### 3.72 $$\int x^{7/2} \sqrt{b x+c x^2} \, dx$$

Optimal. Leaf size=136 $\frac{256 b^4 \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac{128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt{x}}+\frac{32 b^2 \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}$

[Out]

(256*b^4*(b*x + c*x^2)^(3/2))/(3465*c^5*x^(3/2)) - (128*b^3*(b*x + c*x^2)^(3/2))/(1155*c^4*Sqrt[x]) + (32*b^2*
Sqrt[x]*(b*x + c*x^2)^(3/2))/(231*c^3) - (16*b*x^(3/2)*(b*x + c*x^2)^(3/2))/(99*c^2) + (2*x^(5/2)*(b*x + c*x^2
)^(3/2))/(11*c)

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Rubi [A]  time = 0.0551495, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {656, 648} $\frac{256 b^4 \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac{128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt{x}}+\frac{32 b^2 \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^(7/2)*Sqrt[b*x + c*x^2],x]

[Out]

(256*b^4*(b*x + c*x^2)^(3/2))/(3465*c^5*x^(3/2)) - (128*b^3*(b*x + c*x^2)^(3/2))/(1155*c^4*Sqrt[x]) + (32*b^2*
Sqrt[x]*(b*x + c*x^2)^(3/2))/(231*c^3) - (16*b*x^(3/2)*(b*x + c*x^2)^(3/2))/(99*c^2) + (2*x^(5/2)*(b*x + c*x^2
)^(3/2))/(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^{7/2} \sqrt{b x+c x^2} \, dx &=\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac{(8 b) \int x^{5/2} \sqrt{b x+c x^2} \, dx}{11 c}\\ &=-\frac{16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac{\left (16 b^2\right ) \int x^{3/2} \sqrt{b x+c x^2} \, dx}{33 c^2}\\ &=\frac{32 b^2 \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac{\left (64 b^3\right ) \int \sqrt{x} \sqrt{b x+c x^2} \, dx}{231 c^3}\\ &=-\frac{128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt{x}}+\frac{32 b^2 \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}+\frac{\left (128 b^4\right ) \int \frac{\sqrt{b x+c x^2}}{\sqrt{x}} \, dx}{1155 c^4}\\ &=\frac{256 b^4 \left (b x+c x^2\right )^{3/2}}{3465 c^5 x^{3/2}}-\frac{128 b^3 \left (b x+c x^2\right )^{3/2}}{1155 c^4 \sqrt{x}}+\frac{32 b^2 \sqrt{x} \left (b x+c x^2\right )^{3/2}}{231 c^3}-\frac{16 b x^{3/2} \left (b x+c x^2\right )^{3/2}}{99 c^2}+\frac{2 x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}\\ \end{align*}

Mathematica [A]  time = 0.0346805, size = 64, normalized size = 0.47 $\frac{2 (x (b+c x))^{3/2} \left (240 b^2 c^2 x^2-192 b^3 c x+128 b^4-280 b c^3 x^3+315 c^4 x^4\right )}{3465 c^5 x^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^(7/2)*Sqrt[b*x + c*x^2],x]

[Out]

(2*(x*(b + c*x))^(3/2)*(128*b^4 - 192*b^3*c*x + 240*b^2*c^2*x^2 - 280*b*c^3*x^3 + 315*c^4*x^4))/(3465*c^5*x^(3
/2))

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Maple [A]  time = 0.046, size = 66, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 315\,{x}^{4}{c}^{4}-280\,b{x}^{3}{c}^{3}+240\,{b}^{2}{x}^{2}{c}^{2}-192\,{b}^{3}xc+128\,{b}^{4} \right ) }{3465\,{c}^{5}}\sqrt{c{x}^{2}+bx}{\frac{1}{\sqrt{x}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(c*x^2+b*x)^(1/2),x)

[Out]

2/3465*(c*x+b)*(315*c^4*x^4-280*b*c^3*x^3+240*b^2*c^2*x^2-192*b^3*c*x+128*b^4)*(c*x^2+b*x)^(1/2)/c^5/x^(1/2)

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Maxima [A]  time = 1.16032, size = 86, normalized size = 0.63 \begin{align*} \frac{2 \,{\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 )} \sqrt{c x + b}}{3465 \, c^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(c*x^2+b*x)^(1/2),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)*sqrt(c*x + b)/c^5

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Fricas [A]  time = 1.88514, size = 170, normalized size = 1.25 \begin{align*} \frac{2 \,{\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 )} \sqrt{c x^{2} + b x}}{3465 \, c^{5} \sqrt{x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[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)*sqrt(c*x^2 + b*x)
/(c^5*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**(7/2)*(c*x**2+b*x)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.16155, size = 95, normalized size = 0.7 \begin{align*} -\frac{256 \, b^{\frac{11}{2}}}{3465 \, c^{5}} + \frac{2 \,{\left (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}\right )}}{3465 \, c^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-256/3465*b^(11/2)/c^5 + 2/3465*(315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 27
72*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5