\(\int \frac {A+B x^2+C x^4}{x^3 (c+d x^2) \sqrt {a+c x^4}} \, dx\) [25]

Optimal result

Integrand size = 36, antiderivative size = 138 \[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=-\frac {A \sqrt {a+c x^4}}{2 a c x^2}-\frac {\left (c^2 C-B c d+A d^2\right ) \text {arctanh}\left (\frac {a d-c^2 x^2}{\sqrt {c^3+a d^2} \sqrt {a+c x^4}}\right )}{2 c^2 \sqrt {c^3+a d^2}}-\frac {(B c-A d) \text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )}{2 \sqrt {a} c^2} \] Output:

-1/2*A*(c*x^4+a)^(1/2)/a/c/x^2-1/2*(A*d^2-B*c*d+C*c^2)*arctanh((-c^2*x^2+a 
*d)/(a*d^2+c^3)^(1/2)/(c*x^4+a)^(1/2))/c^2/(a*d^2+c^3)^(1/2)-1/2*(-A*d+B*c 
)*arctanh((c*x^4+a)^(1/2)/a^(1/2))/a^(1/2)/c^2
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=-\frac {\frac {A c \sqrt {a+c x^4}}{a x^2}+\frac {2 \left (c^2 C-B c d+A d^2\right ) \arctan \left (\frac {c^{3/2}+\sqrt {c} d x^2-d \sqrt {a+c x^4}}{\sqrt {-c^3-a d^2}}\right )}{\sqrt {-c^3-a d^2}}-\frac {2 (B c-A d) \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 c^2} \] Input:

Integrate[(A + B*x^2 + C*x^4)/(x^3*(c + d*x^2)*Sqrt[a + c*x^4]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2249, 2009}

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.

\(\displaystyle \int \frac {A+B x^2+C x^4}{x^3 \sqrt {a+c x^4} \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 2249

\(\displaystyle \int \left (\frac {x \left (A d^2-B c d+c^2 C\right )}{c^2 \sqrt {a+c x^4} \left (c+d x^2\right )}+\frac {B c-A d}{c^2 x \sqrt {a+c x^4}}+\frac {A}{c x^3 \sqrt {a+c x^4}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) (B c-A d)}{2 \sqrt {a} c^2}-\frac {\text {arctanh}\left (\frac {a d-c^2 x^2}{\sqrt {a d^2+c^3} \sqrt {a+c x^4}}\right ) \left (A d^2-B c d+c^2 C\right )}{2 c^2 \sqrt {a d^2+c^3}}-\frac {A \sqrt {a+c x^4}}{2 a c x^2}\)

Input:

Int[(A + B*x^2 + C*x^4)/(x^3*(c + d*x^2)*Sqrt[a + c*x^4]),x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2249

Int[(Px_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_) 
^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[a + c*x^4], Px*(f*x)^m*(d 
 + e*x^2)^q*(a + c*x^4)^(p + 1/2), x], x] /; FreeQ[{a, c, d, e, f, m}, x] & 
& PolyQ[Px, x] && IntegerQ[p + 1/2] && IntegerQ[q]
 

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.51

method	result	size
default	\(-\frac {A \sqrt {c \,x^{4}+a}}{2 a c \,x^{2}}+\frac {\left (A d -B c \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{4}+a}}{x^{2}}\right )}{2 c^{2} \sqrt {a}}-\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 c^{3}}{d^{2}}-\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}\, \sqrt {\left (x^{2}+\frac {c}{d}\right )^{2} c -\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+c^{3}}{d^{2}}}}{x^{2}+\frac {c}{d}}\right )}{2 c^{2} d \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}}\)	\(209\)
elliptic	\(-\frac {A \sqrt {c \,x^{4}+a}}{2 a c \,x^{2}}+\frac {\left (A d -B c \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{4}+a}}{x^{2}}\right )}{2 c^{2} \sqrt {a}}-\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 c^{3}}{d^{2}}-\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}\, \sqrt {\left (x^{2}+\frac {c}{d}\right )^{2} c -\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+c^{3}}{d^{2}}}}{x^{2}+\frac {c}{d}}\right )}{2 c^{2} d \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}}\)	\(209\)
risch	\(-\frac {A \sqrt {c \,x^{4}+a}}{2 a c \,x^{2}}-\frac {-\frac {\left (A d -B c \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{4}+a}}{x^{2}}\right )}{2 c \sqrt {a}}+\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 c^{3}}{d^{2}}-\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}\, \sqrt {\left (x^{2}+\frac {c}{d}\right )^{2} c -\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+c^{3}}{d^{2}}}}{x^{2}+\frac {c}{d}}\right )}{2 c d \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}}}{c}\)	\(215\)

