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

Optimal. Leaf size=310 \[ -\frac {1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-128 a^2 B c^2+2 c x \left (-120 a A c^2-28 a b B c-10 A b^2 c+3 b^3 B\right )-440 a A b c^2-28 a b^2 B c-10 A b^3 c+3 b^4 B\right )}{128 c^2}+\frac {\left (480 a^2 A c^3+240 a^2 b B c^2+240 a A b^2 c^2-40 a b^3 B c-10 A b^4 c+3 b^5 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} \left (16 a B c+6 c x (10 A c+b B)+70 A b c+3 b^2 B\right )}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x} \]

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Rubi [A]  time = 0.31, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {812, 814, 843, 621, 206, 724} \begin {gather*} -\frac {\sqrt {a+b x+c x^2} \left (-128 a^2 B c^2+2 c x \left (-120 a A c^2-28 a b B c-10 A b^2 c+3 b^3 B\right )-440 a A b c^2-28 a b^2 B c-10 A b^3 c+3 b^4 B\right )}{128 c^2}+\frac {\left (480 a^2 A c^3+240 a^2 b B c^2+240 a A b^2 c^2-40 a b^3 B c-10 A b^4 c+3 b^5 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}-\frac {1}{2} a^{3/2} (2 a B+5 A b) \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 (16 a B c+6 c x (10 A c+b B)+70 A b c+3 b^2 B\right )}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b^4*B - 10*A*b^3*c - 28*a*b^2*B*c - 440*a*A*b*c^2 - 128*a^2*B*c^2 + 2*c*(3*b^3*B - 10*A*b^2*c - 28*a*b*B*
c - 120*a*A*c^2)*x)*Sqrt[a + b*x + c*x^2])/(128*c^2) + ((3*b^2*B + 70*A*b*c + 16*a*B*c + 6*c*(b*B + 10*A*c)*x)
*(a + b*x + c*x^2)^(3/2))/(48*c) - ((5*A - B*x)*(a + b*x + c*x^2)^(5/2))/(5*x) - (a^(3/2)*(5*A*b + 2*a*B)*ArcT
anh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/2 + ((3*b^5*B - 10*A*b^4*c - 40*a*b^3*B*c + 240*a*A*b^2*c^
2 + 240*a^2*b*B*c^2 + 480*a^2*A*c^3)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(5/2))

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 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 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]

Rule 812

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

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]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^2} \, dx &=-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x}-\frac {1}{2} \int \frac {(-5 A b-2 a B-(b B+10 A c) x) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx\\ &=\frac {\left (3 b^2 B+70 A b c+16 a B c+6 c (b B+10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x}+\frac {\int \frac {\left (8 a (5 A b+2 a B) c-\frac {1}{2} \left (3 b^3 B-10 A b^2 c-28 a b B c-120 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{x} \, dx}{16 c}\\ &=-\frac {\left (3 b^4 B-10 A b^3 c-28 a b^2 B c-440 a A b c^2-128 a^2 B c^2+2 c \left (3 b^3 B-10 A b^2 c-28 a b B c-120 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{128 c^2}+\frac {\left (3 b^2 B+70 A b c+16 a B c+6 c (b B+10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x}-\frac {\int \frac {-32 a^2 (5 A b+2 a B) c^2-\frac {1}{4} \left (3 b^5 B-10 A b^4 c-40 a b^3 B c+240 a A b^2 c^2+240 a^2 b B c^2+480 a^2 A c^3\right ) x}{x \sqrt {a+b x+c x^2}} \, dx}{64 c^2}\\ &=-\frac {\left (3 b^4 B-10 A b^3 c-28 a b^2 B c-440 a A b c^2-128 a^2 B c^2+2 c \left (3 b^3 B-10 A b^2 c-28 a b B c-120 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{128 c^2}+\frac {\left (3 b^2 B+70 A b c+16 a B c+6 c (b B+10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x}+\frac {1}{2} \left (a^2 (5 A b+2 a B)\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx+\frac {\left (3 b^5 B-10 A b^4 c-40 a b^3 B c+240 a A b^2 c^2+240 a^2 b B c^2+480 a^2 A c^3\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^2}\\ &=-\frac {\left (3 b^4 B-10 A b^3 c-28 a b^2 B c-440 a A b c^2-128 a^2 B c^2+2 c \left (3 b^3 B-10 A b^2 c-28 a b B c-120 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{128 c^2}+\frac {\left (3 b^2 B+70 A b c+16 a B c+6 c (b B+10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x}-\left (a^2 (5 A b+2 a B)\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 {\left (3 b^5 B-10 A b^4 c-40 a b^3 B c+240 a A b^2 c^2+240 a^2 b B c^2+480 a^2 A c^3\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 )}{128 c^2}\\ &=-\frac {\left (3 b^4 B-10 A b^3 c-28 a b^2 B c-440 a A b c^2-128 a^2 B c^2+2 c \left (3 b^3 B-10 A b^2 c-28 a b B c-120 a A c^2\right ) x\right ) \sqrt {a+b x+c x^2}}{128 c^2}+\frac {\left (3 b^2 B+70 A b c+16 a B c+6 c (b B+10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c}-\frac {(5 A-B x) \left (a+b x+c x^2\right )^{5/2}}{5 x}-\frac {1}{2} a^{3/2} (5 A b+2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )+\frac {\left (3 b^5 B-10 A b^4 c-40 a b^3 B c+240 a A b^2 c^2+240 a^2 b B c^2+480 a^2 A c^3\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 288, normalized size = 0.93 \begin {gather*} -\frac {1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )+\frac {\sqrt {a+x (b+c x)} \left (-128 a^2 c^2 (15 A-23 B x)+4 a c x \left (2 b c (695 A+311 B x)+4 c^2 x (135 A+88 B x)+135 b^2 B\right )+x \left (30 b^3 c (5 A+B x)+4 b^2 c^2 x (295 A+186 B x)+16 b c^3 x^2 (85 A+63 B x)+96 c^4 x^3 (5 A+4 B x)-45 b^4 B\right )\right )}{1920 c^2 x}+\frac {\left (480 a^2 A c^3+240 a^2 b B c^2+240 a A b^2 c^2-40 a b^3 B c-10 A b^4 c+3 b^5 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{256 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*(-128*a^2*c^2*(15*A - 23*B*x) + x*(-45*b^4*B + 30*b^3*c*(5*A + B*x) + 96*c^4*x^3*(5*A +
4*B*x) + 16*b*c^3*x^2*(85*A + 63*B*x) + 4*b^2*c^2*x*(295*A + 186*B*x)) + 4*a*c*x*(135*b^2*B + 4*c^2*x*(135*A
+ 88*B*x) + 2*b*c*(695*A + 311*B*x))))/(1920*c^2*x) - (a^(3/2)*(5*A*b + 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*
Sqrt[a + x*(b + c*x)])])/2 + ((3*b^5*B - 10*A*b^4*c - 40*a*b^3*B*c + 240*a*A*b^2*c^2 + 240*a^2*b*B*c^2 + 480*a
^2*A*c^3)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(256*c^(5/2))

