\(\int \frac {x^{15/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [145]

Optimal result

Integrand size = 26, antiderivative size = 233 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {(b B-A c) \sqrt {x}}{4 c^2 \left (b+c x^2\right )^2}-\frac {(9 b B-A c) \sqrt {x}}{16 b c^2 \left (b+c x^2\right )}-\frac {(5 b B+3 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{7/4} c^{9/4}}+\frac {(5 b B+3 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{7/4} c^{9/4}}+\frac {(5 b B+3 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{7/4} c^{9/4}} \] Output:

1/4*(-A*c+B*b)*x^(1/2)/c^2/(c*x^2+b)^2-1/16*(-A*c+9*B*b)*x^(1/2)/b/c^2/(c* 
x^2+b)-1/64*(3*A*c+5*B*b)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2 
)/b^(7/4)/c^(9/4)+1/64*(3*A*c+5*B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1 
/4))*2^(1/2)/b^(7/4)/c^(9/4)+1/64*(3*A*c+5*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^(7/4)/c^(9/4)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.74 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 b^{3/4} \sqrt [4]{c} \sqrt {x} \left (5 b^2 B-A c^2 x^2+3 b c \left (A+3 B x^2\right )\right )}{\left (b+c x^2\right )^2}-\sqrt {2} (5 b B+3 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+\sqrt {2} (5 b B+3 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{64 b^{7/4} c^{9/4}} \] Input:

Integrate[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

((-4*b^(3/4)*c^(1/4)*Sqrt[x]*(5*b^2*B - A*c^2*x^2 + 3*b*c*(A + 3*B*x^2)))/ 
(b + c*x^2)^2 - Sqrt[2]*(5*b*B + 3*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt 
[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + Sqrt[2]*(5*b*B + 3*A*c)*ArcTanh[(Sqrt[2]*b 
^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(64*b^(7/4)*c^(9/4))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.26, 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, 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.

\(\displaystyle \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle \frac {(3 A c+5 b B) \int \frac {x^{3/2}}{\left (c x^2+b\right )^2}dx}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{4 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\int \frac {1}{c x^2+b}d\sqrt {x}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(3 A c+5 b B) \left (\frac {\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}}{2 c}-\frac {\sqrt {x}}{2 c \left (b+c x^2\right )}\right )}{8 b c}-\frac {x^{5/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2}\)

Input:

Int[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*((b*B - A*c)*x^(5/2))/(b*c*(b + c*x^2)^2) + ((5*b*B + 3*A*c)*(-1/2*Sq 
rt[x]/(c*(b + c*x^2)) + ((-(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)) + 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[b]))/(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 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 = 0.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {\frac {\left (A c -9 B b \right ) x^{\frac {5}{2}}}{16 b c}-\frac {\left (3 A c +5 B b \right ) \sqrt {x}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (3 A c +5 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} b^{2}}\)	\(167\)
default	\(\frac {\frac {\left (A c -9 B b \right ) x^{\frac {5}{2}}}{16 b c}-\frac {\left (3 A c +5 B b \right ) \sqrt {x}}{16 c^{2}}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (3 A c +5 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 c^{2} b^{2}}\)	\(167\)

Input:

int(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2*(1/32*(A*c-9*B*b)/b/c*x^(5/2)-1/32*(3*A*c+5*B*b)/c^2*x^(1/2))/(c*x^2+b)^ 
2+1/128*(3*A*c+5*B*b)/c^2/b^2*(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*arc 
tan(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 = 763, normalized size of antiderivative = 3.27 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

