3.32 \(\int \frac {(a x+b x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=91 \[ 2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )-\frac {2 b^2 \sqrt {a x+b x^2}}{x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5}-\frac {2 b \left (a x+b x^2\right )^{3/2}}{3 x^3} \]

[Out]

-2/3*b*(b*x^2+a*x)^(3/2)/x^3-2/5*(b*x^2+a*x)^(5/2)/x^5+2*b^(5/2)*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))-2*b^2*(b
*x^2+a*x)^(1/2)/x

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Rubi [A]  time = 0.04, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {662, 620, 206} \[ -\frac {2 b^2 \sqrt {a x+b x^2}}{x}+2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )-\frac {2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^6,x]

[Out]

(-2*b^2*Sqrt[a*x + b*x^2])/x - (2*b*(a*x + b*x^2)^(3/2))/(3*x^3) - (2*(a*x + b*x^2)^(5/2))/(5*x^5) + 2*b^(5/2)
*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6} \, dx &=-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5}+b \int \frac {\left (a x+b x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac {2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5}+b^2 \int \frac {\sqrt {a x+b x^2}}{x^2} \, dx\\ &=-\frac {2 b^2 \sqrt {a x+b x^2}}{x}-\frac {2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5}+b^3 \int \frac {1}{\sqrt {a x+b x^2}} \, dx\\ &=-\frac {2 b^2 \sqrt {a x+b x^2}}{x}-\frac {2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5}+\left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )\\ &=-\frac {2 b^2 \sqrt {a x+b x^2}}{x}-\frac {2 b \left (a x+b x^2\right )^{3/2}}{3 x^3}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 x^5}+2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.55 \[ -\frac {2 a^2 \sqrt {x (a+b x)} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {b x}{a}\right )}{5 x^3 \sqrt {\frac {b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^6,x]

[Out]

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

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fricas [A]  time = 0.98, size = 144, normalized size = 1.58 \[ \left [\frac {15 \, b^{\frac {5}{2}} x^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{15 \, x^{3}}, -\frac {2 \, {\left (15 \, \sqrt {-b} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt {b x^{2} + a x}\right )}}{15 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/15*(15*b^(5/2)*x^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(23*b^2*x^2 + 11*a*b*x + 3*a^2)*sqrt(b*
x^2 + a*x))/x^3, -2/15*(15*sqrt(-b)*b^2*x^3*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (23*b^2*x^2 + 11*a*b*x
+ 3*a^2)*sqrt(b*x^2 + a*x))/x^3]

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giac [B]  time = 0.26, size = 175, normalized size = 1.92 \[ -b^{\frac {5}{2}} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right ) + \frac {2 \, {\left (45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} a b^{2} + 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} a^{2} b^{\frac {3}{2}} + 35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{3} b + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{4} \sqrt {b} + 3 \, a^{5}\right )}}{15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^6,x, algorithm="giac")

[Out]

-b^(5/2)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + 2/15*(45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4
*a*b^2 + 45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a^2*b^(3/2) + 35*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^3*b + 15*(s
qrt(b)*x - sqrt(b*x^2 + a*x))*a^4*sqrt(b) + 3*a^5)/(sqrt(b)*x - sqrt(b*x^2 + a*x))^5

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maple [B]  time = 0.04, size = 232, normalized size = 2.55 \[ b^{\frac {5}{2}} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {4 \sqrt {b \,x^{2}+a x}\, b^{4} x}{a^{2}}-\frac {2 \sqrt {b \,x^{2}+a x}\, b^{3}}{a}+\frac {32 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{5} x}{3 a^{4}}+\frac {16 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{4}}{3 a^{3}}+\frac {256 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} b^{5}}{15 a^{5}}-\frac {256 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b^{4}}{15 a^{5} x^{2}}+\frac {32 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b^{3}}{5 a^{4} x^{3}}-\frac {16 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b^{2}}{15 a^{3} x^{4}}-\frac {4 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b}{15 a^{2} x^{5}}-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{5 a \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a*x)^(5/2)/x^6,x)

[Out]

-2/5/a/x^6*(b*x^2+a*x)^(7/2)-4/15/a^2*b/x^5*(b*x^2+a*x)^(7/2)-16/15/a^3*b^2/x^4*(b*x^2+a*x)^(7/2)+32/5/a^4*b^3
/x^3*(b*x^2+a*x)^(7/2)-256/15/a^5*b^4/x^2*(b*x^2+a*x)^(7/2)+256/15/a^5*b^5*(b*x^2+a*x)^(5/2)+32/3/a^4*b^5*(b*x
^2+a*x)^(3/2)*x+16/3/a^3*b^4*(b*x^2+a*x)^(3/2)-4/a^2*b^4*(b*x^2+a*x)^(1/2)*x-2/a*b^3*(b*x^2+a*x)^(1/2)+b^(5/2)
*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))

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maxima [A]  time = 1.42, size = 134, normalized size = 1.47 \[ b^{\frac {5}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {38 \, \sqrt {b x^{2} + a x} b^{2}}{15 \, x} - \frac {7 \, \sqrt {b x^{2} + a x} a b}{30 \, x^{2}} + \frac {3 \, \sqrt {b x^{2} + a x} a^{2}}{10 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b}{3 \, x^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{2 \, x^{4}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^6,x, algorithm="maxima")

[Out]

b^(5/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 38/15*sqrt(b*x^2 + a*x)*b^2/x - 7/30*sqrt(b*x^2 + a*x)*
a*b/x^2 + 3/10*sqrt(b*x^2 + a*x)*a^2/x^3 - 1/3*(b*x^2 + a*x)^(3/2)*b/x^3 - 1/2*(b*x^2 + a*x)^(3/2)*a/x^4 - 1/5
*(b*x^2 + a*x)^(5/2)/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2)/x^6,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a*x)**(5/2)/x**6,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**6, x)

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