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

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

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Rubi [A]  time = 0.24, antiderivative size = 165, 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 = {2020, 2025, 2008, 206} \begin {gather*} -\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} \end {gather*}

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 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 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]

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*}

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Mathematica [C]  time = 0.03, size = 42, normalized size = 0.25 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 16.40, size = 121, normalized size = 0.73 \begin {gather*} \frac {\left (x^2 (a+b x)\right )^{3/2} \left (\frac {3 b^5 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{7/2}}+\frac {\sqrt {a+b x} \left (15 a^4-70 a^3 (a+b x)-128 a^2 (a+b x)^2+70 a (a+b x)^3-15 (a+b x)^4\right )}{640 a^3 x^5}\right )}{x^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(a + b*x))^(3/2)*((Sqrt[a + b*x]*(15*a^4 - 70*a^3*(a + b*x) - 128*a^2*(a + b*x)^2 + 70*a*(a + b*x)^3 - 1
5*(a + b*x)^4))/(640*a^3*x^5) + (3*b^5*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(7/2))))/(x^3*(a + b*x)^(3/2))

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fricas [A]  time = 0.41, size = 219, normalized size = 1.33 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.27, size = 126, normalized size = 0.76 \begin {gather*} -\frac {\frac {15 \, b^{6} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {9}{2}} b^{6} \mathrm {sgn}\relax (x) - 70 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{6} \mathrm {sgn}\relax (x) + 128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{6} \mathrm {sgn}\relax (x) + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{6} \mathrm {sgn}\relax (x) - 15 \, \sqrt {b x + a} a^{4} b^{6} \mathrm {sgn}\relax (x)}{a^{3} b^{5} x^{5}}}{640 \, b} \end {gather*}

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

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maple [A]  time = 0.05, size = 113, normalized size = 0.68 \begin {gather*} -\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (-15 a^{3} b^{5} x^{5} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-15 \sqrt {b x +a}\, a^{\frac {15}{2}}+70 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {13}{2}}+128 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {11}{2}}-70 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {9}{2}}+15 \left (b x +a \right )^{\frac {9}{2}} a^{\frac {7}{2}}\right )}{640 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {13}{2}} x^{8}} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{9}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^9} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(3/2)/x^9,x)

[Out]

int((a*x^2 + b*x^3)^(3/2)/x^9, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{9}}\, dx \end {gather*}

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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