\(\int \frac {1}{\sqrt {x} (b x+c x^2)^{3/2}} \, dx\) [109]

Optimal result

Integrand size = 19, antiderivative size = 81 \[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c \sqrt {x}}{b^2 \sqrt {b x+c x^2}}+\frac {3 c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{5/2}} \]

[Out]

3*c*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(5/2)-1/b/x^(1/2)/(c*x^2+b*x)^(1/2)-3*c*x^(1/2)/b^2/(c*x^2+b*
x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {686, 680, 674, 213} \[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {3 c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{5/2}}-\frac {3 c \sqrt {x}}{b^2 \sqrt {b x+c x^2}}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}} \]

[In]

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

[Out]

-(1/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) - (3*c*Sqrt[x])/(b^2*Sqrt[b*x + c*x^2]) + (3*c*ArcTanh[Sqrt[b*x + c*x^2]/(S
qrt[b]*Sqrt[x])])/b^(5/2)

Rule 213

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

Rule 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

Rule 680

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*c*d - b*e)*(d + e
*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] - Dist[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(
b^2 - 4*a*c))), 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] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 686

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e))),
Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}-\frac {(3 c) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{2 b} \\ & = -\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c \sqrt {x}}{b^2 \sqrt {b x+c x^2}}-\frac {(3 c) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{2 b^2} \\ & = -\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c \sqrt {x}}{b^2 \sqrt {b x+c x^2}}-\frac {(3 c) \text {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{b^2} \\ & = -\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}-\frac {3 c \sqrt {x}}{b^2 \sqrt {b x+c x^2}}+\frac {3 c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {-\sqrt {b} (b+3 c x)+3 c x \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{b^{5/2} \sqrt {x} \sqrt {x (b+c x)}} \]

[In]

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

[Out]

(-(Sqrt[b]*(b + 3*c*x)) + 3*c*x*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(b^(5/2)*Sqrt[x]*Sqrt[x*(b + c*x
)])

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, c x -3 c x \sqrt {b}-b^{\frac {3}{2}}\right )}{x^{\frac {3}{2}} \left (c x +b \right ) b^{\frac {5}{2}}}\)	\(60\)
risch	\(-\frac {c x +b}{b^{2} \sqrt {x}\, \sqrt {x \left (c x +b \right )}}-\frac {c \left (\frac {4}{\sqrt {c x +b}}-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{2 b^{2} \sqrt {x \left (c x +b \right )}}\)	\(76\)

[In]

int(1/x^(1/2)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

(x*(c*x+b))^(1/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*c*x-3*c*x*b^(1/2)-b^(3/2))/x^(3/2)/(c*x+b)/b
^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.32 \[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (c^{2} x^{3} + b c x^{2}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (3 \, b c x + b^{2}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{2 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}}, -\frac {3 \, {\left (c^{2} x^{3} + b c x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, b c x + b^{2}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{b^{3} c x^{3} + b^{4} x^{2}}\right ] \]

[In]

integrate(1/x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {x} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/x**(1/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(1/(sqrt(x)*(x*(b + c*x))**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} \sqrt {x}} \,d x } \]

[In]

integrate(1/x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x)^(3/2)*sqrt(x)), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {3 \, c \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} - \frac {3 \, {\left (c x + b\right )} c - 2 \, b c}{{\left ({\left (c x + b\right )}^{\frac {3}{2}} - \sqrt {c x + b} b\right )} b^{2}} \]

[In]

integrate(1/x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {x}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

[In]

int(1/(x^(1/2)*(b*x + c*x^2)^(3/2)),x)

[Out]

int(1/(x^(1/2)*(b*x + c*x^2)^(3/2)), x)