\(\int \frac {x^4}{(c+d x) \sqrt {a+b x^2}} \, dx\) [1182]

Optimal result

Integrand size = 22, antiderivative size = 187 \[ \int \frac {x^4}{(c+d x) \sqrt {a+b x^2}} \, dx=\frac {\left (3 b c^2-2 a d^2\right ) \sqrt {a+b x^2}}{3 b^2 d^3}-\frac {c x \sqrt {a+b x^2}}{2 b d^2}+\frac {x^2 \sqrt {a+b x^2}}{3 b d}-\frac {c \left (2 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2} d^4}-\frac {c^4 \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{d^4 \sqrt {b c^2+a d^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.86 \[ \int \frac {x^4}{(c+d x) \sqrt {a+b x^2}} \, dx=\frac {\frac {d \sqrt {a+b x^2} \left (-4 a d^2+b \left (6 c^2-3 c d x+2 d^2 x^2\right )\right )}{b^2}-\frac {12 c^4 \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )}{\sqrt {-b c^2-a d^2}}+\frac {3 c \left (2 b c^2-a d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}}}{6 d^4} \] Input:

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

Output:

((d*Sqrt[a + b*x^2]*(-4*a*d^2 + b*(6*c^2 - 3*c*d*x + 2*d^2*x^2)))/b^2 - (1 
2*c^4*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a*d^2 
]])/Sqrt[-(b*c^2) - a*d^2] + (3*c*(2*b*c^2 - a*d^2)*Log[-(Sqrt[b]*x) + Sqr 
t[a + b*x^2]])/b^(3/2))/(6*d^4)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {604, 25, 2185, 25, 2185, 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.

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

\(\Big \downarrow \) 604

\(\displaystyle \frac {\int -\frac {7 b c d^3 x^3+d^2 \left (5 b c^2+2 a d^2\right ) x^2+c d \left (b c^2+4 a d^2\right ) x+2 a c^2 d^2}{(c+d x) \sqrt {b x^2+a}}dx}{3 b d^4}+\frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\int \frac {7 b c d^3 x^3+d^2 \left (5 b c^2+2 a d^2\right ) x^2+c d \left (b c^2+4 a d^2\right ) x+2 a c^2 d^2}{(c+d x) \sqrt {b x^2+a}}dx}{3 b d^4}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {\int -\frac {3 a b c^2 d^5+b \left (11 b c^2-4 a d^2\right ) x^2 d^5+b c \left (5 b c^2-a d^2\right ) x d^4}{(c+d x) \sqrt {b x^2+a}}dx}{2 b d^3}+\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)}{3 b d^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {\int \frac {3 a b c^2 d^5+b \left (11 b c^2-4 a d^2\right ) x^2 d^5+b c \left (5 b c^2-a d^2\right ) x d^4}{(c+d x) \sqrt {b x^2+a}}dx}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {\frac {\int \frac {3 b^2 c d^6 \left (a c d-\left (2 b c^2-a d^2\right ) x\right )}{(c+d x) \sqrt {b x^2+a}}dx}{b d^2}+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {3 b c d^4 \int \frac {a c d-\left (2 b c^2-a d^2\right ) x}{(c+d x) \sqrt {b x^2+a}}dx+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {3 b c d^4 \left (\frac {2 b c^3 \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\left (2 b c^2-a d^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{d}\right )+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {3 b c d^4 \left (\frac {2 b c^3 \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\left (2 b c^2-a d^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{d}\right )+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {3 b c d^4 \left (\frac {2 b c^3 \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (2 b c^2-a d^2\right )}{\sqrt {b} d}\right )+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {3 b c d^4 \left (-\frac {2 b c^3 \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}}{d}-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (2 b c^2-a d^2\right )}{\sqrt {b} d}\right )+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a+b x^2} (c+d x)^2}{3 b d^3}-\frac {\frac {7}{2} c d \sqrt {a+b x^2} (c+d x)-\frac {3 b c d^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (2 b c^2-a d^2\right )}{\sqrt {b} d}-\frac {2 b c^3 \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{d \sqrt {a d^2+b c^2}}\right )+d^4 \sqrt {a+b x^2} \left (11 b c^2-4 a d^2\right )}{2 b d^3}}{3 b d^4}\)

Input:

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

Output:

((c + d*x)^2*Sqrt[a + b*x^2])/(3*b*d^3) - ((7*c*d*(c + d*x)*Sqrt[a + b*x^2 
])/2 - (d^4*(11*b*c^2 - 4*a*d^2)*Sqrt[a + b*x^2] + 3*b*c*d^4*(-(((2*b*c^2 
- a*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(Sqrt[b]*d)) - (2*b*c^3*Arc 
Tanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(d*Sqrt[b*c^2 + 
 a*d^2])))/(2*b*d^3))/(3*b*d^4)
 

