3.1756 \(\int (d+e x)^m (a^2+2 a b x+b^2 x^2)^{5+p} \, dx\)

Optimal. Leaf size=97 \[ \frac {(b d-a e)^{10} \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{m+1} \left (-\frac {e (a+b x)}{b d-a e}\right )^{-2 p} \, _2F_1\left (m+1,-2 (p+5);m+2;\frac {b (d+e x)}{b d-a e}\right )}{e^{11} (m+1)} \]

[Out]

(-a*e+b*d)^10*(e*x+d)^(1+m)*(b^2*x^2+2*a*b*x+a^2)^p*hypergeom([1+m, -10-2*p],[2+m],b*(e*x+d)/(-a*e+b*d))/e^11/
(1+m)/((-e*(b*x+a)/(-a*e+b*d))^(2*p))

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Rubi [A]  time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {646, 70, 69} \[ \frac {(b d-a e)^{10} \left (a^2+2 a b x+b^2 x^2\right )^p (d+e x)^{m+1} \left (-\frac {e (a+b x)}{b d-a e}\right )^{-2 p} \, _2F_1\left (m+1,-2 (p+5);m+2;\frac {b (d+e x)}{b d-a e}\right )}{e^{11} (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m*(a^2 + 2*a*b*x + b^2*x^2)^(5 + p),x]

[Out]

((b*d - a*e)^10*(d + e*x)^(1 + m)*(a^2 + 2*a*b*x + b^2*x^2)^p*Hypergeometric2F1[1 + m, -2*(5 + p), 2 + m, (b*(
d + e*x))/(b*d - a*e)])/(e^11*(1 + m)*(-((e*(a + b*x))/(b*d - a*e)))^(2*p))

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]))

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 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int (d+e x)^m \left (a^2+2 a b x+b^2 x^2\right )^{5+p} \, 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 )^{2 (5+p)} (d+e x)^m \, dx}{b^{10}}\\ &=\frac {\left (\left (-b^2 d+a b e\right )^{10} \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 (5+p)} \, dx}{b^{10} e^{10}}\\ &=\frac {(b d-a e)^{10} \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 (5+p);2+m;\frac {b (d+e x)}{b d-a e}\right )}{e^{11} (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 87, normalized size = 0.90 \[ \frac {(b d-a e)^{10} \left ((a+b x)^2\right )^p (d+e x)^{m+1} \left (\frac {e (a+b x)}{a e-b d}\right )^{-2 p} \, _2F_1\left (m+1,-2 (p+5);m+2;\frac {b (d+e x)}{b d-a e}\right )}{e^{11} (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^m*(a^2 + 2*a*b*x + b^2*x^2)^(5 + p),x]

[Out]

((b*d - a*e)^10*((a + b*x)^2)^p*(d + e*x)^(1 + m)*Hypergeometric2F1[1 + m, -2*(5 + p), 2 + m, (b*(d + e*x))/(b
*d - a*e)])/(e^11*(1 + m)*((e*(a + b*x))/(-(b*d) + a*e))^(2*p))

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p + 5} {\left (e x + d\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(5+p),x, algorithm="fricas")

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2)^(p + 5)*(e*x + d)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p + 5} {\left (e x + d\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(5+p),x, algorithm="giac")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^(p + 5)*(e*x + d)^m, x)

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maple [F]  time = 1.35, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{m} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p +5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(5+p),x)

[Out]

int((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(5+p),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p + 5} {\left (e x + d\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(b^2*x^2+2*a*b*x+a^2)^(5+p),x, algorithm="maxima")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2)^(p + 5)*(e*x + d)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d+e\,x\right )}^m\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{p+5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m*(a^2 + b^2*x^2 + 2*a*b*x)^(p + 5),x)

[Out]

int((d + e*x)^m*(a^2 + b^2*x^2 + 2*a*b*x)^(p + 5), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(b**2*x**2+2*a*b*x+a**2)**(5+p),x)

[Out]

Exception raised: HeuristicGCDFailed

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