3.3.5 \(\int \frac {A+B x^2}{\sqrt {x} (b x^2+c x^4)^2} \, dx\) [205]

Optimal. Leaf size=310 \[ \frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}+\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}} \]

[Out]

1/14*(-11*A*c+7*B*b)/b^2/c/x^(7/2)+1/6*(11*A*c-7*B*b)/b^3/x^(3/2)+1/2*(A*c-B*b)/b/c/x^(7/2)/(c*x^2+b)+1/8*c^(3
/4)*(-11*A*c+7*B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(15/4)*2^(1/2)-1/8*c^(3/4)*(-11*A*c+7*B*b)*arc
tan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(15/4)*2^(1/2)+1/16*c^(3/4)*(-11*A*c+7*B*b)*ln(b^(1/2)+x*c^(1/2)-b^(1
/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(15/4)*2^(1/2)-1/16*c^(3/4)*(-11*A*c+7*B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4
)*2^(1/2)*x^(1/2))/b^(15/4)*2^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 310, 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 = {1598, 468, 331, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {c^{3/4} (7 b B-11 A c) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{15/4}}+\frac {c^{3/4} (7 b B-11 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}+\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b*B - 11*A*c)/(14*b^2*c*x^(7/2)) - (7*b*B - 11*A*c)/(6*b^3*x^(3/2)) - (b*B - A*c)/(2*b*c*x^(7/2)*(b + c*x^2
)) + (c^(3/4)*(7*b*B - 11*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(15/4)) - (c^(3/4)*
(7*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(15/4)) + (c^(3/4)*(7*b*B - 11*A*
c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(15/4)) - (c^(3/4)*(7*b*B - 11*A*c
)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(15/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

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 331

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 335

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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 631

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 642

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 1176

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 1179

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 1598

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}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx &=\int \frac {A+B x^2}{x^{9/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {\left (-\frac {7 b B}{2}+\frac {11 A c}{2}\right ) \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {(7 b B-11 A c) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {(c (7 b B-11 A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 b^3}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {(c (7 b B-11 A c)) \text {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 b^3}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {(c (7 b B-11 A c)) \text {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^{7/2}}-\frac {(c (7 b B-11 A c)) \text {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^{7/2}}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {\left (\sqrt {c} (7 b B-11 A c)\right ) \text {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 )}{8 b^{7/2}}-\frac {\left (\sqrt {c} (7 b B-11 A c)\right ) \text {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 )}{8 b^{7/2}}+\frac {\left (c^{3/4} (7 b B-11 A c)\right ) \text {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 )}{8 \sqrt {2} b^{15/4}}+\frac {\left (c^{3/4} (7 b B-11 A c)\right ) \text {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 )}{8 \sqrt {2} b^{15/4}}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {\left (c^{3/4} (7 b B-11 A c)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}+\frac {\left (c^{3/4} (7 b B-11 A c)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}+\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 187, normalized size = 0.60 \begin {gather*} \frac {-\frac {4 b^{3/4} \left (7 b B x^2 \left (4 b+7 c x^2\right )+A \left (12 b^2-44 b c x^2-77 c^2 x^4\right )\right )}{x^{7/2} \left (b+c x^2\right )}+21 \sqrt {2} c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+21 \sqrt {2} c^{3/4} (-7 b B+11 A c) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{168 b^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b^(3/4)*(7*b*B*x^2*(4*b + 7*c*x^2) + A*(12*b^2 - 44*b*c*x^2 - 77*c^2*x^4)))/(x^(7/2)*(b + c*x^2)) + 21*Sq
rt[2]*c^(3/4)*(7*b*B - 11*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 21*Sqrt[2]*c^
(3/4)*(-7*b*B + 11*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(168*b^(15/4))

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Maple [A]
time = 0.40, size = 170, normalized size = 0.55

