3.3.68 \(\int \frac {1}{x^2 (a x^2+b x^3)^{3/2}} \, dx\) [268]

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

[Out]

-315/64*b^4*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(11/2)+2/a/x^3/(b*x^3+a*x^2)^(1/2)-9/4*(b*x^3+a*x^2)^(1/2
)/a^2/x^5+21/8*b*(b*x^3+a*x^2)^(1/2)/a^3/x^4-105/32*b^2*(b*x^3+a*x^2)^(1/2)/a^4/x^3+315/64*b^3*(b*x^3+a*x^2)^(
1/2)/a^5/x^2

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Rubi [A]
time = 0.28, 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 = {2048, 2050, 2033, 212} \begin {gather*} -\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}} \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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

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 2048

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 2050

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 ) \text {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 [A]
time = 0.15, size = 106, normalized size = 0.64 \begin {gather*} \frac {\sqrt {a} \left (-16 a^4+24 a^3 b x-42 a^2 b^2 x^2+105 a b^3 x^3+315 b^4 x^4\right )-315 b^4 x^4 \sqrt {a+b x} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{11/2} x^3 \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]

(Sqrt[a]*(-16*a^4 + 24*a^3*b*x - 42*a^2*b^2*x^2 + 105*a*b^3*x^3 + 315*b^4*x^4) - 315*b^4*x^4*Sqrt[a + b*x]*Arc
Tanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(11/2)*x^3*Sqrt[x^2*(a + b*x)])

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Maple [A]
time = 0.46, size = 100, normalized size = 0.60

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (315 \sqrt {b x +a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}-24 a^{\frac {7}{2}} b x +42 a^{\frac {5}{2}} b^{2} x^{2}-105 a^{\frac {3}{2}} b^{3} x^{3}-315 b^{4} x^{4} \sqrt {a}+16 a^{\frac {9}{2}}\right )}{64 x \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {11}{2}}}\)	\(100\)
risch	\(-\frac {\left (b x +a \right ) \left (-187 b^{3} x^{3}+82 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right )}{64 a^{5} x^{3} \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{4} \left (-\frac {630 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {256}{\sqrt {b x +a}}\right ) \sqrt {b x +a}\, x}{128 a^{5} \sqrt {x^{2} \left (b x +a \right )}}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(b*x+a)*(315*(b*x+a)^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*b^4*x^4-24*a^(7/2)*b*x+42*a^(5/2)*b^2*x^2-105*
a^(3/2)*b^3*x^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [A]
time = 1.60, 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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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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Giac [A]
time = 1.13, size = 122, normalized size = 0.73 \begin {gather*} \frac {315 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{64 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{4}}{\sqrt {b x + a} a^{5} \mathrm {sgn}\left (x\right )} + \frac {187 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} - 643 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} + 765 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} - 325 \, \sqrt {b x + a} a^{3} b^{4}}{64 \, a^{5} b^{4} x^{4} \mathrm {sgn}\left (x\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="giac")

[Out]

315/64*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5*sgn(x)) + 2*b^4/(sqrt(b*x + a)*a^5*sgn(x)) + 1/64*(187
*(b*x + a)^(7/2)*b^4 - 643*(b*x + a)^(5/2)*a*b^4 + 765*(b*x + a)^(3/2)*a^2*b^4 - 325*sqrt(b*x + a)*a^3*b^4)/(a
^5*b^4*x^4*sgn(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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