3.1093 \(\int \frac{(e x)^m (A+B x)}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=281 \[ \frac{A (e x)^{m+1} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{5/2} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{5/2} F_1\left (m+1;\frac{5}{2},\frac{5}{2};m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (a+b x+c x^2\right )^{5/2}}+\frac{B (e x)^{m+2} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{5/2} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{5/2} F_1\left (m+2;\frac{5}{2},\frac{5}{2};m+3;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (m+2) \left (a+b x+c x^2\right )^{5/2}} \]

[Out]

(A*(e*x)^(1 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^(5/2)*(1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c]))^(5/2)*Appel
lF1[1 + m, 5/2, 5/2, 2 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e*(1 + m)*(a
+ b*x + c*x^2)^(5/2)) + (B*(e*x)^(2 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^(5/2)*(1 + (2*c*x)/(b + Sqrt[b
^2 - 4*a*c]))^(5/2)*AppellF1[2 + m, 5/2, 5/2, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2
- 4*a*c])])/(e^2*(2 + m)*(a + b*x + c*x^2)^(5/2))

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Rubi [A]  time = 0.411276, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {843, 759, 133} \[ \frac{A (e x)^{m+1} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{5/2} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{5/2} F_1\left (m+1;\frac{5}{2},\frac{5}{2};m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (a+b x+c x^2\right )^{5/2}}+\frac{B (e x)^{m+2} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{5/2} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{5/2} F_1\left (m+2;\frac{5}{2},\frac{5}{2};m+3;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (m+2) \left (a+b x+c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(e*x)^(1 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^(5/2)*(1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c]))^(5/2)*Appel
lF1[1 + m, 5/2, 5/2, 2 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e*(1 + m)*(a
+ b*x + c*x^2)^(5/2)) + (B*(e*x)^(2 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^(5/2)*(1 + (2*c*x)/(b + Sqrt[b
^2 - 4*a*c]))^(5/2)*AppellF1[2 + m, 5/2, 5/2, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2
- 4*a*c])])/(e^2*(2 + m)*(a + b*x + c*x^2)^(5/2))

Rule 843

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

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{(e x)^m (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=A \int \frac{(e x)^m}{\left (a+b x+c x^2\right )^{5/2}} \, dx+\frac{B \int \frac{(e x)^{1+m}}{\left (a+b x+c x^2\right )^{5/2}} \, dx}{e}\\ &=\frac{\left (B \left (1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )^{5/2} \left (1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{\left (1+\frac{2 c x}{\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^{5/2} \left (1+\frac{2 c x}{\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^{5/2}} \, dx,x,e x\right )}{e^2 \left (a+b x+c x^2\right )^{5/2}}+\frac{\left (A \left (1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )^{5/2} \left (1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{x^m}{\left (1+\frac{2 c x}{\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^{5/2} \left (1+\frac{2 c x}{\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^{5/2}} \, dx,x,e x\right )}{e \left (a+b x+c x^2\right )^{5/2}}\\ &=\frac{A (e x)^{1+m} \left (1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )^{5/2} \left (1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )^{5/2} F_1\left (1+m;\frac{5}{2},\frac{5}{2};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 )}{e (1+m) \left (a+b x+c x^2\right )^{5/2}}+\frac{B (e x)^{2+m} \left (1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )^{5/2} \left (1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )^{5/2} F_1\left (2+m;\frac{5}{2},\frac{5}{2};3+m;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (2+m) \left (a+b x+c x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.328764, size = 237, normalized size = 0.84 \[ \frac{x (e x)^m \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}} \left (A (m+2) F_1\left (m+1;\frac{5}{2},\frac{5}{2};m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+B (m+1) x F_1\left (m+2;\frac{5}{2},\frac{5}{2};m+3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )\right )}{a^2 (m+1) (m+2) \sqrt{a+x (b+c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.071, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( Bx+A \right ) \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)^m*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)

[Out]

int((e*x)^m*(B*x+A)/(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 (B x + A\right )} \left (e x\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)^m*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^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 (B x + A\right )} \left (e x\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)^m*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)*(e*x)^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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x+A)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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

[Out]

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