1/64*((b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(625*B^4*b^4 + 1500*A*B^3*b^ 
3*c + 1350*A^2*B^2*b^2*c^2 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^(1/4 
)*log(b^2*c^2*(-(625*B^4*b^4 + 1500*A*B^3*b^3*c + 1350*A^2*B^2*b^2*c^2 + 5 
40*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^(1/4) + (5*B*b + 3*A*c)*sqrt(x)) - 
 (-I*b*c^4*x^4 - 2*I*b^2*c^3*x^2 - I*b^3*c^2)*(-(625*B^4*b^4 + 1500*A*B^3* 
b^3*c + 1350*A^2*B^2*b^2*c^2 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^(1 
/4)*log(I*b^2*c^2*(-(625*B^4*b^4 + 1500*A*B^3*b^3*c + 1350*A^2*B^2*b^2*c^2 
 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^(1/4) + (5*B*b + 3*A*c)*sqrt(x 
)) - (I*b*c^4*x^4 + 2*I*b^2*c^3*x^2 + I*b^3*c^2)*(-(625*B^4*b^4 + 1500*A*B 
^3*b^3*c + 1350*A^2*B^2*b^2*c^2 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9)) 
^(1/4)*log(-I*b^2*c^2*(-(625*B^4*b^4 + 1500*A*B^3*b^3*c + 1350*A^2*B^2*b^2 
*c^2 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^(1/4) + (5*B*b + 3*A*c)*sq 
rt(x)) - (b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2)*(-(625*B^4*b^4 + 1500*A*B^3 
*b^3*c + 1350*A^2*B^2*b^2*c^2 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^( 
1/4)*log(-b^2*c^2*(-(625*B^4*b^4 + 1500*A*B^3*b^3*c + 1350*A^2*B^2*b^2*c^2 
 + 540*A^3*B*b*c^3 + 81*A^4*c^4)/(b^7*c^9))^(1/4) + (5*B*b + 3*A*c)*sqrt(x 
)) - 4*(5*B*b^2 + 3*A*b*c + (9*B*b*c - A*c^2)*x^2)*sqrt(x))/(b*c^4*x^4 + 2 
*b^2*c^3*x^2 + b^3*c^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(15/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.20 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {{\left (9 \, B b c - A c^{2}\right )} x^{\frac {5}{2}} + {\left (5 \, B b^{2} + 3 \, A b c\right )} \sqrt {x}}{16 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (5 \, B b + 3 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (5 \, B b + 3 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (5 \, B b + 3 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, B b + 3 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{128 \, b c^{2}} \] Input:

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

-1/16*((9*B*b*c - A*c^2)*x^(5/2) + (5*B*b^2 + 3*A*b*c)*sqrt(x))/(b*c^4*x^4 
 + 2*b^2*c^3*x^2 + b^3*c^2) + 1/128*(2*sqrt(2)*(5*B*b + 3*A*c)*arctan(1/2* 
sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c) 
))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(5*B*b + 3*A*c)*arctan(-1/2 
*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c 
)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(5*B*b + 3*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) 
*(5*B*b + 3*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^2)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.28 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{3}} + \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{2} c^{3}} + \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{3}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b + 3 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{2} c^{3}} - \frac {9 \, B b c x^{\frac {5}{2}} - A c^{2} x^{\frac {5}{2}} + 5 \, B b^{2} \sqrt {x} + 3 \, A b c \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b c^{2}} \] Input:

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

1/64*sqrt(2)*(5*(b*c^3)^(1/4)*B*b + 3*(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))/(b^2*c^3) + 1/64*sqrt(2)* 
(5*(b*c^3)^(1/4)*B*b + 3*(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))/(b^2*c^3) + 1/128*sqrt(2)*(5*(b*c^3)^ 
(1/4)*B*b + 3*(b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqr 
t(b/c))/(b^2*c^3) - 1/128*sqrt(2)*(5*(b*c^3)^(1/4)*B*b + 3*(b*c^3)^(1/4)*A 
*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^3) - 1/16*(9* 
B*b*c*x^(5/2) - A*c^2*x^(5/2) + 5*B*b^2*sqrt(x) + 3*A*b*c*sqrt(x))/((c*x^2 
 + b)^2*b*c^2)
 

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 799, normalized size of antiderivative = 3.43 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int((x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

