\(\int \frac {(b x+c x^2)^{3/2}}{x^4} \, dx\) [18]

Optimal result

Integrand size = 17, antiderivative size = 68 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}+2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {676, 634, 212} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 c \sqrt {b x+c x^2}}{x}-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

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 676

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*} \text {integral}& = -\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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (\sqrt {b+c x} (b+4 c x)+3 c^{3/2} x^{3/2} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{3 x^2 \sqrt {b+c x}} \]

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {6 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) x^{2}-2 \sqrt {x \left (c x +b \right )}\, \left (4 c x +b \right )}{3 x^{2}}\)	\(49\)
risch	\(-\frac {2 \left (c x +b \right ) \left (4 c x +b \right )}{3 x \sqrt {x \left (c x +b \right )}}+c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )\)	\(55\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{3 b \,x^{4}}+\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\right )}{3 b}\)	\(151\)

[In]

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

[Out]

1/3*(6*c^(3/2)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*x^2-2*(x*(c*x+b))^(1/2)*(4*c*x+b))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.71 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, c^{\frac {3}{2}} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} {\left (4 \, c x + b\right )}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, \sqrt {-c} c x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} {\left (4 \, c x + b\right )}\right )}}{3 \, x^{2}}\right ] \]

[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]

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{2} + b x} c}{3 \, x} - \frac {\sqrt {c x^{2} + b x} b}{3 \, x^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{3 \, x^{3}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (56) = 112\).

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.69 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=-c^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \frac {2 \, {\left (6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + b^{3}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^4} \,d x \]

[In]

int((b*x + c*x^2)^(3/2)/x^4,x)

[Out]

int((b*x + c*x^2)^(3/2)/x^4, x)