3.18 \(\int \frac {(b x+c x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=68 \[ 2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 68, 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 = {662, 620, 206} \[ 2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^4,x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}+c \int \frac {\sqrt {b x+c x^2}}{x^2} \, dx\\ &=-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}+c^2 \int \frac {1}{\sqrt {b x+c x^2}} \, dx\\ &=-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}+\left (2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )\\ &=-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}+2 c^{3/2} \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 = 48, normalized size = 0.71 \[ -\frac {2 b \sqrt {x (b+c x)} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x}{b}\right )}{3 x^2 \sqrt {\frac {c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^4,x]

[Out]

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

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fricas [A]  time = 0.83, size = 116, normalized size = 1.71 \[ \left [\frac {3 \, c^{\frac {3}{2}} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} {\left (4 \, c x + b\right )}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, \sqrt {-c} c x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} {\left (4 \, c x + b\right )}\right )}}{3 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.22, size = 115, normalized size = 1.69 \[ -c^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \frac {2 \, {\left (6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + b^{3}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="giac")

[Out]

-c^(3/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + 2/3*(6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b
*c + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^2*sqrt(c) + b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^3

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maple [B]  time = 0.05, size = 149, normalized size = 2.19 \[ c^{\frac {3}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-\frac {4 \sqrt {c \,x^{2}+b x}\, c^{3} x}{b^{2}}-\frac {2 \sqrt {c \,x^{2}+b x}\, c^{2}}{b}-\frac {16 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}{3 b^{3}}+\frac {16 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} c^{2}}{3 b^{3} x^{2}}-\frac {4 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} c}{3 b^{2} x^{3}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{3 b \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^4,x)

[Out]

-2/3/b/x^4*(c*x^2+b*x)^(5/2)-4/3/b^2*c/x^3*(c*x^2+b*x)^(5/2)+16/3/b^3*c^2/x^2*(c*x^2+b*x)^(5/2)-16/3/b^3*c^3*(
c*x^2+b*x)^(3/2)-4/b^2*c^3*(c*x^2+b*x)^(1/2)*x-2/b*c^2*(c*x^2+b*x)^(1/2)+c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2
+b*x)^(1/2))

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maxima [A]  time = 1.42, size = 78, normalized size = 1.15 \[ c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{2} + b x} c}{3 \, x} - \frac {\sqrt {c x^{2} + b x} b}{3 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="maxima")

[Out]

c^(3/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 7/3*sqrt(c*x^2 + b*x)*c/x - 1/3*sqrt(c*x^2 + b*x)*b/x^2
- 1/3*(c*x^2 + b*x)^(3/2)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^4,x)

[Out]

int((b*x + c*x^2)^(3/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**4,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**4, x)

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