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

Optimal. Leaf size=108 \[ -\frac{32 b^3 \left (b x+c x^2\right )^{5/2}}{1155 c^4 x^{5/2}}+\frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac{4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt{x}}+\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2}}{11 c} \]

[Out]

(-32*b^3*(b*x + c*x^2)^(5/2))/(1155*c^4*x^(5/2)) + (16*b^2*(b*x + c*x^2)^(5/2))/(231*c^3*x^(3/2)) - (4*b*(b*x
+ c*x^2)^(5/2))/(33*c^2*Sqrt[x]) + (2*Sqrt[x]*(b*x + c*x^2)^(5/2))/(11*c)

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Rubi [A]  time = 0.0429493, 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} \[ -\frac{32 b^3 \left (b x+c x^2\right )^{5/2}}{1155 c^4 x^{5/2}}+\frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac{4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt{x}}+\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(b*x + c*x^2)^(3/2),x]

[Out]

(-32*b^3*(b*x + c*x^2)^(5/2))/(1155*c^4*x^(5/2)) + (16*b^2*(b*x + c*x^2)^(5/2))/(231*c^3*x^(3/2)) - (4*b*(b*x
+ c*x^2)^(5/2))/(33*c^2*Sqrt[x]) + (2*Sqrt[x]*(b*x + c*x^2)^(5/2))/(11*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} \left (b x+c x^2\right )^{3/2} \, dx &=\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac{(6 b) \int \sqrt{x} \left (b x+c x^2\right )^{3/2} \, dx}{11 c}\\ &=-\frac{4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt{x}}+\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2}}{11 c}+\frac{\left (8 b^2\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{\sqrt{x}} \, dx}{33 c^2}\\ &=\frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac{4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt{x}}+\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac{\left (16 b^3\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{231 c^3}\\ &=-\frac{32 b^3 \left (b x+c x^2\right )^{5/2}}{1155 c^4 x^{5/2}}+\frac{16 b^2 \left (b x+c x^2\right )^{5/2}}{231 c^3 x^{3/2}}-\frac{4 b \left (b x+c x^2\right )^{5/2}}{33 c^2 \sqrt{x}}+\frac{2 \sqrt{x} \left (b x+c x^2\right )^{5/2}}{11 c}\\ \end{align*}

Mathematica [A]  time = 0.0324024, size = 53, normalized size = 0.49 \[ \frac{2 (x (b+c x))^{5/2} \left (40 b^2 c x-16 b^3-70 b c^2 x^2+105 c^3 x^3\right )}{1155 c^4 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(-16*b^3 + 40*b^2*c*x - 70*b*c^2*x^2 + 105*c^3*x^3))/(1155*c^4*x^(5/2))

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Maple [A]  time = 0.047, size = 55, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -105\,{x}^{3}{c}^{3}+70\,b{x}^{2}{c}^{2}-40\,{b}^{2}xc+16\,{b}^{3} \right ) }{1155\,{c}^{4}} \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^(3/2)*(c*x^2+b*x)^(3/2),x)

[Out]

-2/1155*(c*x+b)*(-105*c^3*x^3+70*b*c^2*x^2-40*b^2*c*x+16*b^3)*(c*x^2+b*x)^(3/2)/c^4/x^(3/2)

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Maxima [A]  time = 1.15566, size = 167, normalized size = 1.55 \begin{align*} \frac{2 \,{\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \,{\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt{c x + b}}{3465 \, c^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*((315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*x^4 + 11*(35*b*c
^4*x^5 + 5*b^2*c^3*x^4 - 6*b^3*c^2*x^3 + 8*b^4*c*x^2 - 16*b^5*x)*x^3)*sqrt(c*x + b)/(c^4*x^4)

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Fricas [A]  time = 1.97309, size = 166, normalized size = 1.54 \begin{align*} \frac{2 \,{\left (105 \, c^{5} x^{5} + 140 \, b c^{4} x^{4} + 5 \, b^{2} c^{3} x^{3} - 6 \, b^{3} c^{2} x^{2} + 8 \, b^{4} c x - 16 \, b^{5}\right )} \sqrt{c x^{2} + b x}}{1155 \, c^{4} \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(105*c^5*x^5 + 140*b*c^4*x^4 + 5*b^2*c^3*x^3 - 6*b^3*c^2*x^2 + 8*b^4*c*x - 16*b^5)*sqrt(c*x^2 + b*x)/(c
^4*sqrt(x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.19372, size = 181, normalized size = 1.68 \begin{align*} -\frac{2}{3465} \, c{\left (\frac{128 \, b^{\frac{11}{2}}}{c^{5}} - \frac{315 \,{\left (c x + b\right )}^{\frac{11}{2}} - 1540 \,{\left (c x + b\right )}^{\frac{9}{2}} b + 2970 \,{\left (c x + b\right )}^{\frac{7}{2}} b^{2} - 2772 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{3} + 1155 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{4}}{c^{5}}\right )} + \frac{2}{315} \, b{\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*c*(128*b^(11/2)/c^5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772
*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5) + 2/315*b*(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)