\(\int \frac {\sqrt [4]{a+b x}}{x^4 \sqrt [4]{c+d x}} \, dx\) [500]

Optimal result

Integrand size = 22, antiderivative size = 266 \[ \int \frac {\sqrt [4]{a+b x}}{x^4 \sqrt [4]{c+d x}} \, dx=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}-\frac {(b c-9 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{24 a c^2 x^2}+\frac {(7 b c-15 a d) (b c+3 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{96 a^2 c^3 x}-\frac {(b c-a d) \left (7 b^2 c^2+10 a b c d+15 a^2 d^2\right ) \arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{64 a^{11/4} c^{13/4}}-\frac {(b c-a d) \left (7 b^2 c^2+10 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{64 a^{11/4} c^{13/4}} \] Output:

-1/3*(b*x+a)^(1/4)*(d*x+c)^(3/4)/c/x^3-1/24*(-9*a*d+b*c)*(b*x+a)^(1/4)*(d* 
x+c)^(3/4)/a/c^2/x^2+1/96*(-15*a*d+7*b*c)*(3*a*d+b*c)*(b*x+a)^(1/4)*(d*x+c 
)^(3/4)/a^2/c^3/x-1/64*(-a*d+b*c)*(15*a^2*d^2+10*a*b*c*d+7*b^2*c^2)*arctan 
(c^(1/4)*(b*x+a)^(1/4)/a^(1/4)/(d*x+c)^(1/4))/a^(11/4)/c^(13/4)-1/64*(-a*d 
+b*c)*(15*a^2*d^2+10*a*b*c*d+7*b^2*c^2)*arctanh(c^(1/4)*(b*x+a)^(1/4)/a^(1 
/4)/(d*x+c)^(1/4))/a^(11/4)/c^(13/4)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 10.06 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt [4]{a+b x}}{x^4 \sqrt [4]{c+d x}} \, dx=\frac {\sqrt [4]{a+b x} \left (a (c+d x) \left (7 b^2 c^2 x^2+2 a b c x (-2 c+3 d x)+a^2 \left (-32 c^2+36 c d x-45 d^2 x^2\right )\right )-3 \left (7 b^3 c^3+3 a b^2 c^2 d+5 a^2 b c d^2-15 a^3 d^3\right ) x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {c (a+b x)}{a (c+d x)}\right )\right )}{96 a^3 c^3 x^3 \sqrt [4]{c+d x}} \] Input:

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

Output:

((a + b*x)^(1/4)*(a*(c + d*x)*(7*b^2*c^2*x^2 + 2*a*b*c*x*(-2*c + 3*d*x) + 
a^2*(-32*c^2 + 36*c*d*x - 45*d^2*x^2)) - 3*(7*b^3*c^3 + 3*a*b^2*c^2*d + 5* 
a^2*b*c*d^2 - 15*a^3*d^3)*x^3*Hypergeometric2F1[1/4, 1, 5/4, (c*(a + b*x)) 
/(a*(c + d*x))]))/(96*a^3*c^3*x^3*(c + d*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {110, 27, 168, 27, 168, 27, 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.

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

\(\Big \downarrow \) 110

\(\displaystyle \frac {\int \frac {b c-9 a d-8 b d x}{4 x^3 (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{3 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b c-9 a d-8 b d x}{x^3 (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {-\frac {\int \frac {(7 b c-15 a d) (b c+3 a d)+4 b d (b c-9 a d) x}{4 x^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{2 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {(7 b c-15 a d) (b c+3 a d)+4 b d (b c-9 a d) x}{x^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {-\frac {-\frac {\int \frac {3 (b c-a d) \left (7 b^2 c^2+10 a b d c+15 a^2 d^2\right )}{4 x (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 b c-15 a d) (3 a d+b c)}{a c x}}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {3 (b c-a d) \left (15 a^2 d^2+10 a b c d+7 b^2 c^2\right ) \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}}dx}{4 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 b c-15 a d) (3 a d+b c)}{a c x}}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {-\frac {-\frac {3 (b c-a d) \left (15 a^2 d^2+10 a b c d+7 b^2 c^2\right ) \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 b c-15 a d) (3 a d+b c)}{a c x}}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {-\frac {-\frac {3 (b c-a d) \left (15 a^2 d^2+10 a b c d+7 b^2 c^2\right ) \left (-\frac {\int \frac {1}{\sqrt {a}-\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c+d x}}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a}+\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c+d x}}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{2 \sqrt {a}}\right )}{a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 b c-15 a d) (3 a d+b c)}{a c x}}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {-\frac {-\frac {3 (b c-a d) \left (15 a^2 d^2+10 a b c d+7 b^2 c^2\right ) \left (-\frac {\int \frac {1}{\sqrt {a}-\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c+d x}}}d\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} \sqrt [4]{c}}\right )}{a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 b c-15 a d) (3 a d+b c)}{a c x}}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {-\frac {3 (b c-a d) \left (15 a^2 d^2+10 a b c d+7 b^2 c^2\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} \sqrt [4]{c}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{2 a^{3/4} \sqrt [4]{c}}\right )}{a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (7 b c-15 a d) (3 a d+b c)}{a c x}}{8 a c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (b c-9 a d)}{2 a c x^2}}{12 c}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{3 c x^3}\)

