3.140 \(\int (a+b x^3)^m (c+d x^3) \, dx\)

Optimal. Leaf size=93 \[ \frac {d x \left (a+b x^3\right )^{m+1}}{b (3 m+4)}-\frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} (a d-b c (3 m+4)) \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )}{b (3 m+4)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {388, 246, 245} \[ x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (c-\frac {a d}{3 b m+4 b}\right ) \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )+\frac {d x \left (a+b x^3\right )^{m+1}}{b (3 m+4)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^m*(c + d*x^3),x]

[Out]

(d*x*(a + b*x^3)^(1 + m))/(b*(4 + 3*m)) + ((c - (a*d)/(4*b + 3*b*m))*x*(a + b*x^3)^m*Hypergeometric2F1[1/3, -m
, 4/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^m

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 388

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

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^m \left (c+d x^3\right ) \, dx &=\frac {d x \left (a+b x^3\right )^{1+m}}{b (4+3 m)}-\left (-c+\frac {a d}{4 b+3 b m}\right ) \int \left (a+b x^3\right )^m \, dx\\ &=\frac {d x \left (a+b x^3\right )^{1+m}}{b (4+3 m)}-\left (\left (-c+\frac {a d}{4 b+3 b m}\right ) \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m}\right ) \int \left (1+\frac {b x^3}{a}\right )^m \, dx\\ &=\frac {d x \left (a+b x^3\right )^{1+m}}{b (4+3 m)}+\left (c-\frac {a d}{4 b+3 b m}\right ) x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 90, normalized size = 0.97 \[ \frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left ((b c (3 m+4)-a d) \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )+d \left (a+b x^3\right ) \left (\frac {b x^3}{a}+1\right )^m\right )}{b (3 m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^3)^m*(d*(a + b*x^3)*(1 + (b*x^3)/a)^m + (-(a*d) + b*c*(4 + 3*m))*Hypergeometric2F1[1/3, -m, 4/3, -
((b*x^3)/a)]))/(b*(4 + 3*m)*(1 + (b*x^3)/a)^m)

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x^3 + c)*(b*x^3 + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \left (d \,x^{3}+c \right ) \left (b \,x^{3}+a \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{3} + c\right )} {\left (b x^{3} + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^m\,\left (d\,x^3+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^m*(c + d*x^3),x)

[Out]

int((a + b*x^3)^m*(c + d*x^3), x)

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sympy [C]  time = 100.31, size = 75, normalized size = 0.81 \[ \frac {a^{m} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - m \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{m} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - m \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**m*c*x*gamma(1/3)*hyper((1/3, -m), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + a**m*d*x**4*gamma(4/3)
*hyper((4/3, -m), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))

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