3.1.52 \(\int \frac {x^4}{(b x+c x^2)^{3/2}} \, dx\) [52]

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

[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 = 0.03, 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 = {682, 684, 654, 634, 212} \begin {gather*} \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}} \end {gather*}

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 212

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

Rule 634

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 654

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 682

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 684

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 ) \text {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 [A]
time = 0.11, size = 89, normalized size = 0.92 \begin {gather*} \frac {\sqrt {c} x \left (-15 b^2-5 b c x+2 c^2 x^2\right )-15 b^2 \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{4 c^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(b*x + c*x^2)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.40, size = 145, normalized size = 1.49

method	result	size
risch	\(-\frac {\left (-2 c x +7 b \right ) x \left (c x +b \right )}{4 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {15 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {7}{2}}}-\frac {2 b^{2} \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{c^{4} \left (\frac {b}{c}+x \right )}\)	\(103\)
default	\(\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 91, normalized size = 0.94 \begin {gather*} \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}}} \end {gather*}

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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Fricas [A]
time = 1.54, size = 182, normalized size = 1.88 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\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(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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Giac [A]
time = 0.89, size = 102, normalized size = 1.05 \begin {gather*} \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}} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{{\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(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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