\(\int \frac {1}{x (a+b x^8) \sqrt {c+d x^8}} \, dx\) [94]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^8}}{\sqrt {c}}\right )}{4 a \sqrt {c}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {b c-a d}}\right )}{4 a \sqrt {b c-a d}} \] Output:

-1/4*arctanh((d*x^8+c)^(1/2)/c^(1/2))/a/c^(1/2)+1/4*b^(1/2)*arctanh(b^(1/2 
)*(d*x^8+c)^(1/2)/(-a*d+b*c)^(1/2))/a/(-a*d+b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=-\frac {\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^8}}{\sqrt {c}}\right )}{\sqrt {c}}}{4 a} \] Input:

Integrate[1/(x*(a + b*x^8)*Sqrt[c + d*x^8]),x]
 

Output:

-1/4*((Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^8])/Sqrt[-(b*c) + a*d]])/Sqrt[ 
-(b*c) + a*d] + ArcTanh[Sqrt[c + d*x^8]/Sqrt[c]]/Sqrt[c])/a
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {948, 97, 73, 221}

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.

\(\displaystyle \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{8} \int \frac {1}{x^8 \left (b x^8+a\right ) \sqrt {d x^8+c}}dx^8\)

\(\Big \downarrow \) 97

\(\displaystyle \frac {1}{8} \left (\frac {\int \frac {1}{x^8 \sqrt {d x^8+c}}dx^8}{a}-\frac {b \int \frac {1}{\left (b x^8+a\right ) \sqrt {d x^8+c}}dx^8}{a}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{8} \left (\frac {2 \int \frac {1}{\frac {x^{16}}{d}-\frac {c}{d}}d\sqrt {d x^8+c}}{a d}-\frac {2 b \int \frac {1}{\frac {b x^{16}}{d}+a-\frac {b c}{d}}d\sqrt {d x^8+c}}{a d}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{8} \left (\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^8}}{\sqrt {c}}\right )}{a \sqrt {c}}\right )\)

Input:

Int[1/(x*(a + b*x^8)*Sqrt[c + d*x^8]),x]
 

Output:

((-2*ArcTanh[Sqrt[c + d*x^8]/Sqrt[c]])/(a*Sqrt[c]) + (2*Sqrt[b]*ArcTanh[(S 
qrt[b]*Sqrt[c + d*x^8])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/8
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[p]
 

rule 221

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

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{8}+c}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {b \arctan \left (\frac {\sqrt {d \,x^{8}+c}\, b}{\sqrt {\left (a d -c b \right ) b}}\right )}{\sqrt {\left (a d -c b \right ) b}}}{4 a}\)	\(65\)

Input:

int(1/x/(b*x^8+a)/(d*x^8+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/4/a*(-arctanh((d*x^8+c)^(1/2)/c^(1/2))/c^(1/2)-b/((a*d-b*c)*b)^(1/2)*arc 
tan((d*x^8+c)^(1/2)*b/((a*d-b*c)*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 385, normalized size of antiderivative = 4.53 \[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\left [\frac {c \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{8} + 2 \, b c - a d + 2 \, \sqrt {d x^{8} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{8} + a}\right ) + \sqrt {c} \log \left (\frac {d x^{8} - 2 \, \sqrt {d x^{8} + c} \sqrt {c} + 2 \, c}{x^{8}}\right )}{8 \, a c}, -\frac {2 \, c \sqrt {-\frac {b}{b c - a d}} \arctan \left (\sqrt {d x^{8} + c} \sqrt {-\frac {b}{b c - a d}}\right ) - \sqrt {c} \log \left (\frac {d x^{8} - 2 \, \sqrt {d x^{8} + c} \sqrt {c} + 2 \, c}{x^{8}}\right )}{8 \, a c}, \frac {c \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{8} + 2 \, b c - a d + 2 \, \sqrt {d x^{8} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{8} + a}\right ) + 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{8} + c}}\right )}{8 \, a c}, -\frac {c \sqrt {-\frac {b}{b c - a d}} \arctan \left (\sqrt {d x^{8} + c} \sqrt {-\frac {b}{b c - a d}}\right ) - \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{8} + c}}\right )}{4 \, a c}\right ] \] Input:

