3.2.89 \(\int \frac {\sqrt {x} (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=235 \[ \frac {(b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{5/4} c^{3/4}}-\frac {(b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{5/4} c^{3/4}}-\frac {(b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{5/4} c^{3/4}}+\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{5/4} c^{3/4}}-\frac {2 A}{b \sqrt {x}} \]

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Rubi [A]  time = 0.19, antiderivative size = 235, normalized size of antiderivative = 1.00, 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, 453, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \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} b^{5/4} c^{3/4}}-\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} b^{5/4} c^{3/4}}-\frac {(b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{5/4} c^{3/4}}+\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{5/4} c^{3/4}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(b*Sqrt[x]) - ((b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(5/4)*c^(3/4)) + (
(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(5/4)*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^(5/4)*c^(3/4)) - ((b*B - A*c)*Log[Sqrt[b] + Sqr
t[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(5/4)*c^(3/4))

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

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Mathematica [A]  time = 0.10, size = 74, normalized size = 0.31 \begin {gather*} \frac {\frac {(b B-A c) \left (\tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+\tanh ^{-1}\left (\frac {b \sqrt [4]{c} \sqrt {x}}{(-b)^{5/4}}\right )\right )}{\sqrt [4]{-b} c^{3/4}}-\frac {2 A}{\sqrt {x}}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*A)/Sqrt[x] + ((b*B - A*c)*(ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)] + ArcTanh[(b*c^(1/4)*Sqrt[x])/(-b)^(5/4)]
))/((-b)^(1/4)*c^(3/4)))/b

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IntegrateAlgebraic [A]  time = 0.20, size = 135, normalized size = 0.57 \begin {gather*} -\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} b^{5/4} c^{3/4}}-\frac {(b B-A c) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{5/4} c^{3/4}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(b*Sqrt[x]) - ((b*B - A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*b^
(5/4)*c^(3/4)) - ((b*B - A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*b^(5/
4)*c^(3/4))

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fricas [B]  time = 0.45, size = 843, normalized size = 3.59 \begin {gather*} \frac {4 \, b x \left (-\frac {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^{5} c^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (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}\right )} x - {\left (B^{4} b^{7} c - 4 \, A B^{3} b^{6} c^{2} + 6 \, A^{2} B^{2} b^{5} c^{3} - 4 \, A^{3} B b^{4} c^{4} + A^{4} b^{3} c^{5}\right )} \sqrt {-\frac {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^{5} c^{3}}}} b c \left (-\frac {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^{5} c^{3}}\right )^{\frac {1}{4}} + {\left (B^{3} b^{4} c - 3 \, A B^{2} b^{3} c^{2} + 3 \, A^{2} B b^{2} c^{3} - A^{3} b c^{4}\right )} \sqrt {x} \left (-\frac {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^{5} c^{3}}\right )^{\frac {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}}\right ) - b x \left (-\frac {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^{5} c^{3}}\right )^{\frac {1}{4}} \log \left (b^{4} c^{2} \left (-\frac {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^{5} c^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{3} - 3 \, A B^{2} b^{2} c + 3 \, A^{2} B b c^{2} - A^{3} c^{3}\right )} \sqrt {x}\right ) + b x \left (-\frac {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^{5} c^{3}}\right )^{\frac {1}{4}} \log \left (-b^{4} c^{2} \left (-\frac {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^{5} c^{3}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{3} - 3 \, A B^{2} b^{2} c + 3 \, A^{2} B b c^{2} - A^{3} c^{3}\right )} \sqrt {x}\right ) - 4 \, A \sqrt {x}}{2 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/2*(4*b*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^5*c^3))^(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^7*c - 4*A*B^3*b^6*c^2 + 6*A^2*B^2*b^5*c^3 - 4*A^3*B*b^4*c^4 + A^4*b^3*c^5)*sqrt(-(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^5*c^3)))*b*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^5*c^3))^(1/4) + (B^3*b^4*c - 3*A*B^2*b^3*c^2 + 3*A^2*B*b^2*c^3
- A^3*b*c^4)*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^5*c^3))^(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)) - b*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^5*c^3))^(1/4)*log(b^4*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^5*c^3))^(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)) + b*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^5*c^3))^(1/4)*log(
-b^4*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^5*c^3))^(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)) - 4*A*sqrt(x))/(b*x)

