\(\int \frac {x^3 (A+B x^2+C x^4)}{(c+d x^2) \sqrt {a+c x^4}} \, dx\) [22]

Optimal result

Integrand size = 36, antiderivative size = 191 \[ \int \frac {x^3 \left (A+B x^2+C x^4\right )}{\left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=-\frac {(c C-B d) \sqrt {a+c x^4}}{2 c d^2}+\frac {C x^2 \sqrt {a+c x^4}}{4 c d}+\frac {\left (2 c^3 C-2 B c^2 d+2 A c d^2-a C d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{4 c^{3/2} d^3}+\frac {c \left (c^2 C-B c d+A d^2\right ) \text {arctanh}\left (\frac {a d-c^2 x^2}{\sqrt {c^3+a d^2} \sqrt {a+c x^4}}\right )}{2 d^3 \sqrt {c^3+a d^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \left (A+B x^2+C x^4\right )}{\left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\frac {\left (-2 c C+2 B d+C d x^2\right ) \sqrt {a+c x^4}}{4 c d^2}+\frac {c \left (c^2 C-B c d+A d^2\right ) \arctan \left (\frac {c^{3/2}+\sqrt {c} d x^2-d \sqrt {a+c x^4}}{\sqrt {-c^3-a d^2}}\right )}{d^3 \sqrt {-c^3-a d^2}}+\frac {\left (-2 c^3 C+2 B c^2 d-2 A c d^2+a C d^2\right ) \log \left (-\sqrt {c} x^2+\sqrt {a+c x^4}\right )}{4 c^{3/2} d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {2237, 27, 2237, 27, 2253, 2239, 25, 27, 719, 224, 219, 488, 219}

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.

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

\(\Big \downarrow \) 2237

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2237

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2253

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

\(\Big \downarrow \) 2239

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

\(\displaystyle \frac {C x^2 \sqrt {a+c x^4}}{4 c d}-\frac {\frac {\frac {2 c^2 \left (A d^2-B c d+c^2 C\right ) \int \frac {1}{\left (d x^2+c\right ) \sqrt {c x^4+a}}dx^2}{d}-\frac {\left (-a C d^2+2 A c d^2-2 B c^2 d+2 c^3 C\right ) \int \frac {1}{\sqrt {c x^4+a}}dx^2}{d}}{2 d}+\frac {\sqrt {a+c x^4} (c C-B d)}{d}}{2 c d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {C x^2 \sqrt {a+c x^4}}{4 c d}-\frac {\frac {\frac {2 c^2 \left (A d^2-B c d+c^2 C\right ) \int \frac {1}{\left (d x^2+c\right ) \sqrt {c x^4+a}}dx^2}{d}-\frac {\left (-a C d^2+2 A c d^2-2 B c^2 d+2 c^3 C\right ) \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+a}}}{d}}{2 d}+\frac {\sqrt {a+c x^4} (c C-B d)}{d}}{2 c d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {C x^2 \sqrt {a+c x^4}}{4 c d}-\frac {\frac {\frac {2 c^2 \left (A d^2-B c d+c^2 C\right ) \int \frac {1}{\left (d x^2+c\right ) \sqrt {c x^4+a}}dx^2}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right ) \left (-a C d^2+2 A c d^2-2 B c^2 d+2 c^3 C\right )}{\sqrt {c} d}}{2 d}+\frac {\sqrt {a+c x^4} (c C-B d)}{d}}{2 c d}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {C x^2 \sqrt {a+c x^4}}{4 c d}-\frac {\frac {-\frac {2 c^2 \left (A d^2-B c d+c^2 C\right ) \int \frac {1}{-x^4+c^3+a d^2}d\frac {a d-c^2 x^2}{\sqrt {c x^4+a}}}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right ) \left (-a C d^2+2 A c d^2-2 B c^2 d+2 c^3 C\right )}{\sqrt {c} d}}{2 d}+\frac {\sqrt {a+c x^4} (c C-B d)}{d}}{2 c d}\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 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 224

