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

Optimal. Leaf size=83 \[ \frac {(a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^2 (p+1)}+\frac {e (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+3)} \]

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Rubi [A]  time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {770, 21, 43} \begin {gather*} \frac {(a+b x)^2 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^2 (p+1)}+\frac {e (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^2 (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.61 \begin {gather*} \frac {\left ((a+b x)^2\right )^{p+1} (-a e+b d (2 p+3)+2 b e (p+1) x)}{2 b^2 (p+1) (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.16, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 142, normalized size = 1.71 \begin {gather*} \frac {{\left (2 \, a^{2} b d p + 3 \, a^{2} b d - a^{3} e + 2 \, {\left (b^{3} e p + b^{3} e\right )} x^{3} + {\left (3 \, b^{3} d + 3 \, a b^{2} e + 2 \, {\left (b^{3} d + 2 \, a b^{2} e\right )} p\right )} x^{2} + 2 \, {\left (3 \, a b^{2} d + {\left (2 \, a b^{2} d + a^{2} b e\right )} p\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \, {\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 353, normalized size = 4.25 \begin {gather*} \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} p x^{3} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d p x^{2} + 4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} p x^{2} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} x^{3} e + 4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d p x + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d x^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b p x e + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} x^{2} e + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d p + 6 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d x + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d - {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{3} e}{2 \, {\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*p*x^3*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*p*x^2 + 4*(b^2*x^2 + 2*a*
b*x + a^2)^p*a*b^2*p*x^2*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*x^3*e + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*p
*x + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*x^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*p*x*e + 3*(b^2*x^2 + 2*a*b*
x + a^2)^p*a*b^2*x^2*e + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d*p + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*x + 3
*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d - (b^2*x^2 + 2*a*b*x + a^2)^p*a^3*e)/(2*b^2*p^2 + 5*b^2*p + 3*b^2)

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maple [A]  time = 0.05, size = 67, normalized size = 0.81 \begin {gather*} -\frac {\left (-2 b e p x -2 b d p -2 b e x +a e -3 b d \right ) \left (b x +a \right )^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{2 \left (2 p^{2}+5 p +3\right ) b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.54, size = 206, normalized size = 2.48 \begin {gather*} \frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p} a d}{b {\left (2 \, p + 1\right )}} + \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )} {\left (b x + a\right )}^{2 \, p} d}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b} + \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )} {\left (b x + a\right )}^{2 \, p} a e}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{3} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + a^{3}\right )} {\left (b x + a\right )}^{2 \, p} e}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)*(b*x + a)^(2*p)*a*d/(b*(2*p + 1)) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*d/((2*
p^2 + 3*p + 1)*b) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*a*e/((2*p^2 + 3*p + 1)*b^2) + ((
2*p^2 + 3*p + 1)*b^3*x^3 + (2*p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + a^3)*(b*x + a)^(2*p)*e/((4*p^3 + 12*p^2 + 11*
p + 3)*b^2)

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mupad [B]  time = 2.15, size = 142, normalized size = 1.71 \begin {gather*} {\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p\,\left (\frac {x^2\,\left (3\,a\,e+3\,b\,d+4\,a\,e\,p+2\,b\,d\,p\right )}{2\,\left (2\,p^2+5\,p+3\right )}+\frac {a^2\,\left (3\,b\,d-a\,e+2\,b\,d\,p\right )}{2\,b^2\,\left (2\,p^2+5\,p+3\right )}+\frac {a\,x\,\left (3\,b\,d+a\,e\,p+2\,b\,d\,p\right )}{b\,\left (2\,p^2+5\,p+3\right )}+\frac {b\,e\,x^3\,\left (p+1\right )}{2\,p^2+5\,p+3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2 + b^2*x^2 + 2*a*b*x)^p*((x^2*(3*a*e + 3*b*d + 4*a*e*p + 2*b*d*p))/(2*(5*p + 2*p^2 + 3)) + (a^2*(3*b*d - a
*e + 2*b*d*p))/(2*b^2*(5*p + 2*p^2 + 3)) + (a*x*(3*b*d + a*e*p + 2*b*d*p))/(b*(5*p + 2*p^2 + 3)) + (b*e*x^3*(p
+ 1))/(5*p + 2*p^2 + 3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} a \left (d x + \frac {e x^{2}}{2}\right ) \left (a^{2}\right )^{p} & \text {for}\: b = 0 \\\int \frac {\left (a + b x\right ) \left (d + e x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {a e \log {\left (\frac {a}{b} + x \right )}}{b^{2}} + \frac {d \log {\left (\frac {a}{b} + x \right )}}{b} + \frac {e x}{b} & \text {for}\: p = -1 \\- \frac {a^{3} e \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {2 a^{2} b d p \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {3 a^{2} b d \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {2 a^{2} b e p x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {4 a b^{2} d p x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {6 a b^{2} d x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {4 a b^{2} e p x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {3 a b^{2} e x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {2 b^{3} d p x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {3 b^{3} d x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {2 b^{3} e p x^{3} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} + \frac {2 b^{3} e x^{3} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{4 b^{2} p^{2} + 10 b^{2} p + 6 b^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*(d*x + e*x**2/2)*(a**2)**p, Eq(b, 0)), (Integral((a + b*x)*(d + e*x)/((a + b*x)**2)**(3/2), x), E
q(p, -3/2)), (-a*e*log(a/b + x)/b**2 + d*log(a/b + x)/b + e*x/b, Eq(p, -1)), (-a**3*e*(a**2 + 2*a*b*x + b**2*x
**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 2*a**2*b*d*p*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b*
*2*p + 6*b**2) + 3*a**2*b*d*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 2*a**2*b*e*p*
x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 4*a*b**2*d*p*x*(a**2 + 2*a*b*x + b**2*x
**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 6*a*b**2*d*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b*
*2*p + 6*b**2) + 4*a*b**2*e*p*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 3*a*b*
*2*e*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 2*b**3*d*p*x**2*(a**2 + 2*a*b*x
+ b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2) + 3*b**3*d*x**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p*
*2 + 10*b**2*p + 6*b**2) + 2*b**3*e*p*x**3*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2)
+ 2*b**3*e*x**3*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**2*p**2 + 10*b**2*p + 6*b**2), True))

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