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

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Rubi [A]  time = 0.22, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

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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fricas [A]  time = 0.64, size = 132, normalized size = 1.39 \[ -\frac {2 \, {\left ({\left (b^{2} - 4 \, a c\right )} d^{\frac {1}{2} \, n - 1} h x^{\frac {1}{2} \, n} + {\left (2 \, c^{2} f - b c g\right )} d^{\frac {1}{2} \, n - 1} x^{n} + {\left (b c f - 2 \, a c g\right )} d^{\frac {1}{2} \, n - 1}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{{\left (b^{2} c - 4 \, a c^{2}\right )} n x^{2 \, n} + {\left (b^{3} - 4 \, a b c\right )} n x^{n} + {\left (a b^{2} - 4 \, a^{2} c\right )} n} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c h x^{2 \, n} + c g x^{\frac {3}{2} \, n} + c f x^{\frac {1}{2} \, n} - a h\right )} \left (d x\right )^{\frac {1}{2} \, n - 1}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (c f \,x^{\frac {n}{2}}+c g \,x^{\frac {3 n}{2}}+c h \,x^{2 n}-a h \right ) \left (d x \right )^{\frac {n}{2}-1}}{\left (b \,x^{n}+c \,x^{2 n}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c h x^{2 \, n} + c g x^{\frac {3}{2} \, n} + c f x^{\frac {1}{2} \, n} - a h\right )} \left (d x\right )^{\frac {1}{2} \, n - 1}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x\right )}^{\frac {n}{2}-1}\,\left (c\,f\,x^{n/2}-a\,h+c\,g\,x^{\frac {3\,n}{2}}+c\,h\,x^{2\,n}\right )}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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