3.378 \(\int (g x)^m (d+e x)^n (a+c x^2) \, dx\)

Optimal. Leaf size=164 \[ \frac{(g x)^{m+1} (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} \left (a e^2 (m+n+2) (m+n+3)+c d^2 (m+1) (m+2)\right ) \, _2F_1\left (m+1,-n;m+2;-\frac{e x}{d}\right )}{e^2 g (m+1) (m+n+2) (m+n+3)}-\frac{c d (m+2) (g x)^{m+1} (d+e x)^{n+1}}{e^2 g (m+n+2) (m+n+3)}+\frac{c (g x)^{m+2} (d+e x)^{n+1}}{e g^2 (m+n+3)} \]

[Out]

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

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Rubi [A]  time = 0.133066, antiderivative size = 150, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {952, 80, 66, 64} \[ \frac{(g x)^{m+1} (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} \left (\frac{a}{m+1}+\frac{c d^2 (m+2)}{e^2 (m+n+2) (m+n+3)}\right ) \, _2F_1\left (m+1,-n;m+2;-\frac{e x}{d}\right )}{g}-\frac{c d (m+2) (g x)^{m+1} (d+e x)^{n+1}}{e^2 g (m+n+2) (m+n+3)}+\frac{c (g x)^{m+2} (d+e x)^{n+1}}{e g^2 (m+n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 952

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(c^p*(d
+ e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0]
&& NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c^IntPart[n]*(c + d*x)^FracPart[n])/(1 + (d
*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0]))
||  !RationalQ[n])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

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

Mathematica [A]  time = 0.0691794, size = 113, normalized size = 0.69 \[ \frac{x (g x)^m (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} \left (\left (a e^2+c d^2\right ) \, _2F_1\left (m+1,-n;m+2;-\frac{e x}{d}\right )+c d^2 \, _2F_1\left (m+1,-n-2;m+2;-\frac{e x}{d}\right )-2 c d^2 \, _2F_1\left (m+1,-n-1;m+2;-\frac{e x}{d}\right )\right )}{e^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*(d + e*x)^n*(c*d^2*Hypergeometric2F1[1 + m, -2 - n, 2 + m, -((e*x)/d)] - 2*c*d^2*Hypergeometric2F1[
1 + m, -1 - n, 2 + m, -((e*x)/d)] + (c*d^2 + a*e^2)*Hypergeometric2F1[1 + m, -n, 2 + m, -((e*x)/d)]))/(e^2*(1
+ m)*(1 + (e*x)/d)^n)

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Maple [F]  time = 0.428, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{m} \left ( ex+d \right ) ^{n} \left ( c{x}^{2}+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + a\right )}{\left (e x + d\right )}^{n} \left (g x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + a\right )}{\left (e x + d\right )}^{n} \left (g x\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 18.9112, size = 82, normalized size = 0.5 \begin{align*} \frac{a d^{n} g^{m} x x^{m} \Gamma \left (m + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - n, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{e x e^{i \pi }}{d}} \right )}}{\Gamma \left (m + 2\right )} + \frac{c d^{n} g^{m} x^{3} x^{m} \Gamma \left (m + 3\right ){{}_{2}F_{1}\left (\begin{matrix} - n, m + 3 \\ m + 4 \end{matrix}\middle |{\frac{e x e^{i \pi }}{d}} \right )}}{\Gamma \left (m + 4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)**m*(e*x+d)**n*(c*x**2+a),x)

[Out]

a*d**n*g**m*x*x**m*gamma(m + 1)*hyper((-n, m + 1), (m + 2,), e*x*exp_polar(I*pi)/d)/gamma(m + 2) + c*d**n*g**m
*x**3*x**m*gamma(m + 3)*hyper((-n, m + 3), (m + 4,), e*x*exp_polar(I*pi)/d)/gamma(m + 4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + a\right )}{\left (e x + d\right )}^{n} \left (g x\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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