\(\int \frac {(b x+c x^2)^{3/2}}{x^3} \, dx\) [17]

Optimal result

Integrand size = 17, antiderivative size = 64 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx=3 c \sqrt {b x+c x^2}-\frac {2 \left (b x+c x^2\right )^{3/2}}{x^2}+3 b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {676, 678, 634, 212} \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx=3 b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{x^2}+3 c \sqrt {b x+c x^2} \]

[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 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]

Rule 678

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {b+c x} \left ((-2 b+c x) \sqrt {b+c x}+6 b \sqrt {c} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{\sqrt {x (b+c x)}} \]

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^3,x]

[Out]

(Sqrt[b + c*x]*((-2*b + c*x)*Sqrt[b + c*x] + 6*b*Sqrt[c]*Sqrt[x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b
+ c*x])]))/Sqrt[x*(b + c*x)]

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-c x +2 b \right )}{\sqrt {x \left (c x +b \right )}}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2}\)	\(56\)
pseudoelliptic	\(\frac {c^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}\, x +3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b c x -2 b \sqrt {c}\, \sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\)	\(60\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\)	\(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.81 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, b \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, \sqrt {c x^{2} + b x} {\left (c x - 2 \, b\right )}}{2 \, x}, -\frac {3 \, b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - \sqrt {c x^{2} + b x} {\left (c x - 2 \, b\right )}}{x}\right ] \]

[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]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {3}{2} \, b \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {3 \, \sqrt {c x^{2} + b x} b}{x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.19 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^3} \, dx=-\frac {3}{2} \, b \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \sqrt {c x^{2} + b x} c + \frac {2 \, b^{2}}{\sqrt {c} x - \sqrt {c x^{2} + b x}} \]

[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))

Mupad [F(-1)]

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

[In]

int((b*x + c*x^2)^(3/2)/x^3,x)

[Out]

int((b*x + c*x^2)^(3/2)/x^3, x)