### 3.35 $$\int \frac{(a x+b x^2)^{5/2}}{x^9} \, dx$$

Optimal. Leaf size=74 $-\frac{16 b^2 \left (a x+b x^2\right )^{7/2}}{693 a^3 x^7}+\frac{8 b \left (a x+b x^2\right )^{7/2}}{99 a^2 x^8}-\frac{2 \left (a x+b x^2\right )^{7/2}}{11 a x^9}$

[Out]

(-2*(a*x + b*x^2)^(7/2))/(11*a*x^9) + (8*b*(a*x + b*x^2)^(7/2))/(99*a^2*x^8) - (16*b^2*(a*x + b*x^2)^(7/2))/(6
93*a^3*x^7)

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Rubi [A]  time = 0.0281318, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {658, 650} $-\frac{16 b^2 \left (a x+b x^2\right )^{7/2}}{693 a^3 x^7}+\frac{8 b \left (a x+b x^2\right )^{7/2}}{99 a^2 x^8}-\frac{2 \left (a x+b x^2\right )^{7/2}}{11 a x^9}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*x + b*x^2)^(5/2)/x^9,x]

[Out]

(-2*(a*x + b*x^2)^(7/2))/(11*a*x^9) + (8*b*(a*x + b*x^2)^(7/2))/(99*a^2*x^8) - (16*b^2*(a*x + b*x^2)^(7/2))/(6
93*a^3*x^7)

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 (a x+b x^2\right )^{5/2}}{x^9} \, dx &=-\frac{2 \left (a x+b x^2\right )^{7/2}}{11 a x^9}-\frac{(4 b) \int \frac{\left (a x+b x^2\right )^{5/2}}{x^8} \, dx}{11 a}\\ &=-\frac{2 \left (a x+b x^2\right )^{7/2}}{11 a x^9}+\frac{8 b \left (a x+b x^2\right )^{7/2}}{99 a^2 x^8}+\frac{\left (8 b^2\right ) \int \frac{\left (a x+b x^2\right )^{5/2}}{x^7} \, dx}{99 a^2}\\ &=-\frac{2 \left (a x+b x^2\right )^{7/2}}{11 a x^9}+\frac{8 b \left (a x+b x^2\right )^{7/2}}{99 a^2 x^8}-\frac{16 b^2 \left (a x+b x^2\right )^{7/2}}{693 a^3 x^7}\\ \end{align*}

Mathematica [A]  time = 0.0126369, size = 47, normalized size = 0.64 $-\frac{2 (a+b x)^3 \sqrt{x (a+b x)} \left (63 a^2-28 a b x+8 b^2 x^2\right )}{693 a^3 x^6}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^9,x]

[Out]

(-2*(a + b*x)^3*Sqrt[x*(a + b*x)]*(63*a^2 - 28*a*b*x + 8*b^2*x^2))/(693*a^3*x^6)

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Maple [A]  time = 0.045, size = 44, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 8\,{b}^{2}{x}^{2}-28\,abx+63\,{a}^{2} \right ) }{693\,{x}^{8}{a}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a*x)^(5/2)/x^9,x)

[Out]

-2/693*(b*x+a)*(8*b^2*x^2-28*a*b*x+63*a^2)*(b*x^2+a*x)^(5/2)/x^8/a^3

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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((b*x^2+a*x)^(5/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.04175, size = 161, normalized size = 2.18 \begin{align*} -\frac{2 \,{\left (8 \, b^{5} x^{5} - 4 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{3} + 113 \, a^{3} b^{2} x^{2} + 161 \, a^{4} b x + 63 \, a^{5}\right )} \sqrt{b x^{2} + a x}}{693 \, a^{3} x^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^9,x, algorithm="fricas")

[Out]

-2/693*(8*b^5*x^5 - 4*a*b^4*x^4 + 3*a^2*b^3*x^3 + 113*a^3*b^2*x^2 + 161*a^4*b*x + 63*a^5)*sqrt(b*x^2 + a*x)/(a
^3*x^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (a + b x\right )\right )^{\frac{5}{2}}}{x^{9}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a*x)**(5/2)/x**9,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**9, x)

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Giac [B]  time = 1.36916, size = 340, normalized size = 4.59 \begin{align*} \frac{2 \,{\left (924 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{8} b^{4} + 4851 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{7} a b^{\frac{7}{2}} + 11781 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{6} a^{2} b^{3} + 16863 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{5} a^{3} b^{\frac{5}{2}} + 15345 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{4} a^{4} b^{2} + 9009 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{3} a^{5} b^{\frac{3}{2}} + 3311 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{2} a^{6} b + 693 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} a^{7} \sqrt{b} + 63 \, a^{8}\right )}}{693 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{11}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a*x)^(5/2)/x^9,x, algorithm="giac")

[Out]

2/693*(924*(sqrt(b)*x - sqrt(b*x^2 + a*x))^8*b^4 + 4851*(sqrt(b)*x - sqrt(b*x^2 + a*x))^7*a*b^(7/2) + 11781*(s
qrt(b)*x - sqrt(b*x^2 + a*x))^6*a^2*b^3 + 16863*(sqrt(b)*x - sqrt(b*x^2 + a*x))^5*a^3*b^(5/2) + 15345*(sqrt(b)
*x - sqrt(b*x^2 + a*x))^4*a^4*b^2 + 9009*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a^5*b^(3/2) + 3311*(sqrt(b)*x - sqr
t(b*x^2 + a*x))^2*a^6*b + 693*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^7*sqrt(b) + 63*a^8)/(sqrt(b)*x - sqrt(b*x^2 +
a*x))^11