\(\int \frac {x^2}{(d+e x)^2 \sqrt {a+c x^2}} \, dx\) [345]

Optimal result

Integrand size = 22, antiderivative size = 137 \[ \int \frac {x^2}{(d+e x)^2 \sqrt {a+c x^2}} \, dx=-\frac {d^2 \sqrt {a+c x^2}}{e \left (c d^2+a e^2\right ) (d+e x)}+\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} e^2}+\frac {d \left (c d^2+2 a e^2\right ) \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^2 \left (c d^2+a e^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1665, 858, 223, 212, 739} \[ \int \frac {x^2}{(d+e x)^2 \sqrt {a+c x^2}} \, dx=\frac {d \left (2 a e^2+c d^2\right ) \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^2 \left (a e^2+c d^2\right )^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} e^2}-\frac {d^2 \sqrt {a+c x^2}}{e (d+e x) \left (a e^2+c d^2\right )} \]

[In]

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

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(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] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1665

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 \sqrt {a+c x^2}}{e \left (c d^2+a e^2\right ) (d+e x)}-\frac {\int \frac {a d-\frac {\left (c d^2+a e^2\right ) x}{e}}{(d+e x) \sqrt {a+c x^2}} \, dx}{c d^2+a e^2} \\ & = -\frac {d^2 \sqrt {a+c x^2}}{e \left (c d^2+a e^2\right ) (d+e x)}+\frac {\int \frac {1}{\sqrt {a+c x^2}} \, dx}{e^2}-\frac {\left (d \left (2 a+\frac {c d^2}{e^2}\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{c d^2+a e^2} \\ & = -\frac {d^2 \sqrt {a+c x^2}}{e \left (c d^2+a e^2\right ) (d+e x)}+\frac {\text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{e^2}+\frac {\left (d \left (2 a+\frac {c d^2}{e^2}\right )\right ) \text {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{c d^2+a e^2} \\ & = -\frac {d^2 \sqrt {a+c x^2}}{e \left (c d^2+a e^2\right ) (d+e x)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} e^2}+\frac {d \left (2 a+\frac {c d^2}{e^2}\right ) \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{\left (c d^2+a e^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.15 \[ \int \frac {x^2}{(d+e x)^2 \sqrt {a+c x^2}} \, dx=\frac {d \left (-\frac {d e \sqrt {a+c x^2}}{\left (c d^2+a e^2\right ) (d+e x)}-\frac {2 \left (c d^2+2 a e^2\right ) \arctan \left (\frac {\sqrt {-c d^2-a e^2} x}{\sqrt {a} (d+e x)-d \sqrt {a+c x^2}}\right )}{\left (-c d^2-a e^2\right )^{3/2}}\right )+\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+c x^2}}\right )}{\sqrt {c}}}{e^2} \]

[In]

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

[Out]

(d*(-((d*e*Sqrt[a + c*x^2])/((c*d^2 + a*e^2)*(d + e*x))) - (2*(c*d^2 + 2*a*e^2)*ArcTan[(Sqrt[-(c*d^2) - a*e^2]
*x)/(Sqrt[a]*(d + e*x) - d*Sqrt[a + c*x^2])])/(-(c*d^2) - a*e^2)^(3/2)) + (2*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + S
qrt[a + c*x^2])])/Sqrt[c])/e^2

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(368\) vs. \(2(123)=246\).

Time = 0.42 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.69

method	result	size
default	\(\frac {\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{e^{2} \sqrt {c}}+\frac {d^{2} \left (-\frac {e^{2} \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {e c d \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{e^{4}}+\frac {2 d \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c -\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{3} \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\)	\(369\)

[In]

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

[Out]

1/e^2*ln(x*c^(1/2)+(c*x^2+a)^(1/2))/c^(1/2)+d^2/e^4*(-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*d/e^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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (124) = 248\).

Time = 5.49 (sec) , antiderivative size = 1260, normalized size of antiderivative = 9.20 \[ \int \frac {x^2}{(d+e x)^2 \sqrt {a+c x^2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/2*((c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(c)*log(-2*c*x^2 - 2
*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (c^2*d^4 + 2*a*c*d^2*e^2 + (c^2*d^3*e + 2*a*c*d*e^3)*x)*sqrt(c*d^2 + a*e^2)*
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 + a*c*d^2*e^3)*sqrt(c*x^2 + a))/(c^3*d^5*e^2 + 2*a*c^2*d
^3*e^4 + a^2*c*d*e^6 + (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x), 1/2*(2*(c^2*d^4 + 2*a*c*d^2*e^2 + (c^2*
d^3*e + 2*a*c*d*e^3)*x)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(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^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 +
a^2*e^5)*x)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(c^2*d^4*e + a*c*d^2*e^3)*sqrt(c*x^2 +
a))/(c^3*d^5*e^2 + 2*a*c^2*d^3*e^4 + a^2*c*d*e^6 + (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x), -1/2*(2*(c
^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(
c*x^2 + a)) - (c^2*d^4 + 2*a*c*d^2*e^2 + (c^2*d^3*e + 2*a*c*d*e^3)*x)*sqrt(c*d^2 + a*e^2)*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 + a*c*d^2*e^3)*sqrt(c*x^2 + a))/(c^3*d^5*e^2 + 2*a*c^2*d^3*e^4 + a^2*c*d*e^6
+ (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x), ((c^2*d^4 + 2*a*c*d^2*e^2 + (c^2*d^3*e + 2*a*c*d*e^3)*x)*sq
rt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(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^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-c)*arcta
n(sqrt(-c)*x/sqrt(c*x^2 + a)) - (c^2*d^4*e + a*c*d^2*e^3)*sqrt(c*x^2 + a))/(c^3*d^5*e^2 + 2*a*c^2*d^3*e^4 + a^
2*c*d*e^6 + (c^3*d^4*e^3 + 2*a*c^2*d^2*e^5 + a^2*c*e^7)*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

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

[In]

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

[Out]

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