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

Optimal. Leaf size=117 \[ \frac{15 c^2 \sqrt{x}}{4 b^3 \sqrt{b x+c x^2}}-\frac{15 c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{7/2}}+\frac{5 c}{4 b^2 \sqrt{x} \sqrt{b x+c x^2}}-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}} \]

[Out]

-1/(2*b*x^(3/2)*Sqrt[b*x + c*x^2]) + (5*c)/(4*b^2*Sqrt[x]*Sqrt[b*x + c*x^2]) + (15*c^2*Sqrt[x])/(4*b^3*Sqrt[b*
x + c*x^2]) - (15*c^2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(7/2))

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Rubi [A]  time = 0.048457, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {672, 666, 660, 207} \[ \frac{15 c^2 \sqrt{x}}{4 b^3 \sqrt{b x+c x^2}}-\frac{15 c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{7/2}}+\frac{5 c}{4 b^2 \sqrt{x} \sqrt{b x+c x^2}}-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*x^(3/2)*Sqrt[b*x + c*x^2]) + (5*c)/(4*b^2*Sqrt[x]*Sqrt[b*x + c*x^2]) + (15*c^2*Sqrt[x])/(4*b^3*Sqrt[b*
x + c*x^2]) - (15*c^2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(7/2))

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 666

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 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^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}}-\frac{(5 c) \int \frac{1}{\sqrt{x} \left (b x+c x^2\right )^{3/2}} \, dx}{4 b}\\ &=-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}}+\frac{5 c}{4 b^2 \sqrt{x} \sqrt{b x+c x^2}}+\frac{\left (15 c^2\right ) \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{8 b^2}\\ &=-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}}+\frac{5 c}{4 b^2 \sqrt{x} \sqrt{b x+c x^2}}+\frac{15 c^2 \sqrt{x}}{4 b^3 \sqrt{b x+c x^2}}+\frac{\left (15 c^2\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{8 b^3}\\ &=-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}}+\frac{5 c}{4 b^2 \sqrt{x} \sqrt{b x+c x^2}}+\frac{15 c^2 \sqrt{x}}{4 b^3 \sqrt{b x+c x^2}}+\frac{\left (15 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^3}\\ &=-\frac{1}{2 b x^{3/2} \sqrt{b x+c x^2}}+\frac{5 c}{4 b^2 \sqrt{x} \sqrt{b x+c x^2}}+\frac{15 c^2 \sqrt{x}}{4 b^3 \sqrt{b x+c x^2}}-\frac{15 c^2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0100784, size = 40, normalized size = 0.34 \[ \frac{2 c^2 \sqrt{x} \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{c x}{b}+1\right )}{b^3 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.19, size = 76, normalized size = 0.7 \begin{align*} -{\frac{1}{4\,cx+4\,b}\sqrt{x \left ( cx+b \right ) } \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{2}{c}^{2}-5\,{b}^{3/2}xc-15\,{x}^{2}{c}^{2}\sqrt{b}+2\,{b}^{5/2} \right ){x}^{-{\frac{5}{2}}}{b}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x^(5/2)*(x*(c*x+b))^(1/2)*(15*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^2*c^2-5*b^(3/2)*x*c-15*x^2*c
^2*b^(1/2)+2*b^(5/2))/(c*x+b)/b^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.14278, size = 491, normalized size = 4.2 \begin{align*} \left [\frac{15 \,{\left (c^{3} x^{4} + b c^{2} x^{3}\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 (15 \, b c^{2} x^{2} + 5 \, b^{2} c x - 2 \, b^{3}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{8 \,{\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, \frac{15 \,{\left (c^{3} x^{4} + b c^{2} x^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (15 \, b c^{2} x^{2} + 5 \, b^{2} c x - 2 \, b^{3}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{4 \,{\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(c^3*x^4 + b*c^2*x^3)*sqrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(15
*b*c^2*x^2 + 5*b^2*c*x - 2*b^3)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3), 1/4*(15*(c^3*x^4 + b*c^2*x^3
)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*b*c^2*x^2 + 5*b^2*c*x - 2*b^3)*sqrt(c*x^2 + b*x)*s
qrt(x))/(b^4*c*x^4 + b^5*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{3}{2}} \left (x \left (b + c x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.31458, size = 97, normalized size = 0.83 \begin{align*} \frac{1}{4} \, c^{2}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} + \frac{8}{\sqrt{c x + b} b^{3}} + \frac{7 \,{\left (c x + b\right )}^{\frac{3}{2}} - 9 \, \sqrt{c x + b} b}{b^{3} c^{2} x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^2*(15*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + 8/(sqrt(c*x + b)*b^3) + (7*(c*x + b)^(3/2) - 9*sqr
t(c*x + b)*b)/(b^3*c^2*x^2))