3.1.100 \(\int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx\) [100]

Optimal. Leaf size=89 \[ -\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}+\frac {3 c \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}-\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}} \]

[Out]

-3/4*c^2*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(5/2)-1/2*(c*x^2+b*x)^(1/2)/b/x^(5/2)+3/4*c*(c*x^2+b*x)^
(1/2)/b^2/x^(3/2)

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Rubi [A]
time = 0.02, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {686, 674, 213} \begin {gather*} -\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}}+\frac {3 c \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*Sqrt[b*x + c*x^2]),x]

[Out]

-1/2*Sqrt[b*x + c*x^2]/(b*x^(5/2)) + (3*c*Sqrt[b*x + c*x^2])/(4*b^2*x^(3/2)) - (3*c^2*ArcTanh[Sqrt[b*x + c*x^2
]/(Sqrt[b]*Sqrt[x])])/(4*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 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*} \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx &=-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}-\frac {(3 c) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{4 b}\\ &=-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}+\frac {3 c \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}+\frac {\left (3 c^2\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{8 b^2}\\ &=-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}+\frac {3 c \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}+\frac {\left (3 c^2\right ) \text {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{4 b^2}\\ &=-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}+\frac {3 c \sqrt {b x+c x^2}}{4 b^2 x^{3/2}}-\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 83, normalized size = 0.93 \begin {gather*} \frac {\sqrt {b} \left (-2 b^2+b c x+3 c^2 x^2\right )-3 c^2 x^2 \sqrt {b+c x} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{4 b^{5/2} x^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*Sqrt[b*x + c*x^2]),x]

[Out]

(Sqrt[b]*(-2*b^2 + b*c*x + 3*c^2*x^2) - 3*c^2*x^2*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/(4*b^(5/2)*x^(
3/2)*Sqrt[x*(b + c*x)])

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Maple [A]
time = 0.54, size = 72, normalized size = 0.81

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-3 c x +2 b \right )}{4 b^{2} x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}-\frac {3 c^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{4 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\)	\(71\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (3 \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}-3 c x \sqrt {b}\, \sqrt {c x +b}+2 b^{\frac {3}{2}} \sqrt {c x +b}\right )}{4 b^{\frac {5}{2}} x^{\frac {5}{2}} \sqrt {c x +b}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(x*(c*x+b))^(1/2)/b^(5/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*c^2*x^2-3*c*x*b^(1/2)*(c*x+b)^(1/2)+2*b^(3/2)
*(c*x+b)^(1/2))/x^(5/2)/(c*x+b)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^2 + b*x)*x^(5/2)), x)

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Fricas [A]
time = 2.26, size = 152, normalized size = 1.71 \begin {gather*} \left [\frac {3 \, \sqrt {b} c^{2} x^{3} \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 - 2 \, b^{2}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, b^{3} x^{3}}, \frac {3 \, \sqrt {-b} c^{2} x^{3} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, b c x - 2 \, b^{2}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, b^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*sqrt(b)*c^2*x^3*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 - 2*b^2)*
sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^3), 1/4*(3*sqrt(-b)*c^2*x^3*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (3*
b*c*x - 2*b^2)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{\frac {5}{2}} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(1/(x**(5/2)*sqrt(x*(b + c*x))), x)

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Giac [A]
time = 1.65, size = 69, normalized size = 0.78 \begin {gather*} \frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{3} - 5 \, \sqrt {c x + b} b c^{3}}{b^{2} c^{2} x^{2}}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/4*(3*c^3*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x + b)^(3/2)*c^3 - 5*sqrt(c*x + b)*b*c^3)/(b^
2*c^2*x^2))/c

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{5/2}\,\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(b*x + c*x^2)^(1/2)),x)

[Out]

int(1/(x^(5/2)*(b*x + c*x^2)^(1/2)), x)

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