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

Optimal. Leaf size=53 \[ \frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 53, 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 = {664, 660, 207} \begin {gather*} \frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/x^(3/2),x]

[Out]

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

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 664

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

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Mathematica [A]  time = 0.02, size = 60, normalized size = 1.13 \begin {gather*} \frac {2 \sqrt {x} \sqrt {b+c x} \left (\sqrt {b+c x}-\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{\sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 53, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 111, normalized size = 2.09 \begin {gather*} \left [\frac {\sqrt {b} x \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} \sqrt {x}}{x}, \frac {2 \, {\left (\sqrt {-b} x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 61, normalized size = 1.15 \begin {gather*} \frac {2 \, b \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {c x + b} - \frac {2 \, {\left (b \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + \sqrt {-b} \sqrt {b}\right )}}{\sqrt {-b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 48, normalized size = 0.91 \begin {gather*} -\frac {2 \sqrt {\left (c x +b \right ) x}\, \left (\sqrt {b}\, \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-\sqrt {c x +b}\right )}{\sqrt {c x +b}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x}}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)/x^(3/2),x)

[Out]

int((b*x + c*x^2)^(1/2)/x^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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