\(\int \frac {x^{7/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [149]

Optimal result

Integrand size = 26, antiderivative size = 243 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {2 A}{3 b^3 x^{3/2}}+\frac {(b B-A c) \sqrt {x}}{4 b^2 \left (b+c x^2\right )^2}+\frac {(7 b B-15 A c) \sqrt {x}}{16 b^3 \left (b+c x^2\right )}-\frac {7 (3 b B-11 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}}+\frac {7 (3 b B-11 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{15/4} \sqrt [4]{c}} \] Output:

-2/3*A/b^3/x^(3/2)+1/4*(-A*c+B*b)*x^(1/2)/b^2/(c*x^2+b)^2+1/16*(-15*A*c+7* 
B*b)*x^(1/2)/b^3/(c*x^2+b)-7/64*(-11*A*c+3*B*b)*arctan(1-2^(1/2)*c^(1/4)*x 
^(1/2)/b^(1/4))*2^(1/2)/b^(15/4)/c^(1/4)+7/64*(-11*A*c+3*B*b)*arctan(1+2^( 
1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(15/4)/c^(1/4)+7/64*(-11*A*c+3*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^( 
15/4)/c^(1/4)
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.77 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 b^{3/4} \left (-3 b B x^2 \left (11 b+7 c x^2\right )+A \left (32 b^2+121 b c x^2+77 c^2 x^4\right )\right )}{x^{3/2} \left (b+c x^2\right )^2}+\frac {21 \sqrt {2} (-3 b B+11 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt [4]{c}}+\frac {21 \sqrt {2} (3 b B-11 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt [4]{c}}}{192 b^{15/4}} \] Input:

Integrate[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

((-4*b^(3/4)*(-3*b*B*x^2*(11*b + 7*c*x^2) + A*(32*b^2 + 121*b*c*x^2 + 77*c 
^2*x^4)))/(x^(3/2)*(b + c*x^2)^2) + (21*Sqrt[2]*(-3*b*B + 11*A*c)*ArcTan[( 
Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/c^(1/4) + (21*Sqr 
t[2]*(3*b*B - 11*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + 
 Sqrt[c]*x)])/c^(1/4))/(192*b^(15/4))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.28, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {9, 362, 253, 264, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle -\frac {(3 b B-11 A c) \int \frac {1}{x^{5/2} \left (c x^2+b\right )^2}dx}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \int \frac {1}{x^{5/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {c \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \int \frac {1}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 755

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {(3 b B-11 A c) \left (\frac {7 \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{4 b}+\frac {1}{2 b x^{3/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}\)

Input:

Int[(x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*(b*B - A*c)/(b*c*x^(3/2)*(b + c*x^2)^2) - ((3*b*B - 11*A*c)*(1/(2*b*x 
^(3/2)*(b + c*x^2)) + (7*(-2/(3*b*x^(3/2)) - (2*c*((-(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])))/b))/(4*b)))/(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 253

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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 = 0.43 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(-\frac {2 A}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 \left (\frac {\left (\frac {15}{32} A \,c^{2}-\frac {7}{32} B b c \right ) x^{\frac {5}{2}}+\frac {b \left (19 A c -11 B b \right ) \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {7 \left (11 A c -3 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{3}}\)	\(173\)
default	\(-\frac {2 A}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 \left (\frac {\left (\frac {15}{32} A \,c^{2}-\frac {7}{32} B b c \right ) x^{\frac {5}{2}}+\frac {b \left (19 A c -11 B b \right ) \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {7 \left (11 A c -3 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{3}}\)	\(173\)
risch	\(-\frac {2 A}{3 b^{3} x^{\frac {3}{2}}}-\frac {\frac {2 \left (\frac {15}{32} A \,c^{2}-\frac {7}{32} B b c \right ) x^{\frac {5}{2}}+\frac {b \left (19 A c -11 B b \right ) \sqrt {x}}{16}}{\left (c \,x^{2}+b \right )^{2}}+\frac {7 \left (11 A c -3 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}}{b^{3}}\)	\(174\)

Input:

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

-2/3*A/b^3/x^(3/2)-2/b^3*(((15/32*A*c^2-7/32*B*b*c)*x^(5/2)+1/32*b*(19*A*c 
-11*B*b)*x^(1/2))/(c*x^2+b)^2+7/256*(11*A*c-3*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*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 = 770, normalized size of antiderivative = 3.17 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

