\(\int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx\) [1183]

Optimal result

Integrand size = 22, antiderivative size = 146 \[ \int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx=-\frac {c \sqrt {a+b x^2}}{b d^2}+\frac {x \sqrt {a+b x^2}}{2 b d}+\frac {\left (2 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2} d^3}+\frac {c^3 \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{d^3 \sqrt {b c^2+a d^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx=\frac {\frac {d (-2 c+d x) \sqrt {a+b x^2}}{b}+\frac {4 c^3 \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )}{\sqrt {-b c^2-a d^2}}+\frac {\left (-2 b c^2+a d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}}}{2 d^3} \] Input:

Integrate[x^3/((c + d*x)*Sqrt[a + b*x^2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {604, 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.

\(\displaystyle \int \frac {x^3}{\sqrt {a+b x^2} (c+d x)} \, dx\)

\(\Big \downarrow \) 604

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 719

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

Int[x^3/((c + d*x)*Sqrt[a + b*x^2]),x]
 

Output:

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

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 = 0.37 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.38

method	result	size
risch	\(-\frac {\left (-d x +2 c \right ) \sqrt {b \,x^{2}+a}}{2 b \,d^{2}}-\frac {\frac {\left (a \,d^{2}-2 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d \sqrt {b}}-\frac {2 c^{3} b \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}}{2 b \,d^{2}}\)	\(201\)
default	\(\frac {c^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d^{3} \sqrt {b}}+\frac {\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}}{d}-\frac {c \sqrt {b \,x^{2}+a}}{b \,d^{2}}+\frac {c^{3} \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{4} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\)	\(216\)

Input:

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

Output:

-1/2*(-d*x+2*c)*(b*x^2+a)^(1/2)/b/d^2-1/2/b/d^2*((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 = 1.54 (sec) , antiderivative size = 924, normalized size of antiderivative = 6.33 \[ \int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/4*(2*sqrt(b*c^2 + a*d^2)*b^2*c^3*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)) - (2*b^2*c^4 + a*b*c^2*d^2 - a^2*d^4) 
*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(2*b^2*c^3*d 
+ 2*a*b*c*d^3 - (b^2*c^2*d^2 + a*b*d^4)*x)*sqrt(b*x^2 + a))/(b^3*c^2*d^3 + 
 a*b^2*d^5), 1/4*(4*sqrt(-b*c^2 - a*d^2)*b^2*c^3*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)) - (2*b^2*c^4 + a*b*c^2*d^2 - a^2*d^4)*sqrt(b)*log(-2*b*x^2 + 2*sqrt( 
b*x^2 + a)*sqrt(b)*x - a) - 2*(2*b^2*c^3*d + 2*a*b*c*d^3 - (b^2*c^2*d^2 + 
a*b*d^4)*x)*sqrt(b*x^2 + a))/(b^3*c^2*d^3 + a*b^2*d^5), 1/2*(sqrt(b*c^2 + 
a*d^2)*b^2*c^3*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)) - (2*b^2*c^4 + a*b*c^2*d^2 - a^2*d^4)*sqrt(-b)*arctan(sqrt 
(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*c^3*d + 2*a*b*c*d^3 - (b^2*c^2*d^2 + a*b* 
d^4)*x)*sqrt(b*x^2 + a))/(b^3*c^2*d^3 + a*b^2*d^5), 1/2*(2*sqrt(-b*c^2 - a 
*d^2)*b^2*c^3*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)) - (2*b^2*c^4 + a*b*c^2*d^2 - 
a^2*d^4)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*c^3*d + 2*a* 
b*c*d^3 - (b^2*c^2*d^2 + a*b*d^4)*x)*sqrt(b*x^2 + a))/(b^3*c^2*d^3 + a*b^2 
*d^5)]
 

Sympy [F]

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

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

Output:

Integral(x**3/(sqrt(a + b*x**2)*(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} x}{2 \, b d} + \frac {c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b} d^{3}} - \frac {a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}} d} - \frac {c^{3} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{\sqrt {a + \frac {b c^{2}}{d^{2}}} d^{4}} - \frac {\sqrt {b x^{2} + a} c}{b d^{2}} \] Input:

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

Output:

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

Giac [F(-2)]

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

integrate(x^3/(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. \[ \int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx=\int \frac {x^3}{\sqrt {b\,x^2+a}\,\left (c+d\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 1247, normalized size of antiderivative = 8.54 \[ \int \frac {x^3}{(c+d x) \sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqr 
t(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**4 + 2*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 + s 
qrt(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**3*d**2 + 2*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**5 + 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**4 - 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**4 + 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**3*d** 
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)*a*b**2*c**3*d**2 - sqr 
t(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**3*d**2 - sqrt(2*s 
qrt(b)*sqrt(a*d**2 + b*c**2)*c + a*d**2 + 2*b*c**2)*log( - sqrt(2*sqrt(...