\(\int \frac {x^{13/2} (A+B x^2)}{b x^2+c x^4} \, dx\) [117]

Optimal result

Integrand size = 26, antiderivative size = 214 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}+\frac {b^{7/4} (b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{15/4}}-\frac {b^{7/4} (b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{15/4}}+\frac {b^{7/4} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{15/4}} \] Output:

2/3*b*(-A*c+B*b)*x^(3/2)/c^3-2/7*(-A*c+B*b)*x^(7/2)/c^2+2/11*B*x^(11/2)/c+ 
1/2*b^(7/4)*(-A*c+B*b)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/c 
^(15/4)-1/2*b^(7/4)*(-A*c+B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2 
^(1/2)/c^(15/4)+1/2*b^(7/4)*(-A*c+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 = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.80 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 x^{3/2} \left (77 b^2 B-77 A b c-33 b B c x^2+33 A c^2 x^2+21 B c^2 x^4\right )}{231 c^3}+\frac {b^{7/4} (b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} c^{15/4}}+\frac {b^{7/4} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{15/4}} \] Input:

Integrate[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]
 

Output:

(2*x^(3/2)*(77*b^2*B - 77*A*b*c - 33*b*B*c*x^2 + 33*A*c^2*x^2 + 21*B*c^2*x 
^4))/(231*c^3) + (b^(7/4)*(b*B - A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2 
]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*c^(15/4)) + (b^(7/4)*(b*B - A*c)*Arc 
Tanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*c^ 
(15/4))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 277, 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, 363, 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.

\(\displaystyle \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 363

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]
 

Output:

(2*B*x^(11/2))/(11*c) - ((b*B - 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))/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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

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 = 0.46 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {2 x^{\frac {3}{2}} \left (-21 B \,c^{2} x^{4}-33 A \,c^{2} x^{2}+33 x^{2} B b c +77 A b c -77 B \,b^{2}\right )}{231 c^{3}}+\frac {b^{2} \left (A c -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(164\)
derivativedivides	\(-\frac {2 \left (-\frac {B \,c^{2} x^{\frac {11}{2}}}{11}+\frac {\left (-A \,c^{2}+B b c \right ) x^{\frac {7}{2}}}{7}+\frac {\left (A b c -B \,b^{2}\right ) x^{\frac {3}{2}}}{3}\right )}{c^{3}}+\frac {b^{2} \left (A c -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(165\)
default	\(-\frac {2 \left (-\frac {B \,c^{2} x^{\frac {11}{2}}}{11}+\frac {\left (-A \,c^{2}+B b c \right ) x^{\frac {7}{2}}}{7}+\frac {\left (A b c -B \,b^{2}\right ) x^{\frac {3}{2}}}{3}\right )}{c^{3}}+\frac {b^{2} \left (A c -B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(165\)

Input:

int(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
 

Output:

-2/231*x^(3/2)*(-21*B*c^2*x^4-33*A*c^2*x^2+33*B*b*c*x^2+77*A*b*c-77*B*b^2) 
/c^3+1/4*b^2*(A*c-B*b)/c^4/(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 = 770, normalized size of antiderivative = 3.60 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx =\text {Too large to display} \] Input:

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")
 

Output:

1/462*(231*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B* 
b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4)*log(c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 
 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(3/4) - (B^3*b^8 
 - 3*A*B^2*b^7*c + 3*A^2*B*b^6*c^2 - A^3*b^5*c^3)*sqrt(x)) - 231*I*c^3*(-( 
B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7* 
c^4)/c^15)^(1/4)*log(I*c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c 
^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(3/4) - (B^3*b^8 - 3*A*B^2*b^7*c 
 + 3*A^2*B*b^6*c^2 - A^3*b^5*c^3)*sqrt(x)) + 231*I*c^3*(-(B^4*b^11 - 4*A*B 
^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4) 
*log(-I*c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^ 
8*c^3 + A^4*b^7*c^4)/c^15)^(3/4) - (B^3*b^8 - 3*A*B^2*b^7*c + 3*A^2*B*b^6* 
c^2 - A^3*b^5*c^3)*sqrt(x)) - 231*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2 
*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4)*log(-c^11*(-(B^4 
*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4 
)/c^15)^(3/4) - (B^3*b^8 - 3*A*B^2*b^7*c + 3*A^2*B*b^6*c^2 - A^3*b^5*c^3)* 
sqrt(x)) + 4*(21*B*c^2*x^5 - 33*(B*b*c - A*c^2)*x^3 + 77*(B*b^2 - A*b*c)*x 
)*sqrt(x))/c^3
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\text {Timed out} \] Input:

