3.1091 \(\int \frac{(e x)^m (A+B x)}{\sqrt{a+b x+c x^2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.406889, 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} \[ \frac{A (e x)^{m+1} \sqrt{\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1} F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \sqrt{a+b x+c x^2}}+\frac{B (e x)^{m+2} \sqrt{\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1} F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (m+2) \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(e*x)^(1 + m)*Sqrt[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[
1 + m, 1/2, 1/2, 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)*Sqrt[a
+ b*x + c*x^2]) + (B*(e*x)^(2 + m)*Sqrt[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x)/(b + Sqrt[b^2 -
4*a*c])]*AppellF1[2 + m, 1/2, 1/2, 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)*Sqrt[a + b*x + c*x^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)}{\sqrt{a+b x+c x^2}} \, dx &=A \int \frac{(e x)^m}{\sqrt{a+b x+c x^2}} \, dx+\frac{B \int \frac{(e x)^{1+m}}{\sqrt{a+b x+c x^2}} \, dx}{e}\\ &=\frac{\left (B \sqrt{1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}}\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{\sqrt{1+\frac{2 c x}{\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1+\frac{2 c x}{\left (b+\sqrt{b^2-4 a c}\right ) e}}} \, dx,x,e x\right )}{e^2 \sqrt{a+b x+c x^2}}+\frac{\left (A \sqrt{1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{\sqrt{1+\frac{2 c x}{\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1+\frac{2 c x}{\left (b+\sqrt{b^2-4 a c}\right ) e}}} \, dx,x,e x\right )}{e \sqrt{a+b x+c x^2}}\\ &=\frac{A (e x)^{1+m} \sqrt{1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}} F_1\left (1+m;\frac{1}{2},\frac{1}{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) \sqrt{a+b x+c x^2}}+\frac{B (e x)^{2+m} \sqrt{1+\frac{2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x}{b+\sqrt{b^2-4 a c}}} F_1\left (2+m;\frac{1}{2},\frac{1}{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) \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.220107, size = 234, normalized size = 0.83 \[ \frac{x (e x)^m \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}} \left (A (m+2) F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+B (m+1) x F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )\right )}{(m+1) (m+2) \sqrt{a+x (b+c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

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Maple [F]  time = 0.072, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( Bx+A \right ){\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}\, 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)^(1/2),x)

[Out]

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(1/2),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}}{\sqrt{c x^{2} + b x + a}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )} \left (e x\right )^{m}}{\sqrt{c x^{2} + b x + a}}, 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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e x\right )^{m} \left (A + B x\right )}{\sqrt{a + b x + c x^{2}}}\, dx \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)**(1/2),x)

[Out]

Integral((e*x)**m*(A + B*x)/sqrt(a + b*x + c*x**2), x)

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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}}{\sqrt{c x^{2} + b x + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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