3.734 \(\int (d+e x) (a+c x^2)^p \, dx\)

Optimal. Leaf size=61 \[ \frac {d x \left (a+c x^2\right )^{p+1} \, _2F_1\left (1,p+\frac {3}{2};\frac {3}{2};-\frac {c x^2}{a}\right )}{a}+\frac {e \left (a+c x^2\right )^{p+1}}{2 c (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {641, 246, 245} \[ d x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {c x^2}{a}\right )+\frac {e \left (a+c x^2\right )^{p+1}}{2 c (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + c*x^2)^p,x]

[Out]

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

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 641

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 98, normalized size = 1.61 \[ \frac {\left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (2 c d (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {c x^2}{a}\right )+c e x^2 \left (\frac {c x^2}{a}+1\right )^p+a e \left (\left (\frac {c x^2}{a}+1\right )^p-1\right )\right )}{2 c (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(a + c*x^2)^p,x]

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x + d\right )} {\left (c x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)*(c*x^2 + a)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)*(c*x^2 + a)^p, x)

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maple [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right ) \left (c \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)*(c*x^2 + a)^p, x)

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mupad [B]  time = 1.11, size = 65, normalized size = 1.07 \[ \frac {e\,{\left (c\,x^2+a\right )}^{p+1}}{2\,c\,\left (p+1\right )}+\frac {d\,x\,{\left (c\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ -\frac {c\,x^2}{a}\right )}{{\left (\frac {c\,x^2}{a}+1\right )}^p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^p*(d + e*x),x)

[Out]

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

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sympy [A]  time = 6.89, size = 61, normalized size = 1.00 \[ a^{p} d x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )} + e \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\begin {cases} \frac {\left (a + c x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + c x^{2} \right )} & \text {otherwise} \end {cases}}{2 c} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*d*x*hyper((1/2, -p), (3/2,), c*x**2*exp_polar(I*pi)/a) + e*Piecewise((a**p*x**2/2, Eq(c, 0)), (Piecewise(
((a + c*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + c*x**2), True))/(2*c), True))

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