3.404 \(\int (d+e x)^3 (a+b x^2)^p \, dx\)

Optimal. Leaf size=176 \[ -\frac{e \left (a+b x^2\right )^{p+1} \left (a e^2-3 b d^2 (p+2)\right )}{2 b^2 (p+1) (p+2)}-\frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2-b d^2 (2 p+3)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b (2 p+3)}+\frac{3 d e^2 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}+\frac{e^3 x^2 \left (a+b x^2\right )^{p+1}}{2 b (p+2)} \]

[Out]

-(e*(a*e^2 - 3*b*d^2*(2 + p))*(a + b*x^2)^(1 + p))/(2*b^2*(1 + p)*(2 + p)) + (3*d*e^2*x*(a + b*x^2)^(1 + p))/(
b*(3 + 2*p)) + (e^3*x^2*(a + b*x^2)^(1 + p))/(2*b*(2 + p)) - (d*(3*a*e^2 - b*d^2*(3 + 2*p))*x*(a + b*x^2)^p*Hy
pergeometric2F1[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 = 0.149465, antiderivative size = 169, normalized size of antiderivative = 0.96, 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 = {743, 780, 246, 245} \[ -\frac{e \left (a+b x^2\right )^{p+1} \left ((2 p+3) \left (a e^2-b d^2 (2 p+5)\right )-2 b d e (p+1) (p+3) x\right )}{2 b^2 (p+2) \left (2 p^2+5 p+3\right )}+d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{3 a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{e (d+e x)^2 \left (a+b x^2\right )^{p+1}}{2 b (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

(e*(d + e*x)^2*(a + b*x^2)^(1 + p))/(2*b*(2 + p)) - (e*((3 + 2*p)*(a*e^2 - b*d^2*(5 + 2*p)) - 2*b*d*e*(1 + p)*
(3 + p)*x)*(a + b*x^2)^(1 + p))/(2*b^2*(2 + p)*(3 + 5*p + 2*p^2)) + (d*(d^2 - (3*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 743

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]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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 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])

Rubi steps

\begin{align*} \int (d+e x)^3 \left (a+b x^2\right )^p \, dx &=\frac{e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}+\frac{\int (d+e x) \left (-2 \left (a e^2-b d^2 (2+p)\right )+2 b d e (3+p) x\right ) \left (a+b x^2\right )^p \, dx}{2 b (2+p)}\\ &=\frac{e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac{e \left ((3+2 p) \left (a e^2-b d^2 (5+2 p)\right )-2 b d e (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+\left (d \left (d^2-\frac{3 a e^2}{3 b+2 b p}\right )\right ) \int \left (a+b x^2\right )^p \, dx\\ &=\frac{e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac{e \left ((3+2 p) \left (a e^2-b d^2 (5+2 p)\right )-2 b d e (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+\left (d \left (d^2-\frac{3 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{e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac{e \left ((3+2 p) \left (a e^2-b d^2 (5+2 p)\right )-2 b d e (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+d \left (d^2-\frac{3 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*}

Mathematica [A]  time = 0.218668, size = 223, normalized size = 1.27 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (e \left (-a^2 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+b^2 x^2 \left (\frac{b x^2}{a}+1\right )^p \left (3 d^2 (p+2)+e^2 (p+1) x^2\right )+2 b^2 d e \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+a b \left (3 d^2 (p+2) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )\right )+2 b^2 d^3 \left (p^2+3 p+2\right ) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )\right )}{2 b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(a + b*x^2)^p,x]

[Out]

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

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Maple [F]  time = 0.535, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(b*x^2+a)^p,x)

[Out]

int((e*x+d)^3*(b*x^2+a)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

integrate((e*x + d)^3*(b*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*(b*x^2 + a)^p, x)

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Sympy [C]  time = 21.1549, size = 468, normalized size = 2.66 \begin{align*} a^{p} d^{3} 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 )} + a^{p} d 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 d^{2} 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 ) + e^{3} \left (\begin{cases} \frac{a^{p} x^{4}}{4} & \text{for}\: b = 0 \\\frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{b x^{2} \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{b x^{2} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text{for}\: p = -2 \\- \frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b^{2}} - \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b^{2}} + \frac{x^{2}}{2 b} & \text{for}\: p = -1 \\- \frac{a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac{a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac{b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac{b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(b*x**2+a)**p,x)

[Out]

a**p*d**3*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*d*e**2*x**3*hyper((3/2, -p), (5/2,), b*x
**2*exp_polar(I*pi)/a) + 3*d**2*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)) + e**3*Piecewise((a**p*x**4/4, Eq(b, 0)), (a*log(-I*sqrt(
a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + a*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*
b**2 + 2*b**3*x**2) + b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(I*sqrt(a)*sqr
t(1/b) + x)/(2*a*b**2 + 2*b**3*x**2), Eq(p, -2)), (-a*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*b**2) - a*log(I*sqrt(a)
*sqrt(1/b) + x)/(2*b**2) + x**2/(2*b), Eq(p, -1)), (-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) +
a*b*p*x**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*p*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b*
*2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((e*x + d)^3*(b*x^2 + a)^p, x)