### 3.73 $$\int x^{5/2} \sqrt{b x+c x^2} \, dx$$

Optimal. Leaf size=108 $-\frac{32 b^3 \left (b x+c x^2\right )^{3/2}}{315 c^4 x^{3/2}}+\frac{16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt{x}}-\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}$

[Out]

(-32*b^3*(b*x + c*x^2)^(3/2))/(315*c^4*x^(3/2)) + (16*b^2*(b*x + c*x^2)^(3/2))/(105*c^3*Sqrt[x]) - (4*b*Sqrt[x
]*(b*x + c*x^2)^(3/2))/(21*c^2) + (2*x^(3/2)*(b*x + c*x^2)^(3/2))/(9*c)

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Rubi [A]  time = 0.0402275, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {656, 648} $-\frac{32 b^3 \left (b x+c x^2\right )^{3/2}}{315 c^4 x^{3/2}}+\frac{16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt{x}}-\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^(5/2)*Sqrt[b*x + c*x^2],x]

[Out]

(-32*b^3*(b*x + c*x^2)^(3/2))/(315*c^4*x^(3/2)) + (16*b^2*(b*x + c*x^2)^(3/2))/(105*c^3*Sqrt[x]) - (4*b*Sqrt[x
]*(b*x + c*x^2)^(3/2))/(21*c^2) + (2*x^(3/2)*(b*x + c*x^2)^(3/2))/(9*c)

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]

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]

Rubi steps

\begin{align*} \int x^{5/2} \sqrt{b x+c x^2} \, dx &=\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac{(2 b) \int x^{3/2} \sqrt{b x+c x^2} \, dx}{3 c}\\ &=-\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}+\frac{\left (8 b^2\right ) \int \sqrt{x} \sqrt{b x+c x^2} \, dx}{21 c^2}\\ &=\frac{16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt{x}}-\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac{\left (16 b^3\right ) \int \frac{\sqrt{b x+c x^2}}{\sqrt{x}} \, dx}{105 c^3}\\ &=-\frac{32 b^3 \left (b x+c x^2\right )^{3/2}}{315 c^4 x^{3/2}}+\frac{16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 \sqrt{x}}-\frac{4 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}{21 c^2}+\frac{2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}\\ \end{align*}

Mathematica [A]  time = 0.028814, size = 53, normalized size = 0.49 $\frac{2 (x (b+c x))^{3/2} \left (24 b^2 c x-16 b^3-30 b c^2 x^2+35 c^3 x^3\right )}{315 c^4 x^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^(5/2)*Sqrt[b*x + c*x^2],x]

[Out]

(2*(x*(b + c*x))^(3/2)*(-16*b^3 + 24*b^2*c*x - 30*b*c^2*x^2 + 35*c^3*x^3))/(315*c^4*x^(3/2))

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Maple [A]  time = 0.046, size = 55, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -35\,{x}^{3}{c}^{3}+30\,b{x}^{2}{c}^{2}-24\,{b}^{2}xc+16\,{b}^{3} \right ) }{315\,{c}^{4}}\sqrt{c{x}^{2}+bx}{\frac{1}{\sqrt{x}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(c*x^2+b*x)^(1/2),x)

[Out]

-2/315*(c*x+b)*(-35*c^3*x^3+30*b*c^2*x^2-24*b^2*c*x+16*b^3)*(c*x^2+b*x)^(1/2)/c^4/x^(1/2)

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Maxima [A]  time = 1.14971, size = 72, normalized size = 0.67 \begin{align*} \frac{2 \,{\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt{c x + b}}{315 \, c^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 16*b^4)*sqrt(c*x + b)/c^4

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Fricas [A]  time = 2.20523, size = 139, normalized size = 1.29 \begin{align*} \frac{2 \,{\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt{c x^{2} + b x}}{315 \, c^{4} \sqrt{x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 16*b^4)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{5}{2}} \sqrt{x \left (b + c x\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**(5/2)*sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.20601, size = 78, normalized size = 0.72 \begin{align*} \frac{32 \, b^{\frac{9}{2}}}{315 \, c^{4}} + \frac{2 \,{\left (35 \,{\left (c x + b\right )}^{\frac{9}{2}} - 135 \,{\left (c x + b\right )}^{\frac{7}{2}} b + 189 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{3}\right )}}{315 \, c^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

32/315*b^(9/2)/c^4 + 2/315*(35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x +
b)^(3/2)*b^3)/c^4