\(\int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx\) [53]

Optimal result

Integrand size = 29, antiderivative size = 196 \[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=-\frac {\left (a c^2-d^2\right ) x}{b^2 c^3}+\frac {x^3}{3 b c}-\frac {d x \sqrt {a+b x^2}}{2 b^2 c^2}+\frac {\left (a c^2-d^2\right )^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {a c^2-d^2} x}{a c+d \sqrt {a+b x^2}}\right )}{b^{5/2} c^4}+\frac {a d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2} c^2}+\frac {d \left (a c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2} c^4} \] Output:

-(a*c^2-d^2)*x/b^2/c^3+1/3*x^3/b/c-1/2*d*x*(b*x^2+a)^(1/2)/b^2/c^2+(a*c^2- 
d^2)^(3/2)*arctan(b^(1/2)*(a*c^2-d^2)^(1/2)*x/(a*c+d*(b*x^2+a)^(1/2)))/b^( 
5/2)/c^4+1/2*a*d*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)/c^2+d*(a*c^2-d 
^2)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)/c^4
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.77 \[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b} c x \left (-6 a c^2+6 d^2+2 b c^2 x^2-3 c d \sqrt {a+b x^2}\right )-12 \left (a c^2-d^2\right )^{3/2} \arctan \left (\frac {d+c \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {a c^2-d^2}}\right )+\left (-9 a c^2 d+6 d^3\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 b^{5/2} c^4} \] Input:

Integrate[x^4/(a*c + b*c*x^2 + d*Sqrt[a + b*x^2]),x]
 

Output:

(Sqrt[b]*c*x*(-6*a*c^2 + 6*d^2 + 2*b*c^2*x^2 - 3*c*d*Sqrt[a + b*x^2]) - 12 
*(a*c^2 - d^2)^(3/2)*ArcTan[(d + c*(-(Sqrt[b]*x) + Sqrt[a + b*x^2]))/Sqrt[ 
a*c^2 - d^2]] + (-9*a*c^2*d + 6*d^3)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/ 
(6*b^(5/2)*c^4)
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2587, 27, 254, 381, 398, 224, 219, 291, 218, 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.

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

\(\Big \downarrow \) 2587

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 254

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

\(\Big \downarrow \) 381

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

\(\Big \downarrow \) 398

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 2009

\(\displaystyle c \left (\frac {\left (a c^2-d^2\right )^{3/2} \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{b^{5/2} c^5}-\frac {x \left (a c^2-d^2\right )}{b^2 c^4}+\frac {x^3}{3 b c^2}\right )-d \left (\frac {x \sqrt {a+b x^2}}{2 b^2 c^2}-\frac {\frac {\left (3 a-\frac {2 d^2}{c^2}\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-\frac {2 \left (a c^2-d^2\right )^{3/2} \arctan \left (\frac {\sqrt {b} d x}{\sqrt {a+b x^2} \sqrt {a c^2-d^2}}\right )}{\sqrt {b} c^2 d}}{2 b^2 c^2}\right )\)

Input:

Int[x^4/(a*c + b*c*x^2 + d*Sqrt[a + b*x^2]),x]
 

Output:

c*(-(((a*c^2 - d^2)*x)/(b^2*c^4)) + x^3/(3*b*c^2) + ((a*c^2 - d^2)^(3/2)*A 
rcTan[(Sqrt[b]*c*x)/Sqrt[a*c^2 - d^2]])/(b^(5/2)*c^5)) - d*((x*Sqrt[a + b* 
x^2])/(2*b^2*c^2) - ((-2*(a*c^2 - d^2)^(3/2)*ArcTan[(Sqrt[b]*d*x)/(Sqrt[a* 
c^2 - d^2]*Sqrt[a + b*x^2])])/(Sqrt[b]*c^2*d) + ((3*a - (2*d^2)/c^2)*ArcTa 
nh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/(2*b^2*c^2))
 

Defintions of rubi rules used

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 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 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

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 381

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
+ 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) 
Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 
2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q 
}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 
, p, q, x]
 

rule 398

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

rule 2009

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

rule 2587

Int[(u_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_ 
Symbol] :> Simp[c   Int[u/(c^2 - a*e^2 + c*d*x^n), x], x] - Simp[a*e   Int[ 
u/((c^2 - a*e^2 + c*d*x^n)*Sqrt[a + b*x^n]), x], x] /; FreeQ[{a, b, c, d, e 
, n}, x] && EqQ[b*c - a*d, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1904\) vs. \(2(166)=332\).

