3.15 \(\int \frac{(d x)^{-1+\frac{n}{2}} (-a h+c f x^{n/2}+c g x^{3 n/2}+c h x^{2 n})}{(a+b x^n+c x^{2 n})^{3/2}} \, dx\)

Optimal. Leaf size=95 \[ -\frac{2 x^{1-\frac{n}{2}} (d x)^{\frac{n-2}{2}} \left (h x^{n/2} \left (b^2-4 a c\right )+c (b f-2 a g)+c x^n (2 c f-b g)\right )}{n \left (b^2-4 a c\right ) \sqrt{a+b x^n+c x^{2 n}}} \]

[Out]

(-2*x^(1 - n/2)*(d*x)^((-2 + n)/2)*(c*(b*f - 2*a*g) + (b^2 - 4*a*c)*h*x^(n/2) + c*(2*c*f - b*g)*x^n))/((b^2 -
4*a*c)*n*Sqrt[a + b*x^n + c*x^(2*n)])

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Rubi [A]  time = 0.217227, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 63, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.032, Rules used = {1754, 1753} \[ -\frac{2 x^{1-\frac{n}{2}} (d x)^{\frac{n-2}{2}} \left (h x^{n/2} \left (b^2-4 a c\right )+c (b f-2 a g)+c x^n (2 c f-b g)\right )}{n \left (b^2-4 a c\right ) \sqrt{a+b x^n+c x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[((d*x)^(-1 + n/2)*(-(a*h) + c*f*x^(n/2) + c*g*x^((3*n)/2) + c*h*x^(2*n)))/(a + b*x^n + c*x^(2*n))^(3/2),x]

[Out]

(-2*x^(1 - n/2)*(d*x)^((-2 + n)/2)*(c*(b*f - 2*a*g) + (b^2 - 4*a*c)*h*x^(n/2) + c*(2*c*f - b*g)*x^n))/((b^2 -
4*a*c)*n*Sqrt[a + b*x^n + c*x^(2*n)])

Rule 1754

Int[(((d_)*(x_))^(m_.)*((e_) + (f_.)*(x_)^(q_.) + (g_.)*(x_)^(r_.) + (h_.)*(x_)^(s_.)))/((a_) + (b_.)*(x_)^(n_
.) + (c_.)*(x_)^(n2_.))^(3/2), x_Symbol] :> Dist[(d*x)^m/x^m, Int[(x^m*(e + f*x^(n/2) + g*x^((3*n)/2) + h*x^(2
*n)))/(a + b*x^n + c*x^(2*n))^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[n2, 2*n] && EqQ
[q, n/2] && EqQ[r, (3*n)/2] && EqQ[s, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*m - n + 2, 0] && EqQ[c*e + a*h, 0]

Rule 1753

Int[((x_)^(m_.)*((e_) + (f_.)*(x_)^(q_.) + (g_.)*(x_)^(r_.) + (h_.)*(x_)^(s_.)))/((a_) + (b_.)*(x_)^(n_.) + (c
_.)*(x_)^(n2_.))^(3/2), x_Symbol] :> -Simp[(2*c*(b*f - 2*a*g) + 2*h*(b^2 - 4*a*c)*x^(n/2) + 2*c*(2*c*f - b*g)*
x^n)/(c*n*(b^2 - 4*a*c)*Sqrt[a + b*x^n + c*x^(2*n)]), x] /; FreeQ[{a, b, c, e, f, g, h, m, n}, x] && EqQ[n2, 2
*n] && EqQ[q, n/2] && EqQ[r, (3*n)/2] && EqQ[s, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*m - n + 2, 0] && EqQ[c*e
+ a*h, 0]

Rubi steps

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

Mathematica [F]  time = 0, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[((d*x)^(-1 + n/2)*(-(a*h) + c*f*x^(n/2) + c*g*x^((3*n)/2) + c*h*x^(2*n)))/(a + b*x^n + c*x^(2*n))^(3
/2),x]

