\(\int \frac {A+B x^2}{x^{7/2} (b x^2+c x^4)} \, dx\) [127]

Optimal result

Integrand size = 26, antiderivative size = 213 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=-\frac {2 A}{9 b x^{9/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}+\frac {2 c (b B-A c)}{b^3 \sqrt {x}}-\frac {c^{5/4} (b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}+\frac {c^{5/4} (b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}-\frac {c^{5/4} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{13/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=\frac {-\frac {4 \sqrt [4]{b} \left (9 b B x^2 \left (b-5 c x^2\right )+A \left (5 b^2-9 b c x^2+45 c^2 x^4\right )\right )}{x^{9/2}}+45 \sqrt {2} c^{5/4} (-b B+A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+45 \sqrt {2} c^{5/4} (-b B+A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{90 b^{13/4}} \] Input:

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

Output:

((-4*b^(1/4)*(9*b*B*x^2*(b - 5*c*x^2) + A*(5*b^2 - 9*b*c*x^2 + 45*c^2*x^4) 
))/x^(9/2) + 45*Sqrt[2]*c^(5/4)*(-(b*B) + A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x 
)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 45*Sqrt[2]*c^(5/4)*(-(b*B) + A*c)*A 
rcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(90*b^(13 
/4))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 274, 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, 359, 264, 264, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 359

\(\displaystyle \frac {(b B-A c) \int \frac {1}{x^{7/2} \left (c x^2+b\right )}dx}{b}-\frac {2 A}{9 b x^{9/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {(b B-A c) \left (-\frac {c \int \frac {1}{x^{3/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2 A}{9 b x^{9/2}}\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

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 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 359

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

rule 826

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {\left (A c -B b \right ) c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}-\frac {2 A}{9 b \,x^{\frac {9}{2}}}-\frac {2 \left (-A c +B b \right )}{5 b^{2} x^{\frac {5}{2}}}-\frac {2 \left (A c -B b \right ) c}{b^{3} \sqrt {x}}\)	\(158\)
default	\(-\frac {\left (A c -B b \right ) c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}-\frac {2 A}{9 b \,x^{\frac {9}{2}}}-\frac {2 \left (-A c +B b \right )}{5 b^{2} x^{\frac {5}{2}}}-\frac {2 \left (A c -B b \right ) c}{b^{3} \sqrt {x}}\)	\(158\)
risch	\(-\frac {2 \left (45 x^{4} A \,c^{2}-45 x^{4} B b c -9 A b c \,x^{2}+9 x^{2} B \,b^{2}+5 b^{2} A \right )}{45 b^{3} x^{\frac {9}{2}}}-\frac {\left (A c -B b \right ) c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{3} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(165\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 784, normalized size of antiderivative = 3.68 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/90*(45*b^3*x^5*(-(B^4*b^4*c^5 - 4*A*B^3*b^3*c^6 + 6*A^2*B^2*b^2*c^7 - 4 
*A^3*B*b*c^8 + A^4*c^9)/b^13)^(1/4)*log(b^10*(-(B^4*b^4*c^5 - 4*A*B^3*b^3* 
c^6 + 6*A^2*B^2*b^2*c^7 - 4*A^3*B*b*c^8 + A^4*c^9)/b^13)^(3/4) - (B^3*b^3* 
c^4 - 3*A*B^2*b^2*c^5 + 3*A^2*B*b*c^6 - A^3*c^7)*sqrt(x)) - 45*I*b^3*x^5*( 
-(B^4*b^4*c^5 - 4*A*B^3*b^3*c^6 + 6*A^2*B^2*b^2*c^7 - 4*A^3*B*b*c^8 + A^4* 
c^9)/b^13)^(1/4)*log(I*b^10*(-(B^4*b^4*c^5 - 4*A*B^3*b^3*c^6 + 6*A^2*B^2*b 
^2*c^7 - 4*A^3*B*b*c^8 + A^4*c^9)/b^13)^(3/4) - (B^3*b^3*c^4 - 3*A*B^2*b^2 
*c^5 + 3*A^2*B*b*c^6 - A^3*c^7)*sqrt(x)) + 45*I*b^3*x^5*(-(B^4*b^4*c^5 - 4 
*A*B^3*b^3*c^6 + 6*A^2*B^2*b^2*c^7 - 4*A^3*B*b*c^8 + A^4*c^9)/b^13)^(1/4)* 
log(-I*b^10*(-(B^4*b^4*c^5 - 4*A*B^3*b^3*c^6 + 6*A^2*B^2*b^2*c^7 - 4*A^3*B 
*b*c^8 + A^4*c^9)/b^13)^(3/4) - (B^3*b^3*c^4 - 3*A*B^2*b^2*c^5 + 3*A^2*B*b 
*c^6 - A^3*c^7)*sqrt(x)) - 45*b^3*x^5*(-(B^4*b^4*c^5 - 4*A*B^3*b^3*c^6 + 6 
*A^2*B^2*b^2*c^7 - 4*A^3*B*b*c^8 + A^4*c^9)/b^13)^(1/4)*log(-b^10*(-(B^4*b 
^4*c^5 - 4*A*B^3*b^3*c^6 + 6*A^2*B^2*b^2*c^7 - 4*A^3*B*b*c^8 + A^4*c^9)/b^ 
13)^(3/4) - (B^3*b^3*c^4 - 3*A*B^2*b^2*c^5 + 3*A^2*B*b*c^6 - A^3*c^7)*sqrt 
(x)) - 4*(45*(B*b*c - A*c^2)*x^4 - 5*A*b^2 - 9*(B*b^2 - A*b*c)*x^2)*sqrt(x 
))/(b^3*x^5)
 

