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

Optimal. Leaf size=257 \[ \frac {c^{3/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}-\frac {c^{3/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {c^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}-\frac {c^{3/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{11/4}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}-\frac {2 A}{7 b x^{7/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 257, 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, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {c^{3/4} (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^{11/4}}-\frac {c^{3/4} (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^{11/4}}+\frac {c^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}-\frac {c^{3/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{11/4}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}-\frac {2 A}{7 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(7*b*x^(7/2)) - (2*(b*B - A*c))/(3*b^2*x^(3/2)) + (c^(3/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt
[x])/b^(1/4)])/(Sqrt[2]*b^(11/4)) - (c^(3/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[
2]*b^(11/4)) + (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^(
11/4)) - (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^(11/4))

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 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^{5/2} \left (b x^2+c x^4\right )} \, dx &=\int \frac {A+B x^2}{x^{9/2} \left (b+c x^2\right )} \, dx\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {\left (2 \left (-\frac {7 b B}{2}+\frac {7 A c}{2}\right )\right ) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{7 b}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}-\frac {(c (b B-A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{b^2}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}-\frac {(2 c (b B-A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}-\frac {(c (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^{5/2}}-\frac {(c (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^{5/2}}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}-\frac {\left (\sqrt {c} (b B-A c)\right ) \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^{5/2}}-\frac {\left (\sqrt {c} (b B-A c)\right ) \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^{5/2}}+\frac {\left (c^{3/4} (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^{11/4}}+\frac {\left (c^{3/4} (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^{11/4}}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}+\frac {c^{3/4} (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^{11/4}}-\frac {c^{3/4} (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^{11/4}}-\frac {\left (c^{3/4} (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^{11/4}}+\frac {\left (c^{3/4} (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^{11/4}}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{3 b^2 x^{3/2}}+\frac {c^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}-\frac {c^{3/4} (b B-A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{11/4}}+\frac {c^{3/4} (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^{11/4}}-\frac {c^{3/4} (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^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 47, normalized size = 0.18 \begin {gather*} \frac {14 x^2 (A c-b B) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\frac {c x^2}{b}\right )-6 A b}{21 b^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*A*b + 14*(-(b*B) + A*c)*x^2*Hypergeometric2F1[-3/4, 1, 1/4, -((c*x^2)/b)])/(21*b^2*x^(7/2))

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IntegrateAlgebraic [A]  time = 0.23, size = 162, normalized size = 0.63 \begin {gather*} \frac {\left (b B c^{3/4}-A c^{7/4}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} b^{11/4}}-\frac {\left (b B c^{3/4}-A c^{7/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{11/4}}-\frac {2 \left (3 A b-7 A c x^2+7 b B x^2\right )}{21 b^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*A*b + 7*b*B*x^2 - 7*A*c*x^2))/(21*b^2*x^(7/2)) + ((b*B*c^(3/4) - A*c^(7/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x
)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*b^(11/4)) - ((b*B*c^(3/4) - A*c^(7/4))*ArcTanh[(Sqrt[2]*b^(1/4)
*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*b^(11/4))

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fricas [B]  time = 0.43, size = 707, normalized size = 2.75 \begin {gather*} \frac {84 \, b^{2} x^{4} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{6} \sqrt {-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}} + {\left (B^{2} b^{2} c^{2} - 2 \, A B b c^{3} + A^{2} c^{4}\right )} x} b^{8} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {3}{4}} + {\left (B b^{9} c - A b^{8} c^{2}\right )} \sqrt {x} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {3}{4}}}{B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}\right ) + 21 \, b^{2} x^{4} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {1}{4}} \log \left (b^{3} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {1}{4}} - {\left (B b c - A c^{2}\right )} \sqrt {x}\right ) - 21 \, b^{2} x^{4} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-b^{3} \left (-\frac {B^{4} b^{4} c^{3} - 4 \, A B^{3} b^{3} c^{4} + 6 \, A^{2} B^{2} b^{2} c^{5} - 4 \, A^{3} B b c^{6} + A^{4} c^{7}}{b^{11}}\right )^{\frac {1}{4}} - {\left (B b c - A c^{2}\right )} \sqrt {x}\right ) - 4 \, {\left (7 \, {\left (B b - A c\right )} x^{2} + 3 \, A b\right )} \sqrt {x}}{42 \, b^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*(84*b^2*x^4*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(1/4)*a
rctan((sqrt(b^6*sqrt(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11) + (B
^2*b^2*c^2 - 2*A*B*b*c^3 + A^2*c^4)*x)*b^8*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^
6 + A^4*c^7)/b^11)^(3/4) + (B*b^9*c - A*b^8*c^2)*sqrt(x)*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5
- 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(3/4))/(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A
^4*c^7)) + 21*b^2*x^4*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(1
/4)*log(b^3*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)^(1/4) - (B*b
*c - A*c^2)*sqrt(x)) - 21*b^2*x^4*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c
^7)/b^11)^(1/4)*log(-b^3*(-(B^4*b^4*c^3 - 4*A*B^3*b^3*c^4 + 6*A^2*B^2*b^2*c^5 - 4*A^3*B*b*c^6 + A^4*c^7)/b^11)
^(1/4) - (B*b*c - A*c^2)*sqrt(x)) - 4*(7*(B*b - A*c)*x^2 + 3*A*b)*sqrt(x))/(b^2*x^4)

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giac [A]  time = 0.19, size = 257, normalized size = 1.00 \begin {gather*} -\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^{3}} - \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^{3}} - \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^{3}} + \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^{3}} - \frac {2 \, {\left (7 \, B b x^{2} - 7 \, A c x^{2} + 3 \, A b\right )}}{21 \, b^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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^3 - 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^3 - 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^3 + 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^3 - 2/21*(7*B*b*x^2 - 7*A*c*x^2 + 3*A*b)/(b^2*x^(7/2))

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maple [A]  time = 0.06, size = 308, normalized size = 1.20 \begin {gather*} \frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 b^{3}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 b^{3}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \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^{3}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 b^{2}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 b^{2}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B 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 b^{2}}+\frac {2 A c}{3 b^{2} x^{\frac {3}{2}}}-\frac {2 B}{3 b \,x^{\frac {3}{2}}}-\frac {2 A}{7 b \,x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c^2/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+1/4*c^2/b^3*(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*c^2/b^3*(b/c)^(1/4
)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-1/2*c/b^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x
^(1/2)-1)-1/4*c/b^2*(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^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-2/7*A/b/x^(7/2)+
2/3/b^2/x^(3/2)*A*c-2/3/b/x^(3/2)*B

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maxima [A]  time = 2.93, size = 247, normalized size = 0.96 \begin {gather*} -\frac {\frac {2 \, \sqrt {2} {\left (B b c - A c^{2}\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 c - A c^{2}\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 c - A c^{2}\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 c - A c^{2}\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 \, b^{2}} - \frac {2 \, {\left (7 \, {\left (B b - A c\right )} x^{2} + 3 \, A b\right )}}{21 \, b^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*(B*b*c - A*c^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(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(B*b*c - A*c^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(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(B*b*c - A*c^2)*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*c - A*c^2)*log(-sqrt(
2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^2 - 2/21*(7*(B*b - A*c)*x^2 + 3*A*b)/(b
^2*x^(7/2))