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IntegrateAlgebraic [A]  time = 1.86, size = 329, normalized size = 1.06 \begin {gather*} \left (5 a^{3/2} A b+2 a^{5/2} B\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )+\frac {\sqrt {a+b x+c x^2} \left (-1920 a^2 A c^2+2944 a^2 B c^2 x+5560 a A b c^2 x+2160 a A c^3 x^2+540 a b^2 B c x+2488 a b B c^2 x^2+1408 a B c^3 x^3+150 A b^3 c x+1180 A b^2 c^2 x^2+1360 A b c^3 x^3+480 A c^4 x^4-45 b^4 B x+30 b^3 B c x^2+744 b^2 B c^2 x^3+1008 b B c^3 x^4+384 B c^4 x^5\right )}{1920 c^2 x}+\frac {\left (-480 a^2 A c^3-240 a^2 b B c^2-240 a A b^2 c^2+40 a b^3 B c+10 A b^4 c-3 b^5 B\right ) \log \left (-2 c^{5/2} \sqrt {a+b x+c x^2}+b c^2+2 c^3 x\right )}{256 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-1920*a^2*A*c^2 - 45*b^4*B*x + 150*A*b^3*c*x + 540*a*b^2*B*c*x + 5560*a*A*b*c^2*x + 29
44*a^2*B*c^2*x + 30*b^3*B*c*x^2 + 1180*A*b^2*c^2*x^2 + 2488*a*b*B*c^2*x^2 + 2160*a*A*c^3*x^2 + 744*b^2*B*c^2*x
^3 + 1360*A*b*c^3*x^3 + 1408*a*B*c^3*x^3 + 1008*b*B*c^3*x^4 + 480*A*c^4*x^4 + 384*B*c^4*x^5))/(1920*c^2*x) + (
5*a^(3/2)*A*b + 2*a^(5/2)*B)*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + b*x + c*x^2]/Sqrt[a]] + ((-3*b^5*B + 10*A*
b^4*c + 40*a*b^3*B*c - 240*a*A*b^2*c^2 - 240*a^2*b*B*c^2 - 480*a^2*A*c^3)*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt
[a + b*x + c*x^2]])/(256*c^(5/2))

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fricas [A]  time = 8.46, size = 1393, normalized size = 4.49

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/7680*(1920*(2*B*a^2 + 5*A*a*b)*sqrt(a)*c^3*x*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*(3*B*b^5 + 480*A*a^2*c^3 + 240*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 +
A*b^4)*c)*sqrt(c)*x*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*
(384*B*c^5*x^5 - 1920*A*a^2*c^3 + 48*(21*B*b*c^4 + 10*A*c^5)*x^4 + 8*(93*B*b^2*c^3 + 2*(88*B*a + 85*A*b)*c^4)*
x^3 + 2*(15*B*b^3*c^2 + 1080*A*a*c^4 + 2*(622*B*a*b + 295*A*b^2)*c^3)*x^2 - (45*B*b^4*c - 8*(368*B*a^2 + 695*A
*a*b)*c^3 - 30*(18*B*a*b^2 + 5*A*b^3)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*x), 1/3840*(960*(2*B*a^2 + 5*A*a*b)*
sqrt(a)*c^3*x*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*(3*B*b^5 + 480*A*a^2*c^3 + 240*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 + A*b^4)*c)*sqrt(-c)*x*arctan(1/2*sq
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(384*B*c^5*x^5 - 1920*A*a^2*c^3 + 48*(21
*B*b*c^4 + 10*A*c^5)*x^4 + 8*(93*B*b^2*c^3 + 2*(88*B*a + 85*A*b)*c^4)*x^3 + 2*(15*B*b^3*c^2 + 1080*A*a*c^4 + 2
*(622*B*a*b + 295*A*b^2)*c^3)*x^2 - (45*B*b^4*c - 8*(368*B*a^2 + 695*A*a*b)*c^3 - 30*(18*B*a*b^2 + 5*A*b^3)*c^
2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*x), 1/7680*(3840*(2*B*a^2 + 5*A*a*b)*sqrt(-a)*c^3*x*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*(3*B*b^5 + 480*A*a^2*c^3 + 240*(B*a^2*b + A*a*b^2)
*c^2 - 10*(4*B*a*b^3 + A*b^4)*c)*sqrt(c)*x*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*(384*B*c^5*x^5 - 1920*A*a^2*c^3 + 48*(21*B*b*c^4 + 10*A*c^5)*x^4 + 8*(93*B*b^2*c^3 + 2*
(88*B*a + 85*A*b)*c^4)*x^3 + 2*(15*B*b^3*c^2 + 1080*A*a*c^4 + 2*(622*B*a*b + 295*A*b^2)*c^3)*x^2 - (45*B*b^4*c
- 8*(368*B*a^2 + 695*A*a*b)*c^3 - 30*(18*B*a*b^2 + 5*A*b^3)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*x), 1/3840*(1
920*(2*B*a^2 + 5*A*a*b)*sqrt(-a)*c^3*x*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*(3*B*b^5 + 480*A*a^2*c^3 + 240*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 + A*b^4)*c)*sqrt(-c)*x*arc
tan(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*(384*B*c^5*x^5 - 1920*A*a^2*c^
3 + 48*(21*B*b*c^4 + 10*A*c^5)*x^4 + 8*(93*B*b^2*c^3 + 2*(88*B*a + 85*A*b)*c^4)*x^3 + 2*(15*B*b^3*c^2 + 1080*A
*a*c^4 + 2*(622*B*a*b + 295*A*b^2)*c^3)*x^2 - (45*B*b^4*c - 8*(368*B*a^2 + 695*A*a*b)*c^3 - 30*(18*B*a*b^2 + 5
*A*b^3)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(c^3*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%{[1,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,0,0,0]%%%}+%%%{%%{[-2,0]:
[1,0,%%%{-1,[1]%%%}]%%},[2,0,1,0,0]%%%}+%%%{%%{[1,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,2,0,0]%%%} / %%%{%%%{1,[1]%%
%},[4,0,0,0,0]%%%}+%%%{%%%{-2,[1]%%%},[2,0,1,0,0]%%%}+%%%{%%%{1,[1]%%%},[0,0,2,0,0]%%%} Error: Bad Argument Va
lue

