3.16 \(\int \frac {(b x+c x^2)^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=72 \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c}}+\frac {3}{4} b \sqrt {b x+c x^2}+\frac {\left (b x+c x^2\right )^{3/2}}{2 x} \]

[Out]

1/2*(c*x^2+b*x)^(3/2)/x+3/4*b^2*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(1/2)+3/4*b*(c*x^2+b*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {664, 620, 206} \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c}}+\frac {3}{4} b \sqrt {b x+c x^2}+\frac {\left (b x+c x^2\right )^{3/2}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^2,x]

[Out]

(3*b*Sqrt[b*x + c*x^2])/4 + (b*x + c*x^2)^(3/2)/(2*x) + (3*b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*Sqrt
[c])

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

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Mathematica [A]  time = 0.10, size = 69, normalized size = 0.96 \[ \frac {1}{4} \sqrt {x (b+c x)} \left (\frac {3 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {c} \sqrt {x} \sqrt {\frac {c x}{b}+1}}+5 b+2 c x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^2,x]

[Out]

(Sqrt[x*(b + c*x)]*(5*b + 2*c*x + (3*b^(3/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*
x)/b])))/4

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fricas [A]  time = 0.79, size = 126, normalized size = 1.75 \[ \left [\frac {3 \, b^{2} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{2} x + 5 \, b c\right )} \sqrt {c x^{2} + b x}}{8 \, c}, -\frac {3 \, b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, c^{2} x + 5 \, b c\right )} \sqrt {c x^{2} + b x}}{4 \, c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*b^2*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(2*c^2*x + 5*b*c)*sqrt(c*x^2 + b*x))/c, -
1/4*(3*b^2*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*c^2*x + 5*b*c)*sqrt(c*x^2 + b*x))/c]

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giac [A]  time = 0.20, size = 60, normalized size = 0.83 \[ -\frac {3 \, b^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, \sqrt {c}} + \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (2 \, c x + 5 \, b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 99, normalized size = 1.38 \[ \frac {3 b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 \sqrt {c}}-\frac {3 \sqrt {c \,x^{2}+b x}\, c x}{2}-\frac {3 \sqrt {c \,x^{2}+b x}\, b}{4}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}{b}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b/x^2*(c*x^2+b*x)^(5/2)-2/b*c*(c*x^2+b*x)^(3/2)-3/2*c*(c*x^2+b*x)^(1/2)*x-3/4*b*(c*x^2+b*x)^(1/2)+3/8*b^2/c^
(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 1.36, size = 62, normalized size = 0.86 \[ \frac {3 \, b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, \sqrt {c}} + \frac {3}{4} \, \sqrt {c x^{2} + b x} b + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^2,x, algorithm="maxima")

[Out]

3/8*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + 3/4*sqrt(c*x^2 + b*x)*b + 1/2*(c*x^2 + b*x)^(3/
2)/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^2,x)

[Out]

int((b*x + c*x^2)^(3/2)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**2,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**2, x)

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