3.2555 \(\int (d+e x)^m (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=189 \[ \frac {\left (a+b x+c x^2\right )^{5/2} (d+e x)^{m+1} F_1\left (m+1;-\frac {5}{2},-\frac {5}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{5/2}} \]

[Out]

(e*x+d)^(1+m)*(c*x^2+b*x+a)^(5/2)*AppellF1(1+m,-5/2,-5/2,2+m,2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*
(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))/e/(1+m)/(1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^(5/2)/(1-2*
c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(5/2)

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Rubi [A]  time = 0.25, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {759, 133} \[ \frac {\left (a+b x+c x^2\right )^{5/2} (d+e x)^{m+1} F_1\left (m+1;-\frac {5}{2},-\frac {5}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (m+1) \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(a + b*x + c*x^2)^(5/2)*AppellF1[1 + m, -5/2, -5/2, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sq
rt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(e*(1 + m)*(1 - (2*c*(d + e*x))/(2*
c*d - (b - Sqrt[b^2 - 4*a*c])*e))^(5/2)*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^(5/2))

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+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac {\left (a+b x+c x^2\right )^{5/2} \operatorname {Subst}\left (\int x^m \left (1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \, dx,x,d+e x\right )}{e \left (1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{5/2} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{5/2}}\\ &=\frac {(d+e x)^{1+m} \left (a+b x+c x^2\right )^{5/2} F_1\left (1+m;-\frac {5}{2},-\frac {5}{2};2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m) \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{5/2} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{5/2}}\\ \end {align*}

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Mathematica [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^{5/2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.34, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt {c x^{2} + b x + a} {\left (e x + d\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*sqrt(c*x^2 + b*x + a)*(e*x + d)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.59, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} \left (e x +d \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: HeuristicGCDFailed

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