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

Optimal. Leaf size=57 \[ \frac{x \left (c+d x^3\right )^q \left (\frac{d x^3}{c}+1\right )^{-q} F_1\left (\frac{1}{3};2,-q;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2} \]

[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)

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Rubi [A]  time = 0.0273022, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {430, 429} \[ \frac{x \left (c+d x^3\right )^q \left (\frac{d x^3}{c}+1\right )^{-q} F_1\left (\frac{1}{3};2,-q;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2} \]

Antiderivative was successfully verified.

[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 430

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])

Rule 429

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])

Rubi steps

\begin{align*} \int \frac{\left (c+d x^3\right )^q}{\left (a+b x^3\right )^2} \, dx &=\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]  time = 0.193763, size = 162, normalized size = 2.84 \[ \frac{4 a c x \left (c+d x^3\right )^q F_1\left (\frac{1}{3};2,-q;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{\left (a+b x^3\right )^2 \left (3 x^3 \left (a d q F_1\left (\frac{4}{3};2,1-q;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )-2 b c F_1\left (\frac{4}{3};3,-q;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )+4 a c F_1\left (\frac{1}{3};2,-q;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[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)])))

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Maple [F]  time = 0.443, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d{x}^{3}+c \right ) ^{q}}{ \left ( b{x}^{3}+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{q}}{{\left (b x^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d x^{3} + c\right )}^{q}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{q}}{{\left (b x^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)