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

Optimal. Leaf size=298 \[ \frac {d x \left (a+b x^3\right )^{m+1} \left (28 a^2 d^2-a b c d (15 m+92)+b^2 c^2 \left (9 m^2+60 m+118\right )\right )}{b^3 (3 m+4) (3 m+7) (3 m+10)}-\frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (28 a^3 d^3-12 a^2 b c d^2 (3 m+10)+3 a b^2 c^2 d \left (9 m^2+51 m+70\right )-b^3 c^3 \left (27 m^3+189 m^2+414 m+280\right )\right ) \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )}{b^3 (3 m+4) (3 m+7) (3 m+10)}-\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b^2 (3 m+7) (3 m+10)}+\frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)} \]

[Out]

d*(28*a^2*d^2-a*b*c*d*(92+15*m)+b^2*c^2*(9*m^2+60*m+118))*x*(b*x^3+a)^(1+m)/b^3/(10+3*m)/(9*m^2+33*m+28)-d*(7*
a*d-b*c*(16+3*m))*x*(b*x^3+a)^(1+m)*(d*x^3+c)/b^2/(9*m^2+51*m+70)+d*x*(b*x^3+a)^(1+m)*(d*x^3+c)^2/b/(10+3*m)-(
28*a^3*d^3-12*a^2*b*c*d^2*(10+3*m)+3*a*b^2*c^2*d*(9*m^2+51*m+70)-b^3*c^3*(27*m^3+189*m^2+414*m+280))*x*(b*x^3+
a)^m*hypergeom([1/3, -m],[4/3],-b*x^3/a)/b^3/(10+3*m)/(9*m^2+33*m+28)/((1+b*x^3/a)^m)

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Rubi [A]  time = 0.30, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {416, 528, 388, 246, 245} \[ -\frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (-12 a^2 b c d^2 (3 m+10)+28 a^3 d^3+3 a b^2 c^2 d \left (9 m^2+51 m+70\right )-b^3 c^3 \left (27 m^3+189 m^2+414 m+280\right )\right ) \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )}{b^3 (3 m+4) (3 m+7) (3 m+10)}+\frac {d x \left (a+b x^3\right )^{m+1} \left (28 a^2 d^2-a b c d (15 m+92)+b^2 c^2 \left (9 m^2+60 m+118\right )\right )}{b^3 (3 m+4) (3 m+7) (3 m+10)}-\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b^2 (3 m+7) (3 m+10)}+\frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^m*(c + d*x^3)^3,x]

[Out]

(d*(28*a^2*d^2 - a*b*c*d*(92 + 15*m) + b^2*c^2*(118 + 60*m + 9*m^2))*x*(a + b*x^3)^(1 + m))/(b^3*(4 + 3*m)*(7
+ 3*m)*(10 + 3*m)) - (d*(7*a*d - b*c*(16 + 3*m))*x*(a + b*x^3)^(1 + m)*(c + d*x^3))/(b^2*(7 + 3*m)*(10 + 3*m))
+ (d*x*(a + b*x^3)^(1 + m)*(c + d*x^3)^2)/(b*(10 + 3*m)) - ((28*a^3*d^3 - 12*a^2*b*c*d^2*(10 + 3*m) + 3*a*b^2
*c^2*d*(70 + 51*m + 9*m^2) - b^3*c^3*(280 + 414*m + 189*m^2 + 27*m^3))*x*(a + b*x^3)^m*Hypergeometric2F1[1/3,
-m, 4/3, -((b*x^3)/a)])/(b^3*(4 + 3*m)*(7 + 3*m)*(10 + 3*m)*(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]

Rule 416

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]

