3.1570 \(\int \frac {(b+2 c x) (a+b x+c x^2)^{5/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=508 \[ \frac {\sqrt {a+b x+c x^2} \left (8 c^2 e^2 \left (8 a^2 e^2-67 a b d e+76 b^2 d^2\right )-2 c e x (2 c d-b e) \left (44 a c e^2+b^2 e^2-48 b c d e+48 c^2 d^2\right )-2 b^2 c e^3 (65 b d-62 a e)-32 c^3 d^2 e (27 b d-14 a e)+b^4 e^4+384 c^4 d^4\right )}{32 c e^6}-\frac {(2 c d-b e) \left (16 c^2 e^2 \left (15 a^2 e^2-40 a b d e+26 b^2 d^2\right )-8 b^2 c e^3 (4 b d-5 a e)-128 c^3 d^2 e (6 b d-5 a e)-b^4 e^4+384 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^7}+\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \left (-4 c e (6 b d-a e)+5 b^2 e^2+24 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^7}+\frac {\left (a+b x+c x^2\right )^{3/2} \left (8 a c e^2+19 b^2 e^2-18 c e x (2 c d-b e)-66 b c d e+48 c^2 d^2\right )}{12 e^4}+\frac {\left (a+b x+c x^2\right )^{5/2} (-5 b e+12 c d+2 c e x)}{5 e^2 (d+e x)} \]

[Out]

1/12*(48*c^2*d^2-66*b*c*d*e+19*b^2*e^2+8*a*c*e^2-18*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(3/2)/e^4+1/5*(2*c*e*x-5
*b*e+12*c*d)*(c*x^2+b*x+a)^(5/2)/e^2/(e*x+d)-1/64*(-b*e+2*c*d)*(384*c^4*d^4-b^4*e^4-8*b^2*c*e^3*(-5*a*e+4*b*d)
-128*c^3*d^2*e*(-5*a*e+6*b*d)+16*c^2*e^2*(15*a^2*e^2-40*a*b*d*e+26*b^2*d^2))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*
x^2+b*x+a)^(1/2))/c^(3/2)/e^7+1/2*(a*e^2-b*d*e+c*d^2)^(3/2)*(24*c^2*d^2+5*b^2*e^2-4*c*e*(-a*e+6*b*d))*arctanh(
1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^7+1/32*(384*c^4*d^4+b^4*e^4-2*
b^2*c*e^3*(-62*a*e+65*b*d)-32*c^3*d^2*e*(-14*a*e+27*b*d)+8*c^2*e^2*(8*a^2*e^2-67*a*b*d*e+76*b^2*d^2)-2*c*e*(-b
*e+2*c*d)*(44*a*c*e^2+b^2*e^2-48*b*c*d*e+48*c^2*d^2)*x)*(c*x^2+b*x+a)^(1/2)/c/e^6