Input:

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

Output:

-1/3*((a + b*x)^(1/4)*(c + d*x)^(3/4))/(c*x^3) + (-1/2*((b*c - 9*a*d)*(a + 
 b*x)^(1/4)*(c + d*x)^(3/4))/(a*c*x^2) - (-(((7*b*c - 15*a*d)*(b*c + 3*a*d 
)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(a*c*x)) - (3*(b*c - a*d)*(7*b^2*c^2 + 
10*a*b*c*d + 15*a^2*d^2)*(-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))))/(a*c))/(8*a*c))/(12*c)
 

Defintions of rubi rules used

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 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 110

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[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[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]

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 1831, normalized size of antiderivative = 6.88 \[ \int \frac {\sqrt [4]{a+b x}}{x^4 \sqrt [4]{c+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/384*(3*a^2*c^3*x^3*((2401*b^12*c^12 + 4116*a*b^11*c^11*d + 9506*a^2*b^1 
0*c^10*d^2 - 11004*a^3*b^9*c^9*d^3 - 15249*a^4*b^8*c^8*d^4 - 48600*a^5*b^7 
*c^7*d^5 + 31580*a^6*b^6*c^6*d^6 + 18600*a^7*b^5*c^5*d^7 + 93775*a^8*b^4*c 
^4*d^8 - 61500*a^9*b^3*c^3*d^9 - 6750*a^10*b^2*c^2*d^10 - 67500*a^11*b*c*d 
^11 + 50625*a^12*d^12)/(a^11*c^13))^(1/4)*log(-((7*b^3*c^3 + 3*a*b^2*c^2*d 
 + 5*a^2*b*c*d^2 - 15*a^3*d^3)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (a^3*c^3* 
d*x + a^3*c^4)*((2401*b^12*c^12 + 4116*a*b^11*c^11*d + 9506*a^2*b^10*c^10* 
d^2 - 11004*a^3*b^9*c^9*d^3 - 15249*a^4*b^8*c^8*d^4 - 48600*a^5*b^7*c^7*d^ 
5 + 31580*a^6*b^6*c^6*d^6 + 18600*a^7*b^5*c^5*d^7 + 93775*a^8*b^4*c^4*d^8 
- 61500*a^9*b^3*c^3*d^9 - 6750*a^10*b^2*c^2*d^10 - 67500*a^11*b*c*d^11 + 5 
0625*a^12*d^12)/(a^11*c^13))^(1/4))/(d*x + c)) - 3*a^2*c^3*x^3*((2401*b^12 
*c^12 + 4116*a*b^11*c^11*d + 9506*a^2*b^10*c^10*d^2 - 11004*a^3*b^9*c^9*d^ 
3 - 15249*a^4*b^8*c^8*d^4 - 48600*a^5*b^7*c^7*d^5 + 31580*a^6*b^6*c^6*d^6 
+ 18600*a^7*b^5*c^5*d^7 + 93775*a^8*b^4*c^4*d^8 - 61500*a^9*b^3*c^3*d^9 - 
6750*a^10*b^2*c^2*d^10 - 67500*a^11*b*c*d^11 + 50625*a^12*d^12)/(a^11*c^13 
))^(1/4)*log(-((7*b^3*c^3 + 3*a*b^2*c^2*d + 5*a^2*b*c*d^2 - 15*a^3*d^3)*(b 
*x + a)^(1/4)*(d*x + c)^(3/4) - (a^3*c^3*d*x + a^3*c^4)*((2401*b^12*c^12 + 
 4116*a*b^11*c^11*d + 9506*a^2*b^10*c^10*d^2 - 11004*a^3*b^9*c^9*d^3 - 152 
49*a^4*b^8*c^8*d^4 - 48600*a^5*b^7*c^7*d^5 + 31580*a^6*b^6*c^6*d^6 + 18600 
*a^7*b^5*c^5*d^7 + 93775*a^8*b^4*c^4*d^8 - 61500*a^9*b^3*c^3*d^9 - 6750...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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