\(\int \frac {A+B x^4}{x^3 (a+b x^4) (c+d x^4)} \, dx\) [9]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {A}{2 a c x^2}-\frac {\sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 a^{3/2} (b c-a d)}-\frac {\sqrt {d} (B c-A d) \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 c^{3/2} (b c-a d)} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.96 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {-\sqrt {b} (A b-a B) c^{3/2} x^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )-\sqrt {b} (A b-a B) c^{3/2} x^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+\sqrt {a} \left (A \sqrt {c} (b c-a d)+a \sqrt {d} (-B c+A d) x^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+a \sqrt {d} (-B c+A d) x^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )\right )}{2 a^{3/2} c^{3/2} (-b c+a d) x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1045, 445, 397, 218}

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.

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

\(\Big \downarrow \) 1045

\(\displaystyle \frac {1}{2} \int \frac {B x^4+A}{x^4 \left (b x^4+a\right ) \left (d x^4+c\right )}dx^2\)

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {\sqrt {b} c (A b-a B) \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {a} (b c-a d)}+\frac {a \sqrt {d} (B c-A d) \arctan \left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}}{a c}-\frac {A}{a c x^2}\right )\)

Input:

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

Output:

(-(A/(a*c*x^2)) - ((Sqrt[b]*(A*b - a*B)*c*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/( 
Sqrt[a]*(b*c - a*d)) + (a*Sqrt[d]*(B*c - A*d)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c] 
])/(Sqrt[c]*(b*c - a*d)))/(a*c))/2
 

Defintions of rubi rules used

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 397

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

rule 445

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p 
+ 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) 
 Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c 
+ a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
 

rule 1045

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Si 
mp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q*(e 
 + f*x^(n/k))^r, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, f, p, 
q, r}, x] && IGtQ[n, 0] && IntegerQ[m]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {\left (A b -B a \right ) b \arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 a \left (a d -c b \right ) \sqrt {a b}}-\frac {A}{2 a c \,x^{2}}-\frac {\left (A d -B c \right ) d \arctan \left (\frac {d \,x^{2}}{\sqrt {c d}}\right )}{2 c \left (a d -c b \right ) \sqrt {c d}}\)	\(94\)
risch	\(\text {Expression too large to display}\)	\(1282\)

Input:

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

Output:

1/2*(A*b-B*a)*b/a/(a*d-b*c)/(a*b)^(1/2)*arctan(b*x^2/(a*b)^(1/2))-1/2*A/a/ 
c/x^2-1/2*(A*d-B*c)*d/c/(a*d-b*c)/(c*d)^(1/2)*arctan(d*x^2/(c*d)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 483, normalized size of antiderivative = 4.43 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\left [\frac {{\left (B a - A b\right )} c x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{4} + 2 \, a x^{2} \sqrt {-\frac {b}{a}} - a}{b x^{4} + a}\right ) + {\left (B a c - A a d\right )} x^{2} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{4} - 2 \, c x^{2} \sqrt {-\frac {d}{c}} - c}{d x^{4} + c}\right ) - 2 \, A b c + 2 \, A a d}{4 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}, \frac {{\left (B a - A b\right )} c x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{4} + 2 \, a x^{2} \sqrt {-\frac {b}{a}} - a}{b x^{4} + a}\right ) - 2 \, {\left (B a c - A a d\right )} x^{2} \sqrt {\frac {d}{c}} \arctan \left (x^{2} \sqrt {\frac {d}{c}}\right ) - 2 \, A b c + 2 \, A a d}{4 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}, \frac {2 \, {\left (B a - A b\right )} c x^{2} \sqrt {\frac {b}{a}} \arctan \left (x^{2} \sqrt {\frac {b}{a}}\right ) + {\left (B a c - A a d\right )} x^{2} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{4} - 2 \, c x^{2} \sqrt {-\frac {d}{c}} - c}{d x^{4} + c}\right ) - 2 \, A b c + 2 \, A a d}{4 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}, \frac {{\left (B a - A b\right )} c x^{2} \sqrt {\frac {b}{a}} \arctan \left (x^{2} \sqrt {\frac {b}{a}}\right ) - {\left (B a c - A a d\right )} x^{2} \sqrt {\frac {d}{c}} \arctan \left (x^{2} \sqrt {\frac {d}{c}}\right ) - A b c + A a d}{2 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}\right ] \] Input:

