\(\int \frac {(c+d x^4)^q}{(a+b x^4)^2} \, dx\) [222]

Optimal result

Integrand size = 19, antiderivative size = 57 \[ \int \frac {\left (c+d x^4\right )^q}{\left (a+b x^4\right )^2} \, dx=\frac {x \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{4},2,-q,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2} \]

[Out]

x*(d*x^4+c)^q*AppellF1(1/4,2,-q,5/4,-b*x^4/a,-d*x^4/c)/a^2/((1+d*x^4/c)^q)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {441, 440} \[ \int \frac {\left (c+d x^4\right )^q}{\left (a+b x^4\right )^2} \, dx=\frac {x \left (c+d x^4\right )^q \left (\frac {d x^4}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{4},2,-q,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2} \]

[In]

Int[(c + d*x^4)^q/(a + b*x^4)^2,x]

[Out]

(x*(c + d*x^4)^q*AppellF1[1/4, 2, -q, 5/4, -((b*x^4)/a), -((d*x^4)/c)])/(a^2*(1 + (d*x^4)/c)^q)

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rubi steps \begin{align*} \text {integral}& = \left (\left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q}\right ) \int \frac {\left (1+\frac {d x^4}{c}\right )^q}{\left (a+b x^4\right )^2} \, dx \\ & = \frac {x \left (c+d x^4\right )^q \left (1+\frac {d x^4}{c}\right )^{-q} F_1\left (\frac {1}{4};2,-q;\frac {5}{4};-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^2} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(162\) vs. \(2(57)=114\).

Time = 0.51 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.84 \[ \int \frac {\left (c+d x^4\right )^q}{\left (a+b x^4\right )^2} \, dx=\frac {5 a c x \left (c+d x^4\right )^q \operatorname {AppellF1}\left (\frac {1}{4},2,-q,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{\left (a+b x^4\right )^2 \left (5 a c \operatorname {AppellF1}\left (\frac {1}{4},2,-q,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+4 x^4 \left (a d q \operatorname {AppellF1}\left (\frac {5}{4},2,1-q,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-2 b c \operatorname {AppellF1}\left (\frac {5}{4},3,-q,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )} \]

[In]

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

[Out]

(5*a*c*x*(c + d*x^4)^q*AppellF1[1/4, 2, -q, 5/4, -((b*x^4)/a), -((d*x^4)/c)])/((a + b*x^4)^2*(5*a*c*AppellF1[1
/4, 2, -q, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + 4*x^4*(a*d*q*AppellF1[5/4, 2, 1 - q, 9/4, -((b*x^4)/a), -((d*x^4
)/c)] - 2*b*c*AppellF1[5/4, 3, -q, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))

Maple [F]

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

[In]

int((d*x^4+c)^q/(b*x^4+a)^2,x)

[Out]

int((d*x^4+c)^q/(b*x^4+a)^2,x)

Fricas [F]

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

[In]

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

[Out]

integral((d*x^4 + c)^q/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

Sympy [F(-1)]

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

[In]

integrate((d*x**4+c)**q/(b*x**4+a)**2,x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate((d*x^4 + c)^q/(b*x^4 + a)^2, x)

Giac [F]

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

[In]

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

[Out]

integrate((d*x^4 + c)^q/(b*x^4 + a)^2, x)

Mupad [F(-1)]

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

[In]

int((c + d*x^4)^q/(a + b*x^4)^2,x)

[Out]

int((c + d*x^4)^q/(a + b*x^4)^2, x)