\(\int \frac {(c+d x^3)^q}{(a+b x^3)^2} \, dx\) [137]

Optimal result

Integrand size = 19, antiderivative size = 57 \[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\frac {x \left (c+d x^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{3},2,-q,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2} \]

[Out]

x*(d*x^3+c)^q*AppellF1(1/3,2,-q,4/3,-b*x^3/a,-d*x^3/c)/a^2/((1+d*x^3/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} \[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\frac {x \left (c+d x^3\right )^q \left (\frac {d x^3}{c}+1\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{3},2,-q,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2} \]

[In]

Int[(c + d*x^3)^q/(a + b*x^3)^2,x]

[Out]

(x*(c + d*x^3)^q*AppellF1[1/3, 2, -q, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a^2*(1 + (d*x^3)/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^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q}\right ) \int \frac {\left (1+\frac {d x^3}{c}\right )^q}{\left (a+b x^3\right )^2} \, dx \\ & = \frac {x \left (c+d x^3\right )^q \left (1+\frac {d x^3}{c}\right )^{-q} F_1\left (\frac {1}{3};2,-q;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{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 = 0.44 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.84 \[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\frac {4 a c x \left (c+d x^3\right )^q \operatorname {AppellF1}\left (\frac {1}{3},2,-q,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{\left (a+b x^3\right )^2 \left (4 a c \operatorname {AppellF1}\left (\frac {1}{3},2,-q,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+3 x^3 \left (a d q \operatorname {AppellF1}\left (\frac {4}{3},2,1-q,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-2 b c \operatorname {AppellF1}\left (\frac {4}{3},3,-q,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )} \]

[In]

Integrate[(c + d*x^3)^q/(a + b*x^3)^2,x]

[Out]

(4*a*c*x*(c + d*x^3)^q*AppellF1[1/3, 2, -q, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/((a + b*x^3)^2*(4*a*c*AppellF1[1
/3, 2, -q, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + 3*x^3*(a*d*q*AppellF1[4/3, 2, 1 - q, 7/3, -((b*x^3)/a), -((d*x^3
)/c)] - 2*b*c*AppellF1[4/3, 3, -q, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))

Maple [F]

\[\int \frac {\left (d \,x^{3}+c \right )^{q}}{\left (b \,x^{3}+a \right )^{2}}d x\]

[In]

int((d*x^3+c)^q/(b*x^3+a)^2,x)

[Out]

int((d*x^3+c)^q/(b*x^3+a)^2,x)

Fricas [F]

\[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{q}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \]

[In]

integrate((d*x^3+c)^q/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

integral((d*x^3 + c)^q/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \]

[In]

integrate((d*x**3+c)**q/(b*x**3+a)**2,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{q}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \]

[In]

integrate((d*x^3+c)^q/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^q/(b*x^3 + a)^2, x)

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{q}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \]

[In]

integrate((d*x^3+c)^q/(b*x^3+a)^2,x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^q/(b*x^3 + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx=\int \frac {{\left (d\,x^3+c\right )}^q}{{\left (b\,x^3+a\right )}^2} \,d x \]

[In]

int((c + d*x^3)^q/(a + b*x^3)^2,x)

[Out]

int((c + d*x^3)^q/(a + b*x^3)^2, x)