3.940 \(\int \frac{(A+B x) (a+b x+c x^2)^{5/2}}{x} \, dx\)

Optimal. Leaf size=350 \[ \frac{\sqrt{a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{512 c^3}+\frac{\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{7/2}}+a^{5/2} (-A) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{192 c^2}+\frac{\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c} \]

[Out]

((512*a^2*A*c^3 + b*(64*a*A*b*c^2 + (b^2 - 4*a*c)*(5*b^2*B - 12*A*b*c - 20*a*B*c)) + 2*c*(64*a*A*b*c^2 + (b^2
- 4*a*c)*(5*b^2*B - 12*A*b*c - 20*a*B*c))*x)*Sqrt[a + b*x + c*x^2])/(512*c^3) - ((5*b^3*B - 12*A*b^2*c - 20*a*
b*B*c - 64*a*A*c^2 + 2*c*(5*b^2*B - 12*A*b*c - 20*a*B*c)*x)*(a + b*x + c*x^2)^(3/2))/(192*c^2) + ((5*b*B + 12*
A*c + 10*B*c*x)*(a + b*x + c*x^2)^(5/2))/(60*c) - a^(5/2)*A*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^
2])] + ((512*a^2*A*b*c^3 - (b^2 - 4*a*c)*(5*b^4*B - 12*A*b^3*c - 40*a*b^2*B*c + 112*a*A*b*c^2 + 80*a^2*B*c^2))
*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.419511, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {814, 843, 621, 206, 724} \[ \frac{\sqrt{a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b^2 c^2\right )}{512 c^3}+\frac{\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 c^{7/2}}+a^{5/2} (-A) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )-\frac{\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{192 c^2}+\frac{\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x,x]

[Out]

((64*a*A*b^2*c^2 + 512*a^2*A*c^3 + b*(b^2 - 4*a*c)*(5*b^2*B - 12*A*b*c - 20*a*B*c) + 2*c*(64*a*A*b*c^2 + (b^2
- 4*a*c)*(5*b^2*B - 12*A*b*c - 20*a*B*c))*x)*Sqrt[a + b*x + c*x^2])/(512*c^3) - ((5*b^3*B - 12*A*b^2*c - 20*a*
b*B*c - 64*a*A*c^2 + 2*c*(5*b^2*B - 12*A*b*c - 20*a*B*c)*x)*(a + b*x + c*x^2)^(3/2))/(192*c^2) + ((5*b*B + 12*
A*c + 10*B*c*x)*(a + b*x + c*x^2)^(5/2))/(60*c) - a^(5/2)*A*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^
2])] + ((512*a^2*A*b*c^3 - (b^2 - 4*a*c)*(5*b^4*B - 12*A*b^3*c - 40*a*b^2*B*c + 112*a*A*b*c^2 + 80*a^2*B*c^2))
*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1024*c^(7/2))

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*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]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx &=\frac{(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}-\frac{\int \frac{\left (-12 a A c-\frac{1}{2} \left (12 A b c-5 B \left (b^2-4 a c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx}{12 c}\\ &=-\frac{\left (5 b^3 B-12 A b^2 c-20 a b B c-64 a A c^2+2 c \left (5 b^2 B-12 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}+\frac{\int \frac{\left (96 a^2 A c^2+\frac{3}{4} \left (64 a A b c^2-\left (b^2-4 a c\right ) \left (12 A b c-5 B \left (b^2-4 a c\right )\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{x} \, dx}{96 c^2}\\ &=\frac{\left (64 a A b^2 c^2+512 a^2 A c^3+b \left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )+2 c \left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{\left (5 b^3 B-12 A b^2 c-20 a b B c-64 a A c^2+2 c \left (5 b^2 B-12 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}-\frac{\int \frac{-384 a^3 A c^3-\frac{3}{8} \left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (5 b^4 B-12 A b^3 c-40 a b^2 B c+112 a A b c^2+80 a^2 B c^2\right )\right ) x}{x \sqrt{a+b x+c x^2}} \, dx}{384 c^3}\\ &=\frac{\left (64 a A b^2 c^2+512 a^2 A c^3+b \left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )+2 c \left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{\left (5 b^3 B-12 A b^2 c-20 a b B c-64 a A c^2+2 c \left (5 b^2 B-12 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}+\left (a^3 A\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx+\frac{\left (320 a^3 B-\frac{b^5 (5 b B-12 A c)}{c^3}+\frac{20 a b^3 (3 b B-8 A c)}{c^2}-\frac{240 a^2 b (b B-4 A c)}{c}\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{1024}\\ &=\frac{\left (64 a A b^2 c^2+512 a^2 A c^3+b \left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )+2 c \left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{\left (5 b^3 B-12 A b^2 c-20 a b B c-64 a A c^2+2 c \left (5 b^2 B-12 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}-\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )+\frac{1}{512} \left (320 a^3 B-\frac{b^5 (5 b B-12 A c)}{c^3}+\frac{20 a b^3 (3 b B-8 A c)}{c^2}-\frac{240 a^2 b (b B-4 A c)}{c}\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )\\ &=\frac{\left (64 a A b^2 c^2+512 a^2 A c^3+b \left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )+2 c \left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{512 c^3}-\frac{\left (5 b^3 B-12 A b^2 c-20 a b B c-64 a A c^2+2 c \left (5 b^2 B-12 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac{(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}-a^{5/2} A \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )+\frac{\left (320 a^3 B-\frac{b^5 (5 b B-12 A c)}{c^3}+\frac{20 a b^3 (3 b B-8 A c)}{c^2}-\frac{240 a^2 b (b B-4 A c)}{c}\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{1024 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.491493, size = 349, normalized size = 1. \[ \frac{\sqrt{a+x (b+c x)} \left (32 a^2 c^3 (16 A+5 B x)+16 a b^2 c^2 (7 A-5 B x)-8 b^3 c (5 a B+3 A c x)+16 a b c^2 (5 a B+14 A c x)-2 b^4 c (6 A-5 B x)+5 b^5 B\right )}{512 c^3}+\frac{\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{1024 c^{7/2}}+a^{5/2} (-A) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )+\frac{(a+x (b+c x))^{3/2} \left (4 b c (5 a B+6 A c x)+8 a c^2 (8 A+5 B x)+2 b^2 c (6 A-5 B x)-5 b^3 B\right )}{192 c^2}+\frac{(a+x (b+c x))^{5/2} (2 c (6 A+5 B x)+5 b B)}{60 c} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^(5/2))/x,x]

[Out]

((5*b*B + 2*c*(6*A + 5*B*x))*(a + x*(b + c*x))^(5/2))/(60*c) + ((a + x*(b + c*x))^(3/2)*(-5*b^3*B + 2*b^2*c*(6
*A - 5*B*x) + 8*a*c^2*(8*A + 5*B*x) + 4*b*c*(5*a*B + 6*A*c*x)))/(192*c^2) + (Sqrt[a + x*(b + c*x)]*(5*b^5*B -
2*b^4*c*(6*A - 5*B*x) + 16*a*b^2*c^2*(7*A - 5*B*x) + 32*a^2*c^3*(16*A + 5*B*x) - 8*b^3*c*(5*a*B + 3*A*c*x) + 1
6*a*b*c^2*(5*a*B + 14*A*c*x)))/(512*c^3) - a^(5/2)*A*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])] +
((512*a^2*A*b*c^3 - (b^2 - 4*a*c)*(5*b^4*B - 12*A*b^3*c - 40*a*b^2*B*c + 112*a*A*b*c^2 + 80*a^2*B*c^2))*ArcTan
h[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(1024*c^(7/2))

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Maple [B]  time = 0.008, size = 694, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^(5/2)/x,x)

[Out]

-5/32*B/c*(c*x^2+b*x+a)^(1/2)*x*a*b^2+1/5*A*(c*x^2+b*x+a)^(5/2)-5/64*B/c^2*(c*x^2+b*x+a)^(1/2)*b^3*a-15/64*B/c
^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a^2+5/256*B/c^2*(c*x^2+b*x+a)^(1/2)*x*b^4+5/32*B/c*(c*x
^2+b*x+a)^(1/2)*b*a^2+15/256*B/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^4*a-A*a^(5/2)*ln((2*a+b*x
+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/3*A*a*(c*x^2+b*x+a)^(3/2)+A*a^2*(c*x^2+b*x+a)^(1/2)+1/6*B*x*(c*x^2+b*x+a)
^(5/2)+7/32*A/c*(c*x^2+b*x+a)^(1/2)*b^2*a+7/16*A*b*(c*x^2+b*x+a)^(1/2)*x*a-3/64*A/c*(c*x^2+b*x+a)^(1/2)*x*b^3+
15/16*A*b/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-5/32*A/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2
+b*x+a)^(1/2))*b^3*a-5/1024*B/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^6+1/12*B/c*(c*x^2+b*x+a)^(
5/2)*b+5/24*B*(c*x^2+b*x+a)^(3/2)*x*a-5/192*B/c^2*(c*x^2+b*x+a)^(3/2)*b^3+5/16*B*(c*x^2+b*x+a)^(1/2)*x*a^2+5/5
12*B/c^3*(c*x^2+b*x+a)^(1/2)*b^5+5/16*B/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^3+1/8*A*b*(c*x^2
+b*x+a)^(3/2)*x+1/16*A/c*(c*x^2+b*x+a)^(3/2)*b^2-3/128*A/c^2*(c*x^2+b*x+a)^(1/2)*b^4+3/256*A/c^(5/2)*ln((1/2*b
+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^5-5/96*B/c*(c*x^2+b*x+a)^(3/2)*x*b^2+5/48*B/c*(c*x^2+b*x+a)^(3/2)*b*a

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 96.4683, size = 3779, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x,x, algorithm="fricas")

[Out]

[1/30720*(15360*A*a^(5/2)*c^4*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a)
+ 8*a^2)/x^2) - 15*(5*B*b^6 - 320*(B*a^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 12*(5*B*a*b^4 +
A*b^5)*c)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(
1280*B*c^6*x^5 + 75*B*b^5*c + 11776*A*a^2*c^4 + 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 240*(11*B*a^2*b + 9*A*a*b^2)
*c^3 + 16*(135*B*b^2*c^4 + 4*(65*B*a + 63*A*b)*c^5)*x^3 - 20*(40*B*a*b^3 + 9*A*b^4)*c^2 + 8*(5*B*b^3*c^3 + 704
*A*a*c^5 + 12*(65*B*a*b + 31*A*b^2)*c^4)*x^2 - 2*(25*B*b^4*c^2 - 16*(165*B*a^2 + 311*A*a*b)*c^4 - 60*(4*B*a*b^
2 + A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/15360*(7680*A*a^(5/2)*c^4*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 -
4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 15*(5*B*b^6 - 320*(B*a^3 + 3*A*a^2*b)*c^3 + 80*(3
*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 12*(5*B*a*b^4 + A*b^5)*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*
sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(1280*B*c^6*x^5 + 75*B*b^5*c + 11776*A*a^2*c^4 + 128*(25*B*b*c^5 + 12*A*
c^6)*x^4 + 240*(11*B*a^2*b + 9*A*a*b^2)*c^3 + 16*(135*B*b^2*c^4 + 4*(65*B*a + 63*A*b)*c^5)*x^3 - 20*(40*B*a*b^
3 + 9*A*b^4)*c^2 + 8*(5*B*b^3*c^3 + 704*A*a*c^5 + 12*(65*B*a*b + 31*A*b^2)*c^4)*x^2 - 2*(25*B*b^4*c^2 - 16*(16
5*B*a^2 + 311*A*a*b)*c^4 - 60*(4*B*a*b^2 + A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/30720*(30720*A*sqrt(-a
)*a^2*c^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 15*(5*B*b^6 - 320*(
B*a^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 12*(5*B*a*b^4 + A*b^5)*c)*sqrt(c)*log(-8*c^2*x^2 -
8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(1280*B*c^6*x^5 + 75*B*b^5*c + 11776
*A*a^2*c^4 + 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 240*(11*B*a^2*b + 9*A*a*b^2)*c^3 + 16*(135*B*b^2*c^4 + 4*(65*B*
a + 63*A*b)*c^5)*x^3 - 20*(40*B*a*b^3 + 9*A*b^4)*c^2 + 8*(5*B*b^3*c^3 + 704*A*a*c^5 + 12*(65*B*a*b + 31*A*b^2)
*c^4)*x^2 - 2*(25*B*b^4*c^2 - 16*(165*B*a^2 + 311*A*a*b)*c^4 - 60*(4*B*a*b^2 + A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x
+ a))/c^4, 1/15360*(15360*A*sqrt(-a)*a^2*c^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 +
a*b*x + a^2)) + 15*(5*B*b^6 - 320*(B*a^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 12*(5*B*a*b^4
+ A*b^5)*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(1280*
B*c^6*x^5 + 75*B*b^5*c + 11776*A*a^2*c^4 + 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 240*(11*B*a^2*b + 9*A*a*b^2)*c^3
+ 16*(135*B*b^2*c^4 + 4*(65*B*a + 63*A*b)*c^5)*x^3 - 20*(40*B*a*b^3 + 9*A*b^4)*c^2 + 8*(5*B*b^3*c^3 + 704*A*a*
c^5 + 12*(65*B*a*b + 31*A*b^2)*c^4)*x^2 - 2*(25*B*b^4*c^2 - 16*(165*B*a^2 + 311*A*a*b)*c^4 - 60*(4*B*a*b^2 + A
*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac{5}{2}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**(5/2)/x,x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(5/2)/x, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x,x, algorithm="giac")

[Out]

Exception raised: TypeError