\(\int \frac {x^{7/2} (A+B x^2)}{b x^2+c x^4} \, dx\) [186]

Optimal result

Integrand size = 26, antiderivative size = 255 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{5/2}}{5 c}-\frac {\sqrt [4]{b} (b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}-\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}} \]

[Out]

2/5*B*x^(5/2)/c-1/2*b^(1/4)*(-A*c+B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(9/4)*2^(1/2)+1/2*b^(1/4)*(
-A*c+B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(9/4)*2^(1/2)-1/4*b^(1/4)*(-A*c+B*b)*ln(b^(1/2)+x*c^(1/2
)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/c^(9/4)*2^(1/2)+1/4*b^(1/4)*(-A*c+B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)
*2^(1/2)*x^(1/2))/c^(9/4)*2^(1/2)-2*(-A*c+B*b)*x^(1/2)/c^2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1598, 470, 327, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {\sqrt [4]{b} (b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{9/4}}-\frac {\sqrt [4]{b} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {\sqrt [4]{b} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}-\frac {2 \sqrt {x} (b B-A c)}{c^2}+\frac {2 B x^{5/2}}{5 c} \]

[In]

Int[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(-2*(b*B - A*c)*Sqrt[x])/c^2 + (2*B*x^(5/2))/(5*c) - (b^(1/4)*(b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])
/b^(1/4)])/(Sqrt[2]*c^(9/4)) + (b^(1/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^
(9/4)) - (b^(1/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(9/4))
+ (b^(1/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(9/4))

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 470

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.59 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {4 \sqrt [4]{c} \sqrt {x} \left (-5 b B+5 A c+B c x^2\right )-5 \sqrt {2} \sqrt [4]{b} (b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+5 \sqrt {2} \sqrt [4]{b} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{10 c^{9/4}} \]

[In]

Integrate[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(4*c^(1/4)*Sqrt[x]*(-5*b*B + 5*A*c + B*c*x^2) - 5*Sqrt[2]*b^(1/4)*(b*B - A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sq
rt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 5*Sqrt[2]*b^(1/4)*(b*B - A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqr
t[b] + Sqrt[c]*x)])/(10*c^(9/4))

Maple [A] (verified)

Time = 1.77 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.54

method	result	size
risch	\(\frac {2 \left (B c \,x^{2}+5 A c -5 B b \right ) \sqrt {x}}{5 c^{2}}-\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(138\)
derivativedivides	\(\frac {\frac {2 B c \,x^{\frac {5}{2}}}{5}+2 A c \sqrt {x}-2 b B \sqrt {x}}{c^{2}}-\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(141\)
default	\(\frac {\frac {2 B c \,x^{\frac {5}{2}}}{5}+2 A c \sqrt {x}-2 b B \sqrt {x}}{c^{2}}-\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(141\)

[In]

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

2/5*(B*c*x^2+5*A*c-5*B*b)*x^(1/2)/c^2-1/4*(A*c-B*b)/c^2*(1/c*b)^(1/4)*2^(1/2)*(ln((x+(1/c*b)^(1/4)*x^(1/2)*2^(
1/2)+(1/c*b)^(1/2))/(x-(1/c*b)^(1/4)*x^(1/2)*2^(1/2)+(1/c*b)^(1/2)))+2*arctan(2^(1/2)/(1/c*b)^(1/4)*x^(1/2)+1)
+2*arctan(2^(1/2)/(1/c*b)^(1/4)*x^(1/2)-1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 597, normalized size of antiderivative = 2.34 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {5 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) + 5 i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 5 i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (-i \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 5 \, c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} \log \left (-c^{2} \left (-\frac {B^{4} b^{5} - 4 \, A B^{3} b^{4} c + 6 \, A^{2} B^{2} b^{3} c^{2} - 4 \, A^{3} B b^{2} c^{3} + A^{4} b c^{4}}{c^{9}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - 4 \, {\left (B c x^{2} - 5 \, B b + 5 \, A c\right )} \sqrt {x}}{10 \, c^{2}} \]

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/10*(5*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*log(c^2*
(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x)
) + 5*I*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*log(I*c^2
*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x
)) - 5*I*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*log(-I*c
^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt
(x)) - 5*c^2*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4)*log(-c^2
*(-(B^4*b^5 - 4*A*B^3*b^4*c + 6*A^2*B^2*b^3*c^2 - 4*A^3*B*b^2*c^3 + A^4*b*c^4)/c^9)^(1/4) - (B*b - A*c)*sqrt(x
)) - 4*(B*c*x^2 - 5*B*b + 5*A*c)*sqrt(x))/c^2