integrate(x**(13/2)*(B*x**2+A)/(c*x**4+b*x**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.11 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {{\left (B b^{3} - A b^{2} c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, c^{3}} + \frac {2 \, {\left (21 \, B c^{2} x^{\frac {11}{2}} - 33 \, {\left (B b c - A c^{2}\right )} x^{\frac {7}{2}} + 77 \, {\left (B b^{2} - A b c\right )} x^{\frac {3}{2}}\right )}}{231 \, c^{3}} \] Input:

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")
 

Output:

-1/4*(B*b^3 - A*b^2*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))*sqr 
t(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c) 
*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2) 
*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/ 
4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/ 
(b^(1/4)*c^(3/4)))/c^3 + 2/231*(21*B*c^2*x^(11/2) - 33*(B*b*c - A*c^2)*x^( 
7/2) + 77*(B*b^2 - A*b*c)*x^(3/2))/c^3
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.39 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{6}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{6}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{6}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{6}} + \frac {2 \, {\left (21 \, B c^{10} x^{\frac {11}{2}} - 33 \, B b c^{9} x^{\frac {7}{2}} + 33 \, A c^{10} x^{\frac {7}{2}} + 77 \, B b^{2} c^{8} x^{\frac {3}{2}} - 77 \, A b c^{9} x^{\frac {3}{2}}\right )}}{231 \, c^{11}} \] Input:

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*arctan(1/2*sqrt(2 
)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^6 - 1/2*sqrt(2)*((b*c^3 
)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1 
/4) - 2*sqrt(x))/(b/c)^(1/4))/c^6 + 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b* 
c^3)^(3/4)*A*b*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 - 1 
/4*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*log(-sqrt(2)*sqrt(x 
)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 2/231*(21*B*c^10*x^(11/2) - 33*B*b*c^ 
9*x^(7/2) + 33*A*c^10*x^(7/2) + 77*B*b^2*c^8*x^(3/2) - 77*A*b*c^9*x^(3/2)) 
/c^11
 

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.54 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=x^{7/2}\,\left (\frac {2\,A}{7\,c}-\frac {2\,B\,b}{7\,c^2}\right )+\frac {2\,B\,x^{11/2}}{11\,c}+\frac {{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )}{c^{15/4}}-\frac {b\,x^{3/2}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{3\,c}+\frac {{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{c^{15/4}} \] Input:

int((x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.56 \[ \int \frac {x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {-462 c^{\frac {5}{4}} b^{\frac {7}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a +462 c^{\frac {1}{4}} b^{\frac {15}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+462 c^{\frac {5}{4}} b^{\frac {7}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a -462 c^{\frac {1}{4}} b^{\frac {15}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+231 c^{\frac {5}{4}} b^{\frac {7}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a -231 c^{\frac {1}{4}} b^{\frac {15}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )-231 c^{\frac {5}{4}} b^{\frac {7}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a +231 c^{\frac {1}{4}} b^{\frac {15}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )-616 \sqrt {x}\, a b \,c^{2} x +264 \sqrt {x}\, a \,c^{3} x^{3}+616 \sqrt {x}\, b^{3} c x -264 \sqrt {x}\, b^{2} c^{2} x^{3}+168 \sqrt {x}\, b \,c^{3} x^{5}}{924 c^{4}} \] Input:

int(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x)
 

Output:

( - 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)))*a*b*c + 462*c**(1/4)*b**(3/4)*sq 
rt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1 
/4)*sqrt(2)))*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)))*a*b*c - 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 + 231*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 - 231*c**( 
1/4)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + 
 sqrt(c)*x)*b**3 - 231*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 + 231*c**(1/4)*b**(3/4)*sqrt(2)* 
log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**3 - 616*sq 
rt(x)*a*b*c**2*x + 264*sqrt(x)*a*c**3*x**3 + 616*sqrt(x)*b**3*c*x - 264*sq 
rt(x)*b**2*c**2*x**3 + 168*sqrt(x)*b*c**3*x**5)/(924*c**4)