3.430 \(\int (g x)^m (d+e x)^3 (a+c x^2)^p \, dx\)

Optimal. Leaf size=276 \[ -\frac {e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (a e^2 (m+2)-3 c d^2 (m+2 p+4)\right ) \, _2F_1\left (\frac {m+2}{2},-p;\frac {m+4}{2};-\frac {c x^2}{a}\right )}{c g^2 (m+2) (m+2 p+4)}-\frac {d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (3 a e^2 (m+1)-c d^2 (m+2 p+3)\right ) \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {c x^2}{a}\right )}{c g (m+1) (m+2 p+3)}+\frac {3 d e^2 (g x)^{m+1} \left (a+c x^2\right )^{p+1}}{c g (m+2 p+3)}+\frac {e^3 (g x)^{m+2} \left (a+c x^2\right )^{p+1}}{c g^2 (m+2 p+4)} \]

[Out]

3*d*e^2*(g*x)^(1+m)*(c*x^2+a)^(1+p)/c/g/(3+m+2*p)+e^3*(g*x)^(2+m)*(c*x^2+a)^(1+p)/c/g^2/(4+m+2*p)-d*(3*a*e^2*(
1+m)-c*d^2*(3+m+2*p))*(g*x)^(1+m)*(c*x^2+a)^p*hypergeom([-p, 1/2+1/2*m],[3/2+1/2*m],-c*x^2/a)/c/g/(1+m)/(3+m+2
*p)/((c*x^2/a+1)^p)-e*(a*e^2*(2+m)-3*c*d^2*(4+m+2*p))*(g*x)^(2+m)*(c*x^2+a)^p*hypergeom([-p, 1+1/2*m],[2+1/2*m
],-c*x^2/a)/c/g^2/(2+m)/(4+m+2*p)/((c*x^2/a+1)^p)

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Rubi [A]  time = 0.47, antiderivative size = 254, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1809, 808, 365, 364} \[ \frac {e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (\frac {3 d^2}{m+2}-\frac {a e^2}{c (m+2 p+4)}\right ) \, _2F_1\left (\frac {m+2}{2},-p;\frac {m+4}{2};-\frac {c x^2}{a}\right )}{g^2}+\frac {d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (\frac {d^2}{m+1}-\frac {3 a e^2}{c (m+2 p+3)}\right ) \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {c x^2}{a}\right )}{g}+\frac {3 d e^2 (g x)^{m+1} \left (a+c x^2\right )^{p+1}}{c g (m+2 p+3)}+\frac {e^3 (g x)^{m+2} \left (a+c x^2\right )^{p+1}}{c g^2 (m+2 p+4)} \]

Antiderivative was successfully verified.

[In]

Int[(g*x)^m*(d + e*x)^3*(a + c*x^2)^p,x]

[Out]

(3*d*e^2*(g*x)^(1 + m)*(a + c*x^2)^(1 + p))/(c*g*(3 + m + 2*p)) + (e^3*(g*x)^(2 + m)*(a + c*x^2)^(1 + p))/(c*g
^2*(4 + m + 2*p)) + (d*(d^2/(1 + m) - (3*a*e^2)/(c*(3 + m + 2*p)))*(g*x)^(1 + m)*(a + c*x^2)^p*Hypergeometric2
F1[(1 + m)/2, -p, (3 + m)/2, -((c*x^2)/a)])/(g*(1 + (c*x^2)/a)^p) + (e*((3*d^2)/(2 + m) - (a*e^2)/(c*(4 + m +
2*p)))*(g*x)^(2 + m)*(a + c*x^2)^p*Hypergeometric2F1[(2 + m)/2, -p, (4 + m)/2, -((c*x^2)/a)])/(g^2*(1 + (c*x^2
)/a)^p)

Rule 364

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int (g x)^m (d+e x)^3 \left (a+c x^2\right )^p \, dx &=\frac {e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\frac {\int (g x)^m \left (a+c x^2\right )^p \left (c d^3 (4+m+2 p)-e \left (a e^2 (2+m)-3 c d^2 (4+m+2 p)\right ) x+3 c d e^2 (4+m+2 p) x^2\right ) \, dx}{c (4+m+2 p)}\\ &=\frac {3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac {e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\frac {\int (g x)^m \left (-c d (4+m+2 p) \left (3 a e^2 (1+m)-c d^2 (3+m+2 p)\right )-c e (3+m+2 p) \left (a e^2 (2+m)-3 c d^2 (4+m+2 p)\right ) x\right ) \left (a+c x^2\right )^p \, dx}{c^2 (3+m+2 p) (4+m+2 p)}\\ &=\frac {3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac {e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\left (d \left (d^2-\frac {3 a e^2 (1+m)}{c (3+m+2 p)}\right )\right ) \int (g x)^m \left (a+c x^2\right )^p \, dx+\frac {\left (e \left (3 d^2-\frac {a e^2 (2+m)}{c (4+m+2 p)}\right )\right ) \int (g x)^{1+m} \left (a+c x^2\right )^p \, dx}{g}\\ &=\frac {3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac {e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\left (d \left (d^2-\frac {3 a e^2 (1+m)}{c (3+m+2 p)}\right ) \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int (g x)^m \left (1+\frac {c x^2}{a}\right )^p \, dx+\frac {\left (e \left (3 d^2-\frac {a e^2 (2+m)}{c (4+m+2 p)}\right ) \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int (g x)^{1+m} \left (1+\frac {c x^2}{a}\right )^p \, dx}{g}\\ &=\frac {3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac {e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\frac {d \left (d^2-\frac {3 a e^2 (1+m)}{c (3+m+2 p)}\right ) (g x)^{1+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {c x^2}{a}\right )}{g (1+m)}+\frac {e \left (3 d^2-\frac {a e^2 (2+m)}{c (4+m+2 p)}\right ) (g x)^{2+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {2+m}{2},-p;\frac {4+m}{2};-\frac {c x^2}{a}\right )}{g^2 (2+m)}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 182, normalized size = 0.66 \[ x (g x)^m \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (\frac {d^3 \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {c x^2}{a}\right )}{m+1}+e x \left (\frac {3 d^2 \, _2F_1\left (\frac {m+2}{2},-p;\frac {m+4}{2};-\frac {c x^2}{a}\right )}{m+2}+e x \left (\frac {3 d \, _2F_1\left (\frac {m+3}{2},-p;\frac {m+5}{2};-\frac {c x^2}{a}\right )}{m+3}+\frac {e x \, _2F_1\left (\frac {m+4}{2},-p;\frac {m+6}{2};-\frac {c x^2}{a}\right )}{m+4}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*(a + c*x^2)^p*((d^3*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, -((c*x^2)/a)])/(1 + m) + e*x*((3*d^
2*Hypergeometric2F1[(2 + m)/2, -p, (4 + m)/2, -((c*x^2)/a)])/(2 + m) + e*x*((3*d*Hypergeometric2F1[(3 + m)/2,
-p, (5 + m)/2, -((c*x^2)/a)])/(3 + m) + (e*x*Hypergeometric2F1[(4 + m)/2, -p, (6 + m)/2, -((c*x^2)/a)])/(4 + m
)))))/(1 + (c*x^2)/a)^p

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} {\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*(c*x^2 + a)^p*(g*x)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{3} {\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^3*(c*x^2 + a)^p*(g*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(e*x+d)^3*(c*x^2+a)^p,x)

[Out]

int((g*x)^m*(e*x+d)^3*(c*x^2+a)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{3} {\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^3*(c*x^2 + a)^p*(g*x)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (g\,x\right )}^m\,{\left (c\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(a + c*x^2)^p*(d + e*x)^3,x)

[Out]

int((g*x)^m*(a + c*x^2)^p*(d + e*x)^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((g*x)**m*(e*x+d)**3*(c*x**2+a)**p,x)

[Out]

Timed out

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