Rule 528

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

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^m \left (c+d x^3\right )^3 \, dx &=\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )^2}{b (10+3 m)}+\frac {\int \left (a+b x^3\right )^m \left (c+d x^3\right ) \left (-c (a d-b c (10+3 m))-d (7 a d-b c (16+3 m)) x^3\right ) \, dx}{b (10+3 m)}\\ &=-\frac {d (7 a d-b c (16+3 m)) x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b^2 (7+3 m) (10+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )^2}{b (10+3 m)}+\frac {\int \left (a+b x^3\right )^m \left (c \left (7 a^2 d^2-a b c d (23+6 m)+b^2 c^2 \left (70+51 m+9 m^2\right )\right )+d \left (28 a^2 d^2-a b c d (92+15 m)+b^2 c^2 \left (118+60 m+9 m^2\right )\right ) x^3\right ) \, dx}{b^2 (7+3 m) (10+3 m)}\\ &=\frac {d \left (28 a^2 d^2-a b c d (92+15 m)+b^2 c^2 \left (118+60 m+9 m^2\right )\right ) x \left (a+b x^3\right )^{1+m}}{b^3 (4+3 m) (7+3 m) (10+3 m)}-\frac {d (7 a d-b c (16+3 m)) x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b^2 (7+3 m) (10+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )^2}{b (10+3 m)}-\frac {\left (28 a^3 d^3-12 a^2 b c d^2 (10+3 m)+3 a b^2 c^2 d \left (70+51 m+9 m^2\right )-b^3 c^3 \left (280+414 m+189 m^2+27 m^3\right )\right ) \int \left (a+b x^3\right )^m \, dx}{b^3 (4+3 m) (7+3 m) (10+3 m)}\\ &=\frac {d \left (28 a^2 d^2-a b c d (92+15 m)+b^2 c^2 \left (118+60 m+9 m^2\right )\right ) x \left (a+b x^3\right )^{1+m}}{b^3 (4+3 m) (7+3 m) (10+3 m)}-\frac {d (7 a d-b c (16+3 m)) x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b^2 (7+3 m) (10+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )^2}{b (10+3 m)}-\frac {\left (\left (28 a^3 d^3-12 a^2 b c d^2 (10+3 m)+3 a b^2 c^2 d \left (70+51 m+9 m^2\right )-b^3 c^3 \left (280+414 m+189 m^2+27 m^3\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^3 (4+3 m) (7+3 m) (10+3 m)}\\ &=\frac {d \left (28 a^2 d^2-a b c d (92+15 m)+b^2 c^2 \left (118+60 m+9 m^2\right )\right ) x \left (a+b x^3\right )^{1+m}}{b^3 (4+3 m) (7+3 m) (10+3 m)}-\frac {d (7 a d-b c (16+3 m)) x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )}{b^2 (7+3 m) (10+3 m)}+\frac {d x \left (a+b x^3\right )^{1+m} \left (c+d x^3\right )^2}{b (10+3 m)}-\frac {\left (28 a^3 d^3-12 a^2 b c d^2 (10+3 m)+3 a b^2 c^2 d \left (70+51 m+9 m^2\right )-b^3 c^3 \left (280+414 m+189 m^2+27 m^3\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^3 (4+3 m) (7+3 m) (10+3 m)}\\ \end {align*}

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Mathematica [A]  time = 5.08, size = 137, normalized size = 0.46 \[ \frac {1}{140} x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (140 c^3 \, _2F_1\left (\frac {1}{3},-m;\frac {4}{3};-\frac {b x^3}{a}\right )+d x^3 \left (105 c^2 \, _2F_1\left (\frac {4}{3},-m;\frac {7}{3};-\frac {b x^3}{a}\right )+2 d x^3 \left (30 c \, _2F_1\left (\frac {7}{3},-m;\frac {10}{3};-\frac {b x^3}{a}\right )+7 d x^3 \, _2F_1\left (\frac {10}{3},-m;\frac {13}{3};-\frac {b x^3}{a}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^3)^m*(140*c^3*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)] + d*x^3*(105*c^2*Hypergeometric2F1[4/3
, -m, 7/3, -((b*x^3)/a)] + 2*d*x^3*(30*c*Hypergeometric2F1[7/3, -m, 10/3, -((b*x^3)/a)] + 7*d*x^3*Hypergeometr
ic2F1[10/3, -m, 13/3, -((b*x^3)/a)]))))/(140*(1 + (b*x^3)/a)^m)

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fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{3} x^{9} + 3 \, c d^{2} x^{6} + 3 \, c^{2} d x^{3} + c^{3}\right )} {\left (b x^{3} + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^3*x^9 + 3*c*d^2*x^6 + 3*c^2*d*x^3 + c^3)*(b*x^3 + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \left (d \,x^{3}+c \right )^{3} \left (b \,x^{3}+a \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^m*(c + d*x^3)^3,x)

[Out]

int((a + b*x^3)^m*(c + d*x^3)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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