3.251 \(\int \frac{(a x^2+b x^3)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=165 \[ -\frac{3 b^4 \sqrt{a x^2+b x^3}}{128 a^3 x^2}+\frac{b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^3}+\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{128 a^{7/2}}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8} \]

[Out]

(-3*b*Sqrt[a*x^2 + b*x^3])/(40*x^5) - (b^2*Sqrt[a*x^2 + b*x^3])/(80*a*x^4) + (b^3*Sqrt[a*x^2 + b*x^3])/(64*a^2
*x^3) - (3*b^4*Sqrt[a*x^2 + b*x^3])/(128*a^3*x^2) - (a*x^2 + b*x^3)^(3/2)/(5*x^8) + (3*b^5*ArcTanh[(Sqrt[a]*x)
/Sqrt[a*x^2 + b*x^3]])/(128*a^(7/2))

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Rubi [A]  time = 0.236105, antiderivative size = 165, 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 = {2020, 2025, 2008, 206} \[ -\frac{3 b^4 \sqrt{a x^2+b x^3}}{128 a^3 x^2}+\frac{b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^3}+\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{128 a^{7/2}}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^9,x]

[Out]

(-3*b*Sqrt[a*x^2 + b*x^3])/(40*x^5) - (b^2*Sqrt[a*x^2 + b*x^3])/(80*a*x^4) + (b^3*Sqrt[a*x^2 + b*x^3])/(64*a^2
*x^3) - (3*b^4*Sqrt[a*x^2 + b*x^3])/(128*a^3*x^2) - (a*x^2 + b*x^3)^(3/2)/(5*x^8) + (3*b^5*ArcTanh[(Sqrt[a]*x)
/Sqrt[a*x^2 + b*x^3]])/(128*a^(7/2))

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^9} \, dx &=-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}+\frac{1}{10} (3 b) \int \frac{\sqrt{a x^2+b x^3}}{x^6} \, dx\\ &=-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}+\frac{1}{80} \left (3 b^2\right ) \int \frac{1}{x^3 \sqrt{a x^2+b x^3}} \, dx\\ &=-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}-\frac{b^3 \int \frac{1}{x^2 \sqrt{a x^2+b x^3}} \, dx}{32 a}\\ &=-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}+\frac{b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^3}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}+\frac{\left (3 b^4\right ) \int \frac{1}{x \sqrt{a x^2+b x^3}} \, dx}{128 a^2}\\ &=-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}+\frac{b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^3}-\frac{3 b^4 \sqrt{a x^2+b x^3}}{128 a^3 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}-\frac{\left (3 b^5\right ) \int \frac{1}{\sqrt{a x^2+b x^3}} \, dx}{256 a^3}\\ &=-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}+\frac{b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^3}-\frac{3 b^4 \sqrt{a x^2+b x^3}}{128 a^3 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}+\frac{\left (3 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{x}{\sqrt{a x^2+b x^3}}\right )}{128 a^3}\\ &=-\frac{3 b \sqrt{a x^2+b x^3}}{40 x^5}-\frac{b^2 \sqrt{a x^2+b x^3}}{80 a x^4}+\frac{b^3 \sqrt{a x^2+b x^3}}{64 a^2 x^3}-\frac{3 b^4 \sqrt{a x^2+b x^3}}{128 a^3 x^2}-\frac{\left (a x^2+b x^3\right )^{3/2}}{5 x^8}+\frac{3 b^5 \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^3}}\right )}{128 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.014527, size = 42, normalized size = 0.25 \[ \frac{2 b^5 \left (x^2 (a+b x)\right )^{5/2} \, _2F_1\left (\frac{5}{2},6;\frac{7}{2};\frac{b x}{a}+1\right )}{5 a^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^9,x]

[Out]

(2*b^5*(x^2*(a + b*x))^(5/2)*Hypergeometric2F1[5/2, 6, 7/2, 1 + (b*x)/a])/(5*a^6*x^5)

