\(\int \frac {x^{15/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [131]

Optimal result

Integrand size = 26, antiderivative size = 233 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {2 (2 b B-A c) \sqrt {x}}{c^3}+\frac {2 B x^{5/2}}{5 c^2}-\frac {b (b B-A c) \sqrt {x}}{2 c^3 \left (b+c x^2\right )}-\frac {\sqrt [4]{b} (9 b B-5 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {\sqrt [4]{b} (9 b B-5 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {\sqrt [4]{b} (9 b B-5 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{13/4}} \] Output:

-2*(-A*c+2*B*b)*x^(1/2)/c^3+2/5*B*x^(5/2)/c^2-1/2*b*(-A*c+B*b)*x^(1/2)/c^3 
/(c*x^2+b)-1/8*b^(1/4)*(-5*A*c+9*B*b)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^( 
1/4))*2^(1/2)/c^(13/4)+1/8*b^(1/4)*(-5*A*c+9*B*b)*arctan(1+2^(1/2)*c^(1/4) 
*x^(1/2)/b^(1/4))*2^(1/2)/c^(13/4)+1/8*b^(1/4)*(-5*A*c+9*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^(13/4)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.79 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{c} \sqrt {x} \left (-45 b^2 B+b c \left (25 A-36 B x^2\right )+4 c^2 x^2 \left (5 A+B x^2\right )\right )}{b+c x^2}-5 \sqrt {2} \sqrt [4]{b} (9 b B-5 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+5 \sqrt {2} \sqrt [4]{b} (9 b B-5 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{40 c^{13/4}} \] Input:

Integrate[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

((4*c^(1/4)*Sqrt[x]*(-45*b^2*B + b*c*(25*A - 36*B*x^2) + 4*c^2*x^2*(5*A + 
B*x^2)))/(b + c*x^2) - 5*Sqrt[2]*b^(1/4)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[b] - 
 Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 5*Sqrt[2]*b^(1/4)*(9*b*B 
- 5*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)]) 
/(40*c^(13/4))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 300, 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, 755, 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.

\(\displaystyle \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle \frac {(9 b B-5 A c) \int \frac {x^{7/2}}{c x^2+b}dx}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \int \frac {x^{3/2}}{c x^2+b}dx}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {b \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \int \frac {1}{c x^2+b}d\sqrt {x}}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(9 b B-5 A c) \left (\frac {2 x^{5/2}}{5 c}-\frac {b \left (\frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\right )}{c}\right )}{4 b c}-\frac {x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

Input:

Int[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

-1/2*((b*B - A*c)*x^(9/2))/(b*c*(b + c*x^2)) + ((9*b*B - 5*A*c)*((2*x^(5/2 
))/(5*c) - (b*((2*Sqrt[x])/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[b]) + (-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)) + L 
og[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[b])))/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 755

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

rule 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 = 169, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {2 \left (B c \,x^{2}+5 A c -10 B b \right ) \sqrt {x}}{5 c^{3}}-\frac {b \left (\frac {2 \left (-\frac {A c}{4}+\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (5 A c -9 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}\right )}{c^{3}}\)	\(169\)
derivativedivides	\(\frac {\frac {2 B c \,x^{\frac {5}{2}}}{5}+2 A c \sqrt {x}-4 B b \sqrt {x}}{c^{3}}-\frac {2 b \left (\frac {\left (-\frac {A c}{4}+\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (5 A c -9 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{c^{3}}\)	\(171\)
default	\(\frac {\frac {2 B c \,x^{\frac {5}{2}}}{5}+2 A c \sqrt {x}-4 B b \sqrt {x}}{c^{3}}-\frac {2 b \left (\frac {\left (-\frac {A c}{4}+\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (5 A c -9 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{c^{3}}\)	\(171\)

Input:

int(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

2/5*(B*c*x^2+5*A*c-10*B*b)*x^(1/2)/c^3-b/c^3*(2*(-1/4*A*c+1/4*B*b)*x^(1/2) 
/(c*x^2+b)+1/16*(5*A*c-9*B*b)*(b/c)^(1/4)/b*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*a 
rctan(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 = 696, normalized size of antiderivative = 2.99 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
 

Output:

-1/40*(5*(c^4*x^2 + b*c^3)*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2 
*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13)^(1/4)*log(c^3*(-( 
6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2* 
c^3 + 625*A^4*b*c^4)/c^13)^(1/4) - (9*B*b - 5*A*c)*sqrt(x)) + 5*(I*c^4*x^2 
 + I*b*c^3)*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 
4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13)^(1/4)*log(I*c^3*(-(6561*B^4*b^5 
- 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4 
*b*c^4)/c^13)^(1/4) - (9*B*b - 5*A*c)*sqrt(x)) + 5*(-I*c^4*x^2 - I*b*c^3)* 
(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b 
^2*c^3 + 625*A^4*b*c^4)/c^13)^(1/4)*log(-I*c^3*(-(6561*B^4*b^5 - 14580*A*B 
^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^1 
3)^(1/4) - (9*B*b - 5*A*c)*sqrt(x)) - 5*(c^4*x^2 + b*c^3)*(-(6561*B^4*b^5 
- 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4 
*b*c^4)/c^13)^(1/4)*log(-c^3*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A 
^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13)^(1/4) - (9*B*b 
- 5*A*c)*sqrt(x)) - 4*(4*B*c^2*x^4 - 45*B*b^2 + 25*A*b*c - 4*(9*B*b*c - 5* 
A*c^2)*x^2)*sqrt(x))/(c^4*x^2 + b*c^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(15/2)*(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 = 271, normalized size of antiderivative = 1.16 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (B b^{2} - A b c\right )} \sqrt {x}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (\frac {2 \, \sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )} b}{16 \, c^{3}} + \frac {2 \, {\left (B c x^{\frac {5}{2}} - 5 \, {\left (2 \, B b - A c\right )} \sqrt {x}\right )}}{5 \, c^{3}} \] Input:

