3.43 \(\int \frac {x^3}{\sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}+\frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c} \]

[Out]

-5/8*b^3*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+5/8*b^2*(c*x^2+b*x)^(1/2)/c^3-5/12*b*x*(c*x^2+b*x)^(1/2)
/c^2+1/3*x^2*(c*x^2+b*x)^(1/2)/c

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Rubi [A]  time = 0.04, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {670, 640, 620, 206} \[ \frac {5 b^2 \sqrt {b x+c x^2}}{8 c^3}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}-\frac {5 b x \sqrt {b x+c x^2}}{12 c^2}+\frac {x^2 \sqrt {b x+c x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[b*x + c*x^2],x]

[Out]

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

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Mathematica [A]  time = 0.16, size = 87, normalized size = 0.85 \[ \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^2-10 b c x+8 c^2 x^2\right )-\frac {15 b^{5/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{24 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^2 - 10*b*c*x + 8*c^2*x^2) - (15*b^(5/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/
(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(24*c^(7/2))

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fricas [A]  time = 0.99, size = 147, normalized size = 1.44 \[ \left [\frac {15 \, b^{3} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, \frac {15 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*b^3*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*c^3*x^2 - 10*b*c^2*x + 15*b^2*c)*sqr
t(c*x^2 + b*x))/c^4, 1/24*(15*b^3*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*c^3*x^2 - 10*b*c^2*x
+ 15*b^2*c)*sqrt(c*x^2 + b*x))/c^4]

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giac [A]  time = 0.22, size = 77, normalized size = 0.75 \[ \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, x}{c} - \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2}}{c^{3}}\right )} + \frac {5 \, b^{3} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*x*(4*x/c - 5*b/c^2) + 15*b^2/c^3) + 5/16*b^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*
x))*sqrt(c) - b))/c^(7/2)

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maple [A]  time = 0.04, size = 90, normalized size = 0.88 \[ \frac {\sqrt {c \,x^{2}+b x}\, x^{2}}{3 c}-\frac {5 b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}-\frac {5 \sqrt {c \,x^{2}+b x}\, b x}{12 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{2}}{8 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^2*(c*x^2+b*x)^(1/2)/c-5/12*b*x*(c*x^2+b*x)^(1/2)/c^2+5/8*b^2*(c*x^2+b*x)^(1/2)/c^3-5/16*b^3/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 = 1.39, size = 88, normalized size = 0.86 \[ \frac {\sqrt {c x^{2} + b x} x^{2}}{3 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} b x}{12 \, c^{2}} - \frac {5 \, b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{2}}{8 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(c*x^2 + b*x)*x^2/c - 5/12*sqrt(c*x^2 + b*x)*b*x/c^2 - 5/16*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq
rt(c))/c^(7/2) + 5/8*sqrt(c*x^2 + b*x)*b^2/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x + c*x^2)^(1/2),x)

[Out]

int(x^3/(b*x + c*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/sqrt(x*(b + c*x)), x)

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