\(\int \frac {(a+b x^4)^{3/4}}{c+d x^4} \, dx\) [104]

Optimal result

Integrand size = 21, antiderivative size = 173 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{c+d x^4} \, dx=\frac {b^{3/4} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 d}-\frac {(b c-a d)^{3/4} \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} d}+\frac {b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 d}-\frac {(b c-a d)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} d} \] Output:

1/2*b^(3/4)*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/d-1/2*(-a*d+b*c)^(3/4)*arcta 
n((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(3/4)/d+1/2*b^(3/4)*arctan 
h(b^(1/4)*x/(b*x^4+a)^(1/4))/d-1/2*(-a*d+b*c)^(3/4)*arctanh((-a*d+b*c)^(1/ 
4)*x/c^(1/4)/(b*x^4+a)^(1/4))/c^(3/4)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.19 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{c+d x^4} \, dx=-\frac {-2 b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {(1+i) \left ((-1+i) b^{3/4} c^{3/4} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+(b c-a d)^{3/4} \arctan \left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}-\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )+(b c-a d)^{3/4} \text {arctanh}\left (\frac {\frac {(1-i) \sqrt [4]{b c-a d} x^2}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}+\frac {(1+i) \sqrt [4]{c} \sqrt [4]{a+b x^4}}{\sqrt [4]{b c-a d}}}{2 x}\right )\right )}{c^{3/4}}}{4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {916, 770, 756, 216, 219, 902, 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.

\(\displaystyle \int \frac {\left (a+b x^4\right )^{3/4}}{c+d x^4} \, dx\)

\(\Big \downarrow \) 916

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 902

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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]
 

rule 902

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su 
bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b 
, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
 

rule 916

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(133)=266\).

Time = 4.21 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.79

method	result	size
pseudoelliptic	\(\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} c \left (2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )-\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )\right ) b^{\frac {3}{4}}+\frac {\sqrt {2}\, \left (a d -b c \right ) \left (2 \arctan \left (-\frac {\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}+1\right )-2 \arctan \left (\frac {\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}+1\right )-\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )\right )}{2}}{4 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} d c}\)	\(309\)

Input:

int((b*x^4+a)^(3/4)/(d*x^4+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 733, normalized size of antiderivative = 4.24 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{c+d x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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