3.100 \(\int \frac{1}{x^{5/2} \sqrt{b x+c x^2}} \, dx\)

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

[Out]

-Sqrt[b*x + c*x^2]/(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))

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Rubi [A]  time = 0.0343466, antiderivative size = 89, normalized size of antiderivative = 1., 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 = {672, 660, 207} \[ -\frac{3 c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{5/2}}+\frac{3 c \sqrt{b x+c x^2}}{4 b^2 x^{3/2}}-\frac{\sqrt{b x+c x^2}}{2 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*Sqrt[b*x + c*x^2]),x]

[Out]

-Sqrt[b*x + c*x^2]/(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 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]

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 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])

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 ) \operatorname{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*}

Mathematica [C]  time = 0.0123243, size = 40, normalized size = 0.45 \[ -\frac{2 c^2 \sqrt{x (b+c x)} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{c x}{b}+1\right )}{b^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^2*Sqrt[x*(b + c*x)]*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c*x)/b])/(b^3*Sqrt[x])

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Maple [A]  time = 0.225, size = 72, normalized size = 0.8 \begin{align*} -{\frac{1}{4}\sqrt{x \left ( cx+b \right ) } \left ( 3\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{2}{c}^{2}-3\,xc\sqrt{cx+b}\sqrt{b}+2\,{b}^{3/2}\sqrt{cx+b} \right ){b}^{-{\frac{5}{2}}}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(x*(c*x+b))^(1/2)/b^(5/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*c^2-3*x*c*(c*x+b)^(1/2)*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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + b x} x^{\frac{5}{2}}}\,{d x} \end{align*}

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 = 2.06695, size = 373, normalized size = 4.19 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{5}{2}} \sqrt{x \left (b + c x\right )}}\, dx \end{align*}

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 = 1.30997, size = 81, normalized size = 0.91 \begin{align*} \frac{1}{4} \, c^{2}{\left (\frac{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}} - 5 \, \sqrt{c x + b} b}{b^{2} c^{2} x^{2}}\right )} \end{align*}

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