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
 

Output:

-1/2*(B*b^2 - A*b*c)*sqrt(x)/(c^4*x^2 + b*c^3) + 1/16*(2*sqrt(2)*(9*B*b - 
5*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sq 
rt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(9*B*b - 
5*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/s 
qrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(9*B*b - 5 
*A*c)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)* 
c^(1/4)) - sqrt(2)*(9*B*b - 5*A*c)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + 
sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))*b/c^3 + 2/5*(B*c*x^(5/2) - 5*(2*B* 
b - A*c)*sqrt(x))/c^3
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.28 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{4}} + \frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{4}} + \frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{4}} - \frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{4}} - \frac {B b^{2} \sqrt {x} - A b c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {2 \, {\left (B c^{8} x^{\frac {5}{2}} - 10 \, B b c^{7} \sqrt {x} + 5 \, A c^{8} \sqrt {x}\right )}}{5 \, c^{10}} \] Input:

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
 

Output:

1/8*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*(b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2) 
*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^4 + 1/8*sqrt(2)*(9*(b*c^ 
3)^(1/4)*B*b - 5*(b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/ 
4) - 2*sqrt(x))/(b/c)^(1/4))/c^4 + 1/16*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*( 
b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 1 
/16*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*(b*c^3)^(1/4)*A*c)*log(-sqrt(2)*sqrt( 
x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 1/2*(B*b^2*sqrt(x) - A*b*c*sqrt(x))/ 
((c*x^2 + b)*c^3) + 2/5*(B*c^8*x^(5/2) - 10*B*b*c^7*sqrt(x) + 5*A*c^8*sqrt 
(x))/c^10
 

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 823, normalized size of antiderivative = 3.53 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int((x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
 

Output:

x^(1/2)*((2*A)/c^2 - (4*B*b)/c^3) + (2*B*x^(5/2))/(5*c^2) - (x^(1/2)*((B*b 
^2)/2 - (A*b*c)/2))/(b*c^3 + c^4*x^2) + ((-b)^(1/4)*atan((((-b)^(1/4)*((x^ 
(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b)^(1/4)*(5*A 
*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c))/(8*c^(13/4)))*(5*A*c - 9*B*b)*1i)/(8* 
c^(13/4)) + ((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^ 
3*c))/c^3 + ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c))/(8*c^(13/ 
4)))*(5*A*c - 9*B*b)*1i)/(8*c^(13/4)))/(((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 
+ 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B* 
b^3 - 40*A*b^2*c))/(8*c^(13/4)))*(5*A*c - 9*B*b))/(8*c^(13/4)) - ((-b)^(1/ 
4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 + ((-b)^(1/ 
4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c))/(8*c^(13/4)))*(5*A*c - 9*B*b)) 
/(8*c^(13/4))))*(5*A*c - 9*B*b)*1i)/(4*c^(13/4)) + ((-b)^(1/4)*atan((((-b) 
^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b) 
^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c)*1i)/(8*c^(13/4)))*(5*A*c - 
9*B*b))/(8*c^(13/4)) + ((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 
- 90*A*B*b^3*c))/c^3 + ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c) 
*1i)/(8*c^(13/4)))*(5*A*c - 9*B*b))/(8*c^(13/4)))/(((-b)^(1/4)*((x^(1/2)*( 
81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b)^(1/4)*(5*A*c - 9* 
B*b)*(72*B*b^3 - 40*A*b^2*c)*1i)/(8*c^(13/4)))*(5*A*c - 9*B*b)*1i)/(8*c^(1 
3/4)) - ((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 639, normalized size of antiderivative = 2.74 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
 

Output:

(50*c**(3/4)*b**(1/4)*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 + 50*c**(3/4)*b**(1/4)*sqrt(2) 
*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*s 
qrt(2)))*a*c**2*x**2 - 90*c**(3/4)*b**(1/4)*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 - 90*c**( 
3/4)*b**(1/4)*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 - 50*c**(3/4)*b**(1/4)*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)))*a*b*c - 50*c**(3/4)*b**(1/4)*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 + 90*c**(3 
/4)*b**(1/4)*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 + 90*c**(3/4)*b**(1/4)*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 + 25*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4) 
*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*b*c + 25*c**(3/4)*b**(1/4)*sqrt(2)*log( 
- sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*c**2*x**2 - 4 
5*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqr 
t(b) + sqrt(c)*x)*b**3 - 45*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1 
/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**2*c*x**2 - 25*c**(3/4)*b**( 
1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*...