3.111 \(\int \frac{1}{x^{5/2} (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac{35 c^3 \sqrt{x}}{8 b^4 \sqrt{b x+c x^2}}-\frac{35 c^2}{24 b^3 \sqrt{x} \sqrt{b x+c x^2}}+\frac{35 c^3 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{9/2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}} \]

[Out]

-1/(3*b*x^(5/2)*Sqrt[b*x + c*x^2]) + (7*c)/(12*b^2*x^(3/2)*Sqrt[b*x + c*x^2]) - (35*c^2)/(24*b^3*Sqrt[x]*Sqrt[
b*x + c*x^2]) - (35*c^3*Sqrt[x])/(8*b^4*Sqrt[b*x + c*x^2]) + (35*c^3*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x
])])/(8*b^(9/2))

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Rubi [A]  time = 0.0660191, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {672, 666, 660, 207} \[ -\frac{35 c^3 \sqrt{x}}{8 b^4 \sqrt{b x+c x^2}}-\frac{35 c^2}{24 b^3 \sqrt{x} \sqrt{b x+c x^2}}+\frac{35 c^3 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{9/2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

-1/(3*b*x^(5/2)*Sqrt[b*x + c*x^2]) + (7*c)/(12*b^2*x^(3/2)*Sqrt[b*x + c*x^2]) - (35*c^2)/(24*b^3*Sqrt[x]*Sqrt[
b*x + c*x^2]) - (35*c^3*Sqrt[x])/(8*b^4*Sqrt[b*x + c*x^2]) + (35*c^3*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x
])])/(8*b^(9/2))

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^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}}-\frac{(7 c) \int \frac{1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx}{6 b}\\ &=-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}+\frac{\left (35 c^2\right ) \int \frac{1}{\sqrt{x} \left (b x+c x^2\right )^{3/2}} \, dx}{24 b^2}\\ &=-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2}{24 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{\left (35 c^3\right ) \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{16 b^3}\\ &=-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2}{24 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{35 c^3 \sqrt{x}}{8 b^4 \sqrt{b x+c x^2}}-\frac{\left (35 c^3\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{16 b^4}\\ &=-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2}{24 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{35 c^3 \sqrt{x}}{8 b^4 \sqrt{b x+c x^2}}-\frac{\left (35 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{8 b^4}\\ &=-\frac{1}{3 b x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c}{12 b^2 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2}{24 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{35 c^3 \sqrt{x}}{8 b^4 \sqrt{b x+c x^2}}+\frac{35 c^3 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{8 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0101033, size = 40, normalized size = 0.28 \[ -\frac{2 c^3 \sqrt{x} \, _2F_1\left (-\frac{1}{2},4;\frac{1}{2};\frac{c x}{b}+1\right )}{b^4 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-2*c^3*Sqrt[x]*Hypergeometric2F1[-1/2, 4, 1/2, 1 + (c*x)/b])/(b^4*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.191, size = 87, normalized size = 0.6 \begin{align*}{\frac{1}{24\,cx+24\,b}\sqrt{x \left ( cx+b \right ) } \left ( 105\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{3}{c}^{3}-105\,{x}^{3}{c}^{3}\sqrt{b}-35\,{b}^{3/2}{x}^{2}{c}^{2}+14\,{b}^{5/2}xc-8\,{b}^{7/2} \right ){x}^{-{\frac{7}{2}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(c*x^2+b*x)^(3/2),x)

[Out]

1/24*(x*(c*x+b))^(1/2)*(105*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^3*c^3-105*x^3*c^3*b^(1/2)-35*b^(3/2
)*x^2*c^2+14*b^(5/2)*x*c-8*b^(7/2))/x^(7/2)/(c*x+b)/b^(9/2)

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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{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x)^(3/2)*x^(5/2)), x)

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Fricas [A]  time = 2.06533, size = 549, normalized size = 3.79 \begin{align*} \left [\frac{105 \,{\left (c^{4} x^{5} + b c^{3} x^{4}\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 (105 \, b c^{3} x^{3} + 35 \, b^{2} c^{2} x^{2} - 14 \, b^{3} c x + 8 \, b^{4}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{48 \,{\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}, -\frac{105 \,{\left (c^{4} x^{5} + b c^{3} x^{4}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (105 \, b c^{3} x^{3} + 35 \, b^{2} c^{2} x^{2} - 14 \, b^{3} c x + 8 \, b^{4}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{24 \,{\left (b^{5} c x^{5} + b^{6} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(c^4*x^5 + b*c^3*x^4)*sqrt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(
105*b*c^3*x^3 + 35*b^2*c^2*x^2 - 14*b^3*c*x + 8*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4), -1/24*(
105*(c^4*x^5 + b*c^3*x^4)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (105*b*c^3*x^3 + 35*b^2*c^2*x^
2 - 14*b^3*c*x + 8*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c*x^5 + b^6*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{5}{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**(5/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(1/(x**(5/2)*(x*(b + c*x))**(3/2)), x)

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Giac [A]  time = 1.29756, size = 113, normalized size = 0.78 \begin{align*} -\frac{1}{24} \, c^{3}{\left (\frac{105 \, \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{4}} + \frac{48}{\sqrt{c x + b} b^{4}} + \frac{57 \,{\left (c x + b\right )}^{\frac{5}{2}} - 136 \,{\left (c x + b\right )}^{\frac{3}{2}} b + 87 \, \sqrt{c x + b} b^{2}}{b^{4} c^{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)^(3/2),x, algorithm="giac")

[Out]

-1/24*c^3*(105*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) + 48/(sqrt(c*x + b)*b^4) + (57*(c*x + b)^(5/2) -
136*(c*x + b)^(3/2)*b + 87*sqrt(c*x + b)*b^2)/(b^4*c^3*x^3))