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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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 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 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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.62, size = 218, normalized size = 1.86 \[ \left [\frac {15 \, {\left (c^{3} x^{4} + b c^{2} x^{3}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (15 \, b c^{2} x^{2} + 5 \, b^{2} c x - 2 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, \frac {15 \, {\left (c^{3} x^{4} + b c^{2} x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, b c^{2} x^{2} + 5 \, b^{2} c x - 2 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \]

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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giac [A]  time = 0.25, size = 80, normalized size = 0.68 \[ \frac {15 \, c^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{3}} + \frac {2 \, c^{2}}{\sqrt {c x + b} b^{3}} + \frac {7 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{2} - 9 \, \sqrt {c x + b} b c^{2}}{4 \, b^{3} c^{2} x^{2}} \]

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]

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

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maple [A]  time = 0.06, size = 76, normalized size = 0.65 \[ -\frac {\sqrt {\left (c x +b \right ) x}\, \left (15 \sqrt {c x +b}\, c^{2} x^{2} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 \sqrt {b}\, c^{2} x^{2}-5 b^{\frac {3}{2}} c x +2 b^{\frac {5}{2}}\right )}{4 \left (c x +b \right ) b^{\frac {7}{2}} x^{\frac {5}{2}}} \]

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)*((c*x+b)*x)^(1/2)*(15*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{3/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {3}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

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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