3.11.90 \(\int (e x)^m (A+B x) \sqrt {a+b x+c x^2} \, dx\) [1090]

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

[Out]

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

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Rubi [A]
time = 0.28, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {857, 773, 138} \begin {gather*} \frac {A (e x)^{m+1} \sqrt {a+b x+c x^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}{b+\sqrt {b^2-4 a c}}\right )}{e (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}}+\frac {B (e x)^{m+2} \sqrt {a+b x+c x^2} 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 {\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}} \end {gather*}

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[a + b*x + c*x^2]*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[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x)/(b + Sq
rt[b^2 - 4*a*c])]) + (B*(e*x)^(2 + m)*Sqrt[a + b*x + c*x^2]*AppellF1[2 + m, -1/2, -1/2, 3 + m, (-2*c*x)/(b - S
qrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e^2*(2 + m)*Sqrt[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*S
qrt[1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c])])

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 (e x)^m (A+B x) \sqrt {a+b x+c x^2} \, dx &=A \int (e x)^m \sqrt {a+b x+c x^2} \, dx+\frac {B \int (e x)^{1+m} \sqrt {a+b x+c x^2} \, dx}{e}\\ &=\frac {\left (B \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int 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 {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}}}}+\frac {\left (A \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int 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 {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}}}}\\ &=\frac {A (e x)^{1+m} \sqrt {a+b x+c x^2} 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 {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}}}}+\frac {B (e x)^{2+m} \sqrt {a+b x+c x^2} 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 {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}}}}\\ \end {align*}

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Mathematica [A]
time = 0.45, size = 234, normalized size = 0.83 \begin {gather*} \frac {x (e x)^m \sqrt {a+x (b+c x)} \left (A (2+m) 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 )+B (1+m) x 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 )\right )}{(1+m) (2+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}}} \end {gather*}

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[a + x*(b + c*x)]*(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[(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])])

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Maple [F]
time = 0.34, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (B x +A \right ) \sqrt {c \,x^{2}+b x +a}\, dx\]

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 = 0.00, 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)^m*(B*x+A)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(B*x + A)*(x*e)^m, 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)^m*(B*x+A)*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e x\right )^{m} \left (A + B 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)**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 = 0.00, 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)^m*(B*x+A)*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(A + B*x)*(a + b*x + c*x^2)^(1/2),x)

[Out]

int((e*x)^m*(A + B*x)*(a + b*x + c*x^2)^(1/2), x)

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