\(\int \frac {1}{(c x)^{5/3} (a+b x^2)} \, dx\) [341]

Optimal result
Mathematica [A] (verified)
Rubi [A] (warning: unable to verify)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [C] (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 = 17, antiderivative size = 189 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=-\frac {3}{2 a c (c x)^{2/3}}+\frac {\sqrt {3} \sqrt [3]{b} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a} c^{2/3}}}{\sqrt {3}}\right )}{2 a^{4/3} c^{5/3}}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a} c^{2/3}+\sqrt [3]{b} (c x)^{2/3}\right )}{2 a^{4/3} c^{5/3}}-\frac {\sqrt [3]{b} \log \left (a^{2/3} c^{4/3}-\sqrt [3]{a} \sqrt [3]{b} c^{2/3} (c x)^{2/3}+b^{2/3} (c x)^{4/3}\right )}{4 a^{4/3} c^{5/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=\frac {x \left (-6 \sqrt [3]{a}+2 \sqrt {3} \sqrt [3]{b} x^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x^{2/3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 \sqrt [3]{b} x^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^{2/3}\right )-\sqrt [3]{b} x^{2/3} \log \left (\sqrt [3]{a}-\sqrt {3} \sqrt [6]{a} \sqrt [6]{b} \sqrt [3]{x}+\sqrt [3]{b} x^{2/3}\right )-\sqrt [3]{b} x^{2/3} \log \left (\sqrt [3]{a}+\sqrt {3} \sqrt [6]{a} \sqrt [6]{b} \sqrt [3]{x}+\sqrt [3]{b} x^{2/3}\right )\right )}{4 a^{4/3} (c x)^{5/3}} \] Input:

Integrate[1/((c*x)^(5/3)*(a + b*x^2)),x]
 

Output:

(x*(-6*a^(1/3) + 2*Sqrt[3]*b^(1/3)*x^(2/3)*ArcTan[(1 - (2*b^(1/3)*x^(2/3)) 
/a^(1/3))/Sqrt[3]] + 2*b^(1/3)*x^(2/3)*Log[a^(1/3) + b^(1/3)*x^(2/3)] - b^ 
(1/3)*x^(2/3)*Log[a^(1/3) - Sqrt[3]*a^(1/6)*b^(1/6)*x^(1/3) + b^(1/3)*x^(2 
/3)] - b^(1/3)*x^(2/3)*Log[a^(1/3) + Sqrt[3]*a^(1/6)*b^(1/6)*x^(1/3) + b^( 
1/3)*x^(2/3)]))/(4*a^(4/3)*(c*x)^(5/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.36 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {264, 266, 27, 807, 821, 16, 1142, 25, 27, 1082, 217, 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.

\(\displaystyle \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {b \int \frac {\sqrt [3]{c x}}{b x^2+a}dx}{a c^2}-\frac {3}{2 a c (c x)^{2/3}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {3 b \int \frac {c^3 x}{b x^2 c^2+a c^2}d\sqrt [3]{c x}}{a c^3}-\frac {3}{2 a c (c x)^{2/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 b \int \frac {c x}{b x^2 c^2+a c^2}d\sqrt [3]{c x}}{a c}-\frac {3}{2 a c (c x)^{2/3}}\)

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

\(\displaystyle -\frac {3 b \left (\frac {\int \frac {\sqrt [3]{a} c^{2/3}+\sqrt [3]{b} (c x)^{2/3}}{a^{2/3} c^{4/3}-\sqrt [3]{a} \sqrt [3]{b} (c x)^{2/3} c^{2/3}+b^{2/3} (c x)^{2/3}}d(c x)^{2/3}}{3 \sqrt [3]{a} \sqrt [3]{b} c^{2/3}}-\frac {\log \left (\sqrt [3]{a} c^{2/3}+\sqrt [3]{b} (c x)^{2/3}\right )}{3 \sqrt [3]{a} b^{2/3} c^{2/3}}\right )}{2 a c}-\frac {3}{2 a c (c x)^{2/3}}\)

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/((c*x)^(5/3)*(a + b*x^2)),x]
 

Output:

-3/(2*a*c*(c*x)^(2/3)) - (3*b*(-1/3*Log[a^(1/3)*c^(2/3) + b^(1/3)*(c*x)^(2 
/3)]/(a^(1/3)*b^(2/3)*c^(2/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*(c*x)^ 
(2/3))/(a^(1/3)*c^(2/3)))/Sqrt[3]])/b^(1/3)) + Log[a^(2/3)*c^(4/3) + b^(2/ 
3)*(c*x)^(2/3) - a^(1/3)*b^(1/3)*c^(2/3)*(c*x)^(2/3)]/(2*b^(1/3)))/(3*a^(1 
/3)*b^(1/3)*c^(2/3))))/(2*a*c)
 

Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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 807
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 821
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 
Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.75

method result size
pseudoelliptic \(-\frac {2 \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (c x \right )^{\frac {2}{3}}}{3 \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}-\frac {\sqrt {3}}{3}\right ) \left (c x \right )^{\frac {2}{3}}-2 \ln \left (\left (c x \right )^{\frac {2}{3}}+\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}\right ) \left (c x \right )^{\frac {2}{3}}+\ln \left (c x \left (c x \right )^{\frac {1}{3}}-\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}} \left (c x \right )^{\frac {2}{3}}+\left (\frac {a \,c^{2}}{b}\right )^{\frac {2}{3}}\right ) \left (c x \right )^{\frac {2}{3}}+6 \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}{4 \left (c x \right )^{\frac {2}{3}} \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}} a c}\) \(141\)
derivativedivides \(3 c \left (-\frac {1}{2 a \,c^{2} \left (c x \right )^{\frac {2}{3}}}-\frac {b \left (-\frac {\ln \left (\left (c x \right )^{\frac {2}{3}}+\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\left (c x \right )^{\frac {4}{3}}-\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}} \left (c x \right )^{\frac {2}{3}}+\left (\frac {a \,c^{2}}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (c x \right )^{\frac {2}{3}}}{\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}\right )}{2 a \,c^{2}}\right )\) \(152\)
default \(3 c \left (-\frac {1}{2 a \,c^{2} \left (c x \right )^{\frac {2}{3}}}-\frac {b \left (-\frac {\ln \left (\left (c x \right )^{\frac {2}{3}}+\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (\left (c x \right )^{\frac {4}{3}}-\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}} \left (c x \right )^{\frac {2}{3}}+\left (\frac {a \,c^{2}}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (c x \right )^{\frac {2}{3}}}{\left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a \,c^{2}}{b}\right )^{\frac {1}{3}}}\right )}{2 a \,c^{2}}\right )\) \(152\)

Input:

int(1/(c*x)^(5/3)/(b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

-1/4*(2*3^(1/2)*arctan(2/3*3^(1/2)*(c*x)^(2/3)/(a*c^2/b)^(1/3)-1/3*3^(1/2) 
)*(c*x)^(2/3)-2*ln((c*x)^(2/3)+(a*c^2/b)^(1/3))*(c*x)^(2/3)+ln(c*x*(c*x)^( 
1/3)-(a*c^2/b)^(1/3)*(c*x)^(2/3)+(a*c^2/b)^(2/3))*(c*x)^(2/3)+6*(a*c^2/b)^ 
(1/3))/(c*x)^(2/3)/(a*c^2/b)^(1/3)/a/c
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=-\frac {2 \, \sqrt {3} c x \left (\frac {b}{a c^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (c x\right )^{\frac {2}{3}} \left (\frac {b}{a c^{2}}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - 2 \, c x \left (\frac {b}{a c^{2}}\right )^{\frac {1}{3}} \log \left (a c^{2} \left (\frac {b}{a c^{2}}\right )^{\frac {2}{3}} + \left (c x\right )^{\frac {2}{3}} b\right ) + c x \left (\frac {b}{a c^{2}}\right )^{\frac {1}{3}} \log \left (-\left (c x\right )^{\frac {2}{3}} a c \left (\frac {b}{a c^{2}}\right )^{\frac {2}{3}} + \left (c x\right )^{\frac {1}{3}} b x + a c \left (\frac {b}{a c^{2}}\right )^{\frac {1}{3}}\right ) + 6 \, \left (c x\right )^{\frac {1}{3}}}{4 \, a c^{2} x} \] Input:

integrate(1/(c*x)^(5/3)/(b*x^2+a),x, algorithm="fricas")
 

Output:

-1/4*(2*sqrt(3)*c*x*(b/(a*c^2))^(1/3)*arctan(2/3*sqrt(3)*(c*x)^(2/3)*(b/(a 
*c^2))^(1/3) - 1/3*sqrt(3)) - 2*c*x*(b/(a*c^2))^(1/3)*log(a*c^2*(b/(a*c^2) 
)^(2/3) + (c*x)^(2/3)*b) + c*x*(b/(a*c^2))^(1/3)*log(-(c*x)^(2/3)*a*c*(b/( 
a*c^2))^(2/3) + (c*x)^(1/3)*b*x + a*c*(b/(a*c^2))^(1/3)) + 6*(c*x)^(1/3))/ 
(a*c^2*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=\frac {\Gamma \left (- \frac {1}{3}\right )}{2 a c^{\frac {5}{3}} x^{\frac {2}{3}} \Gamma \left (\frac {2}{3}\right )} - \frac {\sqrt [3]{b} e^{- \frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {2}{3}} e^{\frac {i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {1}{3}\right )}{6 a^{\frac {4}{3}} c^{\frac {5}{3}} \Gamma \left (\frac {2}{3}\right )} - \frac {\sqrt [3]{b} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {2}{3}} e^{i \pi }}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {1}{3}\right )}{6 a^{\frac {4}{3}} c^{\frac {5}{3}} \Gamma \left (\frac {2}{3}\right )} - \frac {\sqrt [3]{b} e^{\frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {2}{3}} e^{\frac {5 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {1}{3}\right )}{6 a^{\frac {4}{3}} c^{\frac {5}{3}} \Gamma \left (\frac {2}{3}\right )} \] Input:

integrate(1/(c*x)**(5/3)/(b*x**2+a),x)
 

Output:

gamma(-1/3)/(2*a*c**(5/3)*x**(2/3)*gamma(2/3)) - b**(1/3)*exp(-2*I*pi/3)*l 
og(1 - b**(1/3)*x**(2/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(-1/3)/(6*a**(4/ 
3)*c**(5/3)*gamma(2/3)) - b**(1/3)*log(1 - b**(1/3)*x**(2/3)*exp_polar(I*p 
i)/a**(1/3))*gamma(-1/3)/(6*a**(4/3)*c**(5/3)*gamma(2/3)) - b**(1/3)*exp(2 
*I*pi/3)*log(1 - b**(1/3)*x**(2/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(-1/ 
3)/(6*a**(4/3)*c**(5/3)*gamma(2/3))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=-\frac {\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (c x\right )^{\frac {2}{3}} - \left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}}}\right )}{\left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}} a} + \frac {\log \left (\left (c x\right )^{\frac {4}{3}} - \left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}} \left (c x\right )^{\frac {2}{3}} + \left (\frac {a c^{2}}{b}\right )^{\frac {2}{3}}\right )}{\left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}} a} - \frac {2 \, \log \left (\left (c x\right )^{\frac {2}{3}} + \left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}}\right )}{\left (\frac {a c^{2}}{b}\right )^{\frac {1}{3}} a} + \frac {6}{\left (c x\right )^{\frac {2}{3}} a}}{4 \, c} \] Input:

integrate(1/(c*x)^(5/3)/(b*x^2+a),x, algorithm="maxima")
 

Output:

-1/4*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(c*x)^(2/3) - (a*c^2/b)^(1/3))/(a*c^ 
2/b)^(1/3))/((a*c^2/b)^(1/3)*a) + log((c*x)^(4/3) - (a*c^2/b)^(1/3)*(c*x)^ 
(2/3) + (a*c^2/b)^(2/3))/((a*c^2/b)^(1/3)*a) - 2*log((c*x)^(2/3) + (a*c^2/ 
b)^(1/3))/((a*c^2/b)^(1/3)*a) + 6/((c*x)^(2/3)*a))/c
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=\frac {\frac {2 \, \left (-\frac {a c^{2}}{b}\right )^{\frac {2}{3}} b \log \left ({\left | \left (c x\right )^{\frac {2}{3}} - \left (-\frac {a c^{2}}{b}\right )^{\frac {1}{3}} \right |}\right )}{a^{2} c^{2}} - \frac {6}{\left (c x\right )^{\frac {2}{3}} a} + \frac {2 \, \sqrt {3} \left (-a b^{2} c^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (c x\right )^{\frac {2}{3}} + \left (-\frac {a c^{2}}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a c^{2}}{b}\right )^{\frac {1}{3}}}\right )}{a^{2} b c^{2}} - \frac {\left (-a b^{2} c^{2}\right )^{\frac {2}{3}} \log \left (\left (c x\right )^{\frac {1}{3}} c x + \left (-\frac {a c^{2}}{b}\right )^{\frac {1}{3}} \left (c x\right )^{\frac {2}{3}} + \left (-\frac {a c^{2}}{b}\right )^{\frac {2}{3}}\right )}{a^{2} b c^{2}}}{4 \, c} \] Input:

integrate(1/(c*x)^(5/3)/(b*x^2+a),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/4*(2*(-a*c^2/b)^(2/3)*b*log(abs((c*x)^(2/3) - (-a*c^2/b)^(1/3)))/(a^2*c^ 
2) - 6/((c*x)^(2/3)*a) + 2*sqrt(3)*(-a*b^2*c^2)^(2/3)*arctan(1/3*sqrt(3)*( 
2*(c*x)^(2/3) + (-a*c^2/b)^(1/3))/(-a*c^2/b)^(1/3))/(a^2*b*c^2) - (-a*b^2* 
c^2)^(2/3)*log((c*x)^(1/3)*c*x + (-a*c^2/b)^(1/3)*(c*x)^(2/3) + (-a*c^2/b) 
^(2/3))/(a^2*b*c^2))/c
 

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=\frac {b^{1/3}\,\ln \left (81\,a^4\,b^6\,c^5+81\,a^{11/3}\,b^{19/3}\,c^{13/3}\,{\left (c\,x\right )}^{2/3}\right )}{2\,a^{4/3}\,c^{5/3}}-\frac {3}{2\,a\,c\,{\left (c\,x\right )}^{2/3}}-\frac {b^{1/3}\,\ln \left (81\,a^4\,b^6\,c^5-81\,a^{11/3}\,b^{19/3}\,c^{13/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (c\,x\right )}^{2/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{4/3}\,c^{5/3}}+\frac {b^{1/3}\,\ln \left (81\,a^4\,b^6\,c^5+162\,a^{11/3}\,b^{19/3}\,c^{13/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (c\,x\right )}^{2/3}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{a^{4/3}\,c^{5/3}} \] Input:

int(1/((c*x)^(5/3)*(a + b*x^2)),x)
 

Output:

(b^(1/3)*log(81*a^4*b^6*c^5 + 81*a^(11/3)*b^(19/3)*c^(13/3)*(c*x)^(2/3)))/ 
(2*a^(4/3)*c^(5/3)) - 3/(2*a*c*(c*x)^(2/3)) - (b^(1/3)*log(81*a^4*b^6*c^5 
- 81*a^(11/3)*b^(19/3)*c^(13/3)*((3^(1/2)*1i)/2 + 1/2)*(c*x)^(2/3))*((3^(1 
/2)*1i)/2 + 1/2))/(2*a^(4/3)*c^(5/3)) + (b^(1/3)*log(81*a^4*b^6*c^5 + 162* 
a^(11/3)*b^(19/3)*c^(13/3)*((3^(1/2)*1i)/4 - 1/4)*(c*x)^(2/3))*((3^(1/2)*1 
i)/4 - 1/4))/(a^(4/3)*c^(5/3))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(c x)^{5/3} \left (a+b x^2\right )} \, dx=\frac {2 x^{\frac {2}{3}} b^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}-2 x^{\frac {1}{3}} b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )+2 x^{\frac {2}{3}} b^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+2 x^{\frac {1}{3}} b^{\frac {1}{3}}}{b^{\frac {1}{6}} a^{\frac {1}{6}}}\right )-6 a^{\frac {1}{3}}+2 x^{\frac {2}{3}} b^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+x^{\frac {2}{3}} b^{\frac {1}{3}}\right )-x^{\frac {2}{3}} b^{\frac {1}{3}} \mathrm {log}\left (-x^{\frac {1}{3}} b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+a^{\frac {1}{3}}+x^{\frac {2}{3}} b^{\frac {1}{3}}\right )-x^{\frac {2}{3}} b^{\frac {1}{3}} \mathrm {log}\left (x^{\frac {1}{3}} b^{\frac {1}{6}} a^{\frac {1}{6}} \sqrt {3}+a^{\frac {1}{3}}+x^{\frac {2}{3}} b^{\frac {1}{3}}\right )}{4 c^{\frac {5}{3}} x^{\frac {2}{3}} a^{\frac {4}{3}}} \] Input:

int(1/(c*x)^(5/3)/(b*x^2+a),x)
 

Output:

(c**(1/3)*(2*x**(2/3)*b**(1/3)*sqrt(3)*atan((b**(1/6)*a**(1/6)*sqrt(3) - 2 
*x**(1/3)*b**(1/3))/(b**(1/6)*a**(1/6))) + 2*x**(2/3)*b**(1/3)*sqrt(3)*ata 
n((b**(1/6)*a**(1/6)*sqrt(3) + 2*x**(1/3)*b**(1/3))/(b**(1/6)*a**(1/6))) - 
 6*a**(1/3) + 2*x**(2/3)*b**(1/3)*log(a**(1/3) + x**(2/3)*b**(1/3)) - x**( 
2/3)*b**(1/3)*log( - x**(1/3)*b**(1/6)*a**(1/6)*sqrt(3) + a**(1/3) + x**(2 
/3)*b**(1/3)) - x**(2/3)*b**(1/3)*log(x**(1/3)*b**(1/6)*a**(1/6)*sqrt(3) + 
 a**(1/3) + x**(2/3)*b**(1/3))))/(4*x**(2/3)*a**(1/3)*a*c**2)