3.250 \(\int \frac{(a x^2+b x^3)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=137 \[ \frac{3 b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^2}-\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{64 a^{5/2}}-\frac{b^2 \sqrt{a x^2+b x^3}}{32 a x^3}-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7} \]

[Out]

-(b*Sqrt[a*x^2 + b*x^3])/(8*x^4) - (b^2*Sqrt[a*x^2 + b*x^3])/(32*a*x^3) + (3*b^3*Sqrt[a*x^2 + b*x^3])/(64*a^2*
x^2) - (a*x^2 + b*x^3)^(3/2)/(4*x^7) - (3*b^4*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(64*a^(5/2))

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Rubi [A]  time = 0.18419, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2008, 206} \[ \frac{3 b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^2}-\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{64 a^{5/2}}-\frac{b^2 \sqrt{a x^2+b x^3}}{32 a x^3}-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^8,x]

[Out]

-(b*Sqrt[a*x^2 + b*x^3])/(8*x^4) - (b^2*Sqrt[a*x^2 + b*x^3])/(32*a*x^3) + (3*b^3*Sqrt[a*x^2 + b*x^3])/(64*a^2*
x^2) - (a*x^2 + b*x^3)^(3/2)/(4*x^7) - (3*b^4*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(64*a^(5/2))

Rule 2020

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

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{\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx &=-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac{1}{8} (3 b) \int \frac{\sqrt{a x^2+b x^3}}{x^5} \, dx\\ &=-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac{1}{16} b^2 \int \frac{1}{x^2 \sqrt{a x^2+b x^3}} \, dx\\ &=-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{b^2 \sqrt{a x^2+b x^3}}{32 a x^3}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac{\left (3 b^3\right ) \int \frac{1}{x \sqrt{a x^2+b x^3}} \, dx}{64 a}\\ &=-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{b^2 \sqrt{a x^2+b x^3}}{32 a x^3}+\frac{3 b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac{\left (3 b^4\right ) \int \frac{1}{\sqrt{a x^2+b x^3}} \, dx}{128 a^2}\\ &=-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{b^2 \sqrt{a x^2+b x^3}}{32 a x^3}+\frac{3 b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac{\left (3 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^2}\\ &=-\frac{b \sqrt{a x^2+b x^3}}{8 x^4}-\frac{b^2 \sqrt{a x^2+b x^3}}{32 a x^3}+\frac{3 b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac{3 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{64 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0145932, size = 42, normalized size = 0.31 \[ -\frac{2 b^4 \left (x^2 (a+b x)\right )^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{b x}{a}+1\right )}{5 a^5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^8,x]

[Out]

(-2*b^4*(x^2*(a + b*x))^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, 1 + (b*x)/a])/(5*a^5*x^5)

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Maple [A]  time = 0.01, size = 101, normalized size = 0.7 \begin{align*}{\frac{1}{64\,{x}^{7}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 3\,{a}^{5/2} \left ( bx+a \right ) ^{7/2}-11\,{a}^{7/2} \left ( bx+a \right ) ^{5/2}-3\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ){a}^{2}{b}^{4}{x}^{4}-11\,{a}^{9/2} \left ( bx+a \right ) ^{3/2}+3\,{a}^{11/2}\sqrt{bx+a} \right ){a}^{-{\frac{9}{2}}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(3/2)/x^8,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a x^{2}\right )}^{\frac{3}{2}}}{x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^8, x)

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Fricas [A]  time = 0.893233, size = 443, normalized size = 3.23 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{4} x^{5} \log \left (\frac{b x^{2} + 2 \, a x - 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{a}}{x^{2}}\right ) + 2 \,{\left (3 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt{b x^{3} + a x^{2}}}{128 \, a^{3} x^{5}}, \frac{3 \, \sqrt{-a} b^{4} x^{5} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) +{\left (3 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt{b x^{3} + a x^{2}}}{64 \, a^{3} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (a + b x\right )\right )^{\frac{3}{2}}}{x^{8}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**8,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**8, x)

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Giac [A]  time = 1.23712, size = 147, normalized size = 1.07 \begin{align*} \frac{\frac{3 \, b^{5} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (x\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x + a\right )}^{\frac{7}{2}} b^{5} \mathrm{sgn}\left (x\right ) - 11 \,{\left (b x + a\right )}^{\frac{5}{2}} a b^{5} \mathrm{sgn}\left (x\right ) - 11 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2} b^{5} \mathrm{sgn}\left (x\right ) + 3 \, \sqrt{b x + a} a^{3} b^{5} \mathrm{sgn}\left (x\right )}{a^{2} b^{4} x^{4}}}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

1/64*(3*b^5*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^2) + (3*(b*x + a)^(7/2)*b^5*sgn(x) - 11*(b*x + a
)^(5/2)*a*b^5*sgn(x) - 11*(b*x + a)^(3/2)*a^2*b^5*sgn(x) + 3*sqrt(b*x + a)*a^3*b^5*sgn(x))/(a^2*b^4*x^4))/b