3.10.50 \(\int \frac {(d+e x)^m (f+g x)}{\sqrt {a+b x+c x^2}} \, dx\) [950]

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

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 388, normalized size of antiderivative = 1.00, 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 = {857, 773, 138} \begin {gather*} \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}} \end {gather*}

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 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

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 857

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]

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 ) \text {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 ) \text {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*}

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Mathematica [F]
time = 0.94, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^m (f+g x)}{\sqrt {a+b x+c x^2}} \, dx \end {gather*}

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 = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (g x +f \right )}{\sqrt {c \,x^{2}+b x +a}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m} \left (f + g x\right )}{\sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (f+g\,x\right )\,{\left (d+e\,x\right )}^m}{\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^m)/(a + b*x + c*x^2)^(1/2),x)

[Out]

int(((f + g*x)*(d + e*x)^m)/(a + b*x + c*x^2)^(1/2), x)

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