3.2.39 \(\int (a+b x^3)^m (c+d x^3)^2 \, dx\) [139]

Optimal. Leaf size=176 \[ -\frac {d (4 a d-b c (10+3 m)) x \left (a+b x^3\right )^{1+m}}{b^2 (4+3 m) (7+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b (7+3 m)}+\frac {\left (4 a^2 d^2-2 a b c d (7+3 m)+b^2 c^2 \left (28+33 m+9 m^2\right )\right ) x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )}{b^2 (4+3 m) (7+3 m)} \]

[Out]

-d*(4*a*d-b*c*(10+3*m))*x*(b*x^3+a)^(1+m)/b^2/(9*m^2+33*m+28)+d*x*(b*x^3+a)^(1+m)*(d*x^3+c)/b/(7+3*m)+(4*a^2*d
^2-2*a*b*c*d*(7+3*m)+b^2*c^2*(9*m^2+33*m+28))*x*(b*x^3+a)^m*hypergeom([1/3, -m],[4/3],-b*x^3/a)/b^2/(9*m^2+33*
m+28)/((1+b*x^3/a)^m)

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Rubi [A]
time = 0.09, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {427, 396, 252, 251} \begin {gather*} \frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (4 a^2 d^2-2 a b c d (3 m+7)+b^2 c^2 \left (9 m^2+33 m+28\right )\right ) \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )}{b^2 (3 m+4) (3 m+7)}-\frac {d x \left (a+b x^3\right )^{m+1} (4 a d-b c (3 m+10))}{b^2 (3 m+4) (3 m+7)}+\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1}}{b (3 m+7)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^m*(c + d*x^3)^2,x]

[Out]

-((d*(4*a*d - b*c*(10 + 3*m))*x*(a + b*x^3)^(1 + m))/(b^2*(4 + 3*m)*(7 + 3*m))) + (d*x*(a + b*x^3)^(1 + m)*(c
+ d*x^3))/(b*(7 + 3*m)) + ((4*a^2*d^2 - 2*a*b*c*d*(7 + 3*m) + b^2*c^2*(28 + 33*m + 9*m^2))*x*(a + b*x^3)^m*Hyp
ergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)])/(b^2*(4 + 3*m)*(7 + 3*m)*(1 + (b*x^3)/a)^m)

Rule 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 396

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]

Rule 427

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*x*(a + b*x^n)^(p + 1)*((c
+ d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^m \left (c+d x^3\right )^2 \, dx &=\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b (7+3 m)}+\frac {\int \left (a+b x^3\right )^m \left (-c (a d-b c (7+3 m))-d (4 a d-b c (10+3 m)) x^3\right ) \, dx}{b (7+3 m)}\\ &=-\frac {d (4 a d-b c (10+3 m)) x \left (a+b x^3\right )^{1+m}}{b^2 (4+3 m) (7+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b (7+3 m)}+\frac {\left (4 a^2 d^2-2 a b c d (7+3 m)+b^2 c^2 \left (28+33 m+9 m^2\right )\right ) \int \left (a+b x^3\right )^m \, dx}{b^2 (4+3 m) (7+3 m)}\\ &=-\frac {d (4 a d-b c (10+3 m)) x \left (a+b x^3\right )^{1+m}}{b^2 (4+3 m) (7+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b (7+3 m)}+\frac {\left (\left (4 a^2 d^2-2 a b c d (7+3 m)+b^2 c^2 \left (28+33 m+9 m^2\right )\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}{b^2 (4+3 m) (7+3 m)}\\ &=-\frac {d (4 a d-b c (10+3 m)) x \left (a+b x^3\right )^{1+m}}{b^2 (4+3 m) (7+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b (7+3 m)}+\frac {\left (4 a^2 d^2-2 a b c d (7+3 m)+b^2 c^2 \left (28+33 m+9 m^2\right )\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 )}{b^2 (4+3 m) (7+3 m)}\\ \end {align*}

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Mathematica [A]
time = 5.68, size = 106, normalized size = 0.60 \begin {gather*} \frac {1}{14} x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \left (14 c^2 \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )+d x^3 \left (7 c \, _2F_1\left (\frac {4}{3},-m;\frac {7}{3};-\frac {b x^3}{a}\right )+2 d x^3 \, _2F_1\left (\frac {7}{3},-m;\frac {10}{3};-\frac {b x^3}{a}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^3)^m*(14*c^2*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)] + d*x^3*(7*c*Hypergeometric2F1[4/3, -m,
7/3, -((b*x^3)/a)] + 2*d*x^3*Hypergeometric2F1[7/3, -m, 10/3, -((b*x^3)/a)])))/(14*(1 + (b*x^3)/a)^m)

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \left (b \,x^{3}+a \right )^{m} \left (d \,x^{3}+c \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 105.62, size = 121, normalized size = 0.69 \begin {gather*} \frac {a^{m} c^{2} 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 {2 a^{m} c 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 )} + \frac {a^{m} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, - m \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**m*c**2*x*gamma(1/3)*hyper((1/3, -m), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + 2*a**m*c*d*x**4*gam
ma(4/3)*hyper((4/3, -m), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**m*d**2*x**7*gamma(7/3)*hyper((7
/3, -m), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^3+a\right )}^m\,{\left (d\,x^3+c\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^m*(c + d*x^3)^2,x)

[Out]

int((a + b*x^3)^m*(c + d*x^3)^2, x)

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