3.22 \(\int \frac{(b x+c x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=100 \[ \frac{32 c^3 \left (b x+c x^2\right )^{5/2}}{1155 b^4 x^5}-\frac{16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}+\frac{4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac{2 \left (b x+c x^2\right )^{5/2}}{11 b x^8} \]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(11*b*x^8) + (4*c*(b*x + c*x^2)^(5/2))/(33*b^2*x^7) - (16*c^2*(b*x + c*x^2)^(5/2))/(2
31*b^3*x^6) + (32*c^3*(b*x + c*x^2)^(5/2))/(1155*b^4*x^5)

________________________________________________________________________________________

Rubi [A]  time = 0.0411665, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {658, 650} \[ \frac{32 c^3 \left (b x+c x^2\right )^{5/2}}{1155 b^4 x^5}-\frac{16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}+\frac{4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac{2 \left (b x+c x^2\right )^{5/2}}{11 b x^8} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^8,x]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(11*b*x^8) + (4*c*(b*x + c*x^2)^(5/2))/(33*b^2*x^7) - (16*c^2*(b*x + c*x^2)^(5/2))/(2
31*b^3*x^6) + (32*c^3*(b*x + c*x^2)^(5/2))/(1155*b^4*x^5)

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), 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 + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{\left (b x+c x^2\right )^{3/2}}{x^8} \, dx &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}-\frac{(6 c) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^7} \, dx}{11 b}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac{4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}+\frac{\left (8 c^2\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{33 b^2}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac{4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac{16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}-\frac{\left (16 c^3\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{231 b^3}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{11 b x^8}+\frac{4 c \left (b x+c x^2\right )^{5/2}}{33 b^2 x^7}-\frac{16 c^2 \left (b x+c x^2\right )^{5/2}}{231 b^3 x^6}+\frac{32 c^3 \left (b x+c x^2\right )^{5/2}}{1155 b^4 x^5}\\ \end{align*}

Mathematica [A]  time = 0.0174287, size = 51, normalized size = 0.51 \[ \frac{2 (x (b+c x))^{5/2} \left (70 b^2 c x-105 b^3-40 b c^2 x^2+16 c^3 x^3\right )}{1155 b^4 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^8,x]

[Out]

(2*(x*(b + c*x))^(5/2)*(-105*b^3 + 70*b^2*c*x - 40*b*c^2*x^2 + 16*c^3*x^3))/(1155*b^4*x^8)

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 55, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -16\,{x}^{3}{c}^{3}+40\,b{x}^{2}{c}^{2}-70\,{b}^{2}xc+105\,{b}^{3} \right ) }{1155\,{x}^{7}{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^8,x)

[Out]

-2/1155*(c*x+b)*(-16*c^3*x^3+40*b*c^2*x^2-70*b^2*c*x+105*b^3)*(c*x^2+b*x)^(3/2)/x^7/b^4

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.96342, size = 161, normalized size = 1.61 \begin{align*} \frac{2 \,{\left (16 \, c^{5} x^{5} - 8 \, b c^{4} x^{4} + 6 \, b^{2} c^{3} x^{3} - 5 \, b^{3} c^{2} x^{2} - 140 \, b^{4} c x - 105 \, b^{5}\right )} \sqrt{c x^{2} + b x}}{1155 \, b^{4} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^8,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{3}{2}}}{x^{8}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**8,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**8, x)

________________________________________________________________________________________

Giac [B]  time = 1.27978, size = 301, normalized size = 3.01 \begin{align*} \frac{2 \,{\left (2310 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} c^{\frac{7}{2}} + 10164 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} b c^{3} + 19635 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} b^{2} c^{\frac{5}{2}} + 21285 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} b^{3} c^{2} + 13860 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} b^{4} c^{\frac{3}{2}} + 5390 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} b^{5} c + 1155 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{6} \sqrt{c} + 105 \, b^{7}\right )}}{1155 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^8,x, algorithm="giac")

[Out]

2/1155*(2310*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*c^(7/2) + 10164*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*b*c^3 + 19635
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^2*c^(5/2) + 21285*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^3*c^2 + 13860*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^3*b^4*c^(3/2) + 5390*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^5*c + 1155*(sqrt(c)*x - sq
rt(c*x^2 + b*x))*b^6*sqrt(c) + 105*b^7)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^11