\(\int \frac {1}{x^3 (a c+b c x^2+d \sqrt {a+b x^2})} \, dx\) [52]

Optimal result

Integrand size = 29, antiderivative size = 151 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {a c-d \sqrt {a+b x^2}}{2 a \left (a c^2-d^2\right ) x^2}-\frac {b d \left (3 a c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2} \left (a c^2-d^2\right )^2}-\frac {b c^3 \log (x)}{\left (a c^2-d^2\right )^2}+\frac {b c^3 \log \left (d+c \sqrt {a+b x^2}\right )}{\left (a c^2-d^2\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {b d \left (-3 a c^2+d^2\right ) x^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\sqrt {a} \left (-\left (\left (a c^2-d^2\right ) \left (a c-d \sqrt {a+b x^2}\right )\right )-a b c^3 x^2 \log \left (b x^2\right )+2 a b c^3 x^2 \log \left (d+c \sqrt {a+b x^2}\right )\right )}{2 a^{3/2} \left (-a c^2+d^2\right )^2 x^2} \] Input:

Integrate[1/(x^3*(a*c + b*c*x^2 + d*Sqrt[a + b*x^2])),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.94 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2586, 7267, 496, 25, 657, 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.

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

\(\Big \downarrow \) 2586

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 496

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

Int[1/(x^3*(a*c + b*c*x^2 + d*Sqrt[a + b*x^2])),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 496

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 
 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a 
 + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 
*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad 
raticQ[a, 0, b, c, d, n, p, x]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

rule 2586

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] 
), x_Symbol] :> Simp[1/n   Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a 
+ b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* 
d, 0] && IntegerQ[(m + 1)/n]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1909\) vs. \(2(137)=274\).

Time = 0.05 (sec) , antiderivative size = 1910, normalized size of antiderivative = 12.65

method	result	size
default	\(\text {Expression too large to display}\)	\(1910\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.53 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\left [\frac {2 \, a^{2} b c^{3} x^{2} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a^{2} b c^{3} x^{2} \log \left (x\right ) + a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} - {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{2}}, \frac {2 \, a^{2} b c^{3} x^{2} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a^{2} b c^{3} x^{2} \log \left (x\right ) + a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} + 2 \, {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + 2 \, {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {b c^{4} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{a^{2} c^{5} - 2 \, a c^{3} d^{2} + c d^{4}} - \frac {b c^{3} \log \left (-b x^{2}\right )}{2 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )}} + \frac {{\left (3 \, a b c^{2} d - b d^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, {\left (a^{3} c^{4} - 2 \, a^{2} c^{2} d^{2} + a d^{4}\right )} \sqrt {-a}} - \frac {a^{2} b c^{3} - a b c d^{2} - {\left (a b c^{2} d - b d^{3}\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a c^{2} - d^{2}\right )}^{2} a b x^{2}} \] Input:

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

Output:

b*c^4*log(abs(sqrt(b*x^2 + a)*c + d))/(a^2*c^5 - 2*a*c^3*d^2 + c*d^4) - 1/ 
2*b*c^3*log(-b*x^2)/(a^2*c^4 - 2*a*c^2*d^2 + d^4) + 1/2*(3*a*b*c^2*d - b*d 
^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/((a^3*c^4 - 2*a^2*c^2*d^2 + a*d^4)*sq 
rt(-a)) - 1/2*(a^2*b*c^3 - a*b*c*d^2 - (a*b*c^2*d - b*d^3)*sqrt(b*x^2 + a) 
)/((a*c^2 - d^2)^2*a*b*x^2)
 

Mupad [B] (verification not implemented)

Time = 25.11 (sec) , antiderivative size = 4602, normalized size of antiderivative = 30.48 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((((12*a^2*b*c^6*d^9 - 28*a^3*b*c^8*d^7 + 32*a^4*b*c^10*d^5 - 18*a 
^5*b*c^12*d^3 - 2*a*b*c^4*d^11 + 4*a^6*b*c^14*d)/(16*(a^5*c^6 - a^2*d^6 + 
3*a^3*c^2*d^4 - 3*a^4*c^4*d^2)) - ((a + b*x^2)^(1/2)*(16*(b^2*d^6 - 6*a*b^ 
2*c^2*d^4 + 9*a^2*b^2*c^4*d^2)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5* 
c^4*d^4 - 4*a^6*c^6*d^2))^(1/2)*(16*a^7*c^14 + 16*a^2*c^4*d^10 - 48*a^3*c^ 
6*d^8 + 32*a^4*c^8*d^6 + 32*a^5*c^10*d^4 - 48*a^6*c^12*d^2))/(512*(a^4*c^4 
 + a^2*d^4 - 2*a^3*c^2*d^2)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4 
*d^4 - 4*a^6*c^6*d^2)))*(16*(b^2*d^6 - 6*a*b^2*c^2*d^4 + 9*a^2*b^2*c^4*d^2 
)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4*d^4 - 4*a^6*c^6*d^2))^(1/ 
2))/(16*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4*d^4 - 4*a^6*c^6*d^2 
)) + ((a + b*x^2)^(1/2)*(b^2*c^6*d^6 - 6*a*b^2*c^8*d^4 + 13*a^2*b^2*c^10*d 
^2))/(32*(a^4*c^4 + a^2*d^4 - 2*a^3*c^2*d^2)))*(16*(b^2*d^6 - 6*a*b^2*c^2* 
d^4 + 9*a^2*b^2*c^4*d^2)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4*d^ 
4 - 4*a^6*c^6*d^2))^(1/2)*1i)/(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c 
^4*d^4 - 4*a^6*c^6*d^2) - (((((12*a^2*b*c^6*d^9 - 28*a^3*b*c^8*d^7 + 32*a^ 
4*b*c^10*d^5 - 18*a^5*b*c^12*d^3 - 2*a*b*c^4*d^11 + 4*a^6*b*c^14*d)/(16*(a 
^5*c^6 - a^2*d^6 + 3*a^3*c^2*d^4 - 3*a^4*c^4*d^2)) + ((a + b*x^2)^(1/2)*(1 
6*(b^2*d^6 - 6*a*b^2*c^2*d^4 + 9*a^2*b^2*c^4*d^2)*(a^7*c^8 + a^3*d^8 - 4*a 
^4*c^2*d^6 + 6*a^5*c^4*d^4 - 4*a^6*c^6*d^2))^(1/2)*(16*a^7*c^14 + 16*a^2*c 
^4*d^10 - 48*a^3*c^6*d^8 + 32*a^4*c^8*d^6 + 32*a^5*c^10*d^4 - 48*a^6*c^...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.41 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {\sqrt {b \,x^{2}+a}\, a^{2} c^{2} d -\sqrt {b \,x^{2}+a}\, a \,d^{3}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d \,x^{2}-\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b \,d^{3} x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d \,x^{2}+\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b \,d^{3} x^{2}-2 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,c^{3} x^{2}-2 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,c^{3} x^{2}+2 \,\mathrm {log}\left (\frac {2 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, c x +2 \sqrt {b \,x^{2}+a}\, d +2 \sqrt {b}\, d x +2 a c +2 b c \,x^{2}}{\sqrt {a}}\right ) a^{2} b \,c^{3} x^{2}-a^{3} c^{3}-2 a^{2} b \,c^{3} x^{2}+a^{2} c \,d^{2}+2 a b c \,d^{2} x^{2}}{2 a^{2} x^{2} \left (a^{2} c^{4}-2 a \,c^{2} d^{2}+d^{4}\right )} \] Input:

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

Output:

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