\(\int \frac {x^{17/2}}{(b x^2+c x^4)^3} \, dx\) [342]

Optimal result

Integrand size = 19, antiderivative size = 242 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}}-\frac {3 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}} \]

[Out]

-1/4*x^(3/2)/c/(c*x^2+b)^2+3/16*x^(3/2)/b/c/(c*x^2+b)-3/64*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(5/4)/c
^(7/4)*2^(1/2)+3/64*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(5/4)/c^(7/4)*2^(1/2)+3/128*ln(b^(1/2)+x*c^(1/
2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(5/4)/c^(7/4)*2^(1/2)-3/128*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)
*x^(1/2))/b^(5/4)/c^(7/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1598, 294, 296, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}}-\frac {3 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2} \]

[In]

Int[x^(17/2)/(b*x^2 + c*x^4)^3,x]

[Out]

-1/4*x^(3/2)/(c*(b + c*x^2)^2) + (3*x^(3/2))/(16*b*c*(b + c*x^2)) - (3*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^
(1/4)])/(32*Sqrt[2]*b^(5/4)*c^(7/4)) + (3*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(5/4)*c
^(7/4)) + (3*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(7/4)) - (3*Log
[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(7/4))

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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 296

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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 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^{5/2}}{\left (b+c x^2\right )^3} \, dx \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 \int \frac {\sqrt {x}}{\left (b+c x^2\right )^2} \, dx}{8 c} \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac {3 \int \frac {\sqrt {x}}{b+c x^2} \, dx}{32 b c} \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac {3 \text {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b c} \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac {3 \text {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b c^{3/2}}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b c^{3/2}} \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac {3 \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 )}{64 b c^2}+\frac {3 \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 )}{64 b c^2}+\frac {3 \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 )}{64 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \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 )}{64 \sqrt {2} b^{5/4} c^{7/4}} \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac {3 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}}-\frac {3 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}} \\ & = -\frac {x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac {3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{7/4}}+\frac {3 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}}-\frac {3 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{5/4} c^{7/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.56 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 \sqrt [4]{b} c^{3/4} x^{3/2} \left (b-3 c x^2\right )}{\left (b+c x^2\right )^2}-3 \sqrt {2} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{64 b^{5/4} c^{7/4}} \]

[In]

Integrate[x^(17/2)/(b*x^2 + c*x^4)^3,x]

[Out]

((-4*b^(1/4)*c^(3/4)*x^(3/2)*(b - 3*c*x^2))/(b + c*x^2)^2 - 3*Sqrt[2]*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^
(1/4)*c^(1/4)*Sqrt[x])] - 3*Sqrt[2]*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(64*b^(5
/4)*c^(7/4))

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.57

method	result	size
derivativedivides	\(\frac {\frac {3 x^{\frac {7}{2}}}{16 b}-\frac {x^{\frac {3}{2}}}{16 c}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} b \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(138\)
default	\(\frac {\frac {3 x^{\frac {7}{2}}}{16 b}-\frac {x^{\frac {3}{2}}}{16 c}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} b \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(138\)

[In]

int(x^(17/2)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

2*(3/32/b*x^(7/2)-1/32*x^(3/2)/c)/(c*x^2+b)^2+3/128/c^2/b/(b/c)^(1/4)*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))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.19 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (i \, b c^{3} x^{4} + 2 i \, b^{2} c^{2} x^{2} + i \, b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (i \, b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (-i \, b c^{3} x^{4} - 2 i \, b^{2} c^{2} x^{2} - i \, b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (-i \, b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) - 3 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (-b^{4} c^{5} \left (-\frac {1}{b^{5} c^{7}}\right )^{\frac {3}{4}} + \sqrt {x}\right ) + 4 \, {\left (3 \, c x^{3} - b x\right )} \sqrt {x}}{64 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )}} \]

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/64*(3*(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)*(-1/(b^5*c^7))^(1/4)*log(b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) -
3*(I*b*c^3*x^4 + 2*I*b^2*c^2*x^2 + I*b^3*c)*(-1/(b^5*c^7))^(1/4)*log(I*b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)
) - 3*(-I*b*c^3*x^4 - 2*I*b^2*c^2*x^2 - I*b^3*c)*(-1/(b^5*c^7))^(1/4)*log(-I*b^4*c^5*(-1/(b^5*c^7))^(3/4) + sq
rt(x)) - 3*(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)*(-1/(b^5*c^7))^(1/4)*log(-b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x
)) + 4*(3*c*x^3 - b*x)*sqrt(x))/(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \]

[In]

integrate(x**(17/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.92 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, c x^{\frac {7}{2}} - b x^{\frac {3}{2}}}{16 \, {\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, b c} \]

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/16*(3*c*x^(7/2) - b*x^(3/2))/(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c) + 3/128*(2*sqrt(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(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*
sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(
-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/(b*c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.88 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, c x^{\frac {7}{2}} - b x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b c} + \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{4}} + \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{4}} - \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{4}} + \frac {3 \, \sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{4}} \]

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/16*(3*c*x^(7/2) - b*x^(3/2))/((c*x^2 + b)^2*b*c) + 3/64*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b
/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^4) + 3/64*sqrt(2)*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^
(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^4) - 3/128*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x +
sqrt(b/c))/(b^2*c^4) + 3/128*sqrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^4)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.35 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {3\,x^{7/2}}{16\,b}-\frac {x^{3/2}}{16\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4}-\frac {3\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{5/4}\,c^{7/4}}+\frac {3\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{5/4}\,c^{7/4}} \]

[In]

int(x^(17/2)/(b*x^2 + c*x^4)^3,x)

[Out]

((3*x^(7/2))/(16*b) - x^(3/2)/(16*c))/(b^2 + c^2*x^4 + 2*b*c*x^2) - (3*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32
*(-b)^(5/4)*c^(7/4)) + (3*atanh((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32*(-b)^(5/4)*c^(7/4))