\(\int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) [94]

Optimal result

Integrand size = 21, antiderivative size = 59 \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c \sqrt [3]{1+\frac {b x^3}{a}}} \]

[Out]

x*(b*x^3+a)^(1/3)*AppellF1(1/3,-1/3,1,4/3,-b*x^3/a,-d*x^3/c)/c/(1+b*x^3/a)^(1/3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {441, 440} \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c \sqrt [3]{\frac {b x^3}{a}+1}} \]

[In]

Int[(a + b*x^3)^(1/3)/(c + d*x^3),x]

[Out]

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

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}& = \frac {\sqrt [3]{a+b x^3} \int \frac {\sqrt [3]{1+\frac {b x^3}{a}}}{c+d x^3} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}} \\ & = \frac {x \sqrt [3]{a+b x^3} F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c \sqrt [3]{1+\frac {b x^3}{a}}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(160\) vs. \(2(59)=118\).

Time = 10.17 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.71 \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {4 a c x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{\left (c+d x^3\right ) \left (4 a c \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^3 \left (-3 a d \operatorname {AppellF1}\left (\frac {4}{3},-\frac {1}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )} \]

[In]

Integrate[(a + b*x^3)^(1/3)/(c + d*x^3),x]

[Out]

(4*a*c*x*(a + b*x^3)^(1/3)*AppellF1[1/3, -1/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/((c + d*x^3)*(4*a*c*Appell
F1[1/3, -1/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + x^3*(-3*a*d*AppellF1[4/3, -1/3, 2, 7/3, -((b*x^3)/a), -((d
*x^3)/c)] + b*c*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))

Maple [F]

\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{d \,x^{3}+c}d x\]

[In]

int((b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(1/3)/(d*x^3+c),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\text {Timed out} \]

[In]

integrate((b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int \frac {\sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \]

[In]

integrate((b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(1/3)/(c + d*x**3), x)

Maxima [F]

\[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c} \,d x } \]

[In]

integrate((b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)/(d*x^3 + c), x)

Giac [F]

\[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c} \,d x } \]

[In]

integrate((b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)/(d*x^3 + c), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/3}}{d\,x^3+c} \,d x \]

[In]

int((a + b*x^3)^(1/3)/(c + d*x^3),x)

[Out]

int((a + b*x^3)^(1/3)/(c + d*x^3), x)