3.14.80 \(\int \frac {(b+2 c x) (a+b x+c x^2)^{5/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=473 \[ \frac {5 \left (16 c^2 e^2 \left (a^2 e^2-8 a b d e+10 b^2 d^2\right )-8 b^2 c e^3 (4 b d-3 a e)-128 c^3 d^2 e (2 b d-a e)+b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{16 e^7 \sqrt {a e^2-b d e+c d^2}}-\frac {5 \sqrt {c} (2 c d-b e) \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 e^7}-\frac {5 \left (a+b x+c x^2\right )^{3/2} \left (-4 c e (3 b d-a e)+b^2 e^2+4 c e x (2 c d-b e)+16 c^2 d^2\right )}{12 e^4 (d+e x)^2}+\frac {5 \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (5 b d-2 a e)+12 b c e^2 (2 b d-a e)-b^3 e^3+64 c^3 d^3\right )}{8 e^6 (d+e x)}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+4 c d+2 c e x)}{3 e^2 (d+e x)^3} \]

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Rubi [A]  time = 0.75, antiderivative size = 473, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {812, 843, 621, 206, 724} \begin {gather*} \frac {5 \left (16 c^2 e^2 \left (a^2 e^2-8 a b d e+10 b^2 d^2\right )-8 b^2 c e^3 (4 b d-3 a e)-128 c^3 d^2 e (2 b d-a e)+b^4 e^4+128 c^4 d^4\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 )}{16 e^7 \sqrt {a e^2-b d e+c d^2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2} \left (-4 c e (3 b d-a e)+b^2 e^2+4 c e x (2 c d-b e)+16 c^2 d^2\right )}{12 e^4 (d+e x)^2}+\frac {5 \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (5 b d-2 a e)+12 b c e^2 (2 b d-a e)-b^3 e^3+64 c^3 d^3\right )}{8 e^6 (d+e x)}-\frac {5 \sqrt {c} (2 c d-b e) \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 e^7}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+4 c d+2 c e x)}{3 e^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

(5*(64*c^3*d^3 - b^3*e^3 - 16*c^2*d*e*(5*b*d - 2*a*e) + 12*b*c*e^2*(2*b*d - a*e) + 2*c*e*(16*c^2*d^2 + 3*b^2*e
^2 - 4*c*e*(4*b*d - a*e))*x)*Sqrt[a + b*x + c*x^2])/(8*e^6*(d + e*x)) - (5*(16*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*
d - a*e) + 4*c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(3/2))/(12*e^4*(d + e*x)^2) + ((4*c*d - b*e + 2*c*e*x)*(a
+ b*x + c*x^2)^(5/2))/(3*e^2*(d + e*x)^3) - (5*Sqrt[c]*(2*c*d - b*e)*(8*c^2*d^2 + b^2*e^2 - 4*c*e*(2*b*d - a*e
))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*e^7) + (5*(128*c^4*d^4 + b^4*e^4 - 8*b^2*c*e^3*(
4*b*d - 3*a*e) - 128*c^3*d^2*e*(2*b*d - