3.4.49 \(\int \frac {1}{x^2 (d+e x)^2 \sqrt {a+c x^2}} \, dx\) [349]

3.4.49.1 Optimal result
3.4.49.2 Mathematica [A] (verified)
3.4.49.3 Rubi [A] (verified)
3.4.49.4 Maple [B] (verified)
3.4.49.5 Fricas [A] (verification not implemented)
3.4.49.6 Sympy [F]
3.4.49.7 Maxima [F]
3.4.49.8 Giac [F]
3.4.49.9 Mupad [F(-1)]

3.4.49.1 Optimal result

Integrand size = 22, antiderivative size = 212 \[ \int \frac {1}{x^2 (d+e x)^2 \sqrt {a+c x^2}} \, dx=-\frac {\sqrt {a+c x^2}}{a d^2 x}-\frac {e^3 \sqrt {a+c x^2}}{d^2 \left (c d^2+a e^2\right ) (d+e x)}-\frac {c e^2 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d \left (c d^2+a e^2\right )^{3/2}}-\frac {2 e^2 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^3 \sqrt {c d^2+a e^2}}+\frac {2 e \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3} \]

output
-c*e^2*arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/d/(a*e^2+ 
c*d^2)^(3/2)+2*e*arctanh((c*x^2+a)^(1/2)/a^(1/2))/d^3/a^(1/2)-2*e^2*arctan 
h((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/d^3/(a*e^2+c*d^2)^(1/2 
)-(c*x^2+a)^(1/2)/a/d^2/x-e^3*(c*x^2+a)^(1/2)/d^2/(a*e^2+c*d^2)/(e*x+d)
 
3.4.49.2 Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\frac {-\frac {d \sqrt {a+c x^2} \left (c d^2 (d+e x)+a e^2 (d+2 e x)\right )}{a \left (c d^2+a e^2\right ) x (d+e x)}+\frac {2 e^2 \left (3 c d^2+2 a e^2\right ) \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{3/2}}-\frac {4 e \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}}{d^3} \]

input
Integrate[1/(x^2*(d + e*x)^2*Sqrt[a + c*x^2]),x]
 
output
(-((d*Sqrt[a + c*x^2]*(c*d^2*(d + e*x) + a*e^2*(d + 2*e*x)))/(a*(c*d^2 + a 
*e^2)*x*(d + e*x))) + (2*e^2*(3*c*d^2 + 2*a*e^2)*ArcTan[(Sqrt[c]*(d + e*x) 
 - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/(-(c*d^2) - a*e^2)^(3/2) - 
(4*e*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/Sqrt[a])/d^3
 
3.4.49.3 Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {617, 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.

\(\displaystyle \int \frac {1}{x^2 \sqrt {a+c x^2} (d+e x)^2} \, dx\)

\(\Big \downarrow \) 617

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

\(\Big \downarrow \) 2009

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

input
Int[1/(x^2*(d + e*x)^2*Sqrt[a + c*x^2]),x]
 
output
-(Sqrt[a + c*x^2]/(a*d^2*x)) - (e^3*Sqrt[a + c*x^2])/(d^2*(c*d^2 + a*e^2)* 
(d + e*x)) - (c*e^2*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c* 
x^2])])/(d*(c*d^2 + a*e^2)^(3/2)) - (2*e^2*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d 
^2 + a*e^2]*Sqrt[a + c*x^2])])/(d^3*Sqrt[c*d^2 + a*e^2]) + (2*e*ArcTanh[Sq 
rt[a + c*x^2]/Sqrt[a]])/(Sqrt[a]*d^3)
 

3.4.49.3.1 Defintions of rubi rules used

rule 617
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[(a + b*x^2)^p, x^m*(c + d*x)^n, x], x] /; FreeQ[{ 
a, b, c, d, p}, x] && ILtQ[n, 0] && IntegerQ[m] && IntegerQ[2*p]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.4.49.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(190)=380\).

