\(\int \frac {\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx\) [499]

Optimal result

Integrand size = 22, antiderivative size = 194 \[ \int \frac {\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx=\frac {(3 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a c^2 x}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}+\frac {(b c-a d) (3 b c+5 a d) \arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}}+\frac {(b c-a d) (3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{7/4} c^{9/4}} \] Output:

1/8*(5*a*d+3*b*c)*(b*x+a)^(1/4)*(d*x+c)^(3/4)/a/c^2/x-1/2*(b*x+a)^(5/4)*(d 
*x+c)^(3/4)/a/c/x^2+1/16*(-a*d+b*c)*(5*a*d+3*b*c)*arctan(c^(1/4)*(b*x+a)^( 
1/4)/a^(1/4)/(d*x+c)^(1/4))/a^(7/4)/c^(9/4)+1/16*(-a*d+b*c)*(5*a*d+3*b*c)* 
arctanh(c^(1/4)*(b*x+a)^(1/4)/a^(1/4)/(d*x+c)^(1/4))/a^(7/4)/c^(9/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx=\frac {\sqrt [4]{a+b x} \left (-a (c+d x) (4 a c+b c x-5 a d x)+\left (3 b^2 c^2+2 a b c d-5 a^2 d^2\right ) x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {c (a+b x)}{a (c+d x)}\right )\right )}{8 a^2 c^2 x^2 \sqrt [4]{c+d x}} \] Input:

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

Output:

((a + b*x)^(1/4)*(-(a*(c + d*x)*(4*a*c + b*c*x - 5*a*d*x)) + (3*b^2*c^2 + 
2*a*b*c*d - 5*a^2*d^2)*x^2*Hypergeometric2F1[1/4, 1, 5/4, (c*(a + b*x))/(a 
*(c + d*x))]))/(8*a^2*c^2*x^2*(c + d*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {107, 105, 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.

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

\(\Big \downarrow \) 107

\(\displaystyle -\frac {(5 a d+3 b c) \int \frac {\sqrt [4]{a+b x}}{x^2 \sqrt [4]{c+d x}}dx}{8 a c}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(5 a d+3 b c) \left (\frac {(b c-a d) \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{4 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}\right )}{8 a c}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {(5 a d+3 b c) \left (\frac {(b c-a d) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}\right )}{8 a c}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {(5 a d+3 b c) \left (\frac {(b c-a d) \left (-\frac {\int \frac {1}{\sqrt {a}-\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c+d x}}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a}+\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c+d x}}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{2 \sqrt {a}}\right )}{c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}\right )}{8 a c}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {(5 a d+3 b c) \left (\frac {(b c-a d) \left (-\frac {\int \frac {1}{\sqrt {a}-\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c+d x}}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} \sqrt [4]{c}}\right )}{c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}\right )}{8 a c}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(5 a d+3 b c) \left (\frac {(b c-a d) \left (-\frac {\arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} \sqrt [4]{c}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} \sqrt [4]{c}}\right )}{c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{c x}\right )}{8 a c}-\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 a c x^2}\)

Input:

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

Output:

-1/2*((a + b*x)^(5/4)*(c + d*x)^(3/4))/(a*c*x^2) - ((3*b*c + 5*a*d)*(-(((a 
 + b*x)^(1/4)*(c + d*x)^(3/4))/(c*x)) + ((b*c - a*d)*(-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))))/ 
c))/(8*a*c)
 

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 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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]

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 1279, normalized size of antiderivative = 6.59 \[ \int \frac {\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*(a*c^2*x^2*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984 
*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^ 
2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4)*log(-((3*b^2* 
c^2 + 2*a*b*c*d - 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (a^2*c^2*d* 
x + a^2*c^3)*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^ 
3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c 
^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4))/(d*x + c)) - a* 
c^2*x^2*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5 
*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^ 
6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4)*log(-((3*b^2*c^2 + 2* 
a*b*c*d - 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (a^2*c^2*d*x + a^2* 
c^3)*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^ 
5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 
 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4))/(d*x + c)) + I*a*c^2*x^ 
2*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d 
^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 10 
00*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*c^9))^(1/4)*log(-((3*b^2*c^2 + 2*a*b*c* 
d - 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (I*a^2*c^2*d*x + I*a^2*c^ 
3)*((81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5* 
d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 ...
 

Sympy [F]

\[ \int \frac {\sqrt [4]{a+b x}}{x^3 \sqrt [4]{c+d x}} \, dx=\int \frac {\sqrt [4]{a + b x}}{x^{3} \sqrt [4]{c + d x}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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