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

Optimal. Leaf size=212 \[ -\frac{e^3 \sqrt{a+c x^2}}{d^2 (d+e x) \left (a e^2+c d^2\right )}-\frac{2 e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^3 \sqrt{a e^2+c d^2}}-\frac{c e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d \left (a e^2+c d^2\right )^{3/2}}+\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^3}-\frac{\sqrt{a+c x^2}}{a d^2 x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16785, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {961, 264, 266, 63, 208, 731, 725, 206} \[ -\frac{e^3 \sqrt{a+c x^2}}{d^2 (d+e x) \left (a e^2+c d^2\right )}-\frac{2 e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^3 \sqrt{a e^2+c d^2}}-\frac{c e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d \left (a e^2+c d^2\right )^{3/2}}+\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^3}-\frac{\sqrt{a+c x^2}}{a d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 725

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.389819, size = 197, normalized size = 0.93 \[ \frac{-d \sqrt{a+c x^2} \left (\frac{e^3}{(d+e x) \left (a e^2+c d^2\right )}+\frac{1}{a x}\right )-\frac{e^2 \left (2 a e^2+3 c d^2\right ) \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}+\frac{e^2 \left (2 a e^2+3 c d^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}+\frac{2 e \log \left (\sqrt{a} \sqrt{a+c x^2}+a\right )}{\sqrt{a}}-\frac{2 e \log (x)}{\sqrt{a}}}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(d*Sqrt[a + c*x^2]*(1/(a*x) + e^3/((c*d^2 + a*e^2)*(d + e*x)))) - (2*e*Log[x])/Sqrt[a] + (e^2*(3*c*d^2 + 2*a
*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^(3/2) + (2*e*Log[a + Sqrt[a]*Sqrt[a + c*x^2]])/Sqrt[a] - (e^2*(3*c*d^2 + 2
*a*e^2)*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2]])/(c*d^2 + a*e^2)^(3/2))/d^3

________________________________________________________________________________________

Maple [B]  time = 0.238, size = 395, normalized size = 1.9 \begin{align*} 2\,{\frac{e}{{d}^{3}\sqrt{a}}\ln \left ({\frac{2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a}}{x}} \right ) }-2\,{\frac{e}{{d}^{3}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{{e}^{2}}{{d}^{2} \left ( a{e}^{2}+c{d}^{2} \right ) }\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \left ({\frac{d}{e}}+x \right ) ^{-1}}-{\frac{ce}{d \left ( a{e}^{2}+c{d}^{2} \right ) }\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{1}{a{d}^{2}x}\sqrt{c{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + a}{\left (e x + d\right )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 4.6874, size = 3093, normalized size = 14.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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)*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^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)*sqrt(-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^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^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)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^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)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (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)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{a + c x^{2}} \left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError