\(\int \frac {x^{7/2}}{b x^2+c x^4} \, dx\) [120]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 147 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=\frac {2 \sqrt {x}}{c}+\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{5/4}}-\frac {\sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{5/4}} \] Output:

2*x^(1/2)/c+1/2*b^(1/4)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/ 
c^(5/4)-1/2*b^(1/4)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/c^(5 
/4)-1/2*b^(1/4)*arctanh(2^(1/2)*b^(1/4)*c^(1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x 
))*2^(1/2)/c^(5/4)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.80 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=\frac {4 \sqrt [4]{c} \sqrt {x}+\sqrt {2} \sqrt [4]{b} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-\sqrt {2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{2 c^{5/4}} \] Input:

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

Output:

(4*c^(1/4)*Sqrt[x] + Sqrt[2]*b^(1/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2] 
*b^(1/4)*c^(1/4)*Sqrt[x])] - Sqrt[2]*b^(1/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1 
/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(2*c^(5/4))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.56, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {9, 262, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {b \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {2 b \int \frac {1}{c x^2+b}d\sqrt {x}}{c}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{c}\)

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 \sqrt {x}}{c}-\frac {2 b \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{c}\)

Input:

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

Output:

(2*Sqrt[x])/c - (2*b*((-(ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sq 
rt[2]*b^(1/4)*c^(1/4))) + ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(S 
qrt[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] + Sq 
rt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2 
*Sqrt[b])))/c
 

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 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 = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.78

method result size
derivativedivides \(\frac {2 \sqrt {x}}{c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c}\) \(115\)
default \(\frac {2 \sqrt {x}}{c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c}\) \(115\)
risch \(\frac {2 \sqrt {x}}{c}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c}\) \(115\)

Input:

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

Output:

2*x^(1/2)/c-1/4/c*(1/c*b)^(1/4)*2^(1/2)*(ln((x+(1/c*b)^(1/4)*x^(1/2)*2^(1/ 
2)+(1/c*b)^(1/2))/(x-(1/c*b)^(1/4)*x^(1/2)*2^(1/2)+(1/c*b)^(1/2)))+2*arcta 
n(2^(1/2)/(1/c*b)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(1/c*b)^(1/4)*x^(1/2)- 
1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.80 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=-\frac {c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \log \left (c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) + i \, c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \log \left (i \, c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - i \, c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \log \left (-i \, c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} \log \left (-c \left (-\frac {b}{c^{5}}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 4 \, \sqrt {x}}{2 \, c} \] Input:

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

Output:

-1/2*(c*(-b/c^5)^(1/4)*log(c*(-b/c^5)^(1/4) + sqrt(x)) + I*c*(-b/c^5)^(1/4 
)*log(I*c*(-b/c^5)^(1/4) + sqrt(x)) - I*c*(-b/c^5)^(1/4)*log(-I*c*(-b/c^5) 
^(1/4) + sqrt(x)) - c*(-b/c^5)^(1/4)*log(-c*(-b/c^5)^(1/4) + sqrt(x)) - 4* 
sqrt(x))/c
 

Sympy [A] (verification not implemented)

Time = 46.44 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=\begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: b = 0 \wedge c = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: c = 0 \\\frac {2 \sqrt {x}}{c} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{c} + \frac {\sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {\sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {\sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*sqrt(x), Eq(b, 0) & Eq(c, 0)), (2*x**(5/2)/(5*b), Eq(c, 0)) 
, (2*sqrt(x)/c, Eq(b, 0)), (2*sqrt(x)/c + (-b/c)**(1/4)*log(sqrt(x) - (-b/ 
c)**(1/4))/(2*c) - (-b/c)**(1/4)*log(sqrt(x) + (-b/c)**(1/4))/(2*c) - (-b/ 
c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4))/c, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.26 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=-\frac {\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}}}{4 \, c} + \frac {2 \, \sqrt {x}}{c} \] Input:

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

Output:

-1/4*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sq 
rt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/sqrt(sqrt(b)*sqrt(c)) + 2*sqrt(2)*sq 
rt(b)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sq 
rt(sqrt(b)*sqrt(c)))/sqrt(sqrt(b)*sqrt(c)) + sqrt(2)*b^(1/4)*log(sqrt(2)*b 
^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4) - sqrt(2)*b^(1/4)*lo 
g(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4))/c + 2*s 
qrt(x)/c
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.21 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=-\frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{2}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{2}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{2}} + \frac {2 \, \sqrt {x}}{c} \] Input:

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

Output:

-1/2*sqrt(2)*(b*c^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqr 
t(x))/(b/c)^(1/4))/c^2 - 1/2*sqrt(2)*(b*c^3)^(1/4)*arctan(-1/2*sqrt(2)*(sq 
rt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^2 - 1/4*sqrt(2)*(b*c^3)^(1/4 
)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^2 + 1/4*sqrt(2)*(b*c^ 
3)^(1/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^2 + 2*sqrt(x) 
/c
 

Mupad [B] (verification not implemented)

Time = 18.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.37 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=\frac {2\,\sqrt {x}}{c}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{c^{5/4}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{c^{5/4}} \] Input:

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

Output:

(2*x^(1/2))/c - ((-b)^(1/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/c^(5/4) - 
((-b)^(1/4)*atanh((c^(1/4)*x^(1/2))/(-b)^(1/4)))/c^(5/4)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99 \[ \int \frac {x^{7/2}}{b x^2+c x^4} \, dx=\frac {2 c^{\frac {3}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )-2 c^{\frac {3}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+c^{\frac {3}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )-c^{\frac {3}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )+8 \sqrt {x}\, c}{4 c^{2}} \] Input:

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

Output:

(2*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*s 
qrt(c))/(c**(1/4)*b**(1/4)*sqrt(2))) - 2*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))) 
 + c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sq 
rt(b) + sqrt(c)*x) - c**(3/4)*b**(1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/ 
4)*sqrt(2) + sqrt(b) + sqrt(c)*x) + 8*sqrt(x)*c)/(4*c**2)