\(\int \frac {1}{x^4 (a c+b c x^2+d \sqrt {a+b x^2})} \, dx\) [57]

Optimal result

Integrand size = 29, antiderivative size = 192 \[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {c}{3 \left (a c^2-d^2\right ) x^3}+\frac {b c^3}{\left (a c^2-d^2\right )^2 x}+\frac {d \sqrt {a+b x^2}}{3 a \left (a c^2-d^2\right ) x^3}-\frac {b d \left (5 a c^2-2 d^2\right ) \sqrt {a+b x^2}}{3 a^2 \left (a c^2-d^2\right )^2 x}+\frac {b^{3/2} c^4 \arctan \left (\frac {\sqrt {b} \sqrt {a c^2-d^2} x}{a c+d \sqrt {a+b x^2}}\right )}{\left (a c^2-d^2\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {-a^3 c^3+2 b d^3 x^2 \sqrt {a+b x^2}-a d \sqrt {a+b x^2} \left (d^2+5 b c^2 x^2\right )+a^2 c \left (d^2+3 b c^2 x^2+c d \sqrt {a+b x^2}\right )}{3 a^2 \left (-a c^2+d^2\right )^2 x^3}-\frac {2 b^{3/2} c^4 \arctan \left (\frac {d+c \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{5/2}} \] Input:

Integrate[1/(x^4*(a*c + b*c*x^2 + d*Sqrt[a + b*x^2])),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.44, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {2587, 27, 264, 264, 218, 382, 25, 27, 445, 27, 291, 218}

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.

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

\(\Big \downarrow \) 2587

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 382

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (\frac {\int -\frac {b \left (2 b x^2 c^2+5 a c^2-2 d^2\right )}{x^2 \sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (-\frac {\int \frac {b \left (2 b x^2 c^2+5 a c^2-2 d^2\right )}{x^2 \sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (-\frac {b \int \frac {2 b x^2 c^2+5 a c^2-2 d^2}{x^2 \sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 445

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (-\frac {b \left (-\frac {\int \frac {3 a^2 b c^4}{\sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2} \left (5 a c^2-2 d^2\right )}{a x \left (a c^2-d^2\right )}\right )}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (-\frac {b \left (-\frac {3 a b c^4 \int \frac {1}{\sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{a c^2-d^2}-\frac {\sqrt {a+b x^2} \left (5 a c^2-2 d^2\right )}{a x \left (a c^2-d^2\right )}\right )}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 291

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (-\frac {b \left (-\frac {3 a b c^4 \int \frac {1}{a c^2-d^2-\frac {\left (b \left (a c^2-d^2\right )-a b c^2\right ) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{a c^2-d^2}-\frac {\sqrt {a+b x^2} \left (5 a c^2-2 d^2\right )}{a x \left (a c^2-d^2\right )}\right )}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle c \left (-\frac {b c^2 \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )}{a c^2-d^2}-\frac {1}{3 x^3 \left (a c^2-d^2\right )}\right )-d \left (-\frac {b \left (-\frac {3 a \sqrt {b} c^4 \arctan \left (\frac {\sqrt {b} d x}{\sqrt {a+b x^2} \sqrt {a c^2-d^2}}\right )}{d \left (a c^2-d^2\right )^{3/2}}-\frac {\sqrt {a+b x^2} \left (5 a c^2-2 d^2\right )}{a x \left (a c^2-d^2\right )}\right )}{3 a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{3 a x^3 \left (a c^2-d^2\right )}\right )\)

Input:

Int[1/(x^4*(a*c + b*c*x^2 + d*Sqrt[a + b*x^2])),x]
 

