3.2.94 \(\int \frac {A+B x^2}{x^{9/2} (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=278 \[ -\frac {c^{7/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{15/4}}-\frac {c^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{15/4}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}-\frac {2 A}{11 b x^{11/2}} \]

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Rubi [A]  time = 0.24, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 14, 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^{7/4} (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^{15/4}}+\frac {c^{7/4} (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^{15/4}}-\frac {c^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{15/4}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}-\frac {2 A}{11 b x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*A)/(11*b*x^(11/2)) - (2*(b*B - A*c))/(7*b^2*x^(7/2)) + (2*c*(b*B - A*c))/(3*b^3*x^(3/2)) - (c^(7/4)*(b*B -
A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(15/4)) + (c^(7/4)*(b*B - A*c)*ArcTan[1 + (Sqr
t[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(15/4)) - (c^(7/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4
)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(15/4)) + (c^(7/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt
[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(15/4))

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^{9/2} \left (b x^2+c x^4\right )} \, dx &=\int \frac {A+B x^2}{x^{13/2} \left (b+c x^2\right )} \, dx\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {\left (2 \left (-\frac {11 b B}{2}+\frac {11 A c}{2}\right )\right ) \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{11 b}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}-\frac {(c (b B-A c)) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{b^2}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}+\frac {\left (c^2 (b B-A c)\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{b^3}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}+\frac {\left (2 c^2 (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}+\frac {\left (c^2 (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^{7/2}}+\frac {\left (c^2 (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^{7/2}}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}+\frac {\left (c^{3/2} (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^{7/2}}+\frac {\left (c^{3/2} (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^{7/2}}-\frac {\left (c^{7/4} (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^{15/4}}-\frac {\left (c^{7/4} (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^{15/4}}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {c^{7/4} (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^{15/4}}+\frac {c^{7/4} (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^{15/4}}+\frac {\left (c^{7/4} (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^{15/4}}-\frac {\left (c^{7/4} (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^{15/4}}\\ &=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {c^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{15/4}}-\frac {c^{7/4} (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^{15/4}}+\frac {c^{7/4} (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^{15/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 47, normalized size = 0.17 \begin {gather*} \frac {-22 x^2 (b B-A c) \, _2F_1\left (-\frac {7}{4},1;-\frac {3}{4};-\frac {c x^2}{b}\right )-14 A b}{77 b^2 x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x]

[Out]

(-14*A*b - 22*(b*B - A*c)*x^2*Hypergeometric2F1[-7/4, 1, -3/4, -((c*x^2)/b)])/(77*b^2*x^(11/2))

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IntegrateAlgebraic [A]  time = 0.24, size = 184, normalized size = 0.66 \begin {gather*} -\frac {\left (b B c^{7/4}-A c^{11/4}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} b^{15/4}}+\frac {\left (b B c^{7/4}-A c^{11/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{15/4}}-\frac {2 \left (21 A b^2-33 A b c x^2+77 A c^2 x^4+33 b^2 B x^2-77 b B c x^4\right )}{231 b^3 x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x]

[Out]

(-2*(21*A*b^2 + 33*b^2*B*x^2 - 33*A*b*c*x^2 - 77*b*B*c*x^4 + 77*A*c^2*x^4))/(231*b^3*x^(11/2)) - ((b*B*c^(7/4)
- A*c^(11/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*b^(15/4)) + ((b*B*c^(
7/4) - A*c^(11/4))*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*b^(15/4))

