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

Optimal. Leaf size=80 \[ \frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{315 c^3 x^{5/2}}-\frac{8 b \left (b x+c x^2\right )^{5/2}}{63 c^2 x^{3/2}}+\frac{2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt{x}} \]

[Out]

(16*b^2*(b*x + c*x^2)^(5/2))/(315*c^3*x^(5/2)) - (8*b*(b*x + c*x^2)^(5/2))/(63*c^2*x^(3/2)) + (2*(b*x + c*x^2)
^(5/2))/(9*c*Sqrt[x])

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Rubi [A]  time = 0.0269342, 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} \[ \frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{315 c^3 x^{5/2}}-\frac{8 b \left (b x+c x^2\right )^{5/2}}{63 c^2 x^{3/2}}+\frac{2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(b*x + c*x^2)^(3/2),x]

[Out]

(16*b^2*(b*x + c*x^2)^(5/2))/(315*c^3*x^(5/2)) - (8*b*(b*x + c*x^2)^(5/2))/(63*c^2*x^(3/2)) + (2*(b*x + c*x^2)
^(5/2))/(9*c*Sqrt[x])

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 \sqrt{x} \left (b x+c x^2\right )^{3/2} \, dx &=\frac{2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt{x}}-\frac{(4 b) \int \frac{\left (b x+c x^2\right )^{3/2}}{\sqrt{x}} \, dx}{9 c}\\ &=-\frac{8 b \left (b x+c x^2\right )^{5/2}}{63 c^2 x^{3/2}}+\frac{2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt{x}}+\frac{\left (8 b^2\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{63 c^2}\\ &=\frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{315 c^3 x^{5/2}}-\frac{8 b \left (b x+c x^2\right )^{5/2}}{63 c^2 x^{3/2}}+\frac{2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0255045, size = 42, normalized size = 0.52 \[ \frac{2 (x (b+c x))^{5/2} \left (8 b^2-20 b c x+35 c^2 x^2\right )}{315 c^3 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(8*b^2 - 20*b*c*x + 35*c^2*x^2))/(315*c^3*x^(5/2))

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Maple [A]  time = 0.046, size = 44, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 35\,{c}^{2}{x}^{2}-20\,bcx+8\,{b}^{2} \right ) }{315\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(c*x+b)*(35*c^2*x^2-20*b*c*x+8*b^2)*(c*x^2+b*x)^(3/2)/c^3/x^(3/2)

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Maxima [A]  time = 1.15089, size = 138, normalized size = 1.72 \begin{align*} \frac{2 \,{\left ({\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 )} x^{3} + 3 \,{\left (15 \, b c^{3} x^{4} + 3 \, b^{2} c^{2} x^{3} - 4 \, b^{3} c x^{2} + 8 \, b^{4} x\right )} x^{2}\right )} \sqrt{c x + b}}{315 \, c^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*(c*x^2+b*x)^(3/2),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)*x^3 + 3*(15*b*c^3*x^4 + 3*b^2*c^2*x^3 -
4*b^3*c*x^2 + 8*b^4*x)*x^2)*sqrt(c*x + b)/(c^3*x^3)

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Fricas [A]  time = 1.90306, size = 139, normalized size = 1.74 \begin{align*} \frac{2 \,{\left (35 \, c^{4} x^{4} + 50 \, b c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} - 4 \, b^{3} c x + 8 \, b^{4}\right )} \sqrt{c x^{2} + b x}}{315 \, c^{3} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*c^4*x^4 + 50*b*c^3*x^3 + 3*b^2*c^2*x^2 - 4*b^3*c*x + 8*b^4)*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 \sqrt{x} \left (x \left (b + c x\right )\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29596, size = 149, normalized size = 1.86 \begin{align*} \frac{2}{315} \, c{\left (\frac{16 \, b^{\frac{9}{2}}}{c^{4}} + \frac{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}}{c^{4}}\right )} - \frac{2}{105} \, b{\left (\frac{8 \, b^{\frac{7}{2}}}{c^{3}} - \frac{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}}{c^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*c*(16*b^(9/2)/c^4 + (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) - 2/105*b*(8*b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b
^2)/c^3)