3.2561 \(\int (d x)^m (a+b x+c x^2)^p \, dx\)

Optimal. Leaf size=137 \[ \frac {(d x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+1;-p,-p;m+2;-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

[Out]

(d*x)^(1+m)*(c*x^2+b*x+a)^p*AppellF1(1+m,-p,-p,2+m,-2*c*x/(b-(-4*a*c+b^2)^(1/2)),-2*c*x/(b+(-4*a*c+b^2)^(1/2))
)/d/(1+m)/((1+2*c*x/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x/(b+(-4*a*c+b^2)^(1/2)))^p)

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Rubi [A]  time = 0.28, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {759, 133} \[ \frac {(d x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+1;-p,-p;m+2;-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a + b*x + c*x^2)^p,x]

[Out]

((d*x)^(1 + m)*(a + b*x + c*x^2)^p*AppellF1[1 + m, -p, -p, 2 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(
b + Sqrt[b^2 - 4*a*c])])/(d*(1 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c
]))^p)

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 759

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

Rubi steps

\begin {align*} \int (d x)^m \left (a+b x+c x^2\right )^p \, dx &=\frac {\left (\left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x+c x^2\right )^p\right ) \operatorname {Subst}\left (\int x^m \left (1+\frac {2 c x}{\left (b-\sqrt {b^2-4 a c}\right ) d}\right )^p \left (1+\frac {2 c x}{\left (b+\sqrt {b^2-4 a c}\right ) d}\right )^p \, dx,x,d x\right )}{d}\\ &=\frac {(d x)^{1+m} \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (1+m;-p,-p;2+m;-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 160, normalized size = 1.17 \[ \frac {x (d x)^m \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )^{-p} (a+x (b+c x))^p F_1\left (m+1;-p,-p;m+2;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )}{m+1} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(d*x)^m*(a + x*(b + c*x))^p*AppellF1[1 + m, -p, -p, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b +
Sqrt[b^2 - 4*a*c])])/((1 + m)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a
*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c]))^p)

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.48, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a + b*x + c*x^2)^p,x)

[Out]

int((d*x)^m*(a + b*x + c*x^2)^p, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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