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

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

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Rubi [A]  time = 0.23, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2023, 2025, 2008, 206} \begin {gather*} \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}} \end {gather*}

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 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 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 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]

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*}

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.23 \begin {gather*} \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)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 6.31, size = 119, normalized size = 0.72 \begin {gather*} \frac {x \sqrt {a+b x} \left (\frac {128 a^4-837 a^3 (a+b x)+1533 a^2 (a+b x)^2-1155 a (a+b x)^3+315 (a+b x)^4}{64 a^5 x^4 \sqrt {a+b x}}-\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2}}\right )}{\sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x]*((128*a^4 - 837*a^3*(a + b*x) + 1533*a^2*(a + b*x)^2 - 1155*a*(a + b*x)^3 + 315*(a + b*x)^4)/
(64*a^5*x^4*Sqrt[a + b*x]) - (315*b^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(11/2))))/Sqrt[x^2*(a + b*x)]

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fricas [A]  time = 0.43, size = 263, normalized size = 1.58 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

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)

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maple [A]  time = 0.06, size = 100, normalized size = 0.60 \begin {gather*} -\frac {\left (b x +a \right ) \left (315 \sqrt {b x +a}\, b^{4} x^{4} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-315 \sqrt {a}\, b^{4} x^{4}-105 a^{\frac {3}{2}} b^{3} x^{3}+42 a^{\frac {5}{2}} b^{2} x^{2}-24 a^{\frac {7}{2}} b x +16 a^{\frac {9}{2}}\right )}{64 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {11}{2}} x} \end {gather*}

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*arctanh((b*x+a)^(1/2)/a^(1/2))*(b*x+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*x^4*b^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

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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mupad [B]  time = 5.68, size = 44, normalized size = 0.27 \begin {gather*} -\frac {2\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {11}{2};\ \frac {13}{2};\ -\frac {a}{b\,x}\right )}{11\,x\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a*x^2 + b*x^3)^(3/2)),x)

[Out]

-(2*(a/(b*x) + 1)^(3/2)*hypergeom([3/2, 11/2], 13/2, -a/(b*x)))/(11*x*(a*x^2 + b*x^3)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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