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

rule 488

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

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 2237

Int[(Px_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> W 
ith[{q = Expon[Px, x]}, Simp[Coeff[Px, x, q]*x^(q - 5)*(Sqrt[a + c*x^4]/(c* 
e*(q - 3))), x] + Simp[1/(c*e*(q - 3))   Int[(c*e*(q - 3)*Px - Coeff[Px, x, 
 q]*x^(q - 6)*(d + e*x^2)*(a*(q - 5) + c*(q - 3)*x^4))/((d + e*x^2)*Sqrt[a 
+ c*x^4]), x], x] /; GtQ[q, 4]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x]
 

rule 2239

Int[(Px_)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[(Px /. x -> Sqrt[x])*(d + e*x)^q*(a + c*x^2) 
^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && PolyQ[Px, x^2]
 

rule 2253

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] 
 :> With[{m = Expon[Px, x, Min]}, Int[x^m*ExpandToSum[Px/x^m, x]*(d + e*x^2 
)^q*(a + c*x^4)^p, x] /; GtQ[m, 0] &&  !MatchQ[Px, x^m*(u_.)]] /; FreeQ[{a, 
 c, d, e, p, q}, x] && PolyQ[Px, x]
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.27

method	result	size
risch	\(\frac {\left (C d \,x^{2}+2 B d -2 C c \right ) \sqrt {c \,x^{4}+a}}{4 c \,d^{2}}+\frac {\frac {\left (2 A c \,d^{2}-2 d \,c^{2} B -C a \,d^{2}+2 C \,c^{3}\right ) \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 d \sqrt {c}}+\frac {c^{2} \left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 c^{3}}{d^{2}}-\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}\, \sqrt {\left (x^{2}+\frac {c}{d}\right )^{2} c -\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+c^{3}}{d^{2}}}}{x^{2}+\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}}}{2 c \,d^{2}}\)	\(243\)
default	\(\frac {\frac {d \left (B d -C c \right ) \sqrt {c \,x^{4}+a}}{2 c}+\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 \sqrt {c}}+C \,d^{2} \left (\frac {x^{2} \sqrt {c \,x^{4}+a}}{4 c}-\frac {a \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{4 c^{\frac {3}{2}}}\right )}{d^{3}}+\frac {c \left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 c^{3}}{d^{2}}-\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}\, \sqrt {\left (x^{2}+\frac {c}{d}\right )^{2} c -\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+c^{3}}{d^{2}}}}{x^{2}+\frac {c}{d}}\right )}{2 d^{4} \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}}\)	\(261\)
elliptic	\(\frac {\frac {A \,d^{2} \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{\sqrt {c}}+C \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )+\frac {d \left (B d -C c \right ) \sqrt {c \,x^{4}+a}}{c}+C \,d^{2} \left (\frac {x^{2} \sqrt {c \,x^{4}+a}}{2 c}-\frac {a \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 c^{\frac {3}{2}}}\right )-B \sqrt {c}\, d \ln \left (\sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+a}\right )}{2 d^{3}}+\frac {c \left (A \,d^{2}-B c d +C \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 c^{3}}{d^{2}}-\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}\, \sqrt {\left (x^{2}+\frac {c}{d}\right )^{2} c -\frac {2 c^{2} \left (x^{2}+\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+c^{3}}{d^{2}}}}{x^{2}+\frac {c}{d}}\right )}{2 d^{4} \sqrt {\frac {a \,d^{2}+c^{3}}{d^{2}}}}\)	\(296\)

Input:

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

Output:

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

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 \left (A+B x^2+C x^4\right )}{\left (c+d x^2\right ) \sqrt {a+c x^4}} \, dx=\left (\int \frac {x^{7}}{\sqrt {c \,x^{4}+a}\, c +\sqrt {c \,x^{4}+a}\, d \,x^{2}}d x \right ) c +\left (\int \frac {x^{5}}{\sqrt {c \,x^{4}+a}\, c +\sqrt {c \,x^{4}+a}\, d \,x^{2}}d x \right ) b +\left (\int \frac {x^{3}}{\sqrt {c \,x^{4}+a}\, c +\sqrt {c \,x^{4}+a}\, d \,x^{2}}d x \right ) a \] Input:

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

Output:

int(x**7/(sqrt(a + c*x**4)*c + sqrt(a + c*x**4)*d*x**2),x)*c + int(x**5/(s 
qrt(a + c*x**4)*c + sqrt(a + c*x**4)*d*x**2),x)*b + int(x**3/(sqrt(a + c*x 
**4)*c + sqrt(a + c*x**4)*d*x**2),x)*a