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

Optimal. Leaf size=111 \[ -\frac{e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^2 \sqrt{a e^2+c d^2}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+c x^2}}{a d x} \]

[Out]

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

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Rubi [A]  time = 0.0953832, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {961, 264, 266, 63, 208, 725, 206} \[ -\frac{e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^2 \sqrt{a e^2+c d^2}}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^2}-\frac{\sqrt{a+c x^2}}{a d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0899508, size = 107, normalized size = 0.96 \[ \frac{-\frac{e^2 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\sqrt{a e^2+c d^2}}-\frac{d \sqrt{a+c x^2}}{a x}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.236, size = 180, normalized size = 1.6 \begin{align*}{\frac{e}{{d}^{2}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{e}{{d}^{2}}\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}{adx}\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)/(c*x^2+a)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.42056, size = 1639, normalized size = 14.77 \begin{align*} \left [\frac{\sqrt{c d^{2} + a e^{2}} a e^{2} x \log \left (\frac{2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} -{\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt{c d^{2} + a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) +{\left (c d^{2} e + a e^{3}\right )} \sqrt{a} x \log \left (-\frac{c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (c d^{3} + a d e^{2}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (a c d^{4} + a^{2} d^{2} e^{2}\right )} x}, -\frac{2 \, \sqrt{-c d^{2} - a e^{2}} a e^{2} x \arctan \left (\frac{\sqrt{-c d^{2} - a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{a c d^{2} + a^{2} e^{2} +{\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) -{\left (c d^{2} e + a e^{3}\right )} \sqrt{a} x \log \left (-\frac{c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (c d^{3} + a d e^{2}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (a c d^{4} + a^{2} d^{2} e^{2}\right )} x}, \frac{\sqrt{c d^{2} + a e^{2}} a e^{2} x \log \left (\frac{2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} -{\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt{c d^{2} + a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \,{\left (c d^{2} e + a e^{3}\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) - 2 \,{\left (c d^{3} + a d e^{2}\right )} \sqrt{c x^{2} + a}}{2 \,{\left (a c d^{4} + a^{2} d^{2} e^{2}\right )} x}, -\frac{\sqrt{-c d^{2} - a e^{2}} a e^{2} x \arctan \left (\frac{\sqrt{-c d^{2} - a e^{2}}{\left (c d x - a e\right )} \sqrt{c x^{2} + a}}{a c d^{2} + a^{2} e^{2} +{\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) +{\left (c d^{2} e + a e^{3}\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) +{\left (c d^{3} + a d e^{2}\right )} \sqrt{c x^{2} + a}}{{\left (a c d^{4} + a^{2} d^{2} e^{2}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(c*d^2 + a*e^2)*a*e^2*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)) + (c*d^2*e + a*e^3)*sqrt(a)*x*log(-(c
*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(c*d^3 + a*d*e^2)*sqrt(c*x^2 + a))/((a*c*d^4 + a^2*d^2*e^2)*x
), -1/2*(2*sqrt(-c*d^2 - a*e^2)*a*e^2*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*d^2*e + a*e^3)*sqrt(a)*x*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a
)/x^2) + 2*(c*d^3 + a*d*e^2)*sqrt(c*x^2 + a))/((a*c*d^4 + a^2*d^2*e^2)*x), 1/2*(sqrt(c*d^2 + a*e^2)*a*e^2*x*lo
g((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*d^2*e + a*e^3)*sqrt(-a)*x*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - 2*(c*
d^3 + a*d*e^2)*sqrt(c*x^2 + a))/((a*c*d^4 + a^2*d^2*e^2)*x), -(sqrt(-c*d^2 - a*e^2)*a*e^2*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*d^2*e + a*e^3)*sqr
t(-a)*x*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (c*d^3 + a*d*e^2)*sqrt(c*x^2 + a))/((a*c*d^4 + a^2*d^2*e^2)*x)]

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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 )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14147, size = 192, normalized size = 1.73 \begin{align*} 2 \, c{\left (\frac{\arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} - a e^{2}}}\right ) e^{2}}{\sqrt{-c d^{2} - a e^{2}} c d^{2}} - \frac{\arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right ) e}{\sqrt{-a} c d^{2}} + \frac{1}{{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )} \sqrt{c} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*(arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^2/(sqrt(-c*d^2 - a*e^2)*c*d
^2) - arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))*e/(sqrt(-a)*c*d^2) + 1/(((sqrt(c)*x - sqrt(c*x^2 + a))^2
- a)*sqrt(c)*d))