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

Optimal. Leaf size=110 \[ -\frac {3 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}+\frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {640, 612, 620, 206} \begin {gather*} \frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {3 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(b*x + c*x^2)^(3/2),x]

[Out]

(3*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^3) - (b*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(16*c^2) + (b*x + c*x^2)
^(5/2)/(5*c) - (3*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*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]

Rubi steps

\begin {align*} \int x \left (b x+c x^2\right )^{3/2} \, dx &=\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {b \int \left (b x+c x^2\right )^{3/2} \, dx}{2 c}\\ &=-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}+\frac {\left (3 b^3\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^2}\\ &=\frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 b^5\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^3}\\ &=\frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {\left (3 b^5\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^3}\\ &=\frac {3 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^3}-\frac {b (b+2 c x) \left (b x+c x^2\right )^{3/2}}{16 c^2}+\frac {\left (b x+c x^2\right )^{5/2}}{5 c}-\frac {3 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 109, normalized size = 0.99 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^4-10 b^3 c x+8 b^2 c^2 x^2+176 b c^3 x^3+128 c^4 x^4\right )-\frac {15 b^{9/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{640 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^4 - 10*b^3*c*x + 8*b^2*c^2*x^2 + 176*b*c^3*x^3 + 128*c^4*x^4) - (15*b^(9/2)*
ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(640*c^(7/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.31, size = 101, normalized size = 0.92 \begin {gather*} \frac {3 b^5 \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{7/2}}+\frac {\sqrt {b x+c x^2} \left (15 b^4-10 b^3 c x+8 b^2 c^2 x^2+176 b c^3 x^3+128 c^4 x^4\right )}{640 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(15*b^4 - 10*b^3*c*x + 8*b^2*c^2*x^2 + 176*b*c^3*x^3 + 128*c^4*x^4))/(640*c^3) + (3*b^5*Log
[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(256*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 191, normalized size = 1.74 \begin {gather*} \left [\frac {15 \, b^{5} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{1280 \, c^{4}}, \frac {15 \, b^{5} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 8 \, b^{2} c^{3} x^{2} - 10 \, b^{3} c^{2} x + 15 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{640 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1280*(15*b^5*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*
c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*sqrt(c*x^2 + b*x))/c^4, 1/640*(15*b^5*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqr
t(-c)/(c*x)) + (128*c^5*x^4 + 176*b*c^4*x^3 + 8*b^2*c^3*x^2 - 10*b^3*c^2*x + 15*b^4*c)*sqrt(c*x^2 + b*x))/c^4]

________________________________________________________________________________________

giac [A]  time = 0.24, size = 95, normalized size = 0.86 \begin {gather*} \frac {3 \, b^{5} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {7}{2}}} + \frac {1}{640} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x + 11 \, b\right )} x + \frac {b^{2}}{c}\right )} x - \frac {5 \, b^{3}}{c^{2}}\right )} x + \frac {15 \, b^{4}}{c^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/256*b^5*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2) + 1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*
(8*c*x + 11*b)*x + b^2/c)*x - 5*b^3/c^2)*x + 15*b^4/c^3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 126, normalized size = 1.15 \begin {gather*} -\frac {3 b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x}\, b^{3} x}{64 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x}\, b^{4}}{128 c^{3}}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} b x}{8 c}-\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2}}{16 c^{2}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c*x^2+b*x)^(5/2)/c-1/8*b/c*x*(c*x^2+b*x)^(3/2)-1/16*b^2/c^2*(c*x^2+b*x)^(3/2)+3/64*b^3/c^2*(c*x^2+b*x)^(1
/2)*x+3/128*b^4/c^3*(c*x^2+b*x)^(1/2)-3/256*b^5/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.29, size = 124, normalized size = 1.13 \begin {gather*} \frac {3 \, \sqrt {c x^{2} + b x} b^{3} x}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x}{8 \, c} - \frac {3 \, b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} + \frac {3 \, \sqrt {c x^{2} + b x} b^{4}}{128 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}}{16 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}}}{5 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*sqrt(c*x^2 + b*x)*b^3*x/c^2 - 1/8*(c*x^2 + b*x)^(3/2)*b*x/c - 3/256*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*
x)*sqrt(c))/c^(7/2) + 3/128*sqrt(c*x^2 + b*x)*b^4/c^3 - 1/16*(c*x^2 + b*x)^(3/2)*b^2/c^2 + 1/5*(c*x^2 + b*x)^(
5/2)/c

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 118, normalized size = 1.07 \begin {gather*} \frac {{\left (c\,x^2+b\,x\right )}^{5/2}}{5\,c}-\frac {b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\frac {\sqrt {c\,x^2+b\,x}\,\left (b+2\,c\,x\right )}{4\,c}-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x + c*x^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \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*(c*x**2+b*x)**(3/2),x)

[Out]

Integral(x*(x*(b + c*x))**(3/2), x)

________________________________________________________________________________________