3.52 \(\int \frac {x^4}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=97 \[ \frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}}-\frac {15 b \sqrt {b x+c x^2}}{4 c^3}+\frac {5 x \sqrt {b x+c x^2}}{2 c^2}-\frac {2 x^3}{c \sqrt {b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {668, 670, 640, 620, 206} \[ \frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}}+\frac {5 x \sqrt {b x+c x^2}}{2 c^2}-\frac {15 b \sqrt {b x+c x^2}}{4 c^3}-\frac {2 x^3}{c \sqrt {b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(b*x + c*x^2)^(3/2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 668

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(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] && GtQ[m, 1] && IntegerQ[2*p]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.52 \[ \frac {2 x^4 \sqrt {\frac {c x}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};-\frac {c x}{b}\right )}{7 b \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(b*x + c*x^2)^(3/2),x]

[Out]

(2*x^4*Sqrt[1 + (c*x)/b]*Hypergeometric2F1[3/2, 7/2, 9/2, -((c*x)/b)])/(7*b*Sqrt[x*(b + c*x)])

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fricas [A]  time = 1.01, size = 182, normalized size = 1.88 \[ \left [\frac {15 \, {\left (b^{2} c x + b^{3}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{3} x^{2} - 5 \, b c^{2} x - 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{5} x + b c^{4}\right )}}, -\frac {15 \, {\left (b^{2} c x + b^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, c^{3} x^{2} - 5 \, b c^{2} x - 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{5} x + b c^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 102, normalized size = 1.05 \[ \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, x}{c^{2}} - \frac {7 \, b}{c^{3}}\right )} - \frac {15 \, b^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, b^{3}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} c + b \sqrt {c}\right )} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*x/c^2 - 7*b/c^3) - 15/8*b^2*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/
c^(7/2) - 2*b^3/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*c + b*sqrt(c))*c^3)

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maple [A]  time = 0.04, size = 93, normalized size = 0.96 \[ \frac {x^{3}}{2 \sqrt {c \,x^{2}+b x}\, c}-\frac {5 b \,x^{2}}{4 \sqrt {c \,x^{2}+b x}\, c^{2}}-\frac {15 b^{2} x}{4 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {15 b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(c*x^2+b*x)^(3/2),x)

[Out]

1/2*x^3/c/(c*x^2+b*x)^(1/2)-5/4*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*b^2/c^3/(c*x^2+b*x)^(1/2)*x+15/8*b^2/c^(7/2)*
ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 1.38, size = 91, normalized size = 0.94 \[ \frac {x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {15 \, b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

1/2*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*b*x^2/(sqrt(c*x^2 + b*x)*c^2) - 15/4*b^2*x/(sqrt(c*x^2 + b*x)*c^3) + 15/8*
b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^4/(b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x**4/(x*(b + c*x))**(3/2), x)

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