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

Output:

[1/4*((B*a - A*b)*c*x^2*sqrt(-b/a)*log((b*x^4 + 2*a*x^2*sqrt(-b/a) - a)/(b 
*x^4 + a)) + (B*a*c - A*a*d)*x^2*sqrt(-d/c)*log((d*x^4 - 2*c*x^2*sqrt(-d/c 
) - c)/(d*x^4 + c)) - 2*A*b*c + 2*A*a*d)/((a*b*c^2 - a^2*c*d)*x^2), 1/4*(( 
B*a - A*b)*c*x^2*sqrt(-b/a)*log((b*x^4 + 2*a*x^2*sqrt(-b/a) - a)/(b*x^4 + 
a)) - 2*(B*a*c - A*a*d)*x^2*sqrt(d/c)*arctan(x^2*sqrt(d/c)) - 2*A*b*c + 2* 
A*a*d)/((a*b*c^2 - a^2*c*d)*x^2), 1/4*(2*(B*a - A*b)*c*x^2*sqrt(b/a)*arcta 
n(x^2*sqrt(b/a)) + (B*a*c - A*a*d)*x^2*sqrt(-d/c)*log((d*x^4 - 2*c*x^2*sqr 
t(-d/c) - c)/(d*x^4 + c)) - 2*A*b*c + 2*A*a*d)/((a*b*c^2 - a^2*c*d)*x^2), 
1/2*((B*a - A*b)*c*x^2*sqrt(b/a)*arctan(x^2*sqrt(b/a)) - (B*a*c - A*a*d)*x 
^2*sqrt(d/c)*arctan(x^2*sqrt(d/c)) - A*b*c + A*a*d)/((a*b*c^2 - a^2*c*d)*x 
^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {{\left (B a b - A b^{2}\right )} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (a b c - a^{2} d\right )} \sqrt {a b}} - \frac {{\left (B c d - A d^{2}\right )} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c^{2} - a c d\right )} \sqrt {c d}} - \frac {A}{2 \, a c x^{2}} \] Input:

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

Output:

1/2*(B*a*b - A*b^2)*arctan(b*x^2/sqrt(a*b))/((a*b*c - a^2*d)*sqrt(a*b)) - 
1/2*(B*c*d - A*d^2)*arctan(d*x^2/sqrt(c*d))/((b*c^2 - a*c*d)*sqrt(c*d)) - 
1/2*A/(a*c*x^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {{\left (B a b - A b^{2}\right )} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (a b c - a^{2} d\right )} \sqrt {a b}} - \frac {{\left (B c d - A d^{2}\right )} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c^{2} - a c d\right )} \sqrt {c d}} - \frac {A}{2 \, a c x^{2}} \] Input:

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

Output:

1/2*(B*a*b - A*b^2)*arctan(b*x^2/sqrt(a*b))/((a*b*c - a^2*d)*sqrt(a*b)) - 
1/2*(B*c*d - A*d^2)*arctan(d*x^2/sqrt(c*d))/((b*c^2 - a*c*d)*sqrt(c*d)) - 
1/2*A/(a*c*x^2)
 

Mupad [B] (verification not implemented)

Time = 6.38 (sec) , antiderivative size = 1537, normalized size of antiderivative = 14.10 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