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Maple [A]  time = 0.013, size = 113, normalized size = 0.7 \begin{align*}{\frac{1}{640\,{x}^{8}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 15\,{a}^{15/2}\sqrt{bx+a}-70\,{a}^{13/2} \left ( bx+a \right ) ^{3/2}-128\,{a}^{11/2} \left ( bx+a \right ) ^{5/2}+70\,{a}^{9/2} \left ( bx+a \right ) ^{7/2}-15\,{a}^{7/2} \left ( bx+a \right ) ^{9/2}+15\,{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ){a}^{3}{b}^{5}{x}^{5} \right ){a}^{-{\frac{13}{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^9,x)

[Out]

1/640*(b*x^3+a*x^2)^(3/2)*(15*a^(15/2)*(b*x+a)^(1/2)-70*a^(13/2)*(b*x+a)^(3/2)-128*a^(11/2)*(b*x+a)^(5/2)+70*a
^(9/2)*(b*x+a)^(7/2)-15*a^(7/2)*(b*x+a)^(9/2)+15*arctanh((b*x+a)^(1/2)/a^(1/2))*a^3*b^5*x^5)/x^8/(b*x+a)^(3/2)
/a^(13/2)

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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^{9}}\,{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^9,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^9, x)

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Fricas [A]  time = 0.909493, size = 504, normalized size = 3.05 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{5} x^{6} \log \left (\frac{b x^{2} + 2 \, a x + 2 \, \sqrt{b x^{3} + a x^{2}} \sqrt{a}}{x^{2}}\right ) - 2 \,{\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} + 8 \, a^{3} b^{2} x^{2} + 176 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt{b x^{3} + a x^{2}}}{1280 \, a^{4} x^{6}}, -\frac{15 \, \sqrt{-a} b^{5} x^{6} \arctan \left (\frac{\sqrt{b x^{3} + a x^{2}} \sqrt{-a}}{a x}\right ) +{\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} + 8 \, a^{3} b^{2} x^{2} + 176 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt{b x^{3} + a x^{2}}}{640 \, a^{4} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="fricas")

[Out]

[1/1280*(15*sqrt(a)*b^5*x^6*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(15*a*b^4*x^4 - 10*a^
2*b^3*x^3 + 8*a^3*b^2*x^2 + 176*a^4*b*x + 128*a^5)*sqrt(b*x^3 + a*x^2))/(a^4*x^6), -1/640*(15*sqrt(-a)*b^5*x^6
*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + (15*a*b^4*x^4 - 10*a^2*b^3*x^3 + 8*a^3*b^2*x^2 + 176*a^4*b*x + 1
28*a^5)*sqrt(b*x^3 + a*x^2))/(a^4*x^6)]

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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^{9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**9,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**9, x)

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Giac [A]  time = 1.32584, size = 170, normalized size = 1.03 \begin{align*} -\frac{\frac{15 \, b^{6} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (x\right )}{\sqrt{-a} a^{3}} + \frac{15 \,{\left (b x + a\right )}^{\frac{9}{2}} b^{6} \mathrm{sgn}\left (x\right ) - 70 \,{\left (b x + a\right )}^{\frac{7}{2}} a b^{6} \mathrm{sgn}\left (x\right ) + 128 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{2} b^{6} \mathrm{sgn}\left (x\right ) + 70 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3} b^{6} \mathrm{sgn}\left (x\right ) - 15 \, \sqrt{b x + a} a^{4} b^{6} \mathrm{sgn}\left (x\right )}{a^{3} b^{5} x^{5}}}{640 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

-1/640*(15*b^6*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^3) + (15*(b*x + a)^(9/2)*b^6*sgn(x) - 70*(b*x
+ a)^(7/2)*a*b^6*sgn(x) + 128*(b*x + a)^(5/2)*a^2*b^6*sgn(x) + 70*(b*x + a)^(3/2)*a^3*b^6*sgn(x) - 15*sqrt(b*
x + a)*a^4*b^6*sgn(x))/(a^3*b^5*x^5))/b