\(\int \frac {(c+d x^n)^{2-\frac {1}{n}}}{(a+b x^n)^4} \, dx\) [336]

Optimal result

Integrand size = 25, antiderivative size = 133 \[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\frac {b x \left (c+d x^n\right )^{3-\frac {1}{n}}}{3 a (b c-a d) n \left (a+b x^n\right )^3}-\frac {c^2 (b c (1-3 n)+3 a d n) x \left (c+d x^n\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{n},1+\frac {1}{n},-\frac {(b c-a d) x^n}{a \left (c+d x^n\right )}\right )}{3 a^4 (b c-a d) n} \]

[Out]

1/3*b*x*(c+d*x^n)^(3-1/n)/a/(-a*d+b*c)/n/(a+b*x^n)^3-1/3*c^2*(b*c*(1-3*n)+3*a*d*n)*x*hypergeom([3, 1/n],[1+1/n
],-(-a*d+b*c)*x^n/a/(c+d*x^n))/a^4/(-a*d+b*c)/n/((c+d*x^n)^(1/n))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {390, 387} \[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\frac {b x \left (c+d x^n\right )^{3-\frac {1}{n}}}{3 a n (b c-a d) \left (a+b x^n\right )^3}-\frac {c^2 x \left (c+d x^n\right )^{-1/n} (3 a d n+b c (1-3 n)) \operatorname {Hypergeometric2F1}\left (3,\frac {1}{n},1+\frac {1}{n},-\frac {(b c-a d) x^n}{a \left (d x^n+c\right )}\right )}{3 a^4 n (b c-a d)} \]

[In]

Int[(c + d*x^n)^(2 - n^(-1))/(a + b*x^n)^4,x]

[Out]

(b*x*(c + d*x^n)^(3 - n^(-1)))/(3*a*(b*c - a*d)*n*(a + b*x^n)^3) - (c^2*(b*c*(1 - 3*n) + 3*a*d*n)*x*Hypergeome
tric2F1[3, n^(-1), 1 + n^(-1), -(((b*c - a*d)*x^n)/(a*(c + d*x^n)))])/(3*a^4*(b*c - a*d)*n*(c + d*x^n)^n^(-1))

Rule 387

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*(x/(c^(p + 1)*(c + d*x^
n)^(1/n)))*Hypergeometric2F1[1/n, -p, 1 + 1/n, (-(b*c - a*d))*(x^n/(a*(c + d*x^n)))], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && ILtQ[p, 0]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b x \left (c+d x^n\right )^{3-\frac {1}{n}}}{3 a (b c-a d) n \left (a+b x^n\right )^3}-\frac {(b c-3 (b c-a d) n) \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^3} \, dx}{3 a (b c-a d) n} \\ & = \frac {b x \left (c+d x^n\right )^{3-\frac {1}{n}}}{3 a (b c-a d) n \left (a+b x^n\right )^3}-\frac {c^2 (b c (1-3 n)+3 a d n) x \left (c+d x^n\right )^{-1/n} \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {(b c-a d) x^n}{a \left (c+d x^n\right )}\right )}{3 a^4 (b c-a d) n} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 36.96 (sec) , antiderivative size = 6405, normalized size of antiderivative = 48.16 \[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\text {Result too large to show} \]

[In]

Integrate[(c + d*x^n)^(2 - n^(-1))/(a + b*x^n)^4,x]

[Out]

Result too large to show

Maple [F]

\[\int \frac {\left (c +d \,x^{n}\right )^{2-\frac {1}{n}}}{\left (a +b \,x^{n}\right )^{4}}d x\]

[In]

int((c+d*x^n)^(2-1/n)/(a+b*x^n)^4,x)

[Out]

int((c+d*x^n)^(2-1/n)/(a+b*x^n)^4,x)

Fricas [F]

\[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{-\frac {1}{n} + 2}}{{\left (b x^{n} + a\right )}^{4}} \,d x } \]

[In]

integrate((c+d*x^n)^(2-1/n)/(a+b*x^n)^4,x, algorithm="fricas")

[Out]

integral((d*x^n + c)^((2*n - 1)/n)/(b^4*x^(4*n) + 4*a*b^3*x^(3*n) + 6*a^2*b^2*x^(2*n) + 4*a^3*b*x^n + a^4), x)

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((c+d*x**n)**(2-1/n)/(a+b*x**n)**4,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{-\frac {1}{n} + 2}}{{\left (b x^{n} + a\right )}^{4}} \,d x } \]

[In]

integrate((c+d*x^n)^(2-1/n)/(a+b*x^n)^4,x, algorithm="maxima")

[Out]

integrate((d*x^n + c)^(-1/n + 2)/(b*x^n + a)^4, x)

Giac [F]

\[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{-\frac {1}{n} + 2}}{{\left (b x^{n} + a\right )}^{4}} \,d x } \]

[In]

integrate((c+d*x^n)^(2-1/n)/(a+b*x^n)^4,x, algorithm="giac")

[Out]

integrate((d*x^n + c)^(-1/n + 2)/(b*x^n + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^n\right )^{2-\frac {1}{n}}}{\left (a+b x^n\right )^4} \, dx=\int \frac {{\left (c+d\,x^n\right )}^{2-\frac {1}{n}}}{{\left (a+b\,x^n\right )}^4} \,d x \]

[In]

int((c + d*x^n)^(2 - 1/n)/(a + b*x^n)^4,x)

[Out]

int((c + d*x^n)^(2 - 1/n)/(a + b*x^n)^4, x)