3.1089 \(\int (e x)^m (A+B x) (a+b x+c x^2)^{3/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.414225, 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} \[ \frac{A (e x)^{m+1} \left (a+b x+c x^2\right )^{3/2} F_1\left (m+1;-\frac{3}{2},-\frac{3}{2};m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1) \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{3/2} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{3/2}}+\frac{B (e x)^{m+2} \left (a+b x+c x^2\right )^{3/2} F_1\left (m+2;-\frac{3}{2},-\frac{3}{2};m+3;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (m+2) \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{3/2} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{3/2}} \]

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

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 (e x)^m (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx &=A \int (e x)^m \left (a+b x+c x^2\right )^{3/2} \, dx+\frac{B \int (e x)^{1+m} \left (a+b x+c x^2\right )^{3/2} \, dx}{e}\\ &=\frac{\left (B \left (a+b x+c x^2\right )^{3/2}\right ) \operatorname{Subst}\left (\int 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 (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}}+\frac{\left (A \left (a+b x+c x^2\right )^{3/2}\right ) \operatorname{Subst}\left (\int 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 (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}}\\ &=\frac{A (e x)^{1+m} \left (a+b x+c x^2\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 (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}}+\frac{B (e x)^{2+m} \left (a+b x+c x^2\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 (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}}\\ \end{align*}

Mathematica [A]  time = 0.486495, size = 405, normalized size = 1.44 \[ \frac{x (e x)^m \sqrt{a+x (b+c x)} \left ((m+1) x \left (\left (m^2+7 m+12\right ) (a B+A b) F_1\left (m+2;-\frac{1}{2},-\frac{1}{2};m+3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+(m+2) x \left ((m+4) (A c+b B) F_1\left (m+3;-\frac{1}{2},-\frac{1}{2};m+4;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+B c (m+3) x F_1\left (m+4;-\frac{1}{2},-\frac{1}{2};m+5;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )\right )\right )+a A \left (m^3+9 m^2+26 m+24\right ) F_1\left (m+1;-\frac{1}{2},-\frac{1}{2};m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )\right )}{(m+1) (m+2) (m+3) (m+4) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}}} \]

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*Sqrt[a + x*(b + c*x)]*(a*A*(24 + 26*m + 9*m^2 + m^3)*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])] + (1 + m)*x*((A*b + a*B)*(12 + 7*m + m^2)*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])] + (2 + m)*x*((b*B
+ A*c)*(4 + m)*AppellF1[3 + m, -1/2, -1/2, 4 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4
*a*c])] + B*c*(3 + m)*x*AppellF1[4 + m, -1/2, -1/2, 5 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqr
t[b^2 - 4*a*c])]))))/((1 + m)*(2 + m)*(3 + m)*(4 + 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 = 0.066, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( Bx+A \right ) \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}\, 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)^(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., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (B x + A\right )} \left (e x\right )^{m}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B c x^{3} +{\left (B b + A c\right )} x^{2} + A a +{\left (B a + A b\right )} x\right )} \sqrt{c x^{2} + b x + a} \left (e x\right )^{m}, 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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{m} \left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac{3}{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)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (B x + A\right )} \left (e x\right )^{m}\,{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)^(3/2),x, algorithm="giac")

[Out]

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