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

Optimal. Leaf size=138 \[ \frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{9/2}}-\frac {35 b^2 \sqrt {a x^2+b x^3}}{8 a^4 x^2}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}} \]

[Out]

35/8*b^3*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(9/2)+2/a/x^2/(b*x^3+a*x^2)^(1/2)-7/3*(b*x^3+a*x^2)^(1/2)/a^
2/x^4+35/12*b*(b*x^3+a*x^2)^(1/2)/a^3/x^3-35/8*b^2*(b*x^3+a*x^2)^(1/2)/a^4/x^2

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Rubi [A]  time = 0.19, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2023, 2025, 2008, 206} \[ -\frac {35 b^2 \sqrt {a x^2+b x^3}}{8 a^4 x^2}+\frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{9/2}}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {2}{a x^2 \sqrt {a x^2+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a*x^2 + b*x^3)^(3/2)),x]

[Out]

2/(a*x^2*Sqrt[a*x^2 + b*x^3]) - (7*Sqrt[a*x^2 + b*x^3])/(3*a^2*x^4) + (35*b*Sqrt[a*x^2 + b*x^3])/(12*a^3*x^3)
- (35*b^2*Sqrt[a*x^2 + b*x^3])/(8*a^4*x^2) + (35*b^3*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*a^(9/2))

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 \left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}+\frac {7 \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx}{a}\\ &=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}-\frac {(35 b) \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx}{6 a^2}\\ &=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}+\frac {\left (35 b^2\right ) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{8 a^3}\\ &=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {35 b^2 \sqrt {a x^2+b x^3}}{8 a^4 x^2}-\frac {\left (35 b^3\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{16 a^4}\\ &=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {35 b^2 \sqrt {a x^2+b x^3}}{8 a^4 x^2}+\frac {\left (35 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{8 a^4}\\ &=\frac {2}{a x^2 \sqrt {a x^2+b x^3}}-\frac {7 \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {35 b \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {35 b^2 \sqrt {a x^2+b x^3}}{8 a^4 x^2}+\frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.28 \[ -\frac {2 b^3 x \, _2F_1\left (-\frac {1}{2},4;\frac {1}{2};\frac {b x}{a}+1\right )}{a^4 \sqrt {x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^3*x*Hypergeometric2F1[-1/2, 4, 1/2, 1 + (b*x)/a])/(a^4*Sqrt[x^2*(a + b*x)])

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fricas [A]  time = 0.40, size = 241, normalized size = 1.75 \[ \left [\frac {105 \, {\left (b^{4} x^{5} + a b^{3} x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}, -\frac {105 \, {\left (b^{4} x^{5} + a b^{3} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(105*(b^4*x^5 + a*b^3*x^4)*sqrt(a)*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(105*a*b
^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt(b*x^3 + a*x^2))/(a^5*b*x^5 + a^6*x^4), -1/24*(105*(b^4*x^5
+ a*b^3*x^4)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + (105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*
x + 8*a^4)*sqrt(b*x^3 + a*x^2))/(a^5*b*x^5 + a^6*x^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 86, normalized size = 0.62 \[ -\frac {\left (b x +a \right ) \left (-105 \sqrt {b x +a}\, b^{3} x^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+105 \sqrt {a}\, b^{3} x^{3}+35 a^{\frac {3}{2}} b^{2} x^{2}-14 a^{\frac {5}{2}} b x +8 a^{\frac {7}{2}}\right )}{24 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(b*x+a)*(-105*arctanh((b*x+a)^(1/2)/a^(1/2))*(b*x+a)^(1/2)*x^3*b^3-14*a^(5/2)*x*b+35*a^(3/2)*x^2*b^2+105
*b^3*x^3*a^(1/2)+8*a^(7/2))/(b*x^3+a*x^2)^(3/2)/a^(9/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^2 + b*x^3)^(3/2)),x)

[Out]

int(1/(x*(a*x^2 + b*x^3)^(3/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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