### 3.74 $$\int x^{3/2} \sqrt{b x+c x^2} \, dx$$

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

[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)

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Rubi [A]  time = 0.0267062, antiderivative size = 80, normalized size of antiderivative = 1., 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} $\frac{16 b^2 \left (b x+c x^2\right )^{3/2}}{105 c^3 x^{3/2}}-\frac{8 b \left (b x+c x^2\right )^{3/2}}{35 c^2 \sqrt{x}}+\frac{2 \sqrt{x} \left (b x+c x^2\right )^{3/2}}{7 c}$

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 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^{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*}

Mathematica [A]  time = 0.0217325, size = 42, normalized size = 0.52 $\frac{2 (x (b+c x))^{3/2} \left (8 b^2-12 b c x+15 c^2 x^2\right )}{105 c^3 x^{3/2}}$

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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Maple [A]  time = 0.049, size = 44, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 15\,{c}^{2}{x}^{2}-12\,bcx+8\,{b}^{2} \right ) }{105\,{c}^{3}}\sqrt{c{x}^{2}+bx}{\frac{1}{\sqrt{x}}}} \end{align*}

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 = 1.15948, size = 57, normalized size = 0.71 \begin{align*} \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{align*}

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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Fricas [A]  time = 2.28884, size = 116, normalized size = 1.45 \begin{align*} \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{align*}

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

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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Giac [A]  time = 1.16809, size = 62, normalized size = 0.78 \begin{align*} -\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{align*}

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