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

Optimal. Leaf size=237 \[ -\frac{(b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{7/4}}+\frac{(b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{7/4}}+\frac{(b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} \sqrt [4]{b} c^{7/4}}-\frac{(b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} \sqrt [4]{b} c^{7/4}}+\frac{2 B x^{3/2}}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.188252, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {1584, 459, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{(b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{7/4}}+\frac{(b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{7/4}}+\frac{(b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} \sqrt [4]{b} c^{7/4}}-\frac{(b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} \sqrt [4]{b} c^{7/4}}+\frac{2 B x^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 459

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

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

Mathematica [A]  time = 0.0748007, size = 95, normalized size = 0.4 \[ \frac{(3 A c-3 b B) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )+3 (b B-A c) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )+2 \sqrt [4]{-b} B c^{3/4} x^{3/2}}{3 \sqrt [4]{-b} c^{7/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 280, normalized size = 1.2 \begin{align*}{\frac{2\,B}{3\,c}{x}^{{\frac{3}{2}}}}+{\frac{\sqrt{2}A}{2\,c}\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}A}{4\,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{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{\sqrt{2}A}{2\,c}\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}Bb}{2\,{c}^{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}Bb}{4\,{c}^{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}Bb}{2\,{c}^{2}}\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(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.335, size = 1698, normalized size = 7.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*B*x^(3/2) - 12*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(1/
4)*arctan((sqrt((B^6*b^6 - 6*A*B^5*b^5*c + 15*A^2*B^4*b^4*c^2 - 20*A^3*B^3*b^3*c^3 + 15*A^4*B^2*b^2*c^4 - 6*A^
5*B*b*c^5 + A^6*c^6)*x - (B^4*b^5*c^3 - 4*A*B^3*b^4*c^4 + 6*A^2*B^2*b^3*c^5 - 4*A^3*B*b^2*c^6 + A^4*b*c^7)*sqr
t(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7)))*c^2*(-(B^4*b^4 - 4*A*B^3*
b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(1/4) + (B^3*b^3*c^2 - 3*A*B^2*b^2*c^3 + 3*A^2*B
*b*c^4 - A^3*c^5)*sqrt(x)*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(
1/4))/(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)) + 3*c*(-(B^4*b^4 - 4*A*B^3*b^3*
c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(1/4)*log(b*c^5*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B
^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(3/4) - (B^3*b^3 - 3*A*B^2*b^2*c + 3*A^2*B*b*c^2 - A^3*c^3)*sqr
t(x)) - 3*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(1/4)*log(-b*c^
5*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b*c^7))^(3/4) - (B^3*b^3 - 3*A*B^
2*b^2*c + 3*A^2*B*b*c^2 - A^3*c^3)*sqrt(x)))/c

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

[Out]

Timed out

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Giac [A]  time = 1.19203, size = 339, normalized size = 1.43 \begin{align*} \frac{2 \, B x^{\frac{3}{2}}}{3 \, c} - \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 c^{4}} - \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 c^{4}} + \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 c^{4}} - \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 c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*B*x^(3/2)/c - 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*c^4) - 1/2*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(s
qrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b*c^4) + 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*c^4) - 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*lo
g(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^4)