\(\int \frac {1}{x^3 (a+b x^4) \sqrt {c+d x^4}} \, dx\) [243]

Optimal result

Integrand size = 24, antiderivative size = 80 \[ \int \frac {1}{x^3 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx=-\frac {\sqrt {c+d x^4}}{2 a c x^2}-\frac {b \arctan \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 a^{3/2} \sqrt {b c-a d}} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^3 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx=-\frac {\sqrt {c+d x^4}}{2 a c x^2}-\frac {b \arctan \left (\frac {a \sqrt {d}+b x^2 \left (\sqrt {d} x^2+\sqrt {c+d x^4}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{3/2} \sqrt {b c-a d}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {965, 382, 25, 27, 291, 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.

\(\displaystyle \int \frac {1}{x^3 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx\)

\(\Big \downarrow \) 965

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

\(\Big \downarrow \) 382

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-(Sqrt[c + d*x^4]/(a*c*x^2)) - (b*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*S 
qrt[c + d*x^4])])/(a^(3/2)*Sqrt[b*c - a*d]))/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 965

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), 
 x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 
 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free 
Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
 

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d \,x^{4}+c}}{x^{2}}-\frac {\operatorname {arctanh}\left (\frac {a \sqrt {d \,x^{4}+c}}{x^{2} \sqrt {a \left (a d -c b \right )}}\right ) b c}{\sqrt {a \left (a d -c b \right )}}}{2 a c}\)	\(67\)
default	\(-\frac {\sqrt {d \,x^{4}+c}}{2 a c \,x^{2}}-\frac {b \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -c b \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -c b}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -c b}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-a b}\, \sqrt {-\frac {a d -c b}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -c b \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -c b}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -c b}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-a b}\, \sqrt {-\frac {a d -c b}{b}}}\right )}{a}\)	\(349\)
risch	\(-\frac {\sqrt {d \,x^{4}+c}}{2 a c \,x^{2}}+\frac {b \ln \left (\frac {-\frac {2 \left (a d -c b \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -c b}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -c b}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 a \sqrt {-a b}\, \sqrt {-\frac {a d -c b}{b}}}-\frac {b \ln \left (\frac {-\frac {2 \left (a d -c b \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -c b}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -c b}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 a \sqrt {-a b}\, \sqrt {-\frac {a d -c b}{b}}}\)	\(350\)
elliptic	\(-\frac {\sqrt {d \,x^{4}+c}}{2 a c \,x^{2}}+\frac {b \ln \left (\frac {-\frac {2 \left (a d -c b \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -c b}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -c b}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 a \sqrt {-a b}\, \sqrt {-\frac {a d -c b}{b}}}-\frac {b \ln \left (\frac {-\frac {2 \left (a d -c b \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -c b}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -c b}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 a \sqrt {-a b}\, \sqrt {-\frac {a d -c b}{b}}}\)	\(350\)

Input:

int(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/a*(-(d*x^4+c)^(1/2)/x^2-1/(a*(a*d-b*c))^(1/2)*arctanh(a*(d*x^4+c)^(1/2 
)/x^2/(a*(a*d-b*c))^(1/2))*b*c)/c
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (64) = 128\).

Time = 0.13 (sec) , antiderivative size = 332, normalized size of antiderivative = 4.15 \[ \int \frac {1}{x^3 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx=\left [-\frac {\sqrt {-a b c + a^{2} d} b c x^{2} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{6} - a c x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) + 4 \, \sqrt {d x^{4} + c} {\left (a b c - a^{2} d\right )}}{8 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x^{2}}, -\frac {\sqrt {a b c - a^{2} d} b c x^{2} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{6} + {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )}}\right ) + 2 \, \sqrt {d x^{4} + c} {\left (a b c - a^{2} d\right )}}{4 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x^{2}}\right ] \] Input:

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="fricas")
 

Output:

[-1/8*(sqrt(-a*b*c + a^2*d)*b*c*x^2*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2) 
*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4 + a^2*c^2 + 4*((b*c - 2*a*d)*x^6 - a* 
c*x^2)*sqrt(d*x^4 + c)*sqrt(-a*b*c + a^2*d))/(b^2*x^8 + 2*a*b*x^4 + a^2)) 
+ 4*sqrt(d*x^4 + c)*(a*b*c - a^2*d))/((a^2*b*c^2 - a^3*c*d)*x^2), -1/4*(sq 
rt(a*b*c - a^2*d)*b*c*x^2*arctan(1/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 
+ c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2*d^2)*x^6 + (a*b*c^2 - a^2*c*d)*x^ 
2)) + 2*sqrt(d*x^4 + c)*(a*b*c - a^2*d))/((a^2*b*c^2 - a^3*c*d)*x^2)]
 

Sympy [F]

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

integrate(1/x**3/(b*x**4+a)/(d*x**4+c)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45 \[ \int \frac {1}{x^3 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx=\frac {1}{2} \, d^{\frac {3}{2}} {\left (\frac {b \arctan \left (\frac {{\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{\sqrt {a b c d - a^{2} d^{2}} a d} + \frac {2}{{\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} - c\right )} a d}\right )} \] Input:

integrate(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(1/(x^3*(a + b*x^4)*(c + d*x^4)^(1/2)),x)
 

Output:

int(1/(x^3*(a + b*x^4)*(c + d*x^4)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 547, normalized size of antiderivative = 6.84 \[ \int \frac {1}{x^3 \left (a+b x^4\right ) \sqrt {c+d x^4}} \, dx=\frac {-\sqrt {d}\, \sqrt {a}\, \sqrt {d \,x^{4}+c}\, \sqrt {a d -b c}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {a}\, \sqrt {a d -b c}-2 a d +b c}+\sqrt {b}\, \sqrt {d \,x^{4}+c}+\sqrt {d}\, \sqrt {b}\, x^{2}\right ) b c \,x^{2}-\sqrt {d}\, \sqrt {a}\, \sqrt {d \,x^{4}+c}\, \sqrt {a d -b c}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {a}\, \sqrt {a d -b c}-2 a d +b c}+\sqrt {b}\, \sqrt {d \,x^{4}+c}+\sqrt {d}\, \sqrt {b}\, x^{2}\right ) b c \,x^{2}+\sqrt {d}\, \sqrt {a}\, \sqrt {d \,x^{4}+c}\, \sqrt {a d -b c}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {a}\, \sqrt {a d -b c}+2 \sqrt {d}\, \sqrt {d \,x^{4}+c}\, b \,x^{2}+2 a d +2 b d \,x^{4}\right ) b c \,x^{2}-\sqrt {a}\, \sqrt {a d -b c}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {a}\, \sqrt {a d -b c}-2 a d +b c}+\sqrt {b}\, \sqrt {d \,x^{4}+c}+\sqrt {d}\, \sqrt {b}\, x^{2}\right ) b c d \,x^{4}-\sqrt {a}\, \sqrt {a d -b c}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {a}\, \sqrt {a d -b c}-2 a d +b c}+\sqrt {b}\, \sqrt {d \,x^{4}+c}+\sqrt {d}\, \sqrt {b}\, x^{2}\right ) b c d \,x^{4}+\sqrt {a}\, \sqrt {a d -b c}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {a}\, \sqrt {a d -b c}+2 \sqrt {d}\, \sqrt {d \,x^{4}+c}\, b \,x^{2}+2 a d +2 b d \,x^{4}\right ) b c d \,x^{4}-4 \sqrt {d \,x^{4}+c}\, a^{2} d^{2} x^{2}+4 \sqrt {d \,x^{4}+c}\, a b c d \,x^{2}-2 \sqrt {d}\, a^{2} c d -4 \sqrt {d}\, a^{2} d^{2} x^{4}+2 \sqrt {d}\, a b \,c^{2}+4 \sqrt {d}\, a b c d \,x^{4}}{4 a^{2} c \,x^{2} \left (\sqrt {d}\, \sqrt {d \,x^{4}+c}\, a d -\sqrt {d}\, \sqrt {d \,x^{4}+c}\, b c +a \,d^{2} x^{2}-b c d \,x^{2}\right )} \] Input:

int(1/x^3/(b*x^4+a)/(d*x^4+c)^(1/2),x)
 

Output:

( - sqrt(d)*sqrt(a)*sqrt(c + d*x**4)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d) 
*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**4) + sqrt( 
d)*sqrt(b)*x**2)*b*c*x**2 - sqrt(d)*sqrt(a)*sqrt(c + d*x**4)*sqrt(a*d - b* 
c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqr 
t(c + d*x**4) + sqrt(d)*sqrt(b)*x**2)*b*c*x**2 + sqrt(d)*sqrt(a)*sqrt(c + 
d*x**4)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt(d)* 
sqrt(c + d*x**4)*b*x**2 + 2*a*d + 2*b*d*x**4)*b*c*x**2 - sqrt(a)*sqrt(a*d 
- b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt 
(b)*sqrt(c + d*x**4) + sqrt(d)*sqrt(b)*x**2)*b*c*d*x**4 - sqrt(a)*sqrt(a*d 
 - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b 
)*sqrt(c + d*x**4) + sqrt(d)*sqrt(b)*x**2)*b*c*d*x**4 + sqrt(a)*sqrt(a*d - 
 b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt(d)*sqrt(c + d*x**4)*b 
*x**2 + 2*a*d + 2*b*d*x**4)*b*c*d*x**4 - 4*sqrt(c + d*x**4)*a**2*d**2*x**2 
 + 4*sqrt(c + d*x**4)*a*b*c*d*x**2 - 2*sqrt(d)*a**2*c*d - 4*sqrt(d)*a**2*d 
**2*x**4 + 2*sqrt(d)*a*b*c**2 + 4*sqrt(d)*a*b*c*d*x**4)/(4*a**2*c*x**2*(sq 
rt(d)*sqrt(c + d*x**4)*a*d - sqrt(d)*sqrt(c + d*x**4)*b*c + a*d**2*x**2 - 
b*c*d*x**2))