Time = 0.04 (sec) , antiderivative size = 1905, normalized size of antiderivative = 9.72

method	result	size
default	\(\text {Expression too large to display}\)	\(1905\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 1475, normalized size of antiderivative = 7.53 \[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/12*(4*b^2*c^3*x^3 - 6*sqrt(b*x^2 + a)*b*c^2*d*x - 3*(3*a*c^2*d - 2*d^3) 
*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 3*(a*b*c^2 - b* 
d^2)*sqrt(-(a*c^2 - d^2)/b)*log((a^4*c^4 - 2*a^3*c^2*d^2 + a^2*d^4 + (a^2* 
b^2*c^4 - 8*a*b^2*c^2*d^2 + 8*b^2*d^4)*x^4 + 2*(a^3*b*c^4 - 5*a^2*b*c^2*d^ 
2 + 4*a*b*d^4)*x^2 - 4*((a*b^2*c^2*d - 2*b^2*d^3)*x^3 + (a^2*b*c^2*d - a*b 
*d^3)*x)*sqrt(b*x^2 + a)*sqrt(-(a*c^2 - d^2)/b))/(b^2*c^4*x^4 + a^2*c^4 - 
2*a*c^2*d^2 + d^4 + 2*(a*b*c^4 - b*c^2*d^2)*x^2)) - 6*(a*b*c^2 - b*d^2)*sq 
rt(-(a*c^2 - d^2)/b)*log((b*c^2*x^2 - 2*b*c*x*sqrt(-(a*c^2 - d^2)/b) - a*c 
^2 + d^2)/(b*c^2*x^2 + a*c^2 - d^2)) - 12*(a*b*c^3 - b*c*d^2)*x)/(b^3*c^4) 
, 1/12*(4*b^2*c^3*x^3 - 6*sqrt(b*x^2 + a)*b*c^2*d*x - 6*(3*a*c^2*d - 2*d^3 
)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*(a*b*c^2 - b*d^2)*sqrt(- 
(a*c^2 - d^2)/b)*log((a^4*c^4 - 2*a^3*c^2*d^2 + a^2*d^4 + (a^2*b^2*c^4 - 8 
*a*b^2*c^2*d^2 + 8*b^2*d^4)*x^4 + 2*(a^3*b*c^4 - 5*a^2*b*c^2*d^2 + 4*a*b*d 
^4)*x^2 - 4*((a*b^2*c^2*d - 2*b^2*d^3)*x^3 + (a^2*b*c^2*d - a*b*d^3)*x)*sq 
rt(b*x^2 + a)*sqrt(-(a*c^2 - d^2)/b))/(b^2*c^4*x^4 + a^2*c^4 - 2*a*c^2*d^2 
 + d^4 + 2*(a*b*c^4 - b*c^2*d^2)*x^2)) - 6*(a*b*c^2 - b*d^2)*sqrt(-(a*c^2 
- d^2)/b)*log((b*c^2*x^2 - 2*b*c*x*sqrt(-(a*c^2 - d^2)/b) - a*c^2 + d^2)/( 
b*c^2*x^2 + a*c^2 - d^2)) - 12*(a*b*c^3 - b*c*d^2)*x)/(b^3*c^4), 1/12*(4*b 
^2*c^3*x^3 - 6*sqrt(b*x^2 + a)*b*c^2*d*x - 12*(a*b*c^2 - b*d^2)*sqrt((a*c^ 
2 - d^2)/b)*arctan(-b*c*x*sqrt((a*c^2 - d^2)/b)/(a*c^2 - d^2)) + 6*(a*b...
 

Sympy [F]

\[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\int \frac {x^{4}}{a c + b c x^{2} + d \sqrt {a + b x^{2}}}\, dx \] Input:

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

Output:

Integral(x**4/(a*c + b*c*x**2 + d*sqrt(a + b*x**2)), x)
 

Maxima [F]

\[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\int { \frac {x^{4}}{b c x^{2} + a c + \sqrt {b x^{2} + a} d} \,d x } \] Input:

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

Output:

integrate(x^4/(b*c*x^2 + a*c + sqrt(b*x^2 + a)*d), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^4/(a*c+b*c*x^2+d*(b*x^2+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. \[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\int \frac {x^4}{a\,c+d\,\sqrt {b\,x^2+a}+b\,c\,x^2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.08 \[ \int \frac {x^4}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {12 \sqrt {b}\, \sqrt {a \,c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\sqrt {b \,x^{2}+a}\, c +\sqrt {b}\, c x +d}{\sqrt {a \,c^{2}-d^{2}}}\right ) a \,c^{2}-12 \sqrt {b}\, \sqrt {a \,c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\sqrt {b \,x^{2}+a}\, c +\sqrt {b}\, c x +d}{\sqrt {a \,c^{2}-d^{2}}}\right ) d^{2}-3 \sqrt {b \,x^{2}+a}\, b \,c^{2} d x +9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,c^{2} d -6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) d^{3}-6 a b \,c^{3} x +2 b^{2} c^{3} x^{3}+6 b c \,d^{2} x}{6 b^{3} c^{4}} \] Input:

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

Output:

(12*sqrt(b)*sqrt(a*c**2 - d**2)*atan((sqrt(a + b*x**2)*c + sqrt(b)*c*x + d 
)/sqrt(a*c**2 - d**2))*a*c**2 - 12*sqrt(b)*sqrt(a*c**2 - d**2)*atan((sqrt( 
a + b*x**2)*c + sqrt(b)*c*x + d)/sqrt(a*c**2 - d**2))*d**2 - 3*sqrt(a + b* 
x**2)*b*c**2*d*x + 9*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a 
*c**2*d - 6*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*d**3 - 6*a 
*b*c**3*x + 2*b**2*c**3*x**3 + 6*b*c*d**2*x)/(6*b**3*c**4)