Output:

c*(-1/3*1/((a*c^2 - d^2)*x^3) - (b*c^2*(-(1/((a*c^2 - d^2)*x)) - (Sqrt[b]* 
c*ArcTan[(Sqrt[b]*c*x)/Sqrt[a*c^2 - d^2]])/(a*c^2 - d^2)^(3/2)))/(a*c^2 - 
d^2)) - d*(-1/3*Sqrt[a + b*x^2]/(a*(a*c^2 - d^2)*x^3) - (b*(-(((5*a*c^2 - 
2*d^2)*Sqrt[a + b*x^2])/(a*(a*c^2 - d^2)*x)) - (3*a*Sqrt[b]*c^4*ArcTan[(Sq 
rt[b]*d*x)/(Sqrt[a*c^2 - d^2]*Sqrt[a + b*x^2])])/(d*(a*c^2 - d^2)^(3/2)))) 
/(3*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 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 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

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 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 445

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

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. \(1982\) vs. \(2(172)=344\).

Time = 0.05 (sec) , antiderivative size = 1983, normalized size of antiderivative = 10.33

method	result	size
default	\(\text {Expression too large to display}\)	\(1983\)

Input:

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

Output:

a*c*(-1/3/a/(a*c^2-d^2)/x^3+b*(2*a*c^2-d^2)/a^2/(a*c^2-d^2)^2/x+1/(a*c^2-d 
^2)^2*b^2*c^5/d^2/(b*(a*c^2-d^2))^(1/2)*arctan(x*b*c/(b*(a*c^2-d^2))^(1/2) 
)-b^2/d^2/a^2/(a*b)^(1/2)*arctan(x*b/(a*b)^(1/2)))-b^2*c^4/(a*c^2-d^2)/d^2 
/(b*(a*c^2-d^2))^(1/2)*arctan(x*b*c/(b*(a*c^2-d^2))^(1/2))+1/a*c*b^2/d^2/( 
a*b)^(1/2)*arctan(x*b/(a*b)^(1/2))-b*c/(a*c^2-d^2)/a/x-d*(-1/3/a^2/(a*c^2- 
d^2)/x^3*(b*x^2+a)^(3/2)-b*(2*a*c^2-d^2)/a^2/(a*c^2-d^2)^2*(-1/a/x*(b*x^2+ 
a)^(3/2)+2*b/a*(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*b^3*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))*(((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^3*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))*(((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^3*c^8/ 
(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))/(-(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)...
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 744, normalized size of antiderivative = 3.88 \[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\left [\frac {3 \, a^{2} b c^{4} x^{3} \sqrt {-\frac {b}{a c^{2} - d^{2}}} \log \left (\frac {a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4} + {\left (a^{2} b^{2} c^{4} - 8 \, a b^{2} c^{2} d^{2} + 8 \, b^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{3} b c^{4} - 5 \, a^{2} b c^{2} d^{2} + 4 \, a b d^{4}\right )} x^{2} + 4 \, {\left ({\left (a^{2} b c^{4} d - 3 \, a b c^{2} d^{3} + 2 \, b d^{5}\right )} x^{3} + {\left (a^{3} c^{4} d - 2 \, a^{2} c^{2} d^{3} + a d^{5}\right )} x\right )} \sqrt {b x^{2} + a} \sqrt {-\frac {b}{a c^{2} - d^{2}}}}{b^{2} c^{4} x^{4} + a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4} + 2 \, {\left (a b c^{4} - b c^{2} d^{2}\right )} x^{2}}\right ) + 6 \, a^{2} b c^{4} x^{3} \sqrt {-\frac {b}{a c^{2} - d^{2}}} \log \left (\frac {b c^{2} x^{2} - a c^{2} + 2 \, {\left (a c^{3} - c d^{2}\right )} x \sqrt {-\frac {b}{a c^{2} - d^{2}}} + d^{2}}{b c^{2} x^{2} + a c^{2} - d^{2}}\right ) + 12 \, a^{2} b c^{3} x^{2} - 4 \, a^{3} c^{3} + 4 \, a^{2} c d^{2} + 4 \, {\left (a^{2} c^{2} d - a d^{3} - {\left (5 \, a b c^{2} d - 2 \, b d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{3}}, \frac {6 \, a^{2} b c^{4} x^{3} \sqrt {\frac {b}{a c^{2} - d^{2}}} \arctan \left (c x \sqrt {\frac {b}{a c^{2} - d^{2}}}\right ) - 3 \, a^{2} b c^{4} x^{3} \sqrt {\frac {b}{a c^{2} - d^{2}}} \arctan \left (-\frac {{\left (a^{2} c^{2} - a d^{2} + {\left (a b c^{2} - 2 \, b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {\frac {b}{a c^{2} - d^{2}}}}{2 \, {\left (b^{2} d x^{3} + a b d x\right )}}\right ) + 6 \, a^{2} b c^{3} x^{2} - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} + 2 \, {\left (a^{2} c^{2} d - a d^{3} - {\left (5 \, a b c^{2} d - 2 \, b d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{3}}\right ] \] Input:

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

Output:

[1/12*(3*a^2*b*c^4*x^3*sqrt(-b/(a*c^2 - d^2))*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^2*b*c^4*d - 3*a*b*c^2*d^3 + 2 
*b*d^5)*x^3 + (a^3*c^4*d - 2*a^2*c^2*d^3 + a*d^5)*x)*sqrt(b*x^2 + a)*sqrt( 
-b/(a*c^2 - d^2)))/(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^2*b*c^4*x^3*sqrt(-b/(a*c^2 - d^2))*log((b*c^2*x^ 
2 - a*c^2 + 2*(a*c^3 - c*d^2)*x*sqrt(-b/(a*c^2 - d^2)) + d^2)/(b*c^2*x^2 + 
 a*c^2 - d^2)) + 12*a^2*b*c^3*x^2 - 4*a^3*c^3 + 4*a^2*c*d^2 + 4*(a^2*c^2*d 
 - a*d^3 - (5*a*b*c^2*d - 2*b*d^3)*x^2)*sqrt(b*x^2 + a))/((a^4*c^4 - 2*a^3 
*c^2*d^2 + a^2*d^4)*x^3), 1/6*(6*a^2*b*c^4*x^3*sqrt(b/(a*c^2 - d^2))*arcta 
n(c*x*sqrt(b/(a*c^2 - d^2))) - 3*a^2*b*c^4*x^3*sqrt(b/(a*c^2 - d^2))*arcta 
n(-1/2*(a^2*c^2 - a*d^2 + (a*b*c^2 - 2*b*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(b/ 
(a*c^2 - d^2))/(b^2*d*x^3 + a*b*d*x)) + 6*a^2*b*c^3*x^2 - 2*a^3*c^3 + 2*a^ 
2*c*d^2 + 2*(a^2*c^2*d - a*d^3 - (5*a*b*c^2*d - 2*b*d^3)*x^2)*sqrt(b*x^2 + 
 a))/((a^4*c^4 - 2*a^3*c^2*d^2 + a^2*d^4)*x^3)]
 

Sympy [F]

\[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\int \frac {1}{x^{4} \left (a c + b c x^{2} + d \sqrt {a + b x^{2}}\right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\int { \frac {1}{{\left (b c x^{2} + a c + \sqrt {b x^{2} + a} d\right )} x^{4}} \,d x } \] Input:

integrate(1/x^4/(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^4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (172) = 344\).

Time = 0.14 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.88 \[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {b^{2} c^{4} \arctan \left (\frac {b c x}{\sqrt {a b c^{2} - b d^{2}}}\right )}{{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \sqrt {a b c^{2} - b d^{2}}} + \frac {1}{3} \, {\left (\frac {3 \, c^{4} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} c^{2} + a c^{2} - 2 \, d^{2}}{2 \, \sqrt {a c^{2} - d^{2}} d}\right )}{{\left (a^{2} b c^{4} - 2 \, a b c^{2} d^{2} + b d^{4}\right )} \sqrt {a c^{2} - d^{2}} d} + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} c^{2} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a c^{2} + 5 \, a^{2} c^{2} + 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d^{2} - 2 \, a d^{2}\right )}}{{\left (a^{2} b c^{4} - 2 \, a b c^{2} d^{2} + b d^{4}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}}\right )} b^{\frac {5}{2}} d + \frac {3 \, b c^{3} x^{2} - a c^{3} + c d^{2}}{3 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} x^{3}} \] Input:

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

Output:

b^2*c^4*arctan(b*c*x/sqrt(a*b*c^2 - b*d^2))/((a^2*c^4 - 2*a*c^2*d^2 + d^4) 
*sqrt(a*b*c^2 - b*d^2)) + 1/3*(3*c^4*arctan(1/2*((sqrt(b)*x - sqrt(b*x^2 + 
 a))^2*c^2 + a*c^2 - 2*d^2)/(sqrt(a*c^2 - d^2)*d))/((a^2*b*c^4 - 2*a*b*c^2 
*d^2 + b*d^4)*sqrt(a*c^2 - d^2)*d) + 2*(3*(sqrt(b)*x - sqrt(b*x^2 + a))^4* 
c^2 - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*c^2 + 5*a^2*c^2 + 6*(sqrt(b)*x 
- sqrt(b*x^2 + a))^2*d^2 - 2*a*d^2)/((a^2*b*c^4 - 2*a*b*c^2*d^2 + b*d^4)*( 
(sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3))*b^(5/2)*d + 1/3*(3*b*c^3*x^2 - a* 
c^3 + c*d^2)/((a^2*c^4 - 2*a*c^2*d^2 + d^4)*x^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\int \frac {1}{x^4\,\left (a\,c+d\,\sqrt {b\,x^2+a}+b\,c\,x^2\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.59 \[ \int \frac {1}{x^4 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {6 \sqrt {b}\, \sqrt {a \,c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\sqrt {b \,x^{2}+a}\, c +\sqrt {b}\, c x +d}{\sqrt {a \,c^{2}-d^{2}}}\right ) a^{2} b \,c^{4} x^{3}+\sqrt {b \,x^{2}+a}\, a^{3} c^{4} d -5 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{4} d \,x^{2}-2 \sqrt {b \,x^{2}+a}\, a^{2} c^{2} d^{3}+7 \sqrt {b \,x^{2}+a}\, a b \,c^{2} d^{3} x^{2}+\sqrt {b \,x^{2}+a}\, a \,d^{5}-2 \sqrt {b \,x^{2}+a}\, b \,d^{5} x^{2}+3 \sqrt {b}\, a^{2} b \,c^{4} d \,x^{3}-5 \sqrt {b}\, a b \,c^{2} d^{3} x^{3}+2 \sqrt {b}\, b \,d^{5} x^{3}-a^{4} c^{5}+3 a^{3} b \,c^{5} x^{2}+2 a^{3} c^{3} d^{2}-3 a^{2} b \,c^{3} d^{2} x^{2}-a^{2} c \,d^{4}}{3 a^{2} x^{3} \left (a^{3} c^{6}-3 a^{2} c^{4} d^{2}+3 a \,c^{2} d^{4}-d^{6}\right )} \] Input:

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

Output:

(6*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**2*b*c**4*x**3 + sqrt(a + b*x**2)*a**3*c**4*d - 5* 
sqrt(a + b*x**2)*a**2*b*c**4*d*x**2 - 2*sqrt(a + b*x**2)*a**2*c**2*d**3 + 
7*sqrt(a + b*x**2)*a*b*c**2*d**3*x**2 + sqrt(a + b*x**2)*a*d**5 - 2*sqrt(a 
 + b*x**2)*b*d**5*x**2 + 3*sqrt(b)*a**2*b*c**4*d*x**3 - 5*sqrt(b)*a*b*c**2 
*d**3*x**3 + 2*sqrt(b)*b*d**5*x**3 - a**4*c**5 + 3*a**3*b*c**5*x**2 + 2*a* 
*3*c**3*d**2 - 3*a**2*b*c**3*d**2*x**2 - a**2*c*d**4)/(3*a**2*x**3*(a**3*c 
**6 - 3*a**2*c**4*d**2 + 3*a*c**2*d**4 - d**6))