method	result	size
derivativedivides	\(-\frac {2 A}{7 b^{2} x^{\frac {7}{2}}}-\frac {2 \left (-2 A c +B b \right )}{3 b^{3} x^{\frac {3}{2}}}+\frac {2 c \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (11 A c -7 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{b^{3}}\)	\(170\)
default	\(-\frac {2 A}{7 b^{2} x^{\frac {7}{2}}}-\frac {2 \left (-2 A c +B b \right )}{3 b^{3} x^{\frac {3}{2}}}+\frac {2 c \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (11 A c -7 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{b^{3}}\)	\(170\)
risch	\(-\frac {2 \left (-14 A c \,x^{2}+7 b B \,x^{2}+3 A b \right )}{21 b^{3} x^{\frac {7}{2}}}+\frac {c^{2} \sqrt {x}\, A}{2 b^{3} \left (c \,x^{2}+b \right )}-\frac {c \sqrt {x}\, B}{2 b^{2} \left (c \,x^{2}+b \right )}+\frac {11 c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b^{4}}+\frac {11 c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b^{4}}+\frac {11 c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )}{16 b^{4}}-\frac {7 c \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b^{3}}-\frac {7 c \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b^{3}}-\frac {7 c \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )}{16 b^{3}}\)	\(347\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*A/b^2/x^(7/2)-2/3*(-2*A*c+B*b)/b^3/x^(3/2)+2/b^3*c*((1/4*A*c-1/4*B*b)*x^(1/2)/(c*x^2+b)+1/32*(11*A*c-7*B*
b)*(b/c)^(1/4)/b*2^(1/2)*(ln((x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^
(1/2)))+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)))

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Maxima [A]
time = 0.52, size = 286, normalized size = 0.92 \begin {gather*} -\frac {7 \, {\left (7 \, B b c - 11 \, A c^{2}\right )} x^{4} + 12 \, A b^{2} + 4 \, {\left (7 \, B b^{2} - 11 \, A b c\right )} x^{2}}{42 \, {\left (b^{3} c x^{\frac {11}{2}} + b^{4} x^{\frac {7}{2}}\right )}} - \frac {\frac {2 \, \sqrt {2} {\left (7 \, B b c - 11 \, 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 (7 \, B b c - 11 \, 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 (7 \, B b c - 11 \, 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 (7 \, B b c - 11 \, 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}}}}{16 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/42*(7*(7*B*b*c - 11*A*c^2)*x^4 + 12*A*b^2 + 4*(7*B*b^2 - 11*A*b*c)*x^2)/(b^3*c*x^(11/2) + b^4*x^(7/2)) - 1/
16*(2*sqrt(2)*(7*B*b*c - 11*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)*(7*B*b*c - 11*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)*(7*B*b*c -
11*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)*(7*B*b*c - 11
*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^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (226) = 452\).
time = 2.09, size = 795, normalized size = 2.56 \begin {gather*} \frac {84 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{8} \sqrt {-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}} + {\left (49 \, B^{2} b^{2} c^{2} - 154 \, A B b c^{3} + 121 \, A^{2} c^{4}\right )} x} b^{11} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {3}{4}} + {\left (7 \, B b^{12} c - 11 \, A b^{11} c^{2}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {3}{4}}}{2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}\right ) + 21 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (b^{4} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} - {\left (7 \, B b c - 11 \, A c^{2}\right )} \sqrt {x}\right ) - 21 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-b^{4} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} - {\left (7 \, B b c - 11 \, A c^{2}\right )} \sqrt {x}\right ) - 4 \, {\left (7 \, {\left (7 \, B b c - 11 \, A c^{2}\right )} x^{4} + 12 \, A b^{2} + 4 \, {\left (7 \, B b^{2} - 11 \, A b c\right )} x^{2}\right )} \sqrt {x}}{168 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(84*(b^3*c*x^6 + b^4*x^4)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*
B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4)*arctan((sqrt(b^8*sqrt(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^
2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15) + (49*B^2*b^2*c^2 - 154*A*B*b*c^3 + 121*A^2*c^4)*x)*b
^11*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^1
5)^(3/4) + (7*B*b^12*c - 11*A*b^11*c^2)*sqrt(x)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*
c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(3/4))/(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*
b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)) + 21*(b^3*c*x^6 + b^4*x^4)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3
*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4)*log(b^4*(-(2401*B^4*b^4*c^3 - 15
092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4) - (7*B*b*c - 11*A*c
^2)*sqrt(x)) - 21*(b^3*c*x^6 + b^4*x^4)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37
268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4)*log(-b^4*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^
2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4) - (7*B*b*c - 11*A*c^2)*sqrt(x)) - 4*(7*(7*B*b*c - 1
1*A*c^2)*x^4 + 12*A*b^2 + 4*(7*B*b^2 - 11*A*b*c)*x^2)*sqrt(x))/(b^3*c*x^6 + b^4*x^4)