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 604

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 
2*p + 1))), x] + Simp[1/(b*d^m*(m + n + 2*p + 1))   Int[(c + d*x)^n*(a + b* 
x^2)^p*ExpandToSum[b*d^m*(m + n + 2*p + 1)*x^m - b*(m + n + 2*p + 1)*(c + d 
*x)^m - (c + d*x)^(m - 2)*(a*d^2*(m + n - 1) - b*c^2*(m + n + 2*p + 1) - 2* 
b*c*d*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && IGtQ[m, 
 1] && NeQ[m + n + 2*p + 1, 0] && IntegerQ[2*p]
 

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 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.19

method	result	size
risch	\(-\frac {\left (-2 b \,x^{2} d^{2}+3 b c d x +4 a \,d^{2}-6 b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{6 b^{2} d^{3}}+\frac {c \left (\frac {\left (a \,d^{2}-2 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d \sqrt {b}}-\frac {2 c^{3} b \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{2 b \,d^{3}}\)	\(222\)
default	\(\frac {\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}}{d}+\frac {c^{2} \sqrt {b \,x^{2}+a}}{d^{3} b}-\frac {c^{4} \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{5} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}-\frac {c^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d^{4} \sqrt {b}}-\frac {c \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{d^{2}}\)	\(258\)

Input:

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

Output:

-1/6*(-2*b*d^2*x^2+3*b*c*d*x+4*a*d^2-6*b*c^2)*(b*x^2+a)^(1/2)/b^2/d^3+1/2* 
c/b/d^3*((a*d^2-2*b*c^2)/d*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)-2*c^3/d^2 
*b/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a 
*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/ 
2))/(x+c/d)))
 

Fricas [A] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 1060, normalized size of antiderivative = 5.67 \[ \int \frac {x^4}{(c+d x) \sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.91 \[ \int \frac {x^4}{(c+d x) \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} x^{2}}{3 \, b d} - \frac {\sqrt {b x^{2} + a} c x}{2 \, b d^{2}} - \frac {c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b} d^{4}} + \frac {a c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}} d^{2}} + \frac {c^{4} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{\sqrt {a + \frac {b c^{2}}{d^{2}}} d^{5}} + \frac {\sqrt {b x^{2} + a} c^{2}}{b d^{3}} - \frac {2 \, \sqrt {b x^{2} + a} a}{3 \, b^{2} d} \] Input:

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

Output:

1/3*sqrt(b*x^2 + a)*x^2/(b*d) - 1/2*sqrt(b*x^2 + a)*c*x/(b*d^2) - c^3*arcs 
inh(b*x/sqrt(a*b))/(sqrt(b)*d^4) + 1/2*a*c*arcsinh(b*x/sqrt(a*b))/(b^(3/2) 
*d^2) + c^4*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d* 
x + c)))/(sqrt(a + b*c^2/d^2)*d^5) + sqrt(b*x^2 + a)*c^2/(b*d^3) - 2/3*sqr 
t(b*x^2 + a)*a/(b^2*d)
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 1318, normalized size of antiderivative = 7.05 \[ \int \frac {x^4}{(c+d x) \sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)* 
sqrt(a*d**2 + b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt( 
b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*b**2*c**5 - 6*sqrt(2*sqrt 
(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*atan((sqrt(a + b*x**2)*d 
+ sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2) 
)*a*b**2*c**4*d**2 - 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2 
*b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d** 
2 + b*c**2)*c - a*d**2 - 2*b*c**2))*b**3*c**6 - 3*sqrt(b)*sqrt(2*sqrt(b)*s 
qrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2)*sqrt(a*d**2 + b*c**2)*log( - s 
qrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2) + sqrt(a + b*x* 
*2)*d + sqrt(b)*d*x)*b**2*c**5 + 3*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b* 
c**2)*c + a*d**2 + 2*b*c**2)*sqrt(a*d**2 + b*c**2)*log(sqrt(2*sqrt(b)*sqrt 
(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2) + sqrt(a + b*x**2)*d + sqrt(b)*d* 
x)*b**2*c**5 - 3*sqrt(a*d**2 + b*c**2)*log( - sqrt(2*sqrt(b)*sqrt(a*d**2 + 
 b*c**2)*c + a*d**2 + 2*b*c**2) + sqrt(a + b*x**2)*d + sqrt(b)*d*x)*a*b**2 
*c**4*d**2 - 3*sqrt(a*d**2 + b*c**2)*log(sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c* 
*2)*c + a*d**2 + 2*b*c**2) + sqrt(a + b*x**2)*d + sqrt(b)*d*x)*a*b**2*c**4 
*d**2 + 3*sqrt(a*d**2 + b*c**2)*log(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c + 2* 
sqrt(b)*sqrt(a + b*x**2)*d**2*x - 2*b*c**2 + 2*b*d**2*x**2)*a*b**2*c**4*d* 
*2 + 3*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2)*log(...