(atan((((3*A*c + 5*B*b)*((x^(1/2)*(9*A^2*c^2 + 25*B^2*b^2 + 30*A*B*b*c))/( 
64*b^2*c) - ((3*A*c^2 + 5*B*b*c)*(3*A*c + 5*B*b))/(64*(-b)^(7/4)*c^(9/4))) 
*1i)/(64*(-b)^(7/4)*c^(9/4)) + ((3*A*c + 5*B*b)*((x^(1/2)*(9*A^2*c^2 + 25* 
B^2*b^2 + 30*A*B*b*c))/(64*b^2*c) + ((3*A*c^2 + 5*B*b*c)*(3*A*c + 5*B*b))/ 
(64*(-b)^(7/4)*c^(9/4)))*1i)/(64*(-b)^(7/4)*c^(9/4)))/(((3*A*c + 5*B*b)*(( 
x^(1/2)*(9*A^2*c^2 + 25*B^2*b^2 + 30*A*B*b*c))/(64*b^2*c) - ((3*A*c^2 + 5* 
B*b*c)*(3*A*c + 5*B*b))/(64*(-b)^(7/4)*c^(9/4))))/(64*(-b)^(7/4)*c^(9/4)) 
- ((3*A*c + 5*B*b)*((x^(1/2)*(9*A^2*c^2 + 25*B^2*b^2 + 30*A*B*b*c))/(64*b^ 
2*c) + ((3*A*c^2 + 5*B*b*c)*(3*A*c + 5*B*b))/(64*(-b)^(7/4)*c^(9/4))))/(64 
*(-b)^(7/4)*c^(9/4))))*(3*A*c + 5*B*b)*1i)/(32*(-b)^(7/4)*c^(9/4)) - ((x^( 
1/2)*(3*A*c + 5*B*b))/(16*c^2) - (x^(5/2)*(A*c - 9*B*b))/(16*b*c))/(b^2 + 
c^2*x^4 + 2*b*c*x^2) + (atan((((3*A*c + 5*B*b)*((x^(1/2)*(9*A^2*c^2 + 25*B 
^2*b^2 + 30*A*B*b*c))/(64*b^2*c) - ((3*A*c^2 + 5*B*b*c)*(3*A*c + 5*B*b)*1i 
)/(64*(-b)^(7/4)*c^(9/4))))/(64*(-b)^(7/4)*c^(9/4)) + ((3*A*c + 5*B*b)*((x 
^(1/2)*(9*A^2*c^2 + 25*B^2*b^2 + 30*A*B*b*c))/(64*b^2*c) + ((3*A*c^2 + 5*B 
*b*c)*(3*A*c + 5*B*b)*1i)/(64*(-b)^(7/4)*c^(9/4))))/(64*(-b)^(7/4)*c^(9/4) 
))/(((3*A*c + 5*B*b)*((x^(1/2)*(9*A^2*c^2 + 25*B^2*b^2 + 30*A*B*b*c))/(64* 
b^2*c) - ((3*A*c^2 + 5*B*b*c)*(3*A*c + 5*B*b)*1i)/(64*(-b)^(7/4)*c^(9/4))) 
*1i)/(64*(-b)^(7/4)*c^(9/4)) - ((3*A*c + 5*B*b)*((x^(1/2)*(9*A^2*c^2 + 25* 
B^2*b^2 + 30*A*B*b*c))/(64*b^2*c) + ((3*A*c^2 + 5*B*b*c)*(3*A*c + 5*B*b...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 943, normalized size of antiderivative = 4.05 \[ \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 6*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**2*c - 12*c**(3/4)*b**(1/4)*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**(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**3*x 
**4 - 10*c**(3/4)*b**(1/4)*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 - 20*c**(3/4)*b**(1/4)*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 - 10*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**2* 
x**4 + 6*c**(3/4)*b**(1/4)*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**(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**2*x**2 + 6*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** 
3*x**4 + 10*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**4 + 20*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*c*x**2 + 10*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...