3.1893 \(\int (A+B x) (d+e x)^m (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=174 \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{m+1} \left (-\frac{e (a+b x)}{b d-a e}\right )^{-2 p} (A b e (m+2 p+2)-B (a e (m+1)+b (2 d p+d))) \, _2F_1\left (m+1,-2 p;m+2;\frac{b (d+e x)}{b d-a e}\right )}{b e^2 (m+1) (m+2 p+2)}+\frac{B (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{m+1}}{b e (m+2 p+2)} \]

[Out]

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

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Rubi [A]  time = 0.186949, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {770, 80, 70, 69} \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{m+1} \left (-\frac{e (a+b x)}{b d-a e}\right )^{-2 p} (A b e (m+2 p+2)-B (a e (m+1)+b (2 d p+d))) \, _2F_1\left (m+1,-2 p;m+2;\frac{b (d+e x)}{b d-a e}\right )}{b e^2 (m+1) (m+2 p+2)}+\frac{B (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{m+1}}{b e (m+2 p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.128592, size = 125, normalized size = 0.72 \[ \frac{\left ((a+b x)^2\right )^p (d+e x)^{m+1} \left (B e (a+b x)-\frac{\left (\frac{e (a+b x)}{a e-b d}\right )^{-2 p} (a B e (m+1)-A b e (m+2 p+2)+b B (2 d p+d)) \, _2F_1\left (m+1,-2 p;m+2;\frac{b (d+e x)}{b d-a e}\right )}{m+1}\right )}{b e^2 (m+2 p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((a + b*x)^2)^p*(d + e*x)^(1 + m)*(B*e*(a + b*x) - ((a*B*e*(1 + m) - A*b*e*(2 + m + 2*p) + b*B*(d + 2*d*p))*H
ypergeometric2F1[1 + m, -2*p, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((1 + m)*((e*(a + b*x))/(-(b*d) + a*e))^(2*p)
)))/(b*e^2*(2 + m + 2*p))

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Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int \left ( Bx+A \right ) \left ( ex+d \right ) ^{m} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(d + e*x)**m*((a + b*x)**2)**p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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