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

Optimal. Leaf size=64 \[ -\frac {2 \left (b x+c x^2\right )^{3/2}}{x^2}+3 c \sqrt {b x+c x^2}+3 b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {662, 664, 620, 206} \[ -\frac {2 \left (b x+c x^2\right )^{3/2}}{x^2}+3 c \sqrt {b x+c x^2}+3 b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^3,x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.72 \[ -\frac {2 b \sqrt {x (b+c x)} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {c x}{b}\right )}{x \sqrt {\frac {c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^3,x]

[Out]

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

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fricas [A]  time = 0.95, size = 116, normalized size = 1.81 \[ \left [\frac {3 \, b \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, \sqrt {c x^{2} + b x} {\left (c x - 2 \, b\right )}}{2 \, x}, -\frac {3 \, b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - \sqrt {c x^{2} + b x} {\left (c x - 2 \, b\right )}}{x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*b*sqrt(c)*x*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*sqrt(c*x^2 + b*x)*(c*x - 2*b))/x, -(3*b*s
qrt(-c)*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - sqrt(c*x^2 + b*x)*(c*x - 2*b))/x]

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giac [A]  time = 0.23, size = 76, normalized size = 1.19 \[ -\frac {3}{2} \, b \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \sqrt {c x^{2} + b x} c + \frac {2 \, b^{2}}{\sqrt {c} x - \sqrt {c x^{2} + b x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*sqrt(c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + sqrt(c*x^2 + b*x)*c + 2*b^2/(sqrt(c)
*x - sqrt(c*x^2 + b*x))

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maple [B]  time = 0.04, size = 124, normalized size = 1.94 \[ \frac {3 b \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}-\frac {6 \sqrt {c \,x^{2}+b x}\, c^{2} x}{b}-3 \sqrt {c \,x^{2}+b x}\, c -\frac {8 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}{b^{2}}+\frac {8 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} c}{b^{2} x^{2}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/b/x^3*(c*x^2+b*x)^(5/2)+8/b^2*c/x^2*(c*x^2+b*x)^(5/2)-8/b^2*c^2*(c*x^2+b*x)^(3/2)-6/b*c^2*(c*x^2+b*x)^(1/2)
*x-3*c*(c*x^2+b*x)^(1/2)+3/2*b*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 = 1.33, size = 62, normalized size = 0.97 \[ \frac {3}{2} \, b \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {3 \, \sqrt {c x^{2} + b x} b}{x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*b*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 3*sqrt(c*x^2 + b*x)*b/x + (c*x^2 + b*x)^(3/2)/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^3,x)

[Out]

int((b*x + c*x^2)^(3/2)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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