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fricas [B]  time = 0.44, size = 734, normalized size = 2.64 \begin {gather*} -\frac {924 \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{8} \sqrt {-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}} + {\left (B^{2} b^{2} c^{4} - 2 \, A B b c^{5} + A^{2} c^{6}\right )} x} b^{11} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {3}{4}} + {\left (B b^{12} c^{2} - A b^{11} c^{3}\right )} \sqrt {x} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {3}{4}}}{B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}\right ) + 231 \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \log \left (b^{4} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} - {\left (B b c^{2} - A c^{3}\right )} \sqrt {x}\right ) - 231 \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-b^{4} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} - {\left (B b c^{2} - A c^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (77 \, {\left (B b c - A c^{2}\right )} x^{4} - 21 \, A b^{2} - 33 \, {\left (B b^{2} - A b c\right )} x^{2}\right )} \sqrt {x}}{462 \, b^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/462*(924*b^3*x^6*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(1
/4)*arctan((sqrt(b^8*sqrt(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^1
5) + (B^2*b^2*c^4 - 2*A*B*b*c^5 + A^2*c^6)*x)*b^11*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^
3*B*b*c^10 + A^4*c^11)/b^15)^(3/4) + (B*b^12*c^2 - A*b^11*c^3)*sqrt(x)*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^
2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(3/4))/(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4
*A^3*B*b*c^10 + A^4*c^11)) + 231*b^3*x^6*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10
+ A^4*c^11)/b^15)^(1/4)*log(b^4*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c
^11)/b^15)^(1/4) - (B*b*c^2 - A*c^3)*sqrt(x)) - 231*b^3*x^6*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c
^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4)*log(-b^4*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*
A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4) - (B*b*c^2 - A*c^3)*sqrt(x)) - 4*(77*(B*b*c - A*c^2)*x^4 - 21*A*b^2 - 33*
(B*b^2 - A*b*c)*x^2)*sqrt(x))/(b^3*x^6)

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giac [A]  time = 0.19, size = 291, normalized size = 1.05 \begin {gather*} \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\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^{4}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\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^{4}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4}} + \frac {2 \, {\left (77 \, B b c x^{4} - 77 \, A c^{2} x^{4} - 33 \, B b^{2} x^{2} + 33 \, A b c x^{2} - 21 \, A b^{2}\right )}}{231 \, b^{3} x^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (b*c^3)^(1/4)*A*c^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(
b/c)^(1/4))/b^4 + 1/2*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (b*c^3)^(1/4)*A*c^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(
1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^4 + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (b*c^3)^(1/4)*A*c^2)*log(sqrt(2)*sqrt(
x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 - 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (b*c^3)^(1/4)*A*c^2)*log(-sqrt(2)*sqr
t(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 + 2/231*(77*B*b*c*x^4 - 77*A*c^2*x^4 - 33*B*b^2*x^2 + 33*A*b*c*x^2 - 21*
A*b^2)/(b^3*x^(11/2))

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maple [A]  time = 0.06, size = 336, normalized size = 1.21 \begin {gather*} -\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 b^{4}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 b^{4}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{3} \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^{4}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,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}\, B \,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}\, B \,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 {2 A \,c^{2}}{3 b^{3} x^{\frac {3}{2}}}+\frac {2 B c}{3 b^{2} x^{\frac {3}{2}}}+\frac {2 A c}{7 b^{2} x^{\frac {7}{2}}}-\frac {2 B}{7 b \,x^{\frac {7}{2}}}-\frac {2 A}{11 b \,x^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x)

[Out]

-1/2*c^3/b^4*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-1/4*c^3/b^4*(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^3/b^4*(b/c)^(1/
4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/2*c^2/b^3*(b/c)^(1/4)*2^(1/2)*B*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)*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^2/b^3*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-2/11*A/b/
x^(11/2)+2/7/b^2/x^(7/2)*A*c-2/7/b/x^(7/2)*B-2/3/b^3*c^2/x^(3/2)*A+2/3/b^2*c/x^(3/2)*B