Time = 0.39 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.86

method result size
default \(-\frac {\sqrt {c \,x^{2}+a}}{a \,d^{2} x}+\frac {2 e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d^{3} \sqrt {a}}+\frac {-\frac {e^{2} \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {e c d \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{d^{2}}-\frac {2 e \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d^{3} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\) \(395\)
risch \(-\frac {\sqrt {c \,x^{2}+a}}{a \,d^{2} x}+\frac {2 e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{d^{3} \sqrt {a}}-\frac {e^{2} \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{d^{2} \left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {e c \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}-\frac {2 e \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d^{3} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\) \(395\)

input
int(1/x^2/(e*x+d)^2/(c*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 
output
-(c*x^2+a)^(1/2)/a/d^2/x+2/d^3*e/a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2) 
)/x)+1/d^2*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)*((x+d/e)^2*c-2/e*c*d*(x+d/e)+(a*e 
^2+c*d^2)/e^2)^(1/2)-e*c*d/(a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*( 
a*e^2+c*d^2)/e^2-2/e*c*d*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c- 
2/e*c*d*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))-2*e/d^3/((a*e^2+c*d^2) 
/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2/e*c*d*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^ 
(1/2)*((x+d/e)^2*c-2/e*c*d*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))
 
3.4.49.5 Fricas [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 1512, normalized size of antiderivative = 7.13 \[ \int \frac {1}{x^2 (d+e x)^2 \sqrt {a+c x^2}} \, dx=\text {Too large to display} \]

input
integrate(1/x^2/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="fricas")
 
output
[1/2*(sqrt(c*d^2 + a*e^2)*((3*a*c*d^2*e^3 + 2*a^2*e^5)*x^2 + (3*a*c*d^3*e^ 
2 + 2*a^2*d*e^4)*x)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + 
a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x 
^2 + 2*d*e*x + d^2)) + 2*((c^2*d^4*e^2 + 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + (c 
^2*d^5*e + 2*a*c*d^3*e^3 + a^2*d*e^5)*x)*sqrt(a)*log(-(c*x^2 + 2*sqrt(c*x^ 
2 + a)*sqrt(a) + 2*a)/x^2) - 2*(c^2*d^6 + 2*a*c*d^4*e^2 + a^2*d^2*e^4 + (c 
^2*d^5*e + 3*a*c*d^3*e^3 + 2*a^2*d*e^5)*x)*sqrt(c*x^2 + a))/((a*c^2*d^7*e 
+ 2*a^2*c*d^5*e^3 + a^3*d^3*e^5)*x^2 + (a*c^2*d^8 + 2*a^2*c*d^6*e^2 + a^3* 
d^4*e^4)*x), -(sqrt(-c*d^2 - a*e^2)*((3*a*c*d^2*e^3 + 2*a^2*e^5)*x^2 + (3* 
a*c*d^3*e^2 + 2*a^2*d*e^4)*x)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sq 
rt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - ((c^2*d^4*e 
^2 + 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + (c^2*d^5*e + 2*a*c*d^3*e^3 + a^2*d*e^5 
)*x)*sqrt(a)*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + (c^2*d^ 
6 + 2*a*c*d^4*e^2 + a^2*d^2*e^4 + (c^2*d^5*e + 3*a*c*d^3*e^3 + 2*a^2*d*e^5 
)*x)*sqrt(c*x^2 + a))/((a*c^2*d^7*e + 2*a^2*c*d^5*e^3 + a^3*d^3*e^5)*x^2 + 
 (a*c^2*d^8 + 2*a^2*c*d^6*e^2 + a^3*d^4*e^4)*x), -1/2*(4*((c^2*d^4*e^2 + 2 
*a*c*d^2*e^4 + a^2*e^6)*x^2 + (c^2*d^5*e + 2*a*c*d^3*e^3 + a^2*d*e^5)*x)*s 
qrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - sqrt(c*d^2 + a*e^2)*((3*a*c*d^2 
*e^3 + 2*a^2*e^5)*x^2 + (3*a*c*d^3*e^2 + 2*a^2*d*e^4)*x)*log((2*a*c*d*e*x 
- a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^...
 
3.4.49.6 Sympy [F]

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

input
integrate(1/x**2/(e*x+d)**2/(c*x**2+a)**(1/2),x)
 
output
Integral(1/(x**2*sqrt(a + c*x**2)*(d + e*x)**2), x)
 
3.4.49.7 Maxima [F]

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

input
integrate(1/x^2/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="maxima")
 
output
integrate(1/(sqrt(c*x^2 + a)*(e*x + d)^2*x^2), x)
 
3.4.49.8 Giac [F]

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

input
integrate(1/x^2/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="giac")
 
output
sage0*x
 
3.4.49.9 Mupad [F(-1)]

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

input
int(1/(x^2*(a + c*x^2)^(1/2)*(d + e*x)^2),x)
 
output
int(1/(x^2*(a + c*x^2)^(1/2)*(d + e*x)^2), x)