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

Optimal. Leaf size=107 \[ -\frac{3 a^3 (a+2 b x) \sqrt{a x+b x^2}}{128 b^2}+\frac{3 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{128 b^{5/2}}+\frac{a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac{1}{5} \left (a x+b x^2\right )^{5/2} \]

[Out]

(-3*a^3*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(128*b^2) + (a*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(16*b) + (a*x + b*x^2)^
(5/2)/5 + (3*a^5*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(128*b^(5/2))

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Rubi [A]  time = 0.0389775, antiderivative size = 107, normalized size of antiderivative = 1., 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 = {664, 612, 620, 206} \[ -\frac{3 a^3 (a+2 b x) \sqrt{a x+b x^2}}{128 b^2}+\frac{3 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{128 b^{5/2}}+\frac{a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac{1}{5} \left (a x+b x^2\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(5/2)/x,x]

[Out]

(-3*a^3*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(128*b^2) + (a*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(16*b) + (a*x + b*x^2)^
(5/2)/5 + (3*a^5*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(128*b^(5/2))

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]

Rule 612

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

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 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])

Rubi steps

\begin{align*} \int \frac{\left (a x+b x^2\right )^{5/2}}{x} \, dx &=\frac{1}{5} \left (a x+b x^2\right )^{5/2}+\frac{1}{2} a \int \left (a x+b x^2\right )^{3/2} \, dx\\ &=\frac{a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac{1}{5} \left (a x+b x^2\right )^{5/2}-\frac{\left (3 a^3\right ) \int \sqrt{a x+b x^2} \, dx}{32 b}\\ &=-\frac{3 a^3 (a+2 b x) \sqrt{a x+b x^2}}{128 b^2}+\frac{a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac{1}{5} \left (a x+b x^2\right )^{5/2}+\frac{\left (3 a^5\right ) \int \frac{1}{\sqrt{a x+b x^2}} \, dx}{256 b^2}\\ &=-\frac{3 a^3 (a+2 b x) \sqrt{a x+b x^2}}{128 b^2}+\frac{a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac{1}{5} \left (a x+b x^2\right )^{5/2}+\frac{\left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )}{128 b^2}\\ &=-\frac{3 a^3 (a+2 b x) \sqrt{a x+b x^2}}{128 b^2}+\frac{a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac{1}{5} \left (a x+b x^2\right )^{5/2}+\frac{3 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{128 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.14581, size = 109, normalized size = 1.02 \[ \frac{\sqrt{x (a+b x)} \left (\sqrt{b} \left (248 a^2 b^2 x^2+10 a^3 b x-15 a^4+336 a b^3 x^3+128 b^4 x^4\right )+\frac{15 a^{9/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{x} \sqrt{\frac{b x}{a}+1}}\right )}{640 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(-15*a^4 + 10*a^3*b*x + 248*a^2*b^2*x^2 + 336*a*b^3*x^3 + 128*b^4*x^4) + (15*a^(9/
2)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[x]*Sqrt[1 + (b*x)/a])))/(640*b^(5/2))

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Maple [A]  time = 0.047, size = 120, normalized size = 1.1 \begin{align*}{\frac{1}{5} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}+{\frac{ax}{8} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}}{16\,b} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-{\frac{3\,x{a}^{3}}{64\,b}\sqrt{b{x}^{2}+ax}}-{\frac{3\,{a}^{4}}{128\,{b}^{2}}\sqrt{b{x}^{2}+ax}}+{\frac{3\,{a}^{5}}{256}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b*x^2+a*x)^(5/2)+1/8*a*(b*x^2+a*x)^(3/2)*x+1/16/b*(b*x^2+a*x)^(3/2)*a^2-3/64/b*a^3*(b*x^2+a*x)^(1/2)*x-3/
128/b^2*a^4*(b*x^2+a*x)^(1/2)+3/256/b^(5/2)*a^5*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0351, size = 458, normalized size = 4.28 \begin{align*} \left [\frac{15 \, a^{5} \sqrt{b} \log \left (2 \, b x + a + 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt{b x^{2} + a x}}{1280 \, b^{3}}, -\frac{15 \, a^{5} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) -{\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt{b x^{2} + a x}}{640 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1280*(15*a^5*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^
2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x^2 + a*x))/b^3, -1/640*(15*a^5*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*
sqrt(-b)/(b*x)) - (128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x^2 + a*x))
/b^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(a + b*x))**(5/2)/x, x)

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Giac [A]  time = 1.30692, size = 130, normalized size = 1.21 \begin{align*} -\frac{3 \, a^{5} \log \left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right )}{256 \, b^{\frac{5}{2}}} - \frac{1}{640} \, \sqrt{b x^{2} + a x}{\left (\frac{15 \, a^{4}}{b^{2}} - 2 \,{\left (\frac{5 \, a^{3}}{b} + 4 \,{\left (31 \, a^{2} + 2 \,{\left (8 \, b^{2} x + 21 \, a b\right )} x\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/256*a^5*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(5/2) - 1/640*sqrt(b*x^2 + a*x)*(15*a^4/
b^2 - 2*(5*a^3/b + 4*(31*a^2 + 2*(8*b^2*x + 21*a*b)*x)*x)*x)