Input:

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

Output:

-1/2*A*(c*x^4+a)^(1/2)/a/c/x^2+1/2*(A*d-B*c)/c^2/a^(1/2)*ln((2*a+2*a^(1/2) 
*(c*x^4+a)^(1/2))/x^2)-1/2*(A*d^2-B*c*d+C*c^2)/c^2/d/((a*d^2+c^3)/d^2)^(1/ 
2)*ln((2*(a*d^2+c^3)/d^2-2*c^2/d*(x^2+c/d)+2*((a*d^2+c^3)/d^2)^(1/2)*((x^2 
+c/d)^2*c-2*c^2/d*(x^2+c/d)+(a*d^2+c^3)/d^2)^(1/2))/(x^2+c/d))
 

Fricas [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 902, normalized size of antiderivative = 6.54 \[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx =\text {Too large to display} \] Input:

integrate((C*x^4+B*x^2+A)/x^3/(d*x^2+c)/(c*x^4+a)^(1/2),x, algorithm="fric 
as")
 

Output:

[1/4*((C*a*c^2 - B*a*c*d + A*a*d^2)*sqrt(c^3 + a*d^2)*x^2*log((2*a*c^2*d*x 
^2 - (2*c^4 + a*c*d^2)*x^4 - a*c^3 - 2*a^2*d^2 - 2*sqrt(c*x^4 + a)*(c^2*x^ 
2 - a*d)*sqrt(c^3 + a*d^2))/(d^2*x^4 + 2*c*d*x^2 + c^2)) - (B*c^4 - A*c^3* 
d + B*a*c*d^2 - A*a*d^3)*sqrt(a)*x^2*log(-(c*x^4 + 2*sqrt(c*x^4 + a)*sqrt( 
a) + 2*a)/x^4) - 2*(A*c^4 + A*a*c*d^2)*sqrt(c*x^4 + a))/((a*c^5 + a^2*c^2* 
d^2)*x^2), -1/4*(2*(C*a*c^2 - B*a*c*d + A*a*d^2)*sqrt(-c^3 - a*d^2)*x^2*ar 
ctan(sqrt(c*x^4 + a)*(c^2*x^2 - a*d)*sqrt(-c^3 - a*d^2)/((c^4 + a*c*d^2)*x 
^4 + a*c^3 + a^2*d^2)) + (B*c^4 - A*c^3*d + B*a*c*d^2 - A*a*d^3)*sqrt(a)*x 
^2*log(-(c*x^4 + 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) + 2*(A*c^4 + A*a*c* 
d^2)*sqrt(c*x^4 + a))/((a*c^5 + a^2*c^2*d^2)*x^2), 1/4*(2*(B*c^4 - A*c^3*d 
 + B*a*c*d^2 - A*a*d^3)*sqrt(-a)*x^2*arctan(sqrt(c*x^4 + a)*sqrt(-a)/a) + 
(C*a*c^2 - B*a*c*d + A*a*d^2)*sqrt(c^3 + a*d^2)*x^2*log((2*a*c^2*d*x^2 - ( 
2*c^4 + a*c*d^2)*x^4 - a*c^3 - 2*a^2*d^2 - 2*sqrt(c*x^4 + a)*(c^2*x^2 - a* 
d)*sqrt(c^3 + a*d^2))/(d^2*x^4 + 2*c*d*x^2 + c^2)) - 2*(A*c^4 + A*a*c*d^2) 
*sqrt(c*x^4 + a))/((a*c^5 + a^2*c^2*d^2)*x^2), -1/2*((C*a*c^2 - B*a*c*d + 
A*a*d^2)*sqrt(-c^3 - a*d^2)*x^2*arctan(sqrt(c*x^4 + a)*(c^2*x^2 - a*d)*sqr 
t(-c^3 - a*d^2)/((c^4 + a*c*d^2)*x^4 + a*c^3 + a^2*d^2)) - (B*c^4 - A*c^3* 
d + B*a*c*d^2 - A*a*d^3)*sqrt(-a)*x^2*arctan(sqrt(c*x^4 + a)*sqrt(-a)/a) + 
 (A*c^4 + A*a*c*d^2)*sqrt(c*x^4 + a))/((a*c^5 + a^2*c^2*d^2)*x^2)]
 

