\(\int \frac {x \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx\) [495]

Optimal result

Integrand size = 20, antiderivative size = 188 \[ \int \frac {x \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=-\frac {(5 b c+3 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 b d^2}+\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 b d}+\frac {(b c-a d) (5 b c+3 a d) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{7/4} d^{9/4}}+\frac {(b c-a d) (5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{7/4} d^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 9.73 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.93 \[ \int \frac {x \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\frac {\sqrt [4]{a+b x} \left (2 b^{3/4} \sqrt [4]{d (a+b x)} (c+d x)^{3/4} (-5 b c+a d+4 b d x)+\left (-5 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d (a+b x)}}\right )+\left (5 b^2 c^2-2 a b c d-3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d (a+b x)}}\right )\right )}{16 b^{7/4} d^2 \sqrt [4]{d (a+b x)}} \] Input:

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

Output:

((a + b*x)^(1/4)*(2*b^(3/4)*(d*(a + b*x))^(1/4)*(c + d*x)^(3/4)*(-5*b*c + 
a*d + 4*b*d*x) + (-5*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTan[(b^(1/4)*(c + 
 d*x)^(1/4))/(d*(a + b*x))^(1/4)] + (5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^2)*Ar 
cTanh[(b^(1/4)*(c + d*x)^(1/4))/(d*(a + b*x))^(1/4)]))/(16*b^(7/4)*d^2*(d* 
(a + b*x))^(1/4))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {90, 60, 73, 770, 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.

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 90

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

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]
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 1262, normalized size of antiderivative = 6.71 \[ \int \frac {x \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*(b*d^2*((625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640* 
a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2* 
c^2*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(b^7*d^9))^(1/4)*log(-((5*b^2*c^2 
- 2*a*b*c*d - 3*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (b^2*d^3*x + b^ 
2*c*d^2)*((625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3 
*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2 
*d^6 + 216*a^7*b*c*d^7 + 81*a^8*d^8)/(b^7*d^9))^(1/4))/(d*x + c)) - b*d^2* 
((625*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5* 
d^3 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 21 
6*a^7*b*c*d^7 + 81*a^8*d^8)/(b^7*d^9))^(1/4)*log(-((5*b^2*c^2 - 2*a*b*c*d 
- 3*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (b^2*d^3*x + b^2*c*d^2)*((6 
25*b^8*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 
 + 646*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a 
^7*b*c*d^7 + 81*a^8*d^8)/(b^7*d^9))^(1/4))/(d*x + c)) + I*b*d^2*((625*b^8* 
c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646* 
a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*b*c* 
d^7 + 81*a^8*d^8)/(b^7*d^9))^(1/4)*log(-((5*b^2*c^2 - 2*a*b*c*d - 3*a^2*d^ 
2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (I*b^2*d^3*x + I*b^2*c*d^2)*((625*b^8 
*c^8 - 1000*a*b^7*c^7*d - 900*a^2*b^6*c^6*d^2 + 1640*a^3*b^5*c^5*d^3 + 646 
*a^4*b^4*c^4*d^4 - 984*a^5*b^3*c^3*d^5 - 324*a^6*b^2*c^2*d^6 + 216*a^7*...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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