### 3.2560 $$\int \frac{(d+e x)^m}{(a+b x+c x^2)^{5/2}} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.110178, antiderivative size = 189, normalized size of antiderivative = 1., 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} $\frac{(d+e x)^{m+1} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{5/2} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{5/2} F_1\left (m+1;\frac{5}{2},\frac{5}{2};m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e (m+1) \left (a+b x+c x^2\right )^{5/2}}$

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

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.354907, size = 225, normalized size = 1.19 $\frac{e^3 (d+e x)^{m+1} \sqrt{\frac{e \left (\sqrt{b^2-4 a c}-b-2 c x\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}} \sqrt{\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt{b^2-4 a c}+b\right )-2 c d}} F_1\left (m+1;\frac{5}{2},\frac{5}{2};m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{(m+1) \sqrt{a+x (b+c x)} \left (e (a e-b d)+c d^2\right )^2}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e^3*Sqrt[(e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[(e*(b + Sqrt[b^2 - 4
*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]*(d + e*x)^(1 + m)*AppellF1[1 + m, 5/2, 5/2, 2 + m, (2*c*
(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/((c*d^2
+ e*(-(b*d) + a*e))^2*(1 + m)*Sqrt[a + x*(b + c*x)])

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Maple [F]  time = 1.259, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x + a}{\left (e x + d\right )}^{m}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \end{align*}

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(sqrt(c*x^2 + b*x + a)*(e*x + d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 +
6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{m}}{\left (a + b x + c x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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