3.23 \(\int \frac{(b x+c x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=126 \[ -\frac{256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}+\frac{128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac{32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac{16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9} \]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(13*b*x^9) + (16*c*(b*x + c*x^2)^(5/2))/(143*b^2*x^8) - (32*c^2*(b*x + c*x^2)^(5/2))/
(429*b^3*x^7) + (128*c^3*(b*x + c*x^2)^(5/2))/(3003*b^4*x^6) - (256*c^4*(b*x + c*x^2)^(5/2))/(15015*b^5*x^5)

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Rubi [A]  time = 0.0574311, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {658, 650} \[ -\frac{256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}+\frac{128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac{32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac{16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^9,x]

[Out]

(-2*(b*x + c*x^2)^(5/2))/(13*b*x^9) + (16*c*(b*x + c*x^2)^(5/2))/(143*b^2*x^8) - (32*c^2*(b*x + c*x^2)^(5/2))/
(429*b^3*x^7) + (128*c^3*(b*x + c*x^2)^(5/2))/(3003*b^4*x^6) - (256*c^4*(b*x + c*x^2)^(5/2))/(15015*b^5*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^9} \, dx &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}-\frac{(8 c) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^8} \, dx}{13 b}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac{16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}+\frac{\left (48 c^2\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^7} \, dx}{143 b^2}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac{16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac{32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}-\frac{\left (64 c^3\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{429 b^3}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac{16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac{32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac{128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}+\frac{\left (128 c^4\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{3003 b^4}\\ &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{13 b x^9}+\frac{16 c \left (b x+c x^2\right )^{5/2}}{143 b^2 x^8}-\frac{32 c^2 \left (b x+c x^2\right )^{5/2}}{429 b^3 x^7}+\frac{128 c^3 \left (b x+c x^2\right )^{5/2}}{3003 b^4 x^6}-\frac{256 c^4 \left (b x+c x^2\right )^{5/2}}{15015 b^5 x^5}\\ \end{align*}

Mathematica [A]  time = 0.0187573, size = 62, normalized size = 0.49 \[ -\frac{2 (x (b+c x))^{5/2} \left (560 b^2 c^2 x^2-840 b^3 c x+1155 b^4-320 b c^3 x^3+128 c^4 x^4\right )}{15015 b^5 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^9,x]

[Out]

(-2*(x*(b + c*x))^(5/2)*(1155*b^4 - 840*b^3*c*x + 560*b^2*c^2*x^2 - 320*b*c^3*x^3 + 128*c^4*x^4))/(15015*b^5*x
^9)

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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 ( 128\,{c}^{4}{x}^{4}-320\,{x}^{3}{c}^{3}b+560\,{c}^{2}{x}^{2}{b}^{2}-840\,cx{b}^{3}+1155\,{b}^{4} \right ) }{15015\,{x}^{8}{b}^{5}} \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^9,x)

[Out]

-2/15015*(c*x+b)*(128*c^4*x^4-320*b*c^3*x^3+560*b^2*c^2*x^2-840*b^3*c*x+1155*b^4)*(c*x^2+b*x)^(3/2)/x^8/b^5

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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^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92996, size = 194, normalized size = 1.54 \begin{align*} -\frac{2 \,{\left (128 \, c^{6} x^{6} - 64 \, b c^{5} x^{5} + 48 \, b^{2} c^{4} x^{4} - 40 \, b^{3} c^{3} x^{3} + 35 \, b^{4} c^{2} x^{2} + 1470 \, b^{5} c x + 1155 \, b^{6}\right )} \sqrt{c x^{2} + b x}}{15015 \, b^{5} x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^9,x, algorithm="fricas")

[Out]

-2/15015*(128*c^6*x^6 - 64*b*c^5*x^5 + 48*b^2*c^4*x^4 - 40*b^3*c^3*x^3 + 35*b^4*c^2*x^2 + 1470*b^5*c*x + 1155*
b^6)*sqrt(c*x^2 + b*x)/(b^5*x^7)

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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^{9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**9,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**9, x)

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Giac [B]  time = 1.21335, size = 340, normalized size = 2.7 \begin{align*} \frac{2 \,{\left (48048 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{8} c^{4} + 240240 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} b c^{\frac{7}{2}} + 531960 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} b^{2} c^{3} + 675675 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} b^{3} c^{\frac{5}{2}} + 535535 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} b^{4} c^{2} + 270270 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} b^{5} c^{\frac{3}{2}} + 84630 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} b^{6} c + 15015 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{7} \sqrt{c} + 1155 \, b^{8}\right )}}{15015 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{13}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^9,x, algorithm="giac")

[Out]

2/15015*(48048*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*c^4 + 240240*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*b*c^(7/2) + 53
1960*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*b^2*c^3 + 675675*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^3*c^(5/2) + 535535
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^4*c^2 + 270270*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^5*c^(3/2) + 84630*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^2*b^6*c + 15015*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^7*sqrt(c) + 1155*b^8)/(sqrt(c)*x
- sqrt(c*x^2 + b*x))^13