Sympy [A] (verification not implemented)

Time = 105.57 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.43 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=A \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {13}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{13 c x^{\frac {13}{2}}} & \text {for}\: b = 0 \\- \frac {2}{9 b x^{\frac {9}{2}}} & \text {for}\: c = 0 \\- \frac {2}{9 b x^{\frac {9}{2}}} + \frac {2 c}{5 b^{2} x^{\frac {5}{2}}} - \frac {c^{2} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{3} \sqrt [4]{- \frac {b}{c}}} + \frac {c^{2} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{3} \sqrt [4]{- \frac {b}{c}}} - \frac {c^{2} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b^{3} \sqrt [4]{- \frac {b}{c}}} - \frac {2 c^{2}}{b^{3} \sqrt {x}} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{9 c x^{\frac {9}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} + \frac {c \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{2} \sqrt [4]{- \frac {b}{c}}} - \frac {c \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{2} \sqrt [4]{- \frac {b}{c}}} + \frac {c \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b^{2} \sqrt [4]{- \frac {b}{c}}} + \frac {2 c}{b^{2} \sqrt {x}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((B*x**2+A)/x**(7/2)/(c*x**4+b*x**2),x)
 

Output:

A*Piecewise((zoo/x**(13/2), Eq(b, 0) & Eq(c, 0)), (-2/(13*c*x**(13/2)), Eq 
(b, 0)), (-2/(9*b*x**(9/2)), Eq(c, 0)), (-2/(9*b*x**(9/2)) + 2*c/(5*b**2*x 
**(5/2)) - c**2*log(sqrt(x) - (-b/c)**(1/4))/(2*b**3*(-b/c)**(1/4)) + c**2 
*log(sqrt(x) + (-b/c)**(1/4))/(2*b**3*(-b/c)**(1/4)) - c**2*atan(sqrt(x)/( 
-b/c)**(1/4))/(b**3*(-b/c)**(1/4)) - 2*c**2/(b**3*sqrt(x)), True)) + B*Pie 
cewise((zoo/x**(9/2), Eq(b, 0) & Eq(c, 0)), (-2/(9*c*x**(9/2)), Eq(b, 0)), 
 (-2/(5*b*x**(5/2)), Eq(c, 0)), (-2/(5*b*x**(5/2)) + c*log(sqrt(x) - (-b/c 
)**(1/4))/(2*b**2*(-b/c)**(1/4)) - c*log(sqrt(x) + (-b/c)**(1/4))/(2*b**2* 
(-b/c)**(1/4)) + c*atan(sqrt(x)/(-b/c)**(1/4))/(b**2*(-b/c)**(1/4)) + 2*c/ 
(b**2*sqrt(x)), True))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, b^{3}} + \frac {2 \, {\left (45 \, {\left (B b c - A c^{2}\right )} x^{4} - 5 \, A b^{2} - 9 \, {\left (B b^{2} - A b c\right )} x^{2}\right )}}{45 \, b^{3} x^{\frac {9}{2}}} \] Input:

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

Output:

1/4*(B*b*c^2 - A*c^3)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/ 
4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt 
(c)) + 2*sqrt(2)*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)))/b^3 + 2/45*(45*(B*b*c - A*c^2)*x^4 - 5*A*b^2 - 9*(B*b^2 
- A*b*c)*x^2)/(b^3*x^(9/2))
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.37 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4} c} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4} c} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4} c} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4} c} + \frac {2 \, {\left (45 \, B b c x^{4} - 45 \, A c^{2} x^{4} - 9 \, B b^{2} x^{2} + 9 \, A b c x^{2} - 5 \, A b^{2}\right )}}{45 \, b^{3} x^{\frac {9}{2}}} \] Input:

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

Output:

1/2*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*arctan(1/2*sqrt(2)*(sq 
rt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^4*c) + 1/2*sqrt(2)*((b*c^3) 
^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 
 2*sqrt(x))/(b/c)^(1/4))/(b^4*c) - 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3 
)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c) + 1/ 
4*sqrt(2)*((b*c^3)^(3/4)*B*b - (b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/ 
c)^(1/4) + x + sqrt(b/c))/(b^4*c) + 2/45*(45*B*b*c*x^4 - 45*A*c^2*x^4 - 9* 
B*b^2*x^2 + 9*A*b*c*x^2 - 5*A*b^2)/(b^3*x^(9/2))
 

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.50 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=\frac {{\left (-c\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (A\,c-B\,b\right )}{b^{13/4}}-\frac {\frac {2\,A}{9\,b}-\frac {2\,x^2\,\left (A\,c-B\,b\right )}{5\,b^2}+\frac {2\,c\,x^4\,\left (A\,c-B\,b\right )}{b^3}}{x^{9/2}}-\frac {{\left (-c\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (A\,c-B\,b\right )}{b^{13/4}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.74 \[ \int \frac {A+B x^2}{x^{7/2} \left (b x^2+c x^4\right )} \, dx=\frac {90 \sqrt {x}\, c^{\frac {9}{4}} b^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a \,x^{4}-90 \sqrt {x}\, c^{\frac {5}{4}} b^{\frac {11}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{4}-90 \sqrt {x}\, c^{\frac {9}{4}} b^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a \,x^{4}+90 \sqrt {x}\, c^{\frac {5}{4}} b^{\frac {11}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{4}-45 \sqrt {x}\, c^{\frac {9}{4}} b^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a \,x^{4}+45 \sqrt {x}\, c^{\frac {5}{4}} b^{\frac {11}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{4}+45 \sqrt {x}\, c^{\frac {9}{4}} b^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a \,x^{4}-45 \sqrt {x}\, c^{\frac {5}{4}} b^{\frac {11}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{4}-40 a \,b^{3}+72 a \,b^{2} c \,x^{2}-360 a b \,c^{2} x^{4}-72 b^{4} x^{2}+360 b^{3} c \,x^{4}}{180 \sqrt {x}\, b^{4} x^{4}} \] Input:

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

Output:

(90*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2* 
sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c**2*x**4 - 90*sqrt(x)*c** 
(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c) 
)/(c**(1/4)*b**(1/4)*sqrt(2)))*b**2*c*x**4 - 90*sqrt(x)*c**(1/4)*b**(3/4)* 
sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b** 
(1/4)*sqrt(2)))*a*c**2*x**4 + 90*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c 
**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2))) 
*b**2*c*x**4 - 45*sqrt(x)*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*c**2*x**4 + 45*sqrt(x)*c**(1/4 
)*b**(3/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sq 
rt(c)*x)*b**2*c*x**4 + 45*sqrt(x)*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*c**2*x**4 - 45*sqrt(x)*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**2*c*x**4 - 40*a*b**3 + 72*a*b**2*c*x**2 - 360*a*b*c**2*x**4 
- 72*b**4*x**2 + 360*b**3*c*x**4)/(180*sqrt(x)*b**4*x**4)