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Rubi [A]  time = 0.75, antiderivative size = 508, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {812, 814, 843, 621, 206, 724} \[ \frac {\sqrt {a+b x+c x^2} \left (8 c^2 e^2 \left (8 a^2 e^2-67 a b d e+76 b^2 d^2\right )-2 c e x (2 c d-b e) \left (44 a c e^2+b^2 e^2-48 b c d e+48 c^2 d^2\right )-2 b^2 c e^3 (65 b d-62 a e)-32 c^3 d^2 e (27 b d-14 a e)+b^4 e^4+384 c^4 d^4\right )}{32 c e^6}-\frac {(2 c d-b e) \left (16 c^2 e^2 \left (15 a^2 e^2-40 a b d e+26 b^2 d^2\right )-8 b^2 c e^3 (4 b d-5 a e)-128 c^3 d^2 e (6 b d-5 a e)-b^4 e^4+384 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^7}+\frac {\left (a+b x+c x^2\right )^{3/2} \left (8 a c e^2+19 b^2 e^2-18 c e x (2 c d-b e)-66 b c d e+48 c^2 d^2\right )}{12 e^4}+\frac {\left (a e^2-b d e+c d^2\right )^{3/2} \left (-4 c e (6 b d-a e)+5 b^2 e^2+24 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^7}+\frac {\left (a+b x+c x^2\right )^{5/2} (-5 b e+12 c d+2 c e x)}{5 e^2 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((384*c^4*d^4 + b^4*e^4 - 2*b^2*c*e^3*(65*b*d - 62*a*e) - 32*c^3*d^2*e*(27*b*d - 14*a*e) + 8*c^2*e^2*(76*b^2*d
^2 - 67*a*b*d*e + 8*a^2*e^2) - 2*c*e*(2*c*d - b*e)*(48*c^2*d^2 - 48*b*c*d*e + b^2*e^2 + 44*a*c*e^2)*x)*Sqrt[a
+ b*x + c*x^2])/(32*c*e^6) + ((48*c^2*d^2 - 66*b*c*d*e + 19*b^2*e^2 + 8*a*c*e^2 - 18*c*e*(2*c*d - b*e)*x)*(a +
b*x + c*x^2)^(3/2))/(12*e^4) + ((12*c*d - 5*b*e + 2*c*e*x)*(a + b*x + c*x^2)^(5/2))/(5*e^2*(d + e*x)) - ((2*c
*d - b*e)*(384*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(4*b*d - 5*a*e) - 128*c^3*d^2*e*(6*b*d - 5*a*e) + 16*c^2*e^2*(2
6*b^2*d^2 - 40*a*b*d*e + 15*a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(3/2)*e^7)
+ ((c*d^2 - b*d*e + a*e^2)^(3/2)*(24*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(6*b*d - a*e))*ArcTanh[(b*d - 2*a*e + (2*c*d
- b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(2*e^7)