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mupad [B]  time = 0.32, size = 555, normalized size = 2.16 \begin {gather*} -\frac {\frac {2\,A}{7\,b}-\frac {2\,x^2\,\left (A\,c-B\,b\right )}{3\,b^2}}{x^{7/2}}+\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^6\,c^7-32\,A\,B\,b^7\,c^6+16\,B^2\,b^8\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^9\,c^5-32\,B\,b^{10}\,c^4\right )\,1{}\mathrm {i}}{2\,b^{11/4}}\right )}{2\,b^{11/4}}+\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^6\,c^7-32\,A\,B\,b^7\,c^6+16\,B^2\,b^8\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^9\,c^5-32\,B\,b^{10}\,c^4\right )\,1{}\mathrm {i}}{2\,b^{11/4}}\right )}{2\,b^{11/4}}}{\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^6\,c^7-32\,A\,B\,b^7\,c^6+16\,B^2\,b^8\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^9\,c^5-32\,B\,b^{10}\,c^4\right )\,1{}\mathrm {i}}{2\,b^{11/4}}\right )\,1{}\mathrm {i}}{2\,b^{11/4}}-\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^6\,c^7-32\,A\,B\,b^7\,c^6+16\,B^2\,b^8\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^9\,c^5-32\,B\,b^{10}\,c^4\right )\,1{}\mathrm {i}}{2\,b^{11/4}}\right )\,1{}\mathrm {i}}{2\,b^{11/4}}}\right )\,\left (A\,c-B\,b\right )}{b^{11/4}}-\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {A^3\,c^8\,\sqrt {x}\,1{}\mathrm {i}-B^3\,b^3\,c^5\,\sqrt {x}\,1{}\mathrm {i}-A^2\,B\,b\,c^7\,\sqrt {x}\,3{}\mathrm {i}+A\,B^2\,b^2\,c^6\,\sqrt {x}\,3{}\mathrm {i}}{b^{1/4}\,{\left (-c\right )}^{19/4}\,\left (c\,\left (c\,\left (A^3\,c-3\,A^2\,B\,b\right )+3\,A\,B^2\,b^2\right )-B^3\,b^3\right )}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{b^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-c)^(3/4)*atan((((-c)^(3/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^6*c^7 + 16*B^2*b^8*c^5 - 32*A*B*b^7*c^6) - ((-c)^
(3/4)*(A*c - B*b)*(32*A*b^9*c^5 - 32*B*b^10*c^4)*1i)/(2*b^(11/4))))/(2*b^(11/4)) + ((-c)^(3/4)*(A*c - B*b)*(x^
(1/2)*(16*A^2*b^6*c^7 + 16*B^2*b^8*c^5 - 32*A*B*b^7*c^6) + ((-c)^(3/4)*(A*c - B*b)*(32*A*b^9*c^5 - 32*B*b^10*c
^4)*1i)/(2*b^(11/4))))/(2*b^(11/4)))/(((-c)^(3/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^6*c^7 + 16*B^2*b^8*c^5 - 32*A
*B*b^7*c^6) - ((-c)^(3/4)*(A*c - B*b)*(32*A*b^9*c^5 - 32*B*b^10*c^4)*1i)/(2*b^(11/4)))*1i)/(2*b^(11/4)) - ((-c
)^(3/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^6*c^7 + 16*B^2*b^8*c^5 - 32*A*B*b^7*c^6) + ((-c)^(3/4)*(A*c - B*b)*(32*
A*b^9*c^5 - 32*B*b^10*c^4)*1i)/(2*b^(11/4)))*1i)/(2*b^(11/4))))*(A*c - B*b))/b^(11/4) - ((2*A)/(7*b) - (2*x^2*
(A*c - B*b))/(3*b^2))/x^(7/2) - ((-c)^(3/4)*atan((A^3*c^8*x^(1/2)*1i - B^3*b^3*c^5*x^(1/2)*1i - A^2*B*b*c^7*x^
(1/2)*3i + A*B^2*b^2*c^6*x^(1/2)*3i)/(b^(1/4)*(-c)^(19/4)*(c*(c*(A^3*c - 3*A^2*B*b) + 3*A*B^2*b^2) - B^3*b^3))
)*(A*c - B*b)*1i)/b^(11/4)

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sympy [A]  time = 109.30, size = 405, normalized size = 1.58 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{b} & \text {for}\: c = 0 \\- \frac {2 A}{7 b x^{\frac {7}{2}}} + \frac {2 A c}{3 b^{2} x^{\frac {3}{2}}} - \frac {\sqrt [4]{-1} A c^{2} \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {11}{4}}} + \frac {\sqrt [4]{-1} A c^{2} \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {11}{4}}} - \frac {\sqrt [4]{-1} A c^{2} \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 {11}{4}}} - \frac {2 B}{3 b x^{\frac {3}{2}}} + \frac {\sqrt [4]{-1} B c \sqrt [4]{\frac {1}{c}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {7}{4}}} - \frac {\sqrt [4]{-1} B c \sqrt [4]{\frac {1}{c}} \log {\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac {1}{c}} + \sqrt {x} \right )}}{2 b^{\frac {7}{4}}} + \frac {\sqrt [4]{-1} B c \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 {7}{4}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(11*x**(11/2)) - 2*B/(7*x**(7/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(11*x**(11/2)) - 2*B/(7*
x**(7/2)))/c, Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(3*x**(3/2)))/b, Eq(c, 0)), (-2*A/(7*b*x**(7/2)) + 2*A*c/(3
*b**2*x**(3/2)) - (-1)**(1/4)*A*c**2*(1/c)**(1/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(11/
4)) + (-1)**(1/4)*A*c**2*(1/c)**(1/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(11/4)) - (-1)**(
1/4)*A*c**2*(1/c)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/b**(11/4) - 2*B/(3*b*x**(3/2)) + (-
1)**(1/4)*B*c*(1/c)**(1/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(7/4)) - (-1)**(1/4)*B*c*(1
/c)**(1/4)*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(7/4)) + (-1)**(1/4)*B*c*(1/c)**(1/4)*atan((
-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/b**(7/4), True))

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