[Out]

$Aborted

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Maple [F]  time = 0.022, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx \right ) ^{-1+{\frac{n}{2}}} \left ( -ah+cf{x}^{{\frac{n}{2}}}+cg{x}^{{\frac{3\,n}{2}}}+ch{x}^{2\,n} \right ) \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(-1+1/2*n)*(-a*h+c*f*x^(1/2*n)+c*g*x^(3/2*n)+c*h*x^(2*n))/(a+b*x^n+c*x^(2*n))^(3/2),x)

[Out]

int((d*x)^(-1+1/2*n)*(-a*h+c*f*x^(1/2*n)+c*g*x^(3/2*n)+c*h*x^(2*n))/(a+b*x^n+c*x^(2*n))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c h x^{2 \, n} + c g x^{\frac{3}{2} \, n} + c f x^{\frac{1}{2} \, n} - a h\right )} \left (d x\right )^{\frac{1}{2} \, n - 1}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(-1+1/2*n)*(-a*h+c*f*x^(1/2*n)+c*g*x^(3/2*n)+c*h*x^(2*n))/(a+b*x^n+c*x^(2*n))^(3/2),x, algorit
hm="maxima")

[Out]

integrate((c*h*x^(2*n) + c*g*x^(3/2*n) + c*f*x^(1/2*n) - a*h)*(d*x)^(1/2*n - 1)/(c*x^(2*n) + b*x^n + a)^(3/2),
x)

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Fricas [A]  time = 1.4377, size = 300, normalized size = 3.16 \begin{align*} -\frac{2 \,{\left ({\left (b^{2} - 4 \, a c\right )} d^{\frac{1}{2} \, n - 1} h x^{\frac{1}{2} \, n} +{\left (2 \, c^{2} f - b c g\right )} d^{\frac{1}{2} \, n - 1} x^{n} +{\left (b c f - 2 \, a c g\right )} d^{\frac{1}{2} \, n - 1}\right )} \sqrt{c x^{2 \, n} + b x^{n} + a}}{{\left (b^{2} c - 4 \, a c^{2}\right )} n x^{2 \, n} +{\left (b^{3} - 4 \, a b c\right )} n x^{n} +{\left (a b^{2} - 4 \, a^{2} c\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(-1+1/2*n)*(-a*h+c*f*x^(1/2*n)+c*g*x^(3/2*n)+c*h*x^(2*n))/(a+b*x^n+c*x^(2*n))^(3/2),x, algorit
hm="fricas")

[Out]

-2*((b^2 - 4*a*c)*d^(1/2*n - 1)*h*x^(1/2*n) + (2*c^2*f - b*c*g)*d^(1/2*n - 1)*x^n + (b*c*f - 2*a*c*g)*d^(1/2*n
- 1))*sqrt(c*x^(2*n) + b*x^n + a)/((b^2*c - 4*a*c^2)*n*x^(2*n) + (b^3 - 4*a*b*c)*n*x^n + (a*b^2 - 4*a^2*c)*n)

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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((d*x)**(-1+1/2*n)*(-a*h+c*f*x**(1/2*n)+c*g*x**(3/2*n)+c*h*x**(2*n))/(a+b*x**n+c*x**(2*n))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c h x^{2 \, n} + c g x^{\frac{3}{2} \, n} + c f x^{\frac{1}{2} \, n} - a h\right )} \left (d x\right )^{\frac{1}{2} \, n - 1}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(-1+1/2*n)*(-a*h+c*f*x^(1/2*n)+c*g*x^(3/2*n)+c*h*x^(2*n))/(a+b*x^n+c*x^(2*n))^(3/2),x, algorit
hm="giac")

[Out]

integrate((c*h*x^(2*n) + c*g*x^(3/2*n) + c*f*x^(1/2*n) - a*h)*(d*x)^(1/2*n - 1)/(c*x^(2*n) + b*x^n + a)^(3/2),
x)