Sympy [A] (verification not implemented)

Time = 98.31 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.08 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {for}\: c = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{c} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x}}{c} + \frac {A \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {A \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {A \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c} - \frac {2 B b \sqrt {x}}{c^{2}} - \frac {B b \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c^{2}} + \frac {B b \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c^{2}} + \frac {B b \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 c} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(7/2)*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(5/2)/5), Eq(b, 0) & Eq(c, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(9/2)/9)/b, Eq(
c, 0)), ((2*A*sqrt(x) + 2*B*x**(5/2)/5)/c, Eq(b, 0)), (2*A*sqrt(x)/c + A*(-b/c)**(1/4)*log(sqrt(x) - (-b/c)**(
1/4))/(2*c) - A*(-b/c)**(1/4)*log(sqrt(x) + (-b/c)**(1/4))/(2*c) - A*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4))
/c - 2*B*b*sqrt(x)/c**2 - B*b*(-b/c)**(1/4)*log(sqrt(x) - (-b/c)**(1/4))/(2*c**2) + B*b*(-b/c)**(1/4)*log(sqrt
(x) + (-b/c)**(1/4))/(2*c**2) + B*b*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4))/c**2 + 2*B*x**(5/2)/(5*c), True)
)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.92 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {{\left (\frac {2 \, \sqrt {2} {\left (B b - A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (B b - A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )} b}{4 \, c^{2}} + \frac {2 \, {\left (B c x^{\frac {5}{2}} - 5 \, {\left (B b - A c\right )} \sqrt {x}\right )}}{5 \, c^{2}} \]

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

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))/sqrt(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)))*b/c^2 + 2/5*(B*c*x^(5/2) - 5*(B*b - A*c)*sqrt(x))/c^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.03 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{3}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{3}} + \frac {2 \, {\left (B c^{4} x^{\frac {5}{2}} - 5 \, B b c^{3} \sqrt {x} + 5 \, A c^{4} \sqrt {x}\right )}}{5 \, c^{5}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.09 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )+\frac {2\,B\,x^{5/2}}{5\,c}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}+\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}}{\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )}{2\,c^{9/4}}-\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )}{2\,c^{9/4}}\right )}{2\,c^{9/4}}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{c^{9/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )}{2\,c^{9/4}}+\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )}{2\,c^{9/4}}}{\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}-\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}-\frac {{\left (-b\right )}^{1/4}\,\left (A\,c-B\,b\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^2\,c^2-2\,A\,B\,b^3\,c+B^2\,b^4\right )}{c}+\frac {{\left (-b\right )}^{1/4}\,\left (32\,A\,b^2\,c^2-32\,B\,b^3\,c\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{9/4}}\right )\,1{}\mathrm {i}}{2\,c^{9/4}}}\right )\,\left (A\,c-B\,b\right )}{c^{9/4}} \]

[In]

int((x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

x^(1/2)*((2*A)/c - (2*B*b)/c^2) + (2*B*x^(5/2))/(5*c) - ((-b)^(1/4)*atan((((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)
*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4)))*
1i)/(2*c^(9/4)) + ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*
(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4)))*1i)/(2*c^(9/4)))/(((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(
B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4))))/(
2*c^(9/4)) - ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A
*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b))/(2*c^(9/4))))/(2*c^(9/4))))*(A*c - B*b)*1i)/c^(9/4) - ((-b)^(1/4)*atan((((
-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B
*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4))))/(2*c^(9/4)) + ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^
2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4))))/(2*c^(9/4)))/(((-b
)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c^2 - 2*A*B*b^3*c))/c - ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b
^3*c)*(A*c - B*b)*1i)/(2*c^(9/4)))*1i)/(2*c^(9/4)) - ((-b)^(1/4)*(A*c - B*b)*((16*x^(1/2)*(B^2*b^4 + A^2*b^2*c
^2 - 2*A*B*b^3*c))/c + ((-b)^(1/4)*(32*A*b^2*c^2 - 32*B*b^3*c)*(A*c - B*b)*1i)/(2*c^(9/4)))*1i)/(2*c^(9/4))))*
(A*c - B*b))/c^(9/4)