Sympy [F]

\[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\int \frac {A + B x^{2} + C x^{4}}{x^{3} \sqrt {a + c x^{4}} \left (c + d x^{2}\right )}\, dx \] Input:

integrate((C*x**4+B*x**2+A)/x**3/(d*x**2+c)/(c*x**4+a)**(1/2),x)
 

Output:

Integral((A + B*x**2 + C*x**4)/(x**3*sqrt(a + c*x**4)*(c + d*x**2)), x)
 

Maxima [F]

\[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\int { \frac {C x^{4} + B x^{2} + A}{\sqrt {c x^{4} + a} {\left (d x^{2} + c\right )} x^{3}} \,d x } \] Input:

integrate((C*x^4+B*x^2+A)/x^3/(d*x^2+c)/(c*x^4+a)^(1/2),x, algorithm="maxi 
ma")
 

Output:

integrate((C*x^4 + B*x^2 + A)/(sqrt(c*x^4 + a)*(d*x^2 + c)*x^3), x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\frac {A}{{\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )}^{2} - a\right )} \sqrt {c}} + \frac {{\left (C c^{2} - B c d + A d^{2}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )} d + c^{\frac {3}{2}}}{\sqrt {-c^{3} - a d^{2}}}\right )}{\sqrt {-c^{3} - a d^{2}} c^{2}} + \frac {{\left (B c - A d\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} c^{2}} \] Input:

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

Output:

A/(((sqrt(c)*x^2 - sqrt(c*x^4 + a))^2 - a)*sqrt(c)) + (C*c^2 - B*c*d + A*d 
^2)*arctan(-((sqrt(c)*x^2 - sqrt(c*x^4 + a))*d + c^(3/2))/sqrt(-c^3 - a*d^ 
2))/(sqrt(-c^3 - a*d^2)*c^2) + (B*c - A*d)*arctan(-(sqrt(c)*x^2 - sqrt(c*x 
^4 + a))/sqrt(-a))/(sqrt(-a)*c^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\int \frac {C\,x^4+B\,x^2+A}{x^3\,\sqrt {c\,x^4+a}\,\left (d\,x^2+c\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x^2+C x^4}{x^3 \left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\left (\int \frac {x}{\sqrt {c \,x^{4}+a}\, c +\sqrt {c \,x^{4}+a}\, d \,x^{2}}d x \right ) c +\left (\int \frac {1}{\sqrt {c \,x^{4}+a}\, c \,x^{3}+\sqrt {c \,x^{4}+a}\, d \,x^{5}}d x \right ) a +\left (\int \frac {1}{\sqrt {c \,x^{4}+a}\, c x +\sqrt {c \,x^{4}+a}\, d \,x^{3}}d x \right ) b \] Input:

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

Output:

int(x/(sqrt(a + c*x**4)*c + sqrt(a + c*x**4)*d*x**2),x)*c + int(1/(sqrt(a 
+ c*x**4)*c*x**3 + sqrt(a + c*x**4)*d*x**5),x)*a + int(1/(sqrt(a + c*x**4) 
*c*x + sqrt(a + c*x**4)*d*x**3),x)*b