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giac [A]  time = 0.19, size = 251, normalized size = 1.07 \begin {gather*} -\frac {2 \, A}{b \sqrt {x}} + \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^{2} c^{3}} + \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^{2} c^{3}} - \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^{2} c^{3}} + \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^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

-2*A/(b*sqrt(x)) + 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^2*c^3) + 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^2*c^3) - 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^2*c^3) + 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^2*c^3)

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maple [A]  time = 0.06, size = 277, normalized size = 1.18 \begin {gather*} -\frac {\sqrt {2}\, A \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}\, A \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}\, A \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 {\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}} c}+\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}} c}+\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}} c}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2),x)

[Out]

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

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maxima [A]  time = 3.01, size = 194, normalized size = 0.83 \begin {gather*} \frac {{\left (B b - A c\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} - \frac {2 \, A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/4*(B*b - A*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)) + 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) + sq
rt(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*A/(b*sqrt(x))

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mupad [B]  time = 0.22, size = 71, normalized size = 0.30 \begin {gather*} \frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )}{{\left (-b\right )}^{5/4}\,c^{3/4}}-\frac {2\,A}{b\,\sqrt {x}}-\frac {\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )}{{\left (-b\right )}^{5/4}\,c^{3/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

(atan((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c - B*b))/((-b)^(5/4)*c^(3/4)) - (2*A)/(b*x^(1/2)) - (atanh((c^(1/4)*x^
(1/2))/(-b)^(1/4))*(A*c - B*b))/((-b)^(5/4)*c^(3/4))

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sympy [A]  time = 18.45, size = 456, normalized size = 1.94 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{\sqrt {x}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{\sqrt {x}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + \frac {2 B x^{\frac {3}{2}}}{3}}{b} & \text {for}\: c = 0 \\- \frac {2 A}{b \sqrt {x}} + \frac {\left (-1\right )^{\frac {3}{4}} A c \left (\frac {1}{c}\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}}} - \frac {\left (-1\right )^{\frac {3}{4}} A c \left (\frac {1}{c}\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}}} + \frac {\left (-1\right )^{\frac {3}{4}} A c \left (\frac {1}{c}\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}}} - \frac {2 \left (-1\right )^{\frac {3}{4}} A \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}}} - \frac {\left (-1\right )^{\frac {3}{4}} B \left (\frac {1}{c}\right )^{\frac {3}{4}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 \sqrt [4]{b}} + \frac {\left (-1\right )^{\frac {3}{4}} B \left (\frac {1}{c}\right )^{\frac {3}{4}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 \sqrt [4]{b}} - \frac {\left (-1\right )^{\frac {3}{4}} B \left (\frac {1}{c}\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 )}}{\sqrt [4]{b}} + \frac {2 \left (-1\right )^{\frac {3}{4}} B \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{\sqrt [4]{b} c \sqrt [4]{\frac {1}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*x**(1/2)/(c*x**4+b*x**2),x)

[Out]

Piecewise((zoo*(-2*A/(5*x**(5/2)) - 2*B/sqrt(x)), Eq(b, 0) & Eq(c, 0)), ((-2*A/(5*x**(5/2)) - 2*B/sqrt(x))/c,
Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*x**(3/2)/3)/b, Eq(c, 0)), (-2*A/(b*sqrt(x)) + (-1)**(3/4)*A*c*(1/c)**(3/4)*log
(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(5/4)) - (-1)**(3/4)*A*c*(1/c)**(3/4)*log((-1)**(1/4)*b**
(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(5/4)) + (-1)**(3/4)*A*c*(1/c)**(3/4)*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(
1/c)**(1/4)))/b**(5/4) - 2*(-1)**(3/4)*A*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/(b**(5/4)*(1/c)**(1
/4)) - (-1)**(3/4)*B*(1/c)**(3/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(1/4)) + (-1)**(3/4)
*B*(1/c)**(3/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(1/4)) - (-1)**(3/4)*B*(1/c)**(3/4)*ata
n((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/b**(1/4) + 2*(-1)**(3/4)*B*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(
1/c)**(1/4)))/(b**(1/4)*c*(1/c)**(1/4)), True))

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