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Sympy [F(-1)] Timed out
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)/(c*x**4+b*x**2)**2/x**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 0.45, size = 292, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \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 )}{8 \, b^{4}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \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 )}{8 \, b^{4}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \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 )}{16 \, b^{4}} + \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \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 )}{16 \, b^{4}} - \frac {B b c \sqrt {x} - A c^{2} \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} b^{3}} - \frac {2 \, {\left (7 \, B b x^{2} - 14 \, A c x^{2} + 3 \, A b\right )}}{21 \, b^{3} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(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^4 - 1/8*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(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^4 - 1/16*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*log(sqrt(2)*
sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 + 1/16*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*log(-sqrt
(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 - 1/2*(B*b*c*sqrt(x) - A*c^2*sqrt(x))/((c*x^2 + b)*b^3) - 2/21*(7
*B*b*x^2 - 14*A*c*x^2 + 3*A*b)/(b^3*x^(7/2))

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Mupad [B]
time = 0.40, size = 595, normalized size = 1.92 \begin {gather*} \frac {\frac {2\,x^2\,\left (11\,A\,c-7\,B\,b\right )}{21\,b^2}-\frac {2\,A}{7\,b}+\frac {c\,x^4\,\left (11\,A\,c-7\,B\,b\right )}{6\,b^3}}{b\,x^{7/2}+c\,x^{11/2}}+\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )}{8\,b^{15/4}}+\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )}{8\,b^{15/4}}}{\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )\,1{}\mathrm {i}}{8\,b^{15/4}}-\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )\,1{}\mathrm {i}}{8\,b^{15/4}}}\right )\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^{15/4}}-\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {A^3\,c^8\,\sqrt {x}\,1331{}\mathrm {i}-B^3\,b^3\,c^5\,\sqrt {x}\,343{}\mathrm {i}-A^2\,B\,b\,c^7\,\sqrt {x}\,2541{}\mathrm {i}+A\,B^2\,b^2\,c^6\,\sqrt {x}\,1617{}\mathrm {i}}{b^{1/4}\,{\left (-c\right )}^{19/4}\,\left (c\,\left (c\,\left (1331\,A^3\,c-2541\,A^2\,B\,b\right )+1617\,A\,B^2\,b^2\right )-343\,B^3\,b^3\right )}\right )\,\left (11\,A\,c-7\,B\,b\right )\,1{}\mathrm {i}}{4\,b^{15/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x^2*(11*A*c - 7*B*b))/(21*b^2) - (2*A)/(7*b) + (c*x^4*(11*A*c - 7*B*b))/(6*b^3))/(b*x^(7/2) + c*x^(11/2))
+ ((-c)^(3/4)*atan((((-c)^(3/4)*(11*A*c - 7*B*b)*(x^(1/2)*(3872*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^1
0*c^6) - ((-c)^(3/4)*(11*A*c - 7*B*b)*(2816*A*b^13*c^5 - 1792*B*b^14*c^4)*1i)/(8*b^(15/4))))/(8*b^(15/4)) + ((
-c)^(3/4)*(11*A*c - 7*B*b)*(x^(1/2)*(3872*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^10*c^6) + ((-c)^(3/4)*(
11*A*c - 7*B*b)*(2816*A*b^13*c^5 - 1792*B*b^14*c^4)*1i)/(8*b^(15/4))))/(8*b^(15/4)))/(((-c)^(3/4)*(11*A*c - 7*
B*b)*(x^(1/2)*(3872*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^10*c^6) - ((-c)^(3/4)*(11*A*c - 7*B*b)*(2816*
A*b^13*c^5 - 1792*B*b^14*c^4)*1i)/(8*b^(15/4)))*1i)/(8*b^(15/4)) - ((-c)^(3/4)*(11*A*c - 7*B*b)*(x^(1/2)*(3872
*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^10*c^6) + ((-c)^(3/4)*(11*A*c - 7*B*b)*(2816*A*b^13*c^5 - 1792*B
*b^14*c^4)*1i)/(8*b^(15/4)))*1i)/(8*b^(15/4))))*(11*A*c - 7*B*b))/(4*b^(15/4)) - ((-c)^(3/4)*atan((A^3*c^8*x^(
1/2)*1331i - B^3*b^3*c^5*x^(1/2)*343i - A^2*B*b*c^7*x^(1/2)*2541i + A*B^2*b^2*c^6*x^(1/2)*1617i)/(b^(1/4)*(-c)
^(19/4)*(c*(c*(1331*A^3*c - 2541*A^2*B*b) + 1617*A*B^2*b^2) - 343*B^3*b^3)))*(11*A*c - 7*B*b)*1i)/(4*b^(15/4))

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