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 {(b+2 c x) \left (a+b x+c x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\frac {(12 c d-5 b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {\int \frac {\left (12 b c d-5 b^2 e-4 a c e+12 c (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx}{2 e^2}\\ &=\frac {\left (48 c^2 d^2-66 b c d e+19 b^2 e^2+8 a c e^2-18 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4}+\frac {(12 c d-5 b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}+\frac {\int \frac {\left (2 c \left (2 e (b d-2 a e) \left (12 b c d-5 b^2 e-4 a c e\right )+6 (2 c d-b e) \left (2 a c d e-b d \left (4 c d-\frac {3 b e}{2}\right )\right )\right )-2 c (2 c d-b e) \left (48 c^2 d^2-48 b c d e+b^2 e^2+44 a c e^2\right ) x\right ) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{16 c e^4}\\ &=\frac {\left (384 c^4 d^4+b^4 e^4-2 b^2 c e^3 (65 b d-62 a e)-32 c^3 d^2 e (27 b d-14 a e)+8 c^2 e^2 \left (76 b^2 d^2-67 a b d e+8 a^2 e^2\right )-2 c e (2 c d-b e) \left (48 c^2 d^2-48 b c d e+b^2 e^2+44 a c e^2\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^6}+\frac {\left (48 c^2 d^2-66 b c d e+19 b^2 e^2+8 a c e^2-18 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4}+\frac {(12 c d-5 b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {\int \frac {c \left (d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right ) \left (48 c^2 d^2-48 b c d e+b^2 e^2+44 a c e^2\right )+4 c e (b d-2 a e) \left (2 e (b d-2 a e) \left (12 b c d-5 b^2 e-4 a c e\right )-3 d (2 c d-b e) \left (8 b c d-3 b^2 e-4 a c e\right )\right )\right )+c (2 c d-b e) \left (384 c^4 d^4-b^4 e^4-8 b^2 c e^3 (4 b d-5 a e)-128 c^3 d^2 e (6 b d-5 a e)+16 c^2 e^2 \left (26 b^2 d^2-40 a b d e+15 a^2 e^2\right )\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{64 c^2 e^6}\\ &=\frac {\left (384 c^4 d^4+b^4 e^4-2 b^2 c e^3 (65 b d-62 a e)-32 c^3 d^2 e (27 b d-14 a e)+8 c^2 e^2 \left (76 b^2 d^2-67 a b d e+8 a^2 e^2\right )-2 c e (2 c d-b e) \left (48 c^2 d^2-48 b c d e+b^2 e^2+44 a c e^2\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^6}+\frac {\left (48 c^2 d^2-66 b c d e+19 b^2 e^2+8 a c e^2-18 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4}+\frac {(12 c d-5 b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}+\frac {\left (\left (c d^2-b d e+a e^2\right )^2 \left (24 c^2 d^2+5 b^2 e^2-4 c e (6 b d-a e)\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 e^7}-\frac {\left ((2 c d-b e) \left (384 c^4 d^4-b^4 e^4-8 b^2 c e^3 (4 b d-5 a e)-128 c^3 d^2 e (6 b d-5 a e)+16 c^2 e^2 \left (26 b^2 d^2-40 a b d e+15 a^2 e^2\right )\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{64 c e^7}\\ &=\frac {\left (384 c^4 d^4+b^4 e^4-2 b^2 c e^3 (65 b d-62 a e)-32 c^3 d^2 e (27 b d-14 a e)+8 c^2 e^2 \left (76 b^2 d^2-67 a b d e+8 a^2 e^2\right )-2 c e (2 c d-b e) \left (48 c^2 d^2-48 b c d e+b^2 e^2+44 a c e^2\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^6}+\frac {\left (48 c^2 d^2-66 b c d e+19 b^2 e^2+8 a c e^2-18 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4}+\frac {(12 c d-5 b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {\left (\left (c d^2-b d e+a e^2\right )^2 \left (24 c^2 d^2+5 b^2 e^2-4 c e (6 b d-a e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{e^7}-\frac {\left ((2 c d-b e) \left (384 c^4 d^4-b^4 e^4-8 b^2 c e^3 (4 b d-5 a e)-128 c^3 d^2 e (6 b d-5 a e)+16 c^2 e^2 \left (26 b^2 d^2-40 a b d e+15 a^2 e^2\right )\right )\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 )}{32 c e^7}\\ &=\frac {\left (384 c^4 d^4+b^4 e^4-2 b^2 c e^3 (65 b d-62 a e)-32 c^3 d^2 e (27 b d-14 a e)+8 c^2 e^2 \left (76 b^2 d^2-67 a b d e+8 a^2 e^2\right )-2 c e (2 c d-b e) \left (48 c^2 d^2-48 b c d e+b^2 e^2+44 a c e^2\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^6}+\frac {\left (48 c^2 d^2-66 b c d e+19 b^2 e^2+8 a c e^2-18 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4}+\frac {(12 c d-5 b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {(2 c d-b e) \left (384 c^4 d^4-b^4 e^4-8 b^2 c e^3 (4 b d-5 a e)-128 c^3 d^2 e (6 b d-5 a e)+16 c^2 e^2 \left (26 b^2 d^2-40 a b d e+15 a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^7}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \left (24 c^2 d^2+5 b^2 e^2-4 c e (6 b d-a e)\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 e^7}\\ \end {align*}

