\(\int \frac {(a+b x^2)^{7/6}}{x^2} \, dx\) [1094]

Optimal result

Integrand size = 15, antiderivative size = 282 \[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=\frac {7}{4} b x \sqrt [6]{a+b x^2}-\frac {\left (a+b x^2\right )^{7/6}}{x}+\frac {7 a^{2/3} \sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{8 \sqrt [4]{3} x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}} \] Output:

7/4*b*x*(b*x^2+a)^(1/6)-(b*x^2+a)^(7/6)/x+7/24*a^(2/3)*(b*x^2+a)^(1/6)*(a^ 
(1/3)-(b*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/ 
(a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))^2)^(1/2)*InverseJacobiAM(arccos((a^( 
1/3)-(1-3^(1/2))*(b*x^2+a)^(1/3))/(a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))),1 
/4*6^(1/2)+1/4*2^(1/2))*3^(3/4)/x/(-(b*x^2+a)^(1/3)*(a^(1/3)-(b*x^2+a)^(1/ 
3))/(a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.62 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.18 \[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=-\frac {a \sqrt [6]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},-\frac {1}{2},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \sqrt [6]{1+\frac {b x^2}{a}}} \] Input:

Integrate[(a + b*x^2)^(7/6)/x^2,x]
 

Output:

-((a*(a + b*x^2)^(1/6)*Hypergeometric2F1[-7/6, -1/2, 1/2, -((b*x^2)/a)])/( 
x*(1 + (b*x^2)/a)^(1/6)))
 

Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {247, 211, 236, 234, 760}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {7}{3} b \int \sqrt [6]{b x^2+a}dx-\frac {\left (a+b x^2\right )^{7/6}}{x}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {7}{3} b \left (\frac {1}{4} a \int \frac {1}{\left (b x^2+a\right )^{5/6}}dx+\frac {3}{4} x \sqrt [6]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{7/6}}{x}\)

\(\Big \downarrow \) 236

\(\displaystyle \frac {7}{3} b \left (\frac {a \int \frac {1}{\left (1-\frac {b x^2}{b x^2+a}\right )^{2/3}}d\frac {x}{\sqrt {b x^2+a}}}{4 \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}+\frac {3}{4} x \sqrt [6]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{7/6}}{x}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {7}{3} b \left (\frac {3}{4} x \sqrt [6]{a+b x^2}-\frac {3 a \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{8 b x \sqrt [3]{\frac {a}{a+b x^2}}}\right )-\frac {\left (a+b x^2\right )^{7/6}}{x}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {7}{3} b \left (\frac {3^{3/4} \sqrt {2-\sqrt {3}} a \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{4 b x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}+\frac {3}{4} x \sqrt [6]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{7/6}}{x}\)

Input:

Int[(a + b*x^2)^(7/6)/x^2,x]
 

Output:

-((a + b*x^2)^(7/6)/x) + (7*b*((3*x*(a + b*x^2)^(1/6))/4 + (3^(3/4)*Sqrt[2 
 - Sqrt[3]]*a*Sqrt[-((b*x^2)/(a + b*x^2))]*(a + b*x^2)^(1/6)*(1 - (1 - (b* 
x^2)/(a + b*x^2))^(1/3))*Sqrt[(1 + x^2/(a + b*x^2) + (1 - (b*x^2)/(a + b*x 
^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2]*EllipticF[A 
rcSin[(1 + Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - 
(b*x^2)/(a + b*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(4*b*x*(a/(a + b*x^2))^(1/3 
)*Sqrt[-1 + x^3/(a + b*x^2)^(3/2)]*Sqrt[-((1 - (1 - (b*x^2)/(a + b*x^2))^( 
1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2)])))/3
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 234

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

rule 236

Int[((a_) + (b_.)*(x_)^2)^(-5/6), x_Symbol] :> Simp[1/((a/(a + b*x^2))^(1/3 
)*(a + b*x^2)^(1/3))   Subst[Int[1/(1 - b*x^2)^(2/3), x], x, x/Sqrt[a + b*x 
^2]], x] /; FreeQ[{a, b}, x]
 

rule 247

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

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {7}{6}}}{x^{2}}d x\]

Input:

int((b*x^2+a)^(7/6)/x^2,x)
 

Output:

int((b*x^2+a)^(7/6)/x^2,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {7}{6}}}{x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^(7/6)/x^2,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(7/6)/x^2, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.94 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.10 \[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=- \frac {a^{\frac {7}{6}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{6}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \] Input:

integrate((b*x**2+a)**(7/6)/x**2,x)
 

Output:

-a**(7/6)*hyper((-7/6, -1/2), (1/2,), b*x**2*exp_polar(I*pi)/a)/x
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {7}{6}}}{x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^(7/6)/x^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(7/6)/x^2, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {7}{6}}}{x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^(7/6)/x^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((b*x^2 + a)^(7/6)/x^2, x)
 

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.14 \[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=\frac {3\,{\left (b\,x^2+a\right )}^{7/6}\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{6},-\frac {2}{3};\ \frac {1}{3};\ -\frac {a}{b\,x^2}\right )}{4\,x\,{\left (\frac {a}{b\,x^2}+1\right )}^{7/6}} \] Input:

int((a + b*x^2)^(7/6)/x^2,x)
 

Output:

(3*(a + b*x^2)^(7/6)*hypergeom([-7/6, -2/3], 1/3, -a/(b*x^2)))/(4*x*(a/(b* 
x^2) + 1)^(7/6))
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{7/6}}{x^2} \, dx=\frac {-\left (b \,x^{2}+a \right )^{\frac {5}{6}} a -\left (b \,x^{2}+a \right )^{\frac {5}{6}} b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {2}{3}} \left (\int \left (b \,x^{2}+a \right )^{\frac {1}{6}}d x \right ) b x}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} x} \] Input:

int((b*x^2+a)^(7/6)/x^2,x)
 

Output:

( - (a + b*x**2)**(5/6)*a - (a + b*x**2)**(5/6)*b*x**2 + (a + b*x**2)**(2/ 
3)*int((a + b*x**2)**(1/6),x)*b*x)/((a + b*x**2)**(2/3)*x)