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]

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

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Rubi [A]  time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 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 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])

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 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 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 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])

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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 1.05, size = 767, normalized size = 6.91 \[ \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 ] \]

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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giac [A]  time = 0.22, size = 142, normalized size = 1.28 \[ 2 \, c {\left (\frac {\arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{2}}{\sqrt {-c d^{2} - a e^{2}} c d^{2}} - \frac {\arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right ) e}{\sqrt {-a} c d^{2}} + \frac {1}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )} \sqrt {c} d}\right )} \]

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))

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maple [A]  time = 0.01, size = 180, normalized size = 1.62 \[ -\frac {e \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{2}}+\frac {e \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}\, d^{2}}-\frac {\sqrt {c \,x^{2}+a}}{a d x} \]

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2} + a} {\left (e x + d\right )} x^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,\sqrt {c\,x^2+a}\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \sqrt {a + c x^{2}} \left (d + e x\right )}\, dx \]

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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