3.2 \(\int x^2 \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=105 \[ -\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}+\frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c} \]

[Out]

-5/24*b*(c*x^2+b*x)^(3/2)/c^2+1/4*x*(c*x^2+b*x)^(3/2)/c-5/64*b^4*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+
5/64*b^2*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 105, 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 = {670, 640, 612, 620, 206} \[ \frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[b*x + c*x^2],x]

[Out]

(5*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - (5*b*(b*x + c*x^2)^(3/2))/(24*c^2) + (x*(b*x + c*x^2)^(3/2))/
(4*c) - (5*b^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 x^2 \sqrt {b x+c x^2} \, dx &=\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {(5 b) \int x \sqrt {b x+c x^2} \, dx}{8 c}\\ &=-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}+\frac {\left (5 b^2\right ) \int \sqrt {b x+c x^2} \, dx}{16 c^2}\\ &=\frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (5 b^4\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (5 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^3}\\ &=\frac {5 b^2 (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}-\frac {5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 98, normalized size = 0.93 \[ \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^3-10 b^2 c x+8 b c^2 x^2+48 c^3 x^3\right )-\frac {15 b^{7/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3 - 10*b^2*c*x + 8*b*c^2*x^2 + 48*c^3*x^3) - (15*b^(7/2)*ArcSinh[(Sqrt[c]*Sq
rt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 169, normalized size = 1.61 \[ \left [\frac {15 \, b^{4} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} - 10 \, b^{2} c^{2} x + 15 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{384 \, c^{4}}, \frac {15 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} - 10 \, b^{2} c^{2} x + 15 \, b^{3} c\right )} \sqrt {c x^{2} + b x}}{192 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*b^4*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(48*c^4*x^3 + 8*b*c^3*x^2 - 10*b^2*c^2
*x + 15*b^3*c)*sqrt(c*x^2 + b*x))/c^4, 1/192*(15*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*c
^4*x^3 + 8*b*c^3*x^2 - 10*b^2*c^2*x + 15*b^3*c)*sqrt(c*x^2 + b*x))/c^4]

________________________________________________________________________________________

giac [A]  time = 0.20, size = 85, normalized size = 0.81 \[ \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, x + \frac {b}{c}\right )} x - \frac {5 \, b^{2}}{c^{2}}\right )} x + \frac {15 \, b^{3}}{c^{3}}\right )} + \frac {5 \, b^{4} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*x + b/c)*x - 5*b^2/c^2)*x + 15*b^3/c^3) + 5/128*b^4*log(abs(-2*(sqrt(c)*x - s
qrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 107, normalized size = 1.02 \[ -\frac {5 b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{2} x}{32 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{3}}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} x}{4 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b}{24 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(c*x^2+b*x)^(3/2)/c-5/24*b*(c*x^2+b*x)^(3/2)/c^2+5/32*b^2/c^2*(c*x^2+b*x)^(1/2)*x+5/64*b^3/c^3*(c*x^2+b*
x)^(1/2)-5/128*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.18, size = 105, normalized size = 1.00 \[ \frac {5 \, \sqrt {c x^{2} + b x} b^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x}{4 \, c} - \frac {5 \, b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{3}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b}{24 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/32*sqrt(c*x^2 + b*x)*b^2*x/c^2 + 1/4*(c*x^2 + b*x)^(3/2)*x/c - 5/128*b^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)
*sqrt(c))/c^(7/2) + 5/64*sqrt(c*x^2 + b*x)*b^3/c^3 - 5/24*(c*x^2 + b*x)^(3/2)*b/c^2

________________________________________________________________________________________

mupad [B]  time = 0.31, size = 93, normalized size = 0.89 \[ \frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x + c*x^2)^(1/2),x)

[Out]

(x*(b*x + c*x^2)^(3/2))/(4*c) - (5*b*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16*c^(5/2)) + ((
b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2)))/(8*c)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(x*(b + c*x)), x)

________________________________________________________________________________________