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

Optimal. Leaf size=89 \[ 5 a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )+5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4} \]

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Rubi [A]  time = 0.04, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {662, 664, 620, 206} \begin {gather*} 5 b^2 \sqrt {a x+b x^2}+5 a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

5*b^2*Sqrt[a*x + b*x^2] - (10*b*(a*x + b*x^2)^(3/2))/(3*x^2) - (2*(a*x + b*x^2)^(5/2))/(3*x^4) + 5*a*b^(3/2)*A
rcTanh[(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]

Rule 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx &=-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\frac {1}{3} (5 b) \int \frac {\left (a x+b x^2\right )^{3/2}}{x^3} \, dx\\ &=-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\left (5 b^2\right ) \int \frac {\sqrt {a x+b x^2}}{x} \, dx\\ &=5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\frac {1}{2} \left (5 a b^2\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx\\ &=5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\left (5 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )\\ &=5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+5 a b^{3/2} \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 = 0.56 \begin {gather*} -\frac {2 a^2 \sqrt {x (a+b x)} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x}{a}\right )}{3 x^2 \sqrt {\frac {b x}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 77, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a x+b x^2} \left (-2 a^2-14 a b x+3 b^2 x^2\right )}{3 x^2}-\frac {5}{2} a b^{3/2} \log \left (-2 \sqrt {b} \sqrt {a x+b x^2}+a+2 b x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

(Sqrt[a*x + b*x^2]*(-2*a^2 - 14*a*b*x + 3*b^2*x^2))/(3*x^2) - (5*a*b^(3/2)*Log[a + 2*b*x - 2*Sqrt[b]*Sqrt[a*x
+ b*x^2]])/2

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fricas [A]  time = 0.41, size = 145, normalized size = 1.63 \begin {gather*} \left [\frac {15 \, a b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{6 \, x^{2}}, -\frac {15 \, a \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a*b^(3/2)*x^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(3*b^2*x^2 - 14*a*b*x - 2*a^2)*sqrt(b*
x^2 + a*x))/x^2, -1/3*(15*a*sqrt(-b)*b*x^2*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (3*b^2*x^2 - 14*a*b*x -
2*a^2)*sqrt(b*x^2 + a*x))/x^2]

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giac [A]  time = 0.37, size = 133, normalized size = 1.49 \begin {gather*} -\frac {5}{2} \, a b^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right ) + \sqrt {b x^{2} + a x} b^{2} + \frac {2 \, {\left (9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{2} b + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{3} \sqrt {b} + a^{4}\right )}}{3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*a*b^(3/2)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + sqrt(b*x^2 + a*x)*b^2 + 2/3*(9*(sqrt
(b)*x - sqrt(b*x^2 + a*x))^2*a^2*b + 3*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^3*sqrt(b) + a^4)/(sqrt(b)*x - sqrt(b*
x^2 + a*x))^3

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maple [B]  time = 0.04, size = 209, normalized size = 2.35 \begin {gather*} \frac {5 a \,b^{\frac {3}{2}} \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2}-\frac {10 \sqrt {b \,x^{2}+a x}\, b^{3} x}{a}+\frac {80 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{4} x}{3 a^{3}}-5 \sqrt {b \,x^{2}+a x}\, b^{2}+\frac {40 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{3}}{3 a^{2}}+\frac {128 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} b^{4}}{3 a^{4}}-\frac {128 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b^{3}}{3 a^{4} x^{2}}+\frac {16 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b^{2}}{a^{3} x^{3}}-\frac {8 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b}{3 a^{2} x^{4}}-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/a/x^5*(b*x^2+a*x)^(7/2)-8/3/a^2*b/x^4*(b*x^2+a*x)^(7/2)+16/a^3*b^2/x^3*(b*x^2+a*x)^(7/2)-128/3/a^4*b^3/x^
2*(b*x^2+a*x)^(7/2)+128/3/a^4*b^4*(b*x^2+a*x)^(5/2)+80/3/a^3*b^4*(b*x^2+a*x)^(3/2)*x+40/3/a^2*b^3*(b*x^2+a*x)^
(3/2)-10/a*b^3*(b*x^2+a*x)^(1/2)*x-5*b^2*(b*x^2+a*x)^(1/2)+5/2*a*b^(3/2)*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 = 99, normalized size = 1.11 \begin {gather*} \frac {5}{2} \, a b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {35 \, \sqrt {b x^{2} + a x} a b}{6 \, x} - \frac {5 \, \sqrt {b x^{2} + a x} a^{2}}{6 \, x^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{6 \, x^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*a*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 35/6*sqrt(b*x^2 + a*x)*a*b/x - 5/6*sqrt(b*x^2 + a
*x)*a^2/x^2 - 5/6*(b*x^2 + a*x)^(3/2)*a/x^3 + (b*x^2 + a*x)^(5/2)/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2)/x^5,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^5, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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