3.2.96 \(\int \frac {x^{17/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [196]

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

[Out]

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

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Rubi [A]
time = 0.16, 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, 327, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {b^{3/4} (11 b B-7 A c) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {x^{3/2} (11 b B-7 A c)}{6 c^3}+\frac {x^{7/2} (11 b B-7 A c)}{14 b c^2}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-1/6*((11*b*B - 7*A*c)*x^(3/2))/c^3 + ((11*b*B - 7*A*c)*x^(7/2))/(14*b*c^2) - ((b*B - A*c)*x^(11/2))/(2*b*c*(b
+ c*x^2)) - (b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (
b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (b^(3/4)*(11*b*
B - 7*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(15/4)) - (b^(3/4)*(11*b*B
- 7*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(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 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 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 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 {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {\left (\frac {11 b B}{2}-\frac {7 A c}{2}\right ) \int \frac {x^{9/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac {(11 b B-7 A c) \int \frac {x^{5/2}}{b+c x^2} \, dx}{4 c^2}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (11 b B-7 A c)) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 c^3}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (11 b B-7 A c)) \text {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^3}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac {(b (11 b B-7 A c)) \text {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^{7/2}}+\frac {(b (11 b B-7 A c)) \text {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^{7/2}}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (11 b B-7 A c)) \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 c^4}+\frac {(b (11 b B-7 A c)) \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 c^4}+\frac {\left (b^{3/4} (11 b B-7 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} c^{15/4}}+\frac {\left (b^{3/4} (11 b B-7 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} c^{15/4}}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}+\frac {\left (b^{3/4} (11 b B-7 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} c^{15/4}}-\frac {\left (b^{3/4} (11 b B-7 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} c^{15/4}}\\ &=-\frac {(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac {(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac {(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{15/4}}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 184, normalized size = 0.59 \begin {gather*} \frac {\frac {4 c^{3/4} x^{3/2} \left (-77 b^2 B+b c \left (49 A-44 B x^2\right )+4 c^2 x^2 \left (7 A+3 B x^2\right )\right )}{b+c x^2}-21 \sqrt {2} b^{3/4} (11 b B-7 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} b^{3/4} (11 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{168 c^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*c^(3/4)*x^(3/2)*(-77*b^2*B + b*c*(49*A - 44*B*x^2) + 4*c^2*x^2*(7*A + 3*B*x^2)))/(b + c*x^2) - 21*Sqrt[2]*
b^(3/4)*(11*b*B - 7*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - 21*Sqrt[2]*b^(3/4)*
(11*b*B - 7*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(168*c^(15/4))

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Maple [A]
time = 0.50, size = 171, normalized size = 0.55

method	result	size
derivativedivides	\(\frac {\frac {2 B c \,x^{\frac {7}{2}}}{7}+\frac {2 \left (A c -2 B b \right ) x^{\frac {3}{2}}}{3}}{c^{3}}-\frac {2 b \left (\frac {\left (-\frac {A c}{4}+\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {7 A c}{4}-\frac {11 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{c^{3}}\)	\(171\)
default	\(\frac {\frac {2 B c \,x^{\frac {7}{2}}}{7}+\frac {2 \left (A c -2 B b \right ) x^{\frac {3}{2}}}{3}}{c^{3}}-\frac {2 b \left (\frac {\left (-\frac {A c}{4}+\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {7 A c}{4}-\frac {11 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{c^{3}}\)	\(171\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (3 B c \,x^{2}+7 A c -14 B b \right )}{21 c^{3}}+\frac {b \,x^{\frac {3}{2}} A}{2 c^{2} \left (c \,x^{2}+b \right )}-\frac {b^{2} x^{\frac {3}{2}} B}{2 c^{3} \left (c \,x^{2}+b \right )}-\frac {7 b \sqrt {2}\, A \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )}{16 c^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}-\frac {7 b \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}-\frac {7 b \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}+\frac {11 b^{2} \sqrt {2}\, B \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )}{16 c^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}+\frac {11 b^{2} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}+\frac {11 b^{2} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(344\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/c^3*(1/7*B*c*x^(7/2)+1/3*(A*c-2*B*b)*x^(3/2))-2*b/c^3*((-1/4*A*c+1/4*B*b)*x^(3/2)/(c*x^2+b)+1/8*(7/4*A*c-11/
4*B*b)/c/(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)))

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Maxima [A]
time = 0.49, size = 247, normalized size = 0.80 \begin {gather*} -\frac {{\left (B b^{2} - A b c\right )} x^{\frac {3}{2}}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (11 \, B b^{2} - 7 \, A b c\right )} {\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 )}}{16 \, c^{3}} + \frac {2 \, {\left (3 \, B c x^{\frac {7}{2}} - 7 \, {\left (2 \, B b - A c\right )} x^{\frac {3}{2}}\right )}}{21 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 989 vs. \(2 (226) = 452\).
time = 2.79, size = 989, normalized size = 3.19 \begin {gather*} \frac {84 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (1771561 \, B^{6} b^{10} - 6764142 \, A B^{5} b^{9} c + 10761135 \, A^{2} B^{4} b^{8} c^{2} - 9130660 \, A^{3} B^{3} b^{7} c^{3} + 4357815 \, A^{4} B^{2} b^{6} c^{4} - 1109262 \, A^{5} B b^{5} c^{5} + 117649 \, A^{6} b^{4} c^{6}\right )} x - {\left (14641 \, B^{4} b^{7} c^{7} - 37268 \, A B^{3} b^{6} c^{8} + 35574 \, A^{2} B^{2} b^{5} c^{9} - 15092 \, A^{3} B b^{4} c^{10} + 2401 \, A^{4} b^{3} c^{11}\right )} \sqrt {-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}}} c^{4} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} + {\left (1331 \, B^{3} b^{5} c^{4} - 2541 \, A B^{2} b^{4} c^{5} + 1617 \, A^{2} B b^{3} c^{6} - 343 \, A^{3} b^{2} c^{7}\right )} \sqrt {x} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}}}{14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}\right ) - 21 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \log \left (c^{11} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {3}{4}} - {\left (1331 \, B^{3} b^{5} - 2541 \, A B^{2} b^{4} c + 1617 \, A^{2} B b^{3} c^{2} - 343 \, A^{3} b^{2} c^{3}\right )} \sqrt {x}\right ) + 21 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \log \left (-c^{11} \left (-\frac {14641 \, B^{4} b^{7} - 37268 \, A B^{3} b^{6} c + 35574 \, A^{2} B^{2} b^{5} c^{2} - 15092 \, A^{3} B b^{4} c^{3} + 2401 \, A^{4} b^{3} c^{4}}{c^{15}}\right )^{\frac {3}{4}} - {\left (1331 \, B^{3} b^{5} - 2541 \, A B^{2} b^{4} c + 1617 \, A^{2} B b^{3} c^{2} - 343 \, A^{3} b^{2} c^{3}\right )} \sqrt {x}\right ) + 4 \, {\left (12 \, B c^{2} x^{5} - 4 \, {\left (11 \, B b c - 7 \, A c^{2}\right )} x^{3} - 7 \, {\left (11 \, B b^{2} - 7 \, A b c\right )} x\right )} \sqrt {x}}{168 \, {\left (c^{4} x^{2} + b c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(84*(c^4*x^2 + b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3
+ 2401*A^4*b^3*c^4)/c^15)^(1/4)*arctan((sqrt((1771561*B^6*b^10 - 6764142*A*B^5*b^9*c + 10761135*A^2*B^4*b^8*c
^2 - 9130660*A^3*B^3*b^7*c^3 + 4357815*A^4*B^2*b^6*c^4 - 1109262*A^5*B*b^5*c^5 + 117649*A^6*b^4*c^6)*x - (1464
1*B^4*b^7*c^7 - 37268*A*B^3*b^6*c^8 + 35574*A^2*B^2*b^5*c^9 - 15092*A^3*B*b^4*c^10 + 2401*A^4*b^3*c^11)*sqrt(-
(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15))*c^
4*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)
^(1/4) + (1331*B^3*b^5*c^4 - 2541*A*B^2*b^4*c^5 + 1617*A^2*B*b^3*c^6 - 343*A^3*b^2*c^7)*sqrt(x)*(-(14641*B^4*b
^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(1/4))/(14641*B
^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)) - 21*(c^4*x^2 +
b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/
c^15)^(1/4)*log(c^11*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401
*A^4*b^3*c^4)/c^15)^(3/4) - (1331*B^3*b^5 - 2541*A*B^2*b^4*c + 1617*A^2*B*b^3*c^2 - 343*A^3*b^2*c^3)*sqrt(x))
+ 21*(c^4*x^2 + b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 24
01*A^4*b^3*c^4)/c^15)^(1/4)*log(-c^11*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3
*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(3/4) - (1331*B^3*b^5 - 2541*A*B^2*b^4*c + 1617*A^2*B*b^3*c^2 - 343*A^3*b
^2*c^3)*sqrt(x)) + 4*(12*B*c^2*x^5 - 4*(11*B*b*c - 7*A*c^2)*x^3 - 7*(11*B*b^2 - 7*A*b*c)*x)*sqrt(x))/(c^4*x^2
+ b*c^3)

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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(x**(17/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.43, size = 299, normalized size = 0.96 \begin {gather*} -\frac {B b^{2} x^{\frac {3}{2}} - A b c x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{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 \, c^{6}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{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 \, c^{6}} - \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{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 \, c^{6}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{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 \, c^{6}} + \frac {2 \, {\left (3 \, B c^{12} x^{\frac {7}{2}} - 14 \, B b c^{11} x^{\frac {3}{2}} + 7 \, A c^{12} x^{\frac {3}{2}}\right )}}{21 \, c^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*b^2*x^(3/2) - A*b*c*x^(3/2))/((c*x^2 + b)*c^3) + 1/8*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A
*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^6 + 1/8*sqrt(2)*(11*(b*c^3)^(3/4)*B*b
- 7*(b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^6 - 1/16*sqrt(2)*(
11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 1/16*sqrt(2
)*(11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 2/21*(3
*B*c^12*x^(7/2) - 14*B*b*c^11*x^(3/2) + 7*A*c^12*x^(3/2))/c^14

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Mupad [B]
time = 0.17, size = 127, normalized size = 0.41 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,c^2}-\frac {4\,B\,b}{3\,c^3}\right )+\frac {2\,B\,x^{7/2}}{7\,c^2}-\frac {x^{3/2}\,\left (\frac {B\,b^2}{2}-\frac {A\,b\,c}{2}\right )}{c^4\,x^2+b\,c^3}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (7\,A\,c-11\,B\,b\right )}{4\,c^{15/4}}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (7\,A\,c-11\,B\,b\right )\,1{}\mathrm {i}}{4\,c^{15/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

x^(3/2)*((2*A)/(3*c^2) - (4*B*b)/(3*c^3)) + (2*B*x^(7/2))/(7*c^2) - (x^(3/2)*((B*b^2)/2 - (A*b*c)/2))/(b*c^3 +
c^4*x^2) + ((-b)^(3/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4))*(7*A*c - 11*B*b))/(4*c^(15/4)) + ((-b)^(3/4)*atan((
c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*(7*A*c - 11*B*b)*1i)/(4*c^(15/4))

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