\(\int \frac {x^{17/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [130]

Optimal result

Integrand size = 26, antiderivative size = 235 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {2 (2 b B-A c) x^{3/2}}{3 c^3}+\frac {2 B x^{7/2}}{7 c^2}-\frac {b (b B-A c) x^{3/2}}{2 c^3 \left (b+c x^2\right )}-\frac {b^{3/4} (11 b B-7 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}+\frac {b^{3/4} (11 b B-7 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{15/4}}-\frac {b^{3/4} (11 b B-7 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{15/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.78 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 c^{3/4} x^{3/2} \left (-77 b^2 B+b c \left (49 A-44 B x^2\right )+4 c^2 x^2 \left (7 A+3 B x^2\right )\right )}{b+c x^2}-21 \sqrt {2} b^{3/4} (11 b B-7 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-21 \sqrt {2} b^{3/4} (11 b B-7 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{168 c^{15/4}} \] Input:

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

Output:

((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)*(1 
1*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))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.29, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {9, 362, 262, 262, 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.

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

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2}dx\)

\(\Big \downarrow \) 362

\(\displaystyle \frac {(11 b B-7 A c) \int \frac {x^{9/2}}{c x^2+b}dx}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \int \frac {x^{5/2}}{c x^2+b}dx}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {b \int \frac {\sqrt {x}}{c x^2+b}dx}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \int \frac {x}{c x^2+b}d\sqrt {x}}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(11 b B-7 A c) \left (\frac {2 x^{7/2}}{7 c}-\frac {b \left (\frac {2 x^{3/2}}{3 c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

Input:

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

Output:

-1/2*((b*B - A*c)*x^(11/2))/(b*c*(b + c*x^2)) + ((11*b*B - 7*A*c)*((2*x^(7 
/2))/(7*c) - (b*((2*x^(3/2))/(3*c) - (2*b*((-(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])))/c))/c))/(4*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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 = 0.47 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {2 x^{\frac {3}{2}} \left (3 B c \,x^{2}+7 A c -14 B b \right )}{21 c^{3}}-\frac {b \left (\frac {2 \left (-\frac {A c}{4}+\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {7 A c}{4}-\frac {11 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{c^{3}}\)	\(170\)
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	\(\frac {\frac {2 B c \,x^{\frac {7}{2}}}{7}+\frac {2 \left (A c -2 B b \right ) x^{\frac {3}{2}}}{3}}{c^{3}}-\frac {2 b \left (\frac {\left (-\frac {A c}{4}+\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {7 A c}{4}-\frac {11 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{c^{3}}\)	\(171\)

Input:

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

Output:

2/21*x^(3/2)*(3*B*c*x^2+7*A*c-14*B*b)/c^3-b/c^3*(2*(-1/4*A*c+1/4*B*b)*x^(3 
/2)/(c*x^2+b)+1/4*(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)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 848, normalized size of antiderivative = 3.61 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/168*(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*(-I*c^4*x^2 - 
 I*b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 1 
5092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(1/4)*log(I*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 + 2 
401*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*(I*c^4*x^2 + I*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(-I*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 + 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)) - 
 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)*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.05 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (B b^{2} - A b c\right )} x^{\frac {3}{2}}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (11 \, B b^{2} - 7 \, A b c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, c^{3}} + \frac {2 \, {\left (3 \, B c x^{\frac {7}{2}} - 7 \, {\left (2 \, B b - A c\right )} x^{\frac {3}{2}}\right )}}{21 \, c^{3}} \] Input:

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

Output:

-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)*arct 
an(-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(-sqr 
t(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
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.27 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {B b^{2} x^{\frac {3}{2}} - A b c x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{6}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{6}} - \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{6}} + \frac {\sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{6}} + \frac {2 \, {\left (3 \, B c^{12} x^{\frac {7}{2}} - 14 \, B b c^{11} x^{\frac {3}{2}} + 7 \, A c^{12} x^{\frac {3}{2}}\right )}}{21 \, c^{14}} \] Input:

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

Output:

-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)*(1 
1*(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
 

Mupad [B] (verification not implemented)

Time = 9.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.54 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,c^2}-\frac {4\,B\,b}{3\,c^3}\right )+\frac {2\,B\,x^{7/2}}{7\,c^2}-\frac {x^{3/2}\,\left (\frac {B\,b^2}{2}-\frac {A\,b\,c}{2}\right )}{c^4\,x^2+b\,c^3}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (7\,A\,c-11\,B\,b\right )}{4\,c^{15/4}}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (7\,A\,c-11\,B\,b\right )\,1{}\mathrm {i}}{4\,c^{15/4}} \] Input:

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

Output:

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))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 641, normalized size of antiderivative = 2.73 \[ \int \frac {x^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(294*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 + 294*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**2*x**2 - 462*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 - 462* 
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*x**2 - 294*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 - 294*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**2*x**2 + 46 
2*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sq 
rt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**3 + 462*c**(1/4)*b**(3/4)*sqrt(2)*a 
tan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqr 
t(2)))*b**2*c*x**2 - 147*c**(1/4)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4) 
*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*b*c - 147*c**(1/4)*b**(3/4)*sqr 
t(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*c** 
2*x**2 + 231*c**(1/4)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sq 
rt(2) + sqrt(b) + sqrt(c)*x)*b**3 + 231*c**(1/4)*b**(3/4)*sqrt(2)*log( - s 
qrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**2*c*x**2 + 147* 
c**(1/4)*b**(3/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(...