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

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

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Rubi [A]  time = 0.18, 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, 459, 329, 211, 1165, 628, 1162, 617, 204} \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^{3/4} c^{5/4}}-\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^{3/4} c^{5/4}}+\frac {(b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{3/4} c^{5/4}}-\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{3/4} c^{5/4}}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.15, size = 166, normalized size = 0.71 \begin {gather*} \frac {(b B-A c) \left (\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )\right )}{2 \sqrt {2} b^{3/4} c^{5/4}}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 134, 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^{3/4} c^{5/4}}-\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^{3/4} c^{5/4}}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 645, normalized size = 2.74 \begin {gather*} \frac {4 \, 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^{3} c^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{2} c^{2} \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^{3} c^{5}}} + {\left (B^{2} b^{2} - 2 \, A B b c + A^{2} c^{2}\right )} x} b^{2} c^{4} \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^{3} c^{5}}\right )^{\frac {3}{4}} + {\left (B b^{3} c^{4} - A b^{2} c^{5}\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^{3} c^{5}}\right )^{\frac {3}{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 ) + 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^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (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^{3} c^{5}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 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^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (-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^{3} c^{5}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) + 4 \, B \sqrt {x}}{2 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 251, normalized size = 1.07 \begin {gather*} \frac {2 \, B \sqrt {x}}{c} - \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 c^{2}} - \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 c^{2}} - \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 c^{2}} + \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 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 277, normalized size = 1.18 \begin {gather*} \frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 b}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 b}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \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 b}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \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 c}+\frac {2 B \sqrt {x}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.16, size = 218, normalized size = 0.93 \begin {gather*} \frac {2 \, B \sqrt {x}}{c} - \frac {\frac {2 \, \sqrt {2} {\left (B b - A c\right )} \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 {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (B b - A c\right )} \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 {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.29, size = 739, normalized size = 3.14 \begin {gather*} \frac {2\,B\,\sqrt {x}}{c}-\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )-\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}+\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )+\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}}{\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )-\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}-\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )+\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{{\left (-b\right )}^{3/4}\,c^{5/4}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )-\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}+\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )+\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}}{\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )-\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}-\frac {\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,c^3-32\,A\,B\,b\,c^2+16\,B^2\,b^2\,c\right )+\frac {\left (32\,B\,b^2\,c^2-32\,A\,b\,c^3\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}\right )\,1{}\mathrm {i}}{2\,{\left (-b\right )}^{3/4}\,c^{5/4}}}\right )\,\left (A\,c-B\,b\right )}{{\left (-b\right )}^{3/4}\,c^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*x^(1/2))/c - (atan((((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) - ((32*B*b^2*c^2 - 3
2*A*b*c^3)*(A*c - B*b))/(2*(-b)^(3/4)*c^(5/4)))*1i)/(2*(-b)^(3/4)*c^(5/4)) + ((A*c - B*b)*(x^(1/2)*(16*A^2*c^3
+ 16*B^2*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b))/(2*(-b)^(3/4)*c^(5/4)))*1i)/(2*(-b
)^(3/4)*c^(5/4)))/(((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) - ((32*B*b^2*c^2 - 32*A*b*
c^3)*(A*c - B*b))/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)) - ((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2
*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b))/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5
/4))))*(A*c - B*b)*1i)/((-b)^(3/4)*c^(5/4)) - (atan((((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B
*b*c^2) - ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b)*1i)/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)) + ((A*c
- B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b)*1i)/(2*
(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)))/(((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2
) - ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b)*1i)/(2*(-b)^(3/4)*c^(5/4)))*1i)/(2*(-b)^(3/4)*c^(5/4)) - ((A*c -
B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b)*1i)/(2*(-b
)^(3/4)*c^(5/4)))*1i)/(2*(-b)^(3/4)*c^(5/4))))*(A*c - B*b))/((-b)^(3/4)*c^(5/4))

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sympy [A]  time = 27.83, size = 355, normalized size = 1.51 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}}{c} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: c = 0 \\- \frac {\sqrt [4]{-1} A \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} A \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {3}{4}}} - \frac {\sqrt [4]{-1} A \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{b^{\frac {3}{4}}} + \frac {\sqrt [4]{-1} B \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} - \frac {\sqrt [4]{-1} B \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 c} + \frac {\sqrt [4]{-1} B \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{b} \sqrt [4]{\frac {1}{c}}} \right )}}{c} + \frac {2 B \sqrt {x}}{c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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