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

Optimal. Leaf size=131 \[ \frac {7 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}-\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \]

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Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \begin {gather*} -\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}+\frac {7 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(-7*b^3*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (7*b^2*(b*x + c*x^2)^(3/2))/(48*c^3) - (7*b*x*(b*x + c*x^2)
^(3/2))/(40*c^2) + (x^2*(b*x + c*x^2)^(3/2))/(5*c) + (7*b^5*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(128*c^(9/
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^3 \sqrt {b x+c x^2} \, dx &=\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {(7 b) \int x^2 \sqrt {b x+c x^2} \, dx}{10 c}\\ &=-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (7 b^2\right ) \int x \sqrt {b x+c x^2} \, dx}{16 c^2}\\ &=\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}-\frac {\left (7 b^3\right ) \int \sqrt {b x+c x^2} \, dx}{32 c^3}\\ &=-\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (7 b^5\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^4}\\ &=-\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (7 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^4}\\ &=-\frac {7 b^3 (b+2 c x) \sqrt {b x+c x^2}}{128 c^4}+\frac {7 b^2 \left (b x+c x^2\right )^{3/2}}{48 c^3}-\frac {7 b x \left (b x+c x^2\right )^{3/2}}{40 c^2}+\frac {x^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac {7 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 109, normalized size = 0.83 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\frac {105 b^{9/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (-105 b^4+70 b^3 c x-56 b^2 c^2 x^2+48 b c^3 x^3+384 c^4 x^4\right )\right )}{1920 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4 + 70*b^3*c*x - 56*b^2*c^2*x^2 + 48*b*c^3*x^3 + 384*c^4*x^4) + (105*b^(9/
2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

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IntegrateAlgebraic [A]  time = 0.25, size = 101, normalized size = 0.77 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-105 b^4+70 b^3 c x-56 b^2 c^2 x^2+48 b c^3 x^3+384 c^4 x^4\right )}{1920 c^4}-\frac {7 b^5 \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{256 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[b*x + c*x^2]*(-105*b^4 + 70*b^3*c*x - 56*b^2*c^2*x^2 + 48*b*c^3*x^3 + 384*c^4*x^4))/(1920*c^4) - (7*b^5*
Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(256*c^(9/2))

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fricas [A]  time = 0.41, size = 192, normalized size = 1.47 \begin {gather*} \left [\frac {105 \, b^{5} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{3840 \, c^{5}}, -\frac {105 \, b^{5} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (384 \, c^{5} x^{4} + 48 \, b c^{4} x^{3} - 56 \, b^{2} c^{3} x^{2} + 70 \, b^{3} c^{2} x - 105 \, b^{4} c\right )} \sqrt {c x^{2} + b x}}{1920 \, c^{5}}\right ] \end {gather*}

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]

[1/3840*(105*b^5*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2
*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*sqrt(c*x^2 + b*x))/c^5, -1/1920*(105*b^5*sqrt(-c)*arctan(sqrt(c*x^2 + b*x
)*sqrt(-c)/(c*x)) - (384*c^5*x^4 + 48*b*c^4*x^3 - 56*b^2*c^3*x^2 + 70*b^3*c^2*x - 105*b^4*c)*sqrt(c*x^2 + b*x)
)/c^5]

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giac [A]  time = 0.20, size = 97, normalized size = 0.74 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x + \frac {b}{c}\right )} x - \frac {7 \, b^{2}}{c^{2}}\right )} x + \frac {35 \, b^{3}}{c^{3}}\right )} x - \frac {105 \, b^{4}}{c^{4}}\right )} - \frac {7 \, 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 {9}{2}}} \end {gather*}

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]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*x + b/c)*x - 7*b^2/c^2)*x + 35*b^3/c^3)*x - 105*b^4/c^4) - 7/256*b^5*log(
abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)

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maple [A]  time = 0.05, size = 129, normalized size = 0.98 \begin {gather*} \frac {7 b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {9}{2}}}-\frac {7 \sqrt {c \,x^{2}+b x}\, b^{3} x}{64 c^{3}}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} x^{2}}{5 c}-\frac {7 \sqrt {c \,x^{2}+b x}\, b^{4}}{128 c^{4}}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b x}{40 c^{2}}+\frac {7 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2}}{48 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^2*(c*x^2+b*x)^(3/2)/c-7/40*b*x*(c*x^2+b*x)^(3/2)/c^2+7/48*b^2*(c*x^2+b*x)^(3/2)/c^3-7/64*b^3/c^3*(c*x^2+
b*x)^(1/2)*x-7/128*b^4/c^4*(c*x^2+b*x)^(1/2)+7/256*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 1.16, size = 127, normalized size = 0.97 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2}}{5 \, c} - \frac {7 \, \sqrt {c x^{2} + b x} b^{3} x}{64 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x}{40 \, c^{2}} + \frac {7 \, b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {9}{2}}} - \frac {7 \, \sqrt {c x^{2} + b x} b^{4}}{128 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}}{48 \, c^{3}} \end {gather*}

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]

1/5*(c*x^2 + b*x)^(3/2)*x^2/c - 7/64*sqrt(c*x^2 + b*x)*b^3*x/c^3 - 7/40*(c*x^2 + b*x)^(3/2)*b*x/c^2 + 7/256*b^
5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) - 7/128*sqrt(c*x^2 + b*x)*b^4/c^4 + 7/48*(c*x^2 + b*x)^
(3/2)*b^2/c^3

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mupad [B]  time = 0.57, size = 119, normalized size = 0.91 \begin {gather*} \frac {x^2\,{\left (c\,x^2+b\,x\right )}^{3/2}}{5\,c}-\frac {7\,b\,\left (\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}\right )}{10\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x + c*x^2)^(1/2),x)

[Out]

(x^2*(b*x + c*x^2)^(3/2))/(5*c) - (7*b*((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)))/(
10*c)

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

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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