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

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

[Out]

-2/5*A/b/x^(5/2)+1/2*c^(1/4)*(-A*c+B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)*2^(1/2)-1/2*c^(1/4)*
(-A*c+B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(9/4)*2^(1/2)-1/4*c^(1/4)*(-A*c+B*b)*ln(b^(1/2)+x*c^(1/
2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(9/4)*2^(1/2)+1/4*c^(1/4)*(-A*c+B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4
)*2^(1/2)*x^(1/2))/b^(9/4)*2^(1/2)-2*(-A*c+B*b)/b^2/x^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 255, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.02, 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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fricas [B]  time = 0.91, size = 883, normalized size = 3.46 \[ -\frac {20 \, b^{2} x^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (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}\right )} x - {\left (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}\right )} \sqrt {-\frac {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} \left (-\frac {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}}\right )^{\frac {1}{4}} + {\left (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}\right )} \sqrt {x} \left (-\frac {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}}\right )^{\frac {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}}\right ) - 5 \, b^{2} x^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (b^{7} \left (-\frac {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}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{3} c - 3 \, A B^{2} b^{2} c^{2} + 3 \, A^{2} B b c^{3} - A^{3} c^{4}\right )} \sqrt {x}\right ) + 5 \, b^{2} x^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (-b^{7} \left (-\frac {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}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{3} c - 3 \, A B^{2} b^{2} c^{2} + 3 \, A^{2} B b c^{3} - A^{3} c^{4}\right )} \sqrt {x}\right ) + 4 \, {\left (5 \, {\left (B b - A c\right )} x^{2} + A b\right )} \sqrt {x}}{10 \, b^{2} x^{3}} \]

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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giac [A]  time = 0.24, size = 268, normalized size = 1.05 \[ -\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3} c^{2}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3} c^{2}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3} c^{2}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3} c^{2}} - \frac {2 \, {\left (5 \, B b x^{2} - 5 \, A c x^{2} + A b\right )}}{5 \, b^{2} x^{\frac {5}{2}}} \]

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))

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maple [A]  time = 0.06, size = 299, normalized size = 1.17 \[ \frac {\sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}+\frac {\sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}+\frac {\sqrt {2}\, A c \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}-\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b}-\frac {\sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b}-\frac {\sqrt {2}\, B \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} b}+\frac {2 A c}{b^{2} \sqrt {x}}-\frac {2 B}{b \sqrt {x}}-\frac {2 A}{5 b \,x^{\frac {5}{2}}} \]

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]

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

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maxima [A]  time = 3.08, size = 213, normalized size = 0.84 \[ -\frac {{\left (B b c - A c^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, b^{2}} - \frac {2 \, {\left (5 \, {\left (B b - A c\right )} x^{2} + A b\right )}}{5 \, b^{2} x^{\frac {5}{2}}} \]

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]

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

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mupad [B]  time = 0.23, size = 90, normalized size = 0.35 \[ \frac {{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (A\,c-B\,b\right )}{b^{9/4}}-\frac {\frac {2\,A}{5\,b}-\frac {2\,x^2\,\left (A\,c-B\,b\right )}{b^2}}{x^{5/2}}-\frac {{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (A\,c-B\,b\right )}{b^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-c)^(1/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4))*(A*c - B*b))/b^(9/4) - ((2*A)/(5*b) - (2*x^2*(A*c - B*b))/b^2)/
x^(5/2) - ((-c)^(1/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4))*(A*c - B*b))/b^(9/4)

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sympy [A]  time = 108.60, size = 366, normalized size = 1.44 \[ A \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{9 c x^{\frac {9}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} + \frac {2 c}{b^{2} \sqrt {x}} - \frac {\left (-1\right )^{\frac {3}{4}} c \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {9}{4}} \sqrt [4]{\frac {1}{c}}} + \frac {\left (-1\right )^{\frac {3}{4}} c \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {9}{4}} \sqrt [4]{\frac {1}{c}}} + \frac {\left (-1\right )^{\frac {3}{4}} c \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{b^{\frac {9}{4}} \sqrt [4]{\frac {1}{c}}} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{5 c x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{b \sqrt {x}} & \text {for}\: c = 0 \\- \frac {2}{b \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} - \frac {\left (-1\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{b^{\frac {5}{4}} \sqrt [4]{\frac {1}{c}}} & \text {otherwise} \end {cases}\right ) \]

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]

A*Piecewise((zoo/x**(9/2), Eq(b, 0) & Eq(c, 0)), (-2/(9*c*x**(9/2)), Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(c, 0)),
(-2/(5*b*x**(5/2)) + 2*c/(b**2*sqrt(x)) - (-1)**(3/4)*c*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*
b**(9/4)*(1/c)**(1/4)) + (-1)**(3/4)*c*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(9/4)*(1/c)**(1/
4)) + (-1)**(3/4)*c*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/(b**(9/4)*(1/c)**(1/4)), True)) + B*Piec
ewise((zoo/x**(5/2), Eq(b, 0) & Eq(c, 0)), (-2/(5*c*x**(5/2)), Eq(b, 0)), (-2/(b*sqrt(x)), Eq(c, 0)), (-2/(b*s
qrt(x)) + (-1)**(3/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(5/4)*(1/c)**(1/4)) - (-1)**(3/4
)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(5/4)*(1/c)**(1/4)) - (-1)**(3/4)*atan((-1)**(3/4)*sq
rt(x)/(b**(1/4)*(1/c)**(1/4)))/(b**(5/4)*(1/c)**(1/4)), True))

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