\(\int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\) [514]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{a^{3/4} \sqrt [4]{c}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{a^{3/4} \sqrt [4]{c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=-\frac {2 \left (\arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )\right )}{a^{3/4} \sqrt [4]{c}} \] Input:

Integrate[1/(x*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {104, 756, 218, 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.

\(\displaystyle \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\)

\(\Big \downarrow \) 104

\(\displaystyle 4 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\)

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(x*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x]
 

Output:

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

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 218

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

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 756

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

Maple [F]

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

Input:

int(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)
 

Output:

int(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.61 \[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=-\left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (\frac {{\left (a d x + a c\right )} \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (-\frac {{\left (a d x + a c\right )} \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} - {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) - i \, \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (\frac {{\left (i \, a d x + i \, a c\right )} \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) + i \, \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} \log \left (\frac {{\left (-i \, a d x - i \, a c\right )} \left (\frac {1}{a^{3} c}\right )^{\frac {1}{4}} + {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{d x + c}\right ) \] Input:

integrate(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="fricas")
 

Output:

-(1/(a^3*c))^(1/4)*log(((a*d*x + a*c)*(1/(a^3*c))^(1/4) + (b*x + a)^(1/4)* 
(d*x + c)^(3/4))/(d*x + c)) + (1/(a^3*c))^(1/4)*log(-((a*d*x + a*c)*(1/(a^ 
3*c))^(1/4) - (b*x + a)^(1/4)*(d*x + c)^(3/4))/(d*x + c)) - I*(1/(a^3*c))^ 
(1/4)*log(((I*a*d*x + I*a*c)*(1/(a^3*c))^(1/4) + (b*x + a)^(1/4)*(d*x + c) 
^(3/4))/(d*x + c)) + I*(1/(a^3*c))^(1/4)*log(((-I*a*d*x - I*a*c)*(1/(a^3*c 
))^(1/4) + (b*x + a)^(1/4)*(d*x + c)^(3/4))/(d*x + c))
 

Sympy [F]

\[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{x \left (a + b x\right )^{\frac {3}{4}} \sqrt [4]{c + d x}}\, dx \] Input:

integrate(1/x/(b*x+a)**(3/4)/(d*x+c)**(1/4),x)
 

Output:

Integral(1/(x*(a + b*x)**(3/4)*(c + d*x)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} x} \,d x } \] Input:

integrate(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(1/4)*x), x)
 

Giac [F]

\[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} x} \,d x } \] Input:

integrate(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(1/4)*x), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{x\,{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \] Input:

int(1/(x*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x)
 

Output:

int(1/(x*(a + b*x)^(3/4)*(c + d*x)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} x}d x \] Input:

int(1/x/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)
 

Output:

int(1/((c + d*x)**(1/4)*(a + b*x)**(3/4)*x),x)