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

Optimal. Leaf size=56 \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {672, 660, 207} \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*Sqrt[b*x + c*x^2]),x]

[Out]

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

Rule 207

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

Rule 660

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 672

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

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Mathematica [A]  time = 0.05, size = 63, normalized size = 1.12 \[ \frac {2 c \sqrt {x (b+c x)} \left (\frac {\tanh ^{-1}\left (\sqrt {\frac {c x}{b}+1}\right )}{2 \sqrt {\frac {c x}{b}+1}}-\frac {b}{2 c x}\right )}{b^2 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.91, size = 126, normalized size = 2.25 \[ \left [\frac {\sqrt {b} c x^{2} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, \sqrt {c x^{2} + b x} b \sqrt {x}}{2 \, b^{2} x^{2}}, -\frac {\sqrt {-b} c x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} b \sqrt {x}}{b^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 47, normalized size = 0.84 \[ -\frac {\frac {c^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {c x + b} c}{b x}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 52, normalized size = 0.93 \[ \frac {\sqrt {\left (c x +b \right ) x}\, \left (c x \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-\sqrt {c x +b}\, \sqrt {b}\right )}{\sqrt {c x +b}\, b^{\frac {3}{2}} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2} + b x} x^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {3}{2}} \sqrt {x \left (b + c x\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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