-1/192*(21*(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^ 
3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15* 
c))^(1/4)*log(7*b^4*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^ 
2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4) - 7*(3*B*b - 11*A*c 
)*sqrt(x)) + 21*(I*b^3*c^2*x^6 + 2*I*b^4*c*x^4 + I*b^5*x^2)*(-(81*B^4*b^4 
- 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4* 
c^4)/(b^15*c))^(1/4)*log(7*I*b^4*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A 
^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4) - 7*(3 
*B*b - 11*A*c)*sqrt(x)) + 21*(-I*b^3*c^2*x^6 - 2*I*b^4*c*x^4 - I*b^5*x^2)* 
(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^ 
3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*log(-7*I*b^4*(-(81*B^4*b^4 - 1188*A*B^3 
*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c 
))^(1/4) - 7*(3*B*b - 11*A*c)*sqrt(x)) - 21*(b^3*c^2*x^6 + 2*b^4*c*x^4 + b 
^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^ 
3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*log(-7*b^4*(-(81*B^4*b^4 - 1188 
*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/( 
b^15*c))^(1/4) - 7*(3*B*b - 11*A*c)*sqrt(x)) - 4*(7*(3*B*b*c - 11*A*c^2)*x 
^4 - 32*A*b^2 + 11*(3*B*b^2 - 11*A*b*c)*x^2)*sqrt(x))/(b^3*c^2*x^6 + 2*b^4 
*c*x^4 + b^5*x^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(7/2)*(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 = 285, normalized size of antiderivative = 1.17 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {7 \, {\left (3 \, B b c - 11 \, A c^{2}\right )} x^{4} - 32 \, A b^{2} + 11 \, {\left (3 \, B b^{2} - 11 \, A b c\right )} x^{2}}{48 \, {\left (b^{3} c^{2} x^{\frac {11}{2}} + 2 \, b^{4} c x^{\frac {7}{2}} + b^{5} x^{\frac {3}{2}}\right )}} + \frac {7 \, {\left (\frac {2 \, \sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, B b - 11 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{128 \, b^{3}} \] Input:

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

1/48*(7*(3*B*b*c - 11*A*c^2)*x^4 - 32*A*b^2 + 11*(3*B*b^2 - 11*A*b*c)*x^2) 
/(b^3*c^2*x^(11/2) + 2*b^4*c*x^(7/2) + b^5*x^(3/2)) + 7/128*(2*sqrt(2)*(3* 
B*b - 11*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)*(3 
*B*b - 11*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sq 
rt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(3 
*B*b - 11*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)*(3*B*b - 11*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^3
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.25 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{4} c} + \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{4} c} + \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{4} c} - \frac {7 \, \sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{4} c} - \frac {2 \, A}{3 \, b^{3} x^{\frac {3}{2}}} + \frac {7 \, B b c x^{\frac {5}{2}} - 15 \, A c^{2} x^{\frac {5}{2}} + 11 \, B b^{2} \sqrt {x} - 19 \, A b c \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{3}} \] Input:

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

7/64*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(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^4*c) + 7/64*sqrt(2)*( 
3*(b*c^3)^(1/4)*B*b - 11*(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^4*c) + 7/128*sqrt(2)*(3*(b*c^3)^(1 
/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt 
(b/c))/(b^4*c) - 7/128*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c 
)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c) - 2/3*A/(b^3*x 
^(3/2)) + 1/16*(7*B*b*c*x^(5/2) - 15*A*c^2*x^(5/2) + 11*B*b^2*sqrt(x) - 19 
*A*b*c*sqrt(x))/((c*x^2 + b)^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 888, normalized size of antiderivative = 3.65 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \] Input:

int((x^(7/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

- ((2*A)/(3*b) + (11*x^2*(11*A*c - 3*B*b))/(48*b^2) + (7*c*x^4*(11*A*c - 3 
*B*b))/(48*b^3))/(b^2*x^(3/2) + c^2*x^(11/2) + 2*b*c*x^(7/2)) - (atan((((1 
1*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 529 
85856*A*B*b^10*c^4) - (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096* 
B*b^14*c^3))/(64*(-b)^(15/4)*c^(1/4)))*7i)/(64*(-b)^(15/4)*c^(1/4)) + ((11 
*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 5298 
5856*A*B*b^10*c^4) + (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B 
*b^14*c^3))/(64*(-b)^(15/4)*c^(1/4)))*7i)/(64*(-b)^(15/4)*c^(1/4)))/((7*(1 
1*A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 529 
85856*A*B*b^10*c^4) - (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096* 
B*b^14*c^3))/(64*(-b)^(15/4)*c^(1/4))))/(64*(-b)^(15/4)*c^(1/4)) - (7*(11* 
A*c - 3*B*b)*(x^(1/2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985 
856*A*B*b^10*c^4) + (7*(11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B* 
b^14*c^3))/(64*(-b)^(15/4)*c^(1/4))))/(64*(-b)^(15/4)*c^(1/4))))*(11*A*c - 
 3*B*b)*7i)/(32*(-b)^(15/4)*c^(1/4)) - (7*atan(((7*(11*A*c - 3*B*b)*(x^(1/ 
2)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985856*A*B*b^10*c^4) - 
 ((11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B*b^14*c^3)*7i)/(64*(-b 
)^(15/4)*c^(1/4))))/(64*(-b)^(15/4)*c^(1/4)) + (7*(11*A*c - 3*B*b)*(x^(1/2 
)*(97140736*A^2*b^9*c^5 + 7225344*B^2*b^11*c^3 - 52985856*A*B*b^10*c^4) + 
((11*A*c - 3*B*b)*(80740352*A*b^13*c^4 - 22020096*B*b^14*c^3)*7i)/(64*(...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1011, normalized size of antiderivative = 4.16 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

(462*sqrt(x)*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*x + 924*sqrt(x)*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**3 + 462*sqrt(x)*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**5 - 126*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*at 
an((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt 
(2)))*b**4*x - 252*sqrt(x)*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**3 - 
 126*sqrt(x)*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**5 - 462*sqrt(x 
)*c**(3/4)*b**(1/4)*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)))*a*b**2*c*x - 924*sqrt(x)*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**3 - 462*sqrt(x)*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**3*x**5 + 126*sqrt(x)*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* 
x + 252*sqrt(x)*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**3 + 126*sqr...