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maple [B]  time = 0.06, size = 615, normalized size = 1.98 \begin {gather*} \frac {15 A \,a^{2} \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8}-\frac {5 A \,a^{\frac {3}{2}} b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2}+\frac {15 A a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 \sqrt {c}}-\frac {5 A \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {3}{2}}}-B \,a^{\frac {5}{2}} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+\frac {15 B \,a^{2} b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 \sqrt {c}}-\frac {5 B a \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {3}{2}}}+\frac {3 B \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {5}{2}}}+\frac {15 \sqrt {c \,x^{2}+b x +a}\, A a c x}{8}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} x}{32}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, B a b x}{16}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3} x}{64 c}+\frac {55 \sqrt {c \,x^{2}+b x +a}\, A a b}{16}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3}}{64 c}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A c x}{4}+\sqrt {c \,x^{2}+b x +a}\, B \,a^{2}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, B a \,b^{2}}{32 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{4}}{128 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b x}{8}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A c x}{a}+\frac {35 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{24}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B a}{3}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2}}{16 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A b}{a}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B}{5}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} A}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*(c*x^2+b*x+a)^(5/2)+15/16*A/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*b^2+15/8*A*(c*x^2+b*x+
a)^(1/2)*x*a*c+A/a*b*(c*x^2+b*x+a)^(5/2)-5/2*A*a^(3/2)*b*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+5/4*A*c
*(c*x^2+b*x+a)^(3/2)*x+5/32*A*(c*x^2+b*x+a)^(1/2)*x*b^2+5/64*A/c*(c*x^2+b*x+a)^(1/2)*b^3-5/128*A/c^(3/2)*ln((c
*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^4+55/16*A*(c*x^2+b*x+a)^(1/2)*b*a-A/a/x*(c*x^2+b*x+a)^(7/2)+15/8*A*c^
(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2+1/8*B*b*(c*x^2+b*x+a)^(3/2)*x+1/16*B/c*(c*x^2+b*x+a)^(3/
2)*b^2-3/128*B/c^2*(c*x^2+b*x+a)^(1/2)*b^4-5/32*B/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^3*a+A*
c/a*(c*x^2+b*x+a)^(5/2)*x+3/256*B/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^5+B*a^2*(c*x^2+b*x+a)^
(1/2)+35/24*A*b*(c*x^2+b*x+a)^(3/2)-B*a^(5/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+1/3*B*a*(c*x^2+b*x
+a)^(3/2)+7/16*B*b*(c*x^2+b*x+a)^(1/2)*x*a-3/64*B/c*(c*x^2+b*x+a)^(1/2)*x*b^3+7/32*B/c*(c*x^2+b*x+a)^(1/2)*b^2
*a+15/16*B*b/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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