\(\int \frac {x^{17/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [144]

Optimal result

Integrand size = 26, antiderivative size = 230 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {(b B-A c) x^{3/2}}{4 c^2 \left (b+c x^2\right )^2}-\frac {(11 b B-3 A c) x^{3/2}}{16 b c^2 \left (b+c x^2\right )}-\frac {3 (7 b B+A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}+\frac {3 (7 b B+A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}}-\frac {3 (7 b B+A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{5/4} c^{11/4}} \] Output:

1/4*(-A*c+B*b)*x^(3/2)/c^2/(c*x^2+b)^2-1/16*(-3*A*c+11*B*b)*x^(3/2)/b/c^2/ 
(c*x^2+b)-3/64*(A*c+7*B*b)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/ 
2)/b^(5/4)/c^(11/4)+3/64*(A*c+7*B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1 
/4))*2^(1/2)/b^(5/4)/c^(11/4)-3/64*(A*c+7*B*b)*arctanh(2^(1/2)*b^(1/4)*c^( 
1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/b^(5/4)/c^(11/4)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.74 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 \sqrt [4]{b} c^{3/4} x^{3/2} \left (7 b^2 B-3 A c^2 x^2+b c \left (A+11 B x^2\right )\right )}{\left (b+c x^2\right )^2}-3 \sqrt {2} (7 b B+A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-3 \sqrt {2} (7 b B+A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{64 b^{5/4} c^{11/4}} \] Input:

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

Output:

((-4*b^(1/4)*c^(3/4)*x^(3/2)*(7*b^2*B - 3*A*c^2*x^2 + b*c*(A + 11*B*x^2))) 
/(b + c*x^2)^2 - 3*Sqrt[2]*(7*b*B + A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqr 
t[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - 3*Sqrt[2]*(7*b*B + A*c)*ArcTanh[(Sqrt[2]* 
b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(64*b^(5/4)*c^(11/4))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.27, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {9, 362, 252, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 362

\(\displaystyle \frac {(A c+7 b B) \int \frac {x^{5/2}}{\left (c x^2+b\right )^2}dx}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \int \frac {\sqrt {x}}{c x^2+b}dx}{4 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \int \frac {x}{c x^2+b}d\sqrt {x}}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(A c+7 b B) \left (\frac {3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{2 c}-\frac {x^{3/2}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{7/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {\frac {\left (3 A c -11 B b \right ) x^{\frac {7}{2}}}{16 b c}-\frac {\left (A c +7 B b \right ) x^{\frac {3}{2}}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \left (A c +7 B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{3} b \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(166\)
default	\(\frac {\frac {\left (3 A c -11 B b \right ) x^{\frac {7}{2}}}{16 b c}-\frac {\left (A c +7 B b \right ) x^{\frac {3}{2}}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {3 \left (A c +7 B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{3} b \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(166\)

Input:

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

Output:

2*(1/32*(3*A*c-11*B*b)/b/c*x^(7/2)-1/32*(A*c+7*B*b)/c^2*x^(3/2))/(c*x^2+b) 
^2+3/128*(A*c+7*B*b)/c^3/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 = 0.11 (sec) , antiderivative size = 871, normalized size of antiderivative = 3.79 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

1/64*(3*(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(2401*B^4*b^4 + 1372*A*B^3 
*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(1/4) 
*log(27*b^4*c^8*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 
 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(3/4) + 27*(343*B^3*b^3 + 147*A*B^2 
*b^2*c + 21*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)) - 3*(I*b*c^4*x^4 + 2*I*b^2*c^3 
*x^2 + I*b^3*c^2)*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 
 + 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(1/4)*log(27*I*b^4*c^8*(-(2401*B^ 
4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4) 
/(b^5*c^11))^(3/4) + 27*(343*B^3*b^3 + 147*A*B^2*b^2*c + 21*A^2*B*b*c^2 + 
A^3*c^3)*sqrt(x)) - 3*(-I*b*c^4*x^4 - 2*I*b^2*c^3*x^2 - I*b^3*c^2)*(-(2401 
*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c 
^4)/(b^5*c^11))^(1/4)*log(-27*I*b^4*c^8*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c 
 + 294*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(3/4) + 27* 
(343*B^3*b^3 + 147*A*B^2*b^2*c + 21*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)) - 3*(b 
*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 2 
94*A^2*B^2*b^2*c^2 + 28*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^11))^(1/4)*log(-27*b 
^4*c^8*(-(2401*B^4*b^4 + 1372*A*B^3*b^3*c + 294*A^2*B^2*b^2*c^2 + 28*A^3*B 
*b*c^3 + A^4*c^4)/(b^5*c^11))^(3/4) + 27*(343*B^3*b^3 + 147*A*B^2*b^2*c + 
21*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)) - 4*((11*B*b*c - 3*A*c^2)*x^3 + (7*B*b^ 
2 + A*b*c)*x)*sqrt(x))/(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.09 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {{\left (11 \, B b c - 3 \, A c^{2}\right )} x^{\frac {7}{2}} + {\left (7 \, B b^{2} + A b c\right )} x^{\frac {3}{2}}}{16 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}} + \frac {3 \, {\left (7 \, B b + A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, b c^{2}} \] Input:

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

Output:

-1/16*((11*B*b*c - 3*A*c^2)*x^(7/2) + (7*B*b^2 + A*b*c)*x^(3/2))/(b*c^4*x^ 
4 + 2*b^2*c^3*x^2 + b^3*c^2) + 3/128*(7*B*b + A*c)*(2*sqrt(2)*arctan(1/2*s 
qrt(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)*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)) - 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)*sqr 
t(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/(b*c^2)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.27 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {11 \, B b c x^{\frac {7}{2}} - 3 \, A c^{2} x^{\frac {7}{2}} + 7 \, B b^{2} x^{\frac {3}{2}} + A b c x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b c^{2}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{5}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{5}} - \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{5}} + \frac {3 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{5}} \] Input:

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

Output:

-1/16*(11*B*b*c*x^(7/2) - 3*A*c^2*x^(7/2) + 7*B*b^2*x^(3/2) + A*b*c*x^(3/2 
))/((c*x^2 + b)^2*b*c^2) + 3/64*sqrt(2)*(7*(b*c^3)^(3/4)*B*b + (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))/ 
(b^2*c^5) + 3/64*sqrt(2)*(7*(b*c^3)^(3/4)*B*b + (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))/(b^2*c^5) - 3/ 
128*sqrt(2)*(7*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)* 
(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^5) + 3/128*sqrt(2)*(7*(b*c^3)^(3/4)*B* 
b + (b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/( 
b^2*c^5)
 

Mupad [B] (verification not implemented)

Time = 9.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.53 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+7\,B\,b\right )}{32\,{\left (-b\right )}^{5/4}\,c^{11/4}}-\frac {3\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+7\,B\,b\right )}{32\,{\left (-b\right )}^{5/4}\,c^{11/4}}-\frac {\frac {x^{3/2}\,\left (A\,c+7\,B\,b\right )}{16\,c^2}-\frac {x^{7/2}\,\left (3\,A\,c-11\,B\,b\right )}{16\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4} \] Input:

int((x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

(3*atanh((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c + 7*B*b))/(32*(-b)^(5/4)*c^(11 
/4)) - (3*atan((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c + 7*B*b))/(32*(-b)^(5/4) 
*c^(11/4)) - ((x^(3/2)*(A*c + 7*B*b))/(16*c^2) - (x^(7/2)*(3*A*c - 11*B*b) 
)/(16*b*c))/(b^2 + c^2*x^4 + 2*b*c*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 945, normalized size of antiderivative = 4.11 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 6*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x 
)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b**2*c - 12*c**(1/4)*b**(3/4)*sq 
rt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1 
/4)*sqrt(2)))*a*b*c**2*x**2 - 6*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b 
**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c**3*x 
**4 - 42*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqr 
t(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**4 - 84*c**(1/4)*b**(3/4)*sqr 
t(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/ 
4)*sqrt(2)))*b**3*c*x**2 - 42*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b** 
(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**2*c**2* 
x**4 + 6*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqr 
t(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b**2*c + 12*c**(1/4)*b**(3/4) 
*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b* 
*(1/4)*sqrt(2)))*a*b*c**2*x**2 + 6*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4 
)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c** 
3*x**4 + 42*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2* 
sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**4 + 84*c**(1/4)*b**(3/4)* 
sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b** 
(1/4)*sqrt(2)))*b**3*c*x**2 + 42*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)* 
b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**2...