3.5.32 \(\int (g x)^m (d+e x) (a+c x^2)^p \, dx\) [432]

Optimal. Leaf size=135 \[ \frac {d (g x)^{1+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {c x^2}{a}\right )}{g (1+m)}+\frac {e (g x)^{2+m} \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {2+m}{2},-p;\frac {4+m}{2};-\frac {c x^2}{a}\right )}{g^2 (2+m)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {822, 372, 371} \begin {gather*} \frac {d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {c x^2}{a}\right )}{g (m+1)}+\frac {e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {m+2}{2},-p;\frac {m+4}{2};-\frac {c x^2}{a}\right )}{g^2 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 822

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]

Rubi steps

\begin {align*} \int (g x)^m (d+e x) \left (a+c x^2\right )^p \, dx &=d \int (g x)^m \left (a+c x^2\right )^p \, dx+\frac {e \int (g x)^{1+m} \left (a+c x^2\right )^p \, dx}{g}\\ &=\left (d \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 (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 {d (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 (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 = 0.05, size = 106, normalized size = 0.79 \begin {gather*} \frac {x (g x)^m \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \left (e (1+m) x \, _2F_1\left (1+\frac {m}{2},-p;2+\frac {m}{2};-\frac {c x^2}{a}\right )+d (2+m) \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {c x^2}{a}\right )\right )}{(1+m) (2+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*(a + c*x^2)^p*(e*(1 + m)*x*Hypergeometric2F1[1 + m/2, -p, 2 + m/2, -((c*x^2)/a)] + d*(2 + m)*Hyperg
eometric2F1[(1 + m)/2, -p, (3 + m)/2, -((c*x^2)/a)]))/((1 + m)*(2 + m)*(1 + (c*x^2)/a)^p)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (g x \right )^{m} \left (e x +d \right ) \left (c \,x^{2}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 40.47, size = 109, normalized size = 0.81 \begin {gather*} \frac {a^{p} d g^{m} x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a^{p} e g^{m} x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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