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

Optimal. Leaf size=80 \[ \frac {16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \]

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {656, 648} \begin {gather*} \frac {16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*Sqrt[b*x + c*x^2],x]

[Out]

(16*b^2*(b*x + c*x^2)^(3/2))/(105*c^3*x^(3/2)) - (8*b*(b*x + c*x^2)^(3/2))/(35*c^2*Sqrt[x]) + (2*Sqrt[x]*(b*x
+ c*x^2)^(3/2))/(7*c)

Rule 648

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] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 656

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[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rubi steps

\begin {align*} \int x^{3/2} \sqrt {b x+c x^2} \, dx &=\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {(4 b) \int \sqrt {x} \sqrt {b x+c x^2} \, dx}{7 c}\\ &=-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}+\frac {\left (8 b^2\right ) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{35 c^2}\\ &=\frac {16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac {8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt {x}}+\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.52 \begin {gather*} \frac {2 (x (b+c x))^{3/2} \left (8 b^2-12 b c x+15 c^2 x^2\right )}{105 c^3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2)*(8*b^2 - 12*b*c*x + 15*c^2*x^2))/(105*c^3*x^(3/2))

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IntegrateAlgebraic [A]  time = 0.09, size = 55, normalized size = 0.69 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (8 b^3-4 b^2 c x+3 b c^2 x^2+15 c^3 x^3\right )}{105 c^3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + c*x^2]*(8*b^3 - 4*b^2*c*x + 3*b*c^2*x^2 + 15*c^3*x^3))/(105*c^3*Sqrt[x])

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fricas [A]  time = 0.38, size = 49, normalized size = 0.61 \begin {gather*} \frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x^{2} + b x}}{105 \, c^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))

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giac [A]  time = 0.17, size = 46, normalized size = 0.58 \begin {gather*} -\frac {16 \, b^{\frac {7}{2}}}{105 \, c^{3}} + \frac {2 \, {\left (15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}\right )}}{105 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/105*b^(7/2)/c^3 + 2/105*(15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)/c^3

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maple [A]  time = 0.05, size = 44, normalized size = 0.55 \begin {gather*} \frac {2 \left (c x +b \right ) \left (15 c^{2} x^{2}-12 b c x +8 b^{2}\right ) \sqrt {c \,x^{2}+b x}}{105 c^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(c*x+b)*(15*c^2*x^2-12*b*c*x+8*b^2)*(c*x^2+b*x)^(1/2)/c^3/x^(1/2)

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maxima [A]  time = 1.45, size = 42, normalized size = 0.52 \begin {gather*} \frac {2 \, {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x + b}}{105 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*sqrt(c*x + b)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x + c*x^2)^(1/2),x)

[Out]

int(x^(3/2)*(b*x + c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{\frac {3}{2}} \sqrt {x \left (b + c x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(3/2)*sqrt(x*(b + c*x)), x)

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