\(\int (d+e x) (a+b x^2)^p \, dx\) [386]

Optimal result

Integrand size = 15, antiderivative size = 70 \[ \int (d+e x) \left (a+b x^2\right )^p \, dx=\frac {e \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+d x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {655, 252, 251} \[ \int (d+e x) \left (a+b x^2\right )^p \, dx=d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+\frac {e \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]

[In]

Int[(d + e*x)*(a + b*x^2)^p,x]

[Out]

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

Rule 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 655

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.40 \[ \int (d+e x) \left (a+b x^2\right )^p \, dx=\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (b e x^2 \left (1+\frac {b x^2}{a}\right )^p+a e \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )+2 b d (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{2 b (1+p)} \]

[In]

Integrate[(d + e*x)*(a + b*x^2)^p,x]

[Out]

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

Maple [F]

\[\int \left (e x +d \right ) \left (b \,x^{2}+a \right )^{p}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \]

[In]

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

[Out]

integral((e*x + d)*(b*x^2 + a)^p, x)

Sympy [A] (verification not implemented)

Time = 2.45 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int (d+e x) \left (a+b x^2\right )^p \, dx=a^{p} d x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + e \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \]

[In]

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

[Out]

integrate((e*x + d)*(b*x^2 + a)^p, x)

Giac [F]

\[ \int (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \]

[In]

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

[Out]

integrate((e*x + d)*(b*x^2 + a)^p, x)

Mupad [B] (verification not implemented)

Time = 12.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int (d+e x) \left (a+b x^2\right )^p \, dx=\frac {e\,{\left (b\,x^2+a\right )}^{p+1}}{2\,b\,\left (p+1\right )}+\frac {d\,x\,{\left (b\,x^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^p} \]

[In]

int((a + b*x^2)^p*(d + e*x),x)

[Out]

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