3.4.95 \(\int (d+e x)^2 (a+b x^2)^p \, dx\) [395]

Optimal. Leaf size=133 \[ \frac {d e (2+p) \left (a+b x^2\right )^{1+p}}{b (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x^2\right )^{1+p}}{b (3+2 p)}-\frac {\left (a e^2-b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b (3+2 p)} \]

[Out]

d*e*(2+p)*(b*x^2+a)^(1+p)/b/(2*p^2+5*p+3)+e*(e*x+d)*(b*x^2+a)^(1+p)/b/(3+2*p)-(a*e^2-b*d^2*(3+2*p))*x*(b*x^2+a
)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/b/(3+2*p)/((1+b*x^2/a)^p)

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Rubi [A]
time = 0.05, antiderivative size = 125, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {757, 655, 252, 251} \begin {gather*} x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (d^2-\frac {a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+\frac {e (d+e x) \left (a+b x^2\right )^{p+1}}{b (2 p+3)}+\frac {d e (p+2) \left (a+b x^2\right )^{p+1}}{b (p+1) (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(a + b*x^2)^p,x]

[Out]

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

Rule 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 133, normalized size = 1.00 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (3 b d^2 (1+p) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+e \left (3 d \left (b x^2 \left (1+\frac {b x^2}{a}\right )^p+a \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+b e (1+p) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\right )\right )}{3 b (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^p*(3*b*d^2*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + e*(3*d*(b*x^2*(1 + (b*x^2)/a
)^p + a*(-1 + (1 + (b*x^2)/a)^p)) + b*e*(1 + p)*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])))/(3*b*(1 +
p)*(1 + (b*x^2)/a)^p)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [A]
time = 5.28, size = 97, normalized size = 0.73 \begin {gather*} a^{p} d^{2} 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 )} + \frac {a^{p} e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + 2 d 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*d**2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*e**2*x**3*hyper((3/2, -p), (5/2,), b*x**
2*exp_polar(I*pi)/a)/3 + 2*d*e*Piecewise((a**p*x**2/2, Eq(b, 0)), (Piecewise(((a + b*x**2)**(p + 1)/(p + 1), N
e(p, -1)), (log(a + b*x**2), True))/(2*b), True))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p*(d + e*x)^2,x)

[Out]

int((a + b*x^2)^p*(d + e*x)^2, x)

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