3.15.35 \(\int (b d+2 c d x)^m (a+b x+c x^2)^p \, dx\) [1435]

Optimal. Leaf size=107 \[ -\frac {2 (b d+2 c d x)^{1+m} \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \, _2F_1\left (1,\frac {1}{2} (3+m+2 p);\frac {3+m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+m)} \]

[Out]

-2*(2*c*d*x+b*d)^(1+m)*(a-1/4*b^2/c+1/4*(2*c*x+b)^2/c)^(1+p)*hypergeom([1, 3/2+1/2*m+p],[3/2+1/2*m],(2*c*x+b)^
2/(-4*a*c+b^2))/(-4*a*c+b^2)/d/(1+m)

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Rubi [A]
time = 0.05, antiderivative size = 102, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {708, 372, 371} \begin {gather*} \frac {\left (a+b x+c x^2\right )^p \left (1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )^{-p} (d (b+2 c x))^{m+1} \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*(b + 2*c*x))^(1 + m)*(a + b*x + c*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (b + 2*c*x)^2/(b^2 -
4*a*c)])/(2*c*d*(1 + m)*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^p)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx &=\frac {\text {Subst}\left (\int x^m \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )^p \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=\frac {\left (2^{-1+2 p} \left (a+b x+c x^2\right )^p \left (4+\frac {(b d+2 c d x)^2}{\left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^{-p}\right ) \text {Subst}\left (\int x^m \left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^p \, dx,x,b d+2 c d x\right )}{c d}\\ &=\frac {2^{-1+2 p} (d (b+2 c x))^{1+m} \left (a+b x+c x^2\right )^p \left (4-\frac {4 (b+2 c x)^2}{b^2-4 a c}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c d (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 113, normalized size = 1.06 \begin {gather*} \frac {2^{-1-2 p} (b+2 c x) (d (b+2 c x))^m (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \, _2F_1\left (\frac {1}{2}+\frac {m}{2},-p;\frac {3}{2}+\frac {m}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1 - 2*p)*(b + 2*c*x)*(d*(b + 2*c*x))^m*(a + x*(b + c*x))^p*HypergeometricPFQ[{1/2 + m/2, -p}, {3/2 + m/2}
, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*(1 + m)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^p)

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Maple [F]
time = 0.47, size = 0, normalized size = 0.00 \[\int \left (2 c d x +b d \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,d+2\,c\,d\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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