3.1092 \(\int \frac {(e x)^m (A+B x)}{(a+b x+c x^2)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.41, 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 = {843, 759, 133} \[ \frac {A (e x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{3/2} F_1\left (m+1;\frac {3}{2},\frac {3}{2};m+2;-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (m+1) \left (a+b x+c x^2\right )^{3/2}}+\frac {B (e x)^{m+2} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{3/2} F_1\left (m+2;\frac {3}{2},\frac {3}{2};m+3;-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e^2 (m+2) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(3/2),x]

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 272, normalized size = 0.97 \[ \frac {x (e x)^m \left (\sqrt {b^2-4 a c}-b-2 c x\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )^{3/2} \left (A (m+2) F_1\left (m+1;\frac {3}{2},\frac {3}{2};m+2;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )+B (m+1) x F_1\left (m+2;\frac {3}{2},\frac {3}{2};m+3;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )\right )}{(m+1) (m+2) \left (\sqrt {b^2-4 a c}-b\right ) (a+x (b+c x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (B x + A\right )} \left (e x\right )^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)*(e*x)^m/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x)^m/(c*x^2 + b*x + a)^(3/2), x)

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maple [F]  time = 1.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) \left (e x \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(3/2),x)

[Out]

int((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} \left (e x\right )^{m}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^m/(c*x^2 + b*x + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(3/2),x)

[Out]

int(((e*x)^m*(A + B*x))/(a + b*x + c*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x\right )^{m} \left (A + B x\right )}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x+A)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((e*x)**m*(A + B*x)/(a + b*x + c*x**2)**(3/2), x)

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