integrate(1/x/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/8*(c*sqrt(b/(b*c - a*d))*log((b*d*x^8 + 2*b*c - a*d + 2*sqrt(d*x^8 + c) 
*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^8 + a)) + sqrt(c)*log((d*x^8 - 2*sq 
rt(d*x^8 + c)*sqrt(c) + 2*c)/x^8))/(a*c), -1/8*(2*c*sqrt(-b/(b*c - a*d))*a 
rctan(sqrt(d*x^8 + c)*sqrt(-b/(b*c - a*d))) - sqrt(c)*log((d*x^8 - 2*sqrt( 
d*x^8 + c)*sqrt(c) + 2*c)/x^8))/(a*c), 1/8*(c*sqrt(b/(b*c - a*d))*log((b*d 
*x^8 + 2*b*c - a*d + 2*sqrt(d*x^8 + c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b 
*x^8 + a)) + 2*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^8 + c)))/(a*c), -1/4*(c*s 
qrt(-b/(b*c - a*d))*arctan(sqrt(d*x^8 + c)*sqrt(-b/(b*c - a*d))) - sqrt(-c 
)*arctan(sqrt(-c)/sqrt(d*x^8 + c)))/(a*c)]
 

Sympy [A] (verification not implemented)

Time = 12.95 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\begin {cases} \frac {2 \left (- \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{8}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{8 a \sqrt {\frac {a d - b c}{b}}} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{8}}}{\sqrt {- c}} \right )}}{8 a \sqrt {- c}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\operatorname {atan}{\left (\frac {2 \left (\frac {a}{2 b} + x^{8}\right )}{\sqrt {- \frac {a^{2}}{b^{2}}}} \right )}}{4 b \sqrt {c} \sqrt {- \frac {a^{2}}{b^{2}}}} & \text {otherwise} \end {cases} \] Input:

integrate(1/x/(b*x**8+a)/(d*x**8+c)**(1/2),x)
 

Output:

Piecewise((2*(-d*atan(sqrt(c + d*x**8)/sqrt((a*d - b*c)/b))/(8*a*sqrt((a*d 
 - b*c)/b)) + d*atan(sqrt(c + d*x**8)/sqrt(-c))/(8*a*sqrt(-c)))/d, Ne(d, 0 
)), (atan(2*(a/(2*b) + x**8)/sqrt(-a**2/b**2))/(4*b*sqrt(c)*sqrt(-a**2/b** 
2)), True))
 

Maxima [F]

\[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\int { \frac {1}{{\left (b x^{8} + a\right )} \sqrt {d x^{8} + c} x} \,d x } \] Input:

integrate(1/x/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^8 + a)*sqrt(d*x^8 + c)*x), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=-\frac {b \arctan \left (\frac {\sqrt {d x^{8} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, \sqrt {-b^{2} c + a b d} a} + \frac {\arctan \left (\frac {\sqrt {d x^{8} + c}}{\sqrt {-c}}\right )}{4 \, a \sqrt {-c}} \] Input:

integrate(1/x/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")
 

Output:

-1/4*b*arctan(sqrt(d*x^8 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d 
)*a) + 1/4*arctan(sqrt(d*x^8 + c)/sqrt(-c))/(a*sqrt(-c))
 

Mupad [B] (verification not implemented)

Time = 3.79 (sec) , antiderivative size = 652, normalized size of antiderivative = 7.67 \[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^8+c}}{\sqrt {c}}\right )}{4\,a\,\sqrt {c}}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (\frac {b^3\,d^2\,\sqrt {d\,x^8+c}}{4}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^8+c}\,\sqrt {b^2\,c-a\,b\,d}}{8\,\left (a^2\,d-a\,b\,c\right )}\right )}{8\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{8\,\left (a^2\,d-a\,b\,c\right )}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (\frac {b^3\,d^2\,\sqrt {d\,x^8+c}}{4}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^8+c}\,\sqrt {b^2\,c-a\,b\,d}}{8\,\left (a^2\,d-a\,b\,c\right )}\right )}{8\,\left (a^2\,d-a\,b\,c\right )}\right )\,1{}\mathrm {i}}{8\,\left (a^2\,d-a\,b\,c\right )}}{\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (\frac {b^3\,d^2\,\sqrt {d\,x^8+c}}{4}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (a^2\,b^2\,d^3-\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^8+c}\,\sqrt {b^2\,c-a\,b\,d}}{8\,\left (a^2\,d-a\,b\,c\right )}\right )}{8\,\left (a^2\,d-a\,b\,c\right )}\right )}{8\,\left (a^2\,d-a\,b\,c\right )}-\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (\frac {b^3\,d^2\,\sqrt {d\,x^8+c}}{4}+\frac {\sqrt {b^2\,c-a\,b\,d}\,\left (a^2\,b^2\,d^3+\frac {\left (8\,a^3\,b^2\,d^3-16\,a^2\,b^3\,c\,d^2\right )\,\sqrt {d\,x^8+c}\,\sqrt {b^2\,c-a\,b\,d}}{8\,\left (a^2\,d-a\,b\,c\right )}\right )}{8\,\left (a^2\,d-a\,b\,c\right )}\right )}{8\,\left (a^2\,d-a\,b\,c\right )}}\right )\,\sqrt {b^2\,c-a\,b\,d}\,1{}\mathrm {i}}{4\,\left (a^2\,d-a\,b\,c\right )} \] Input:

int(1/(x*(a + b*x^8)*(c + d*x^8)^(1/2)),x)
 

Output:

- atanh((c + d*x^8)^(1/2)/c^(1/2))/(4*a*c^(1/2)) - (atan((((b^2*c - a*b*d) 
^(1/2)*((b^3*d^2*(c + d*x^8)^(1/2))/4 - ((b^2*c - a*b*d)^(1/2)*(a^2*b^2*d^ 
3 - ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x^8)^(1/2)*(b^2*c - a*b*d)^ 
(1/2))/(8*(a^2*d - a*b*c))))/(8*(a^2*d - a*b*c)))*1i)/(8*(a^2*d - a*b*c)) 
+ ((b^2*c - a*b*d)^(1/2)*((b^3*d^2*(c + d*x^8)^(1/2))/4 + ((b^2*c - a*b*d) 
^(1/2)*(a^2*b^2*d^3 + ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2)*(c + d*x^8)^(1/2 
)*(b^2*c - a*b*d)^(1/2))/(8*(a^2*d - a*b*c))))/(8*(a^2*d - a*b*c)))*1i)/(8 
*(a^2*d - a*b*c)))/(((b^2*c - a*b*d)^(1/2)*((b^3*d^2*(c + d*x^8)^(1/2))/4 
- ((b^2*c - a*b*d)^(1/2)*(a^2*b^2*d^3 - ((8*a^3*b^2*d^3 - 16*a^2*b^3*c*d^2 
)*(c + d*x^8)^(1/2)*(b^2*c - a*b*d)^(1/2))/(8*(a^2*d - a*b*c))))/(8*(a^2*d 
 - a*b*c))))/(8*(a^2*d - a*b*c)) - ((b^2*c - a*b*d)^(1/2)*((b^3*d^2*(c + d 
*x^8)^(1/2))/4 + ((b^2*c - a*b*d)^(1/2)*(a^2*b^2*d^3 + ((8*a^3*b^2*d^3 - 1 
6*a^2*b^3*c*d^2)*(c + d*x^8)^(1/2)*(b^2*c - a*b*d)^(1/2))/(8*(a^2*d - a*b* 
c))))/(8*(a^2*d - a*b*c))))/(8*(a^2*d - a*b*c))))*(b^2*c - a*b*d)^(1/2)*1i 
)/(4*(a^2*d - a*b*c))
 

Reduce [F]

\[ \int \frac {1}{x \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx=\frac {\sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{8}+c}-\sqrt {c}\right )-\sqrt {c}\, \mathrm {log}\left (\sqrt {d \,x^{8}+c}+\sqrt {c}\right )-8 \left (\int \frac {\sqrt {d \,x^{8}+c}\, x^{7}}{b d \,x^{16}+a d \,x^{8}+b c \,x^{8}+a c}d x \right ) b c}{8 a c} \] Input:

int(1/x/(b*x^8+a)/(d*x^8+c)^(1/2),x)
 

Output:

(sqrt(c)*log(sqrt(c + d*x**8) - sqrt(c)) - sqrt(c)*log(sqrt(c + d*x**8) + 
sqrt(c)) - 8*int((sqrt(c + d*x**8)*x**7)/(a*c + a*d*x**8 + b*c*x**8 + b*d* 
x**16),x)*b*c)/(8*a*c)