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

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

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Rubi [A]  time = 0.07, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {672, 666, 660, 207} \begin {gather*} -\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}} \end {gather*}

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 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^{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*}

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.28 \begin {gather*} -\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)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 1.26, size = 100, normalized size = 0.69 \begin {gather*} \frac {35 c^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{9/2}}+\frac {\sqrt {b x+c x^2} \left (-8 b^3+14 b^2 c x-35 b c^2 x^2-105 c^3 x^3\right )}{24 b^4 x^{7/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-8*b^3 + 14*b^2*c*x - 35*b*c^2*x^2 - 105*c^3*x^3))/(24*b^4*x^(7/2)*(b + c*x)) + (35*c^3*Ar
cTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(8*b^(9/2))

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fricas [A]  time = 0.42, size = 240, normalized size = 1.66 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.26, size = 95, normalized size = 0.66 \begin {gather*} -\frac {35 \, c^{3} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4}} - \frac {2 \, c^{3}}{\sqrt {c x + b} b^{4}} - \frac {57 \, {\left (c x + b\right )}^{\frac {5}{2}} c^{3} - 136 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{3} + 87 \, \sqrt {c x + b} b^{2} c^{3}}{24 \, b^{4} c^{3} x^{3}} \end {gather*}

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]

-35/8*c^3*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) - 2*c^3/(sqrt(c*x + b)*b^4) - 1/24*(57*(c*x + b)^(5/2)
*c^3 - 136*(c*x + b)^(3/2)*b*c^3 + 87*sqrt(c*x + b)*b^2*c^3)/(b^4*c^3*x^3)

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maple [A]  time = 0.06, size = 87, normalized size = 0.60 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (105 \sqrt {c x +b}\, c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-105 \sqrt {b}\, c^{3} x^{3}-35 b^{\frac {3}{2}} c^{2} x^{2}+14 b^{\frac {5}{2}} c x -8 b^{\frac {7}{2}}\right )}{24 \left (c x +b \right ) b^{\frac {9}{2}} x^{\frac {7}{2}}} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {5}{2}}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{5/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{\frac {5}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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