3.950 \(\int \frac{(d+e x)^m (f+g x)}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=388 \[ \frac{(e f-d g) (d+e x)^{m+1} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e^2 (m+1) \sqrt{a+b x+c x^2}}+\frac{g (d+e x)^{m+2} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e^2 (m+2) \sqrt{a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.346535, antiderivative size = 388, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {843, 759, 133} \[ \frac{(e f-d g) (d+e x)^{m+1} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e^2 (m+1) \sqrt{a+b x+c x^2}}+\frac{g (d+e x)^{m+2} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e^2 (m+2) \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*f - d*g)*(d + e*x)^(1 + m)*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d
+ e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*AppellF1[1 + m, 1/2, 1/2, 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^2*(1 + m)*Sqrt[a + b*x + c*x^2])
+ (g*(d + e*x)^(2 + m)*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))
/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*AppellF1[2 + m, 1/2, 1/2, 3 + 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^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{(d+e x)^m (f+g x)}{\sqrt{a+b x+c x^2}} \, dx &=\frac{g \int \frac{(d+e x)^{1+m}}{\sqrt{a+b x+c x^2}} \, dx}{e}+\frac{(e f-d g) \int \frac{(d+e x)^m}{\sqrt{a+b x+c x^2}} \, dx}{e}\\ &=\frac{\left (g \sqrt{1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}} \sqrt{1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}}\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{\sqrt{1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}} \, dx,x,d+e x\right )}{e^2 \sqrt{a+b x+c x^2}}+\frac{\left ((e f-d g) \sqrt{1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}} \sqrt{1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{\sqrt{1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}} \, dx,x,d+e x\right )}{e^2 \sqrt{a+b x+c x^2}}\\ &=\frac{(e f-d g) (d+e x)^{1+m} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} F_1\left (1+m;\frac{1}{2},\frac{1}{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^2 (1+m) \sqrt{a+b x+c x^2}}+\frac{g (d+e x)^{2+m} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} F_1\left (2+m;\frac{1}{2},\frac{1}{2};3+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^2 (2+m) \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [F]  time = 0.742372, size = 0, normalized size = 0. \[ \int \frac{(d+e x)^m (f+g x)}{\sqrt{a+b x+c x^2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.243, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( gx+f \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+d)^m*(g*x+f)/(c*x^2+b*x+a)^(1/2),x)

[Out]

int((e*x+d)^m*(g*x+f)/(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 (g x + f\right )}{\left (e x + d\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+d)^m*(g*x+f)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)*(e*x + d)^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 (g x + f\right )}{\left (e x + d\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+d)^m*(g*x+f)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((g*x + f)*(e*x + d)^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 (d + e x\right )^{m} \left (f + g 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+d)**m*(g*x+f)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d + e*x)**m*(f + g*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 (g x + f\right )}{\left (e x + d\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+d)^m*(g*x+f)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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