- (atan((A^4*a^2*d^5*x^2*(-a^3*b)^(7/2)*1i + A^4*b^6*c^5*x^2*(-a^3*b)^(5/2 
)*1i - B^4*a*b*c^5*x^2*(-a^3*b)^(7/2)*1i + A*B^3*b^2*c^5*x^2*(-a^3*b)^(7/2 
)*4i + B^4*a^2*c^4*d*x^2*(-a^3*b)^(7/2)*1i + A^4*b^2*c^2*d^3*x^2*(-a^3*b)^ 
(7/2)*1i + A^4*a*b^5*c^4*d*x^2*(-a^3*b)^(5/2)*1i - A*B^3*a^2*c^3*d^2*x^2*( 
-a^3*b)^(7/2)*4i - A^2*B^2*b^2*c^4*d*x^2*(-a^3*b)^(7/2)*4i + A^2*B^2*a^2*b 
^4*c^5*x^2*(-a^3*b)^(5/2)*6i + A^2*B^2*a^2*c^2*d^3*x^2*(-a^3*b)^(7/2)*6i + 
 A^4*a*b*c*d^4*x^2*(-a^3*b)^(7/2)*1i + A^4*a^2*b^4*c^3*d^2*x^2*(-a^3*b)^(5 
/2)*1i - A^3*B*a*b^5*c^5*x^2*(-a^3*b)^(5/2)*4i - A^3*B*a^2*c*d^4*x^2*(-a^3 
*b)^(7/2)*4i - A^3*B*a*b*c^2*d^3*x^2*(-a^3*b)^(7/2)*4i + A^2*B^2*a*b*c^3*d 
^2*x^2*(-a^3*b)^(7/2)*4i - A^3*B*a^2*b^4*c^4*d*x^2*(-a^3*b)^(5/2)*4i)/(a^6 
*b^2*(a^3*b*(A^4*a^4*d^5 - a^3*b*(B^4*c^5 - A^4*c*d^4 - 4*A^2*B^2*c^3*d^2 
+ 4*A^3*B*c^2*d^3) + 4*A^3*B*b^4*c^5 - A^4*b^4*c^4*d + B^4*a^4*c^4*d + A^4 
*a^2*b^2*c^2*d^3 - 4*A^3*B*a^4*c*d^4 + 4*A*B^3*a^2*b^2*c^5 - 6*A^2*B^2*a*b 
^3*c^5 - 4*A*B^3*a^4*c^3*d^2 - A^4*a*b^3*c^3*d^2 + 6*A^2*B^2*a^4*c^2*d^3 - 
 4*A^2*B^2*a^2*b^2*c^4*d + 4*A^3*B*a*b^3*c^4*d) - A^4*a^2*b^6*c^5)))*(A*b 
- B*a)*(-a^3*b)^(1/2)*1i)/(2*(a^4*d - a^3*b*c)) - (atan((A^4*b^5*c^2*x^2*( 
-c^3*d)^(7/2)*1i + A^4*a^5*d^6*x^2*(-c^3*d)^(5/2)*1i + A^4*a^2*b^3*d^2*x^2 
*(-c^3*d)^(7/2)*1i - B^4*a^5*c*d*x^2*(-c^3*d)^(7/2)*1i + A*B^3*a^5*d^2*x^2 
*(-c^3*d)^(7/2)*4i + B^4*a^4*b*c^2*x^2*(-c^3*d)^(7/2)*1i + A^4*a^4*b*c*d^5 
*x^2*(-c^3*d)^(5/2)*1i - A*B^3*a^3*b^2*c^2*x^2*(-c^3*d)^(7/2)*4i - A^2*...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x^4}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}-2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}+2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) x^{2}-c}{2 c^{2} x^{2}} \] Input:

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

Output:

(sqrt(d)*sqrt(c)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)* 
c**(1/4)*sqrt(2)))*x**2 + sqrt(d)*sqrt(c)*atan((d**(1/4)*c**(1/4)*sqrt(2) 
+ 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*x**2 - c)/(2*c**2*x**2)