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maxima [A]  time = 3.12, size = 276, normalized size = 0.99 \begin {gather*} \frac {\frac {2 \, \sqrt {2} {\left (B b c^{2} - A c^{3}\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^{2} - A c^{3}\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^{2} - A c^{3}\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^{2} - A c^{3}\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^{3}} + \frac {2 \, {\left (77 \, {\left (B b c - A c^{2}\right )} x^{4} - 21 \, A b^{2} - 33 \, {\left (B b^{2} - A b c\right )} x^{2}\right )}}{231 \, b^{3} x^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/4*(2*sqrt(2)*(B*b*c^2 - A*c^3)*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^2 - A*c^3)*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^2 - A*c^3
)*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^2 - A*c^3)*log
(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^3 + 2/231*(77*(B*b*c - A*c^2)*x^
4 - 21*A*b^2 - 33*(B*b^2 - A*b*c)*x^2)/(b^3*x^(11/2))

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mupad [B]  time = 0.35, size = 563, normalized size = 2.03 \begin {gather*} \frac {{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {A^3\,c^{10}\,\sqrt {x}-B^3\,b^3\,c^7\,\sqrt {x}-3\,A^2\,B\,b\,c^9\,\sqrt {x}+3\,A\,B^2\,b^2\,c^8\,\sqrt {x}}{b^{1/4}\,{\left (-c\right )}^{27/4}\,\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 )}{b^{15/4}}-\frac {\frac {2\,A}{11\,b}-\frac {2\,x^2\,\left (A\,c-B\,b\right )}{7\,b^2}+\frac {2\,c\,x^4\,\left (A\,c-B\,b\right )}{3\,b^3}}{x^{11/2}}-\frac {{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )-\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )\,1{}\mathrm {i}}{2\,b^{15/4}}+\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )+\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )\,1{}\mathrm {i}}{2\,b^{15/4}}}{\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )-\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )}{2\,b^{15/4}}-\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )+\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )}{2\,b^{15/4}}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{b^{15/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x)

[Out]

((-c)^(7/4)*atan((A^3*c^10*x^(1/2) - B^3*b^3*c^7*x^(1/2) - 3*A^2*B*b*c^9*x^(1/2) + 3*A*B^2*b^2*c^8*x^(1/2))/(b
^(1/4)*(-c)^(27/4)*(c*(c*(A^3*c - 3*A^2*B*b) + 3*A*B^2*b^2) - B^3*b^3)))*(A*c - B*b))/b^(15/4) - ((-c)^(7/4)*a
tan((((-c)^(7/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^9*c^9 + 16*B^2*b^11*c^7 - 32*A*B*b^10*c^8) - ((-c)^(7/4)*(A*c
- B*b)*(32*A*b^13*c^6 - 32*B*b^14*c^5))/(2*b^(15/4)))*1i)/(2*b^(15/4)) + ((-c)^(7/4)*(A*c - B*b)*(x^(1/2)*(16*
A^2*b^9*c^9 + 16*B^2*b^11*c^7 - 32*A*B*b^10*c^8) + ((-c)^(7/4)*(A*c - B*b)*(32*A*b^13*c^6 - 32*B*b^14*c^5))/(2
*b^(15/4)))*1i)/(2*b^(15/4)))/(((-c)^(7/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^9*c^9 + 16*B^2*b^11*c^7 - 32*A*B*b^1
0*c^8) - ((-c)^(7/4)*(A*c - B*b)*(32*A*b^13*c^6 - 32*B*b^14*c^5))/(2*b^(15/4))))/(2*b^(15/4)) - ((-c)^(7/4)*(A
*c - B*b)*(x^(1/2)*(16*A^2*b^9*c^9 + 16*B^2*b^11*c^7 - 32*A*B*b^10*c^8) + ((-c)^(7/4)*(A*c - B*b)*(32*A*b^13*c
^6 - 32*B*b^14*c^5))/(2*b^(15/4))))/(2*b^(15/4))))*(A*c - B*b)*1i)/b^(15/4) - ((2*A)/(11*b) - (2*x^2*(A*c - B*
b))/(7*b^2) + (2*c*x^4*(A*c - B*b))/(3*b^3))/x^(11/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**(9/2)/(c*x**4+b*x**2),x)

[Out]

Timed out

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