3.2.70 \(\int \frac {(a x^2+b x^3)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=109 \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{3/2}}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6} \]

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Rubi [A]  time = 0.13, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{3/2}}-\frac {b^2 \sqrt {a x^2+b x^3}}{8 a x^2}-\frac {b \sqrt {a x^2+b x^3}}{4 x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^7,x]

[Out]

-(b*Sqrt[a*x^2 + b*x^3])/(4*x^3) - (b^2*Sqrt[a*x^2 + b*x^3])/(8*a*x^2) - (a*x^2 + b*x^3)^(3/2)/(3*x^6) + (b^3*
ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*a^(3/2))

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

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Mathematica [C]  time = 0.02, size = 42, normalized size = 0.39 \begin {gather*} \frac {2 b^3 \left (x^2 (a+b x)\right )^{5/2} \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {b x}{a}+1\right )}{5 a^4 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^7,x]

[Out]

(2*b^3*(x^2*(a + b*x))^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, 1 + (b*x)/a])/(5*a^4*x^5)

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IntegrateAlgebraic [A]  time = 15.70, size = 97, normalized size = 0.89 \begin {gather*} \frac {\left (x^2 (a+b x)\right )^{3/2} \left (\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {\sqrt {a+b x} \left (3 a^2-8 a (a+b x)-3 (a+b x)^2\right )}{24 a x^3}\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^7,x]

[Out]

((x^2*(a + b*x))^(3/2)*((Sqrt[a + b*x]*(3*a^2 - 8*a*(a + b*x) - 3*(a + b*x)^2))/(24*a*x^3) + (b^3*ArcTanh[Sqrt
[a + b*x]/Sqrt[a]])/(8*a^(3/2))))/(x^3*(a + b*x)^(3/2))

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fricas [A]  time = 0.43, size = 175, normalized size = 1.61 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{3} x^{4} \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^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{2} x^{4}}, -\frac {3 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 92, normalized size = 0.84 \begin {gather*} -\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-a} a} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\relax (x) + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} \mathrm {sgn}\relax (x) - 3 \, \sqrt {b x + a} a^{2} b^{4} \mathrm {sgn}\relax (x)}{a b^{3} x^{3}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="giac")

[Out]

-1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a) + (3*(b*x + a)^(5/2)*b^4*sgn(x) + 8*(b*x + a)^
(3/2)*a*b^4*sgn(x) - 3*sqrt(b*x + a)*a^2*b^4*sgn(x))/(a*b^3*x^3))/b

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maple [A]  time = 0.06, size = 87, normalized size = 0.80 \begin {gather*} -\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (-3 a \,b^{3} x^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-3 \sqrt {b x +a}\, a^{\frac {7}{2}}+8 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}}+3 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {3}{2}}\right )}{24 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(3/2)/x^7,x)

[Out]

-1/24*(b*x^3+a*x^2)^(3/2)*(3*(b*x+a)^(5/2)*a^(3/2)-3*arctanh((b*x+a)^(1/2)/a^(1/2))*x^3*a*b^3+8*(b*x+a)^(3/2)*
a^(5/2)-3*(b*x+a)^(1/2)*a^(7/2))/x^6/(b*x+a)^(3/2)/a^(5/2)

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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^{7}}\,{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^7,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^7, 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^7} \,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^7,x)

[Out]

int((a*x^2 + b*x^3)^(3/2)/x^7, 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^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**7,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**7, x)

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