3.2162 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^p}{d+e x} \, dx\)

Optimal. Leaf size=74 \[ \frac{(a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (1,2 (p+1);2 p+3;-\frac{e (a+b x)}{b d-a e}\right )}{2 (p+1) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0509769, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {770, 21, 68} \[ \frac{(a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, _2F_1\left (1,2 (p+1);2 p+3;-\frac{e (a+b x)}{b d-a e}\right )}{2 (p+1) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^p*Hypergeometric2F1[1, 2*(1 + p), 3 + 2*p, -((e*(a + b*x))/(b*d - a*e))
])/(2*(b*d - a*e)*(1 + 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 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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.0193485, size = 59, normalized size = 0.8 \[ \frac{\left ((a+b x)^2\right )^{p+1} \, _2F_1\left (1,2 (p+1);2 p+3;\frac{e (a+b x)}{a e-b d}\right )}{2 (p+1) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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