3.333 \(\int \frac{1}{x^3 (d+e x) \sqrt{a+c x^2}} \, dx\)

Optimal. Leaf size=168 \[ \frac{c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{2 a^{3/2} d}+\frac{e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^3 \sqrt{a e^2+c d^2}}-\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^3}+\frac{e \sqrt{a+c x^2}}{a d^2 x}-\frac{\sqrt{a+c x^2}}{2 a d x^2} \]

[Out]

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

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Rubi [A]  time = 0.140467, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {961, 266, 51, 63, 208, 264, 725, 206} \[ \frac{c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{2 a^{3/2} d}+\frac{e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{d^3 \sqrt{a e^2+c d^2}}-\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^3}+\frac{e \sqrt{a+c x^2}}{a d^2 x}-\frac{\sqrt{a+c x^2}}{2 a d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[a + c*x^2]/(2*a*d*x^2) + (e*Sqrt[a + c*x^2])/(a*d^2*x) + (e^3*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]) + (c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2)*d) - (e^2*ArcT
anh[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 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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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^3 (d+e x) \sqrt{a+c x^2}} \, dx &=\int \left (\frac{1}{d x^3 \sqrt{a+c x^2}}-\frac{e}{d^2 x^2 \sqrt{a+c x^2}}+\frac{e^2}{d^3 x \sqrt{a+c x^2}}-\frac{e^3}{d^3 (d+e x) \sqrt{a+c x^2}}\right ) \, dx\\ &=\frac{\int \frac{1}{x^3 \sqrt{a+c x^2}} \, dx}{d}-\frac{e \int \frac{1}{x^2 \sqrt{a+c x^2}} \, dx}{d^2}+\frac{e^2 \int \frac{1}{x \sqrt{a+c x^2}} \, dx}{d^3}-\frac{e^3 \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{d^3}\\ &=\frac{e \sqrt{a+c x^2}}{a d^2 x}+\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+c x}} \, dx,x,x^2\right )}{2 d}+\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{2 d^3}+\frac{e^3 \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{\sqrt{a+c x^2}}{2 a d x^2}+\frac{e \sqrt{a+c x^2}}{a d^2 x}+\frac{e^3 \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{c \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{4 a d}+\frac{e^2 \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{\sqrt{a+c x^2}}{2 a d x^2}+\frac{e \sqrt{a+c x^2}}{a d^2 x}+\frac{e^3 \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{e^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^3}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{2 a d}\\ &=-\frac{\sqrt{a+c x^2}}{2 a d x^2}+\frac{e \sqrt{a+c x^2}}{a d^2 x}+\frac{e^3 \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{c \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{2 a^{3/2} d}-\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a} d^3}\\ \end{align*}

Mathematica [A]  time = 0.716823, size = 163, normalized size = 0.97 \[ \frac{\frac{2 e^3 \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{\sqrt{a e^2+c d^2}}+\frac{d \left (c d x^2 \sqrt{\frac{c x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{c x^2}{a}+1}\right )-\left (a+c x^2\right ) (d-2 e x)\right )}{a x^2 \sqrt{a+c x^2}}-\frac{2 e^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.246, size = 236, normalized size = 1.4 \begin{align*} -{\frac{{e}^{2}}{{d}^{3}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{{e}^{2}}{{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{1}{2\,ad{x}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{c}{2\,d}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}+{\frac{e}{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^3/(e*x+d)/(c*x^2+a)^(1/2),x)

[Out]

-e^2/d^3/a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)+e^2/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))
-1/2*(c*x^2+a)^(1/2)/a/d/x^2+1/2/d*c/a^(3/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)+e*(c*x^2+a)^(1/2)/a/d^2/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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^2 + a)*(e*x + d)*x^3), x)

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Fricas [A]  time = 2.80162, size = 1987, normalized size = 11.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(c*d^2 + a*e^2)*a^2*e^3*x^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)) - (c^2*d^4 - a*c*d^2*e^2 - 2*a^
2*e^4)*sqrt(a)*x^2*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(a*c*d^4 + a^2*d^2*e^2 - 2*(a*c*d^3
*e + a^2*d*e^3)*x)*sqrt(c*x^2 + a))/((a^2*c*d^5 + a^3*d^3*e^2)*x^2), 1/4*(4*sqrt(-c*d^2 - a*e^2)*a^2*e^3*x^2*a
rctan(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 - a*c*d^2*e^2 - 2*a^2*e^4)*sqrt(a)*x^2*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(a*c*d^4 +
a^2*d^2*e^2 - 2*(a*c*d^3*e + a^2*d*e^3)*x)*sqrt(c*x^2 + a))/((a^2*c*d^5 + a^3*d^3*e^2)*x^2), 1/2*(sqrt(c*d^2
+ a*e^2)*a^2*e^3*x^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)) - (c^2*d^4 - a*c*d^2*e^2 - 2*a^2*e^4)*sqrt(-a)*x^2
*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (a*c*d^4 + a^2*d^2*e^2 - 2*(a*c*d^3*e + a^2*d*e^3)*x)*sqrt(c*x^2 + a))/((a
^2*c*d^5 + a^3*d^3*e^2)*x^2), 1/2*(2*sqrt(-c*d^2 - a*e^2)*a^2*e^3*x^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^4 - a*c*d^2*e^2 - 2*a^2*e^4)*sqrt(-a
)*x^2*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (a*c*d^4 + a^2*d^2*e^2 - 2*(a*c*d^3*e + a^2*d*e^3)*x)*sqrt(c*x^2 + a)
)/((a^2*c*d^5 + a^3*d^3*e^2)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \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**3/(e*x+d)/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + c*x**2)*(d + e*x)), x)

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Giac [A]  time = 1.17297, size = 323, normalized size = 1.92 \begin{align*} -c^{\frac{3}{2}}{\left (\frac{2 \, \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^{3}}{\sqrt{-c d^{2} - a e^{2}} c^{\frac{3}{2}} d^{3}} + \frac{{\left (c d^{2} - 2 \, a e^{2}\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a c^{\frac{3}{2}} d^{3}} - \frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} \sqrt{c} d - 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a e +{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a \sqrt{c} d + 2 \, a^{2} e}{{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )}^{2} a c d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x+d)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-c^(3/2)*(2*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^3/(sqrt(-c*d^2 - a*e
^2)*c^(3/2)*d^3) + (c*d^2 - 2*a*e^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a*c^(3/2)*d^3)
- ((sqrt(c)*x - sqrt(c*x^2 + a))^3*sqrt(c)*d - 2*(sqrt(c)*x - sqrt(c*x^2 + a))^2*a*e + (sqrt(c)*x - sqrt(c*x^2
+ a))*a*sqrt(c)*d + 2*a^2*e)/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^2*a*c*d^2))