size = 1793, normalized size = 7.03 \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^(3/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/10*(20*b^2*x^3*(-(B^4*b^4*c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^9)^(1/4)*arc
tan((sqrt((B^6*b^6*c^2 - 6*A*B^5*b^5*c^3 + 15*A^2*B^4*b^4*c^4 - 20*A^3*B^3*b^3*c^5 + 15*A^4*B^2*b^2*c^6 - 6*A^
5*B*b*c^7 + A^6*c^8)*x - (B^4*b^9*c - 4*A*B^3*b^8*c^2 + 6*A^2*B^2*b^7*c^3 - 4*A^3*B*b^6*c^4 + A^4*b^5*c^5)*sqr
t(-(B^4*b^4*c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^9))*b^2*(-(B^4*b^4*c - 4*A*B^
3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^9)^(1/4) + (B^3*b^5*c - 3*A*B^2*b^4*c^2 + 3*A^2*B*b
^3*c^3 - A^3*b^2*c^4)*sqrt(x)*(-(B^4*b^4*c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^
9)^(1/4))/(B^4*b^4*c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)) - 5*b^2*x^3*(-(B^4*b^4*
c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^9)^(1/4)*log(b^7*(-(B^4*b^4*c - 4*A*B^3*b
^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^9)^(3/4) - (B^3*b^3*c - 3*A*B^2*b^2*c^2 + 3*A^2*B*b*c^
3 - A^3*c^4)*sqrt(x)) + 5*b^2*x^3*(-(B^4*b^4*c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5
)/b^9)^(1/4)*log(-b^7*(-(B^4*b^4*c - 4*A*B^3*b^3*c^2 + 6*A^2*B^2*b^2*c^3 - 4*A^3*B*b*c^4 + A^4*c^5)/b^9)^(3/4)
- (B^3*b^3*c - 3*A*B^2*b^2*c^2 + 3*A^2*B*b*c^3 - A^3*c^4)*sqrt(x)) + 4*(5*(B*b - A*c)*x^2 + A*b)*sqrt(x))/(b^
2*x^3)

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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**(3/2)/(c*x**4+b*x**2),x)

[Out]

Timed out

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Giac [A]  time = 1.17888, size = 362, normalized size = 1.42 \begin{align*} -\frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b - \left (b c^{3}\right )^{\frac{3}{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} c^{2}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b - \left (b c^{3}\right )^{\frac{3}{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} c^{2}} + \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b - \left (b c^{3}\right )^{\frac{3}{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} c^{2}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b - \left (b c^{3}\right )^{\frac{3}{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} c^{2}} - \frac{2 \,{\left (5 \, B b x^{2} - 5 \, A c x^{2} + A b\right )}}{5 \, b^{2} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(3/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c
)^(1/4))/(b^3*c^2) - 1/2*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1
/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^3*c^2) + 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^2) - 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*log(-sqrt(2)*
sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^2) - 2/5*(5*B*b*x^2 - 5*A*c*x^2 + A*b)/(b^2*x^(5/2))