3.961 \(\int \frac{x^8 (a+b x)^n}{(c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{a^3 x (a+b x)^{n+1}}{b^4 c^2 (n+1) \sqrt{c x^2}}+\frac{3 a^2 x (a+b x)^{n+2}}{b^4 c^2 (n+2) \sqrt{c x^2}}-\frac{3 a x (a+b x)^{n+3}}{b^4 c^2 (n+3) \sqrt{c x^2}}+\frac{x (a+b x)^{n+4}}{b^4 c^2 (n+4) \sqrt{c x^2}} \]

[Out]

-((a^3*x*(a + b*x)^(1 + n))/(b^4*c^2*(1 + n)*Sqrt[c*x^2])) + (3*a^2*x*(a + b*x)^(2 + n))/(b^4*c^2*(2 + n)*Sqrt
[c*x^2]) - (3*a*x*(a + b*x)^(3 + n))/(b^4*c^2*(3 + n)*Sqrt[c*x^2]) + (x*(a + b*x)^(4 + n))/(b^4*c^2*(4 + n)*Sq
rt[c*x^2])

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Rubi [A]  time = 0.0460681, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 43} \[ -\frac{a^3 x (a+b x)^{n+1}}{b^4 c^2 (n+1) \sqrt{c x^2}}+\frac{3 a^2 x (a+b x)^{n+2}}{b^4 c^2 (n+2) \sqrt{c x^2}}-\frac{3 a x (a+b x)^{n+3}}{b^4 c^2 (n+3) \sqrt{c x^2}}+\frac{x (a+b x)^{n+4}}{b^4 c^2 (n+4) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(a + b*x)^n)/(c*x^2)^(5/2),x]

[Out]

-((a^3*x*(a + b*x)^(1 + n))/(b^4*c^2*(1 + n)*Sqrt[c*x^2])) + (3*a^2*x*(a + b*x)^(2 + n))/(b^4*c^2*(2 + n)*Sqrt
[c*x^2]) - (3*a*x*(a + b*x)^(3 + n))/(b^4*c^2*(3 + n)*Sqrt[c*x^2]) + (x*(a + b*x)^(4 + n))/(b^4*c^2*(4 + n)*Sq
rt[c*x^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^8 (a+b x)^n}{\left (c x^2\right )^{5/2}} \, dx &=\frac{x \int x^3 (a+b x)^n \, dx}{c^2 \sqrt{c x^2}}\\ &=\frac{x \int \left (-\frac{a^3 (a+b x)^n}{b^3}+\frac{3 a^2 (a+b x)^{1+n}}{b^3}-\frac{3 a (a+b x)^{2+n}}{b^3}+\frac{(a+b x)^{3+n}}{b^3}\right ) \, dx}{c^2 \sqrt{c x^2}}\\ &=-\frac{a^3 x (a+b x)^{1+n}}{b^4 c^2 (1+n) \sqrt{c x^2}}+\frac{3 a^2 x (a+b x)^{2+n}}{b^4 c^2 (2+n) \sqrt{c x^2}}-\frac{3 a x (a+b x)^{3+n}}{b^4 c^2 (3+n) \sqrt{c x^2}}+\frac{x (a+b x)^{4+n}}{b^4 c^2 (4+n) \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0331838, size = 99, normalized size = 0.73 \[ \frac{x (a+b x)^{n+1} \left (6 a^2 b (n+1) x-6 a^3-3 a b^2 \left (n^2+3 n+2\right ) x^2+b^3 \left (n^3+6 n^2+11 n+6\right ) x^3\right )}{b^4 c^2 (n+1) (n+2) (n+3) (n+4) \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(a + b*x)^n)/(c*x^2)^(5/2),x]

[Out]

(x*(a + b*x)^(1 + n)*(-6*a^3 + 6*a^2*b*(1 + n)*x - 3*a*b^2*(2 + 3*n + n^2)*x^2 + b^3*(6 + 11*n + 6*n^2 + n^3)*
x^3))/(b^4*c^2*(1 + n)*(2 + n)*(3 + n)*(4 + n)*Sqrt[c*x^2])

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Maple [A]  time = 0.005, size = 136, normalized size = 1. \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+n}{x}^{5} \left ( -{b}^{3}{n}^{3}{x}^{3}-6\,{b}^{3}{n}^{2}{x}^{3}+3\,a{b}^{2}{n}^{2}{x}^{2}-11\,{b}^{3}n{x}^{3}+9\,a{b}^{2}n{x}^{2}-6\,{b}^{3}{x}^{3}-6\,{a}^{2}bnx+6\,a{b}^{2}{x}^{2}-6\,{a}^{2}bx+6\,{a}^{3} \right ) }{{b}^{4} \left ({n}^{4}+10\,{n}^{3}+35\,{n}^{2}+50\,n+24 \right ) } \left ( c{x}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b*x+a)^n/(c*x^2)^(5/2),x)

[Out]

-(b*x+a)^(1+n)*x^5*(-b^3*n^3*x^3-6*b^3*n^2*x^3+3*a*b^2*n^2*x^2-11*b^3*n*x^3+9*a*b^2*n*x^2-6*b^3*x^3-6*a^2*b*n*
x+6*a*b^2*x^2-6*a^2*b*x+6*a^3)/(c*x^2)^(5/2)/b^4/(n^4+10*n^3+35*n^2+50*n+24)

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Maxima [A]  time = 1.03958, size = 140, normalized size = 1.04 \begin{align*} \frac{{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} +{\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \,{\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )}{\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4} c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x+a)^n/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

((n^3 + 6*n^2 + 11*n + 6)*b^4*x^4 + (n^3 + 3*n^2 + 2*n)*a*b^3*x^3 - 3*(n^2 + n)*a^2*b^2*x^2 + 6*a^3*b*n*x - 6*
a^4)*(b*x + a)^n/((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^4*c^(5/2))

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Fricas [A]  time = 1.4507, size = 340, normalized size = 2.52 \begin{align*} \frac{{\left (6 \, a^{3} b n x +{\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} +{\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \,{\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \sqrt{c x^{2}}{\left (b x + a\right )}^{n}}{{\left (b^{4} c^{3} n^{4} + 10 \, b^{4} c^{3} n^{3} + 35 \, b^{4} c^{3} n^{2} + 50 \, b^{4} c^{3} n + 24 \, b^{4} c^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x+a)^n/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

(6*a^3*b*n*x + (b^4*n^3 + 6*b^4*n^2 + 11*b^4*n + 6*b^4)*x^4 - 6*a^4 + (a*b^3*n^3 + 3*a*b^3*n^2 + 2*a*b^3*n)*x^
3 - 3*(a^2*b^2*n^2 + a^2*b^2*n)*x^2)*sqrt(c*x^2)*(b*x + a)^n/((b^4*c^3*n^4 + 10*b^4*c^3*n^3 + 35*b^4*c^3*n^2 +
50*b^4*c^3*n + 24*b^4*c^3)*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**8*(b*x+a)**n/(c*x**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{n} x^{8}}{\left (c x^{2}\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x+a)^n/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)^n*x^8/(c*x^2)^(5/2), x)