3.30 \(\int \frac {(a x+b x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=92 \[ \frac {15}{4} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}+\frac {15}{4} a b \sqrt {a x+b x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3} \]

[Out]

5/2*b*(b*x^2+a*x)^(3/2)/x-2*(b*x^2+a*x)^(5/2)/x^3+15/4*a^2*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))*b^(1/2)+15/4*a
*b*(b*x^2+a*x)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 92, 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} \[ \frac {15}{4} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}+\frac {15}{4} a b \sqrt {a x+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x^4,x]

[Out]

(15*a*b*Sqrt[a*x + b*x^2])/4 + (5*b*(a*x + b*x^2)^(3/2))/(2*x) - (2*(a*x + b*x^2)^(5/2))/x^3 + (15*a^2*Sqrt[b]
*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/4

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^4} \, dx &=-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+(5 b) \int \frac {\left (a x+b x^2\right )^{3/2}}{x^2} \, dx\\ &=\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{4} (15 a b) \int \frac {\sqrt {a x+b x^2}}{x} \, dx\\ &=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{8} \left (15 a^2 b\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx\\ &=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {1}{4} \left (15 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )\\ &=\frac {15}{4} a b \sqrt {a x+b x^2}+\frac {5 b \left (a x+b x^2\right )^{3/2}}{2 x}-\frac {2 \left (a x+b x^2\right )^{5/2}}{x^3}+\frac {15}{4} a^2 \sqrt {b} \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 = 48, normalized size = 0.52 \[ -\frac {2 a^2 \sqrt {x (a+b x)} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x}{a}\right )}{x \sqrt {\frac {b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^4,x]

[Out]

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

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fricas [A]  time = 0.81, size = 144, normalized size = 1.57 \[ \left [\frac {15 \, a^{2} \sqrt {b} x \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{8 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{4 \, x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt(b*x
^2 + a*x))/x, -1/4*(15*a^2*sqrt(-b)*x*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (2*b^2*x^2 + 9*a*b*x - 8*a^2)
*sqrt(b*x^2 + a*x))/x]

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giac [A]  time = 0.23, size = 89, normalized size = 0.97 \[ -\frac {15}{8} \, a^{2} \sqrt {b} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right ) + \frac {2 \, a^{3}}{\sqrt {b} x - \sqrt {b x^{2} + a x}} + \frac {1}{4} \, {\left (2 \, b^{2} x + 9 \, a b\right )} \sqrt {b x^{2} + a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="giac")

[Out]

-15/8*a^2*sqrt(b)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + 2*a^3/(sqrt(b)*x - sqrt(b*x^2 + a
*x)) + 1/4*(2*b^2*x + 9*a*b)*sqrt(b*x^2 + a*x)

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maple [B]  time = 0.04, size = 185, normalized size = 2.01 \[ \frac {15 a^{2} \sqrt {b}\, \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8}-\frac {15 \sqrt {b \,x^{2}+a x}\, b^{2} x}{2}-\frac {15 \sqrt {b \,x^{2}+a x}\, a b}{4}+\frac {20 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{3} x}{a^{2}}+\frac {10 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{2}}{a}+\frac {32 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} b^{3}}{a^{3}}-\frac {32 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b^{2}}{a^{3} x^{2}}+\frac {12 \left (b \,x^{2}+a x \right )^{\frac {7}{2}} b}{a^{2} x^{3}}-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a*x)^(5/2)/x^4,x)

[Out]

-2/a/x^4*(b*x^2+a*x)^(7/2)+12/a^2*b/x^3*(b*x^2+a*x)^(7/2)-32/a^3*b^2/x^2*(b*x^2+a*x)^(7/2)+32/a^3*b^3*(b*x^2+a
*x)^(5/2)+20/a^2*b^3*(b*x^2+a*x)^(3/2)*x+10/a*b^2*(b*x^2+a*x)^(3/2)-15/2*b^2*(b*x^2+a*x)^(1/2)*x-15/4*a*b*(b*x
^2+a*x)^(1/2)+15/8*a^2*b^(1/2)*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))

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maxima [A]  time = 1.38, size = 84, normalized size = 0.91 \[ \frac {15}{8} \, a^{2} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {15 \, \sqrt {b x^{2} + a x} a^{2}}{4 \, x} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{4 \, x^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{2 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^4,x, algorithm="maxima")

[Out]

15/8*a^2*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 15/4*sqrt(b*x^2 + a*x)*a^2/x + 5/4*(b*x^2 + a*
x)^(3/2)*a/x^2 + 1/2*(b*x^2 + a*x)^(5/2)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^2)^(5/2)/x^4,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a*x)**(5/2)/x**4,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**4, x)

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