3.268 \(\int \frac{1}{x^2 (a x^2+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=166 \[ \frac{315 b^3 \sqrt{a x^2+b x^3}}{64 a^5 x^2}-\frac{105 b^2 \sqrt{a x^2+b x^3}}{32 a^4 x^3}-\frac{315 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{64 a^{11/2}}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{2}{a x^3 \sqrt{a x^2+b x^3}} \]

[Out]

2/(a*x^3*Sqrt[a*x^2 + b*x^3]) - (9*Sqrt[a*x^2 + b*x^3])/(4*a^2*x^5) + (21*b*Sqrt[a*x^2 + b*x^3])/(8*a^3*x^4) -
(105*b^2*Sqrt[a*x^2 + b*x^3])/(32*a^4*x^3) + (315*b^3*Sqrt[a*x^2 + b*x^3])/(64*a^5*x^2) - (315*b^4*ArcTanh[(S
qrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(64*a^(11/2))

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Rubi [A]  time = 0.232448, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2023, 2025, 2008, 206} \[ \frac{315 b^3 \sqrt{a x^2+b x^3}}{64 a^5 x^2}-\frac{105 b^2 \sqrt{a x^2+b x^3}}{32 a^4 x^3}-\frac{315 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{64 a^{11/2}}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{2}{a x^3 \sqrt{a x^2+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a*x^2 + b*x^3)^(3/2)),x]

[Out]

2/(a*x^3*Sqrt[a*x^2 + b*x^3]) - (9*Sqrt[a*x^2 + b*x^3])/(4*a^2*x^5) + (21*b*Sqrt[a*x^2 + b*x^3])/(8*a^3*x^4) -
(105*b^2*Sqrt[a*x^2 + b*x^3])/(32*a^4*x^3) + (315*b^3*Sqrt[a*x^2 + b*x^3])/(64*a^5*x^2) - (315*b^4*ArcTanh[(S
qrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(64*a^(11/2))

Rule 2023

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

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 \left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}+\frac{9 \int \frac{1}{x^4 \sqrt{a x^2+b x^3}} \, dx}{a}\\ &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}-\frac{(63 b) \int \frac{1}{x^3 \sqrt{a x^2+b x^3}} \, dx}{8 a^2}\\ &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}+\frac{\left (105 b^2\right ) \int \frac{1}{x^2 \sqrt{a x^2+b x^3}} \, dx}{16 a^3}\\ &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}-\frac{105 b^2 \sqrt{a x^2+b x^3}}{32 a^4 x^3}-\frac{\left (315 b^3\right ) \int \frac{1}{x \sqrt{a x^2+b x^3}} \, dx}{64 a^4}\\ &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}-\frac{105 b^2 \sqrt{a x^2+b x^3}}{32 a^4 x^3}+\frac{315 b^3 \sqrt{a x^2+b x^3}}{64 a^5 x^2}+\frac{\left (315 b^4\right ) \int \frac{1}{\sqrt{a x^2+b x^3}} \, dx}{128 a^5}\\ &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}-\frac{105 b^2 \sqrt{a x^2+b x^3}}{32 a^4 x^3}+\frac{315 b^3 \sqrt{a x^2+b x^3}}{64 a^5 x^2}-\frac{\left (315 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{x}{\sqrt{a x^2+b x^3}}\right )}{64 a^5}\\ &=\frac{2}{a x^3 \sqrt{a x^2+b x^3}}-\frac{9 \sqrt{a x^2+b x^3}}{4 a^2 x^5}+\frac{21 b \sqrt{a x^2+b x^3}}{8 a^3 x^4}-\frac{105 b^2 \sqrt{a x^2+b x^3}}{32 a^4 x^3}+\frac{315 b^3 \sqrt{a x^2+b x^3}}{64 a^5 x^2}-\frac{315 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{64 a^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0099285, size = 38, normalized size = 0.23 \[ \frac{2 b^4 x \, _2F_1\left (-\frac{1}{2},5;\frac{1}{2};\frac{b x}{a}+1\right )}{a^5 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a*x^2 + b*x^3)^(3/2)),x]

[Out]

(2*b^4*x*Hypergeometric2F1[-1/2, 5, 1/2, 1 + (b*x)/a])/(a^5*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.014, size = 100, normalized size = 0.6 \begin{align*} -{\frac{bx+a}{64\,x} \left ( 315\,\sqrt{bx+a}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ){x}^{4}{b}^{4}-24\,{a}^{7/2}xb+42\,{a}^{5/2}{x}^{2}{b}^{2}-105\,{a}^{3/2}{x}^{3}{b}^{3}-315\,{b}^{4}{x}^{4}\sqrt{a}+16\,{a}^{9/2} \right ) \left ( b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}{a}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^3+a*x^2)^(3/2),x)

[Out]

-1/64*(b*x+a)*(315*(b*x+a)^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*x^4*b^4-24*a^(7/2)*x*b+42*a^(5/2)*x^2*b^2-105*
a^(3/2)*x^3*b^3-315*b^4*x^4*a^(1/2)+16*a^(9/2))/x/(b*x^3+a*x^2)^(3/2)/a^(11/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a x^{2}\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^3 + a*x^2)^(3/2)*x^2), x)

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Fricas [A]  time = 0.821706, size = 575, normalized size = 3.46 \begin{align*} \left [\frac{315 \,{\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt{a} \log \left (\frac{b x^{2} + 2 \, a x - 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{a}}{x^{2}}\right ) + 2 \,{\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt{b x^{3} + a x^{2}}}{128 \,{\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}, \frac{315 \,{\left (b^{5} x^{6} + a b^{4} x^{5}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) +{\left (315 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} - 42 \, a^{3} b^{2} x^{2} + 24 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt{b x^{3} + a x^{2}}}{64 \,{\left (a^{6} b x^{6} + a^{7} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/128*(315*(b^5*x^6 + a*b^4*x^5)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(315*a*
b^4*x^4 + 105*a^2*b^3*x^3 - 42*a^3*b^2*x^2 + 24*a^4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2))/(a^6*b*x^6 + a^7*x^5),
1/64*(315*(b^5*x^6 + a*b^4*x^5)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + (315*a*b^4*x^4 + 105*a^2
*b^3*x^3 - 42*a^3*b^2*x^2 + 24*a^4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2))/(a^6*b*x^6 + a^7*x^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (x^{2} \left (a + b x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(1/(x**2*(x**2*(a + b*x))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a x^{2}\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^3 + a*x^2)^(3/2)*x^2), x)