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Mathematica [A]  time = 2.45, size = 659, normalized size = 1.30 \[ \frac {\frac {-2 c^2 e \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right ) \left (4 c^2 e^2 \left (16 a^2 e^2+2 a b e (11 e x-67 d)+b^2 d (152 d-25 e x)\right )+2 b^2 c e^3 (62 a e-65 b d+b e x)-16 c^3 d e (a e (11 e x-28 d)+18 b d (3 d-e x))+b^4 e^4+192 c^4 d^3 (2 d-e x)\right )+c^{3/2} (2 c d-b e) \left (16 c^3 d^2 e^2 \left (55 a^2 e^2-128 a b d e+74 b^2 d^2\right )+b^2 c e^4 \left (40 a^2 e^2-72 a b d e+31 b^2 d^2\right )+8 c^2 e^3 \left (30 a^3 e^3-110 a^2 b d e^2+137 a b^2 d^2 e-56 b^3 d^3\right )+b^4 e^5 (b d-a e)-128 c^4 d^4 e (9 b d-8 a e)+384 c^5 d^6\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+32 c^3 \left (4 c e (a e-6 b d)+5 b^2 e^2+24 c^2 d^2\right ) \left (e (a e-b d)+c d^2\right )^{5/2} \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{64 c^3 e^7}-\frac {(a+x (b+c x))^{3/2} \left (e (a e-b d)+c d^2\right ) \left (2 c e (4 a e-33 b d+9 b e x)+19 b^2 e^2+12 c^2 d (4 d-3 e x)\right )}{12 e^4}+\frac {(a+x (b+c x))^{5/2} \left (c e (-2 a e+17 b d-5 b e x)-5 b^2 e^2-2 c^2 d (6 d-5 e x)\right )}{5 e^2}+\frac {(a+x (b+c x))^{7/2} (b e-2 c d)}{d+e x}}{e (b d-a e)-c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*c*d + b*e)*(a + x*(b + c*x))^(7/2))/(d + e*x) + ((a + x*(b + c*x))^(5/2)*(-5*b^2*e^2 - 2*c^2*d*(6*d - 5*
e*x) + c*e*(17*b*d - 2*a*e - 5*b*e*x)))/(5*e^2) - ((c*d^2 + e*(-(b*d) + a*e))*(a + x*(b + c*x))^(3/2)*(19*b^2*
e^2 + 12*c^2*d*(4*d - 3*e*x) + 2*c*e*(-33*b*d + 4*a*e + 9*b*e*x)))/(12*e^4) + (-2*c^2*e*(c*d^2 + e*(-(b*d) + a
*e))*Sqrt[a + x*(b + c*x)]*(b^4*e^4 + 192*c^4*d^3*(2*d - e*x) + 2*b^2*c*e^3*(-65*b*d + 62*a*e + b*e*x) + 4*c^2
*e^2*(16*a^2*e^2 + b^2*d*(152*d - 25*e*x) + 2*a*b*e*(-67*d + 11*e*x)) - 16*c^3*d*e*(18*b*d*(3*d - e*x) + a*e*(
-28*d + 11*e*x))) + c^(3/2)*(2*c*d - b*e)*(384*c^5*d^6 - 128*c^4*d^4*e*(9*b*d - 8*a*e) + b^4*e^5*(b*d - a*e) +
b^2*c*e^4*(31*b^2*d^2 - 72*a*b*d*e + 40*a^2*e^2) + 16*c^3*d^2*e^2*(74*b^2*d^2 - 128*a*b*d*e + 55*a^2*e^2) + 8
*c^2*e^3*(-56*b^3*d^3 + 137*a*b^2*d^2*e - 110*a^2*b*d*e^2 + 30*a^3*e^3))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ x*(b + c*x)])] + 32*c^3*(24*c^2*d^2 + 5*b^2*e^2 + 4*c*e*(-6*b*d + a*e))*(c*d^2 + e*(-(b*d) + a*e))^(5/2)*Ar
cTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(64*c^3*e^
7))/(-(c*d^2) + e*(b*d - a*e))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.07, size = 13167, normalized size = 25.92 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(c*x^2+b*x+a)^(5/2)/(e*x+d)^2,x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)^(5/2)/(e*x+d)^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(a*e^2-b*d*e>0)', see `assume?`
for more details)Is a*e^2-b*d*e                            +c*d^2 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^2,x)

[Out]

int(((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b + 2 c x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x**2+b*x+a)**(5/2)/(e*x+d)**2,x)

[Out]

Integral((b + 2*c*x)*(a + b*x + c*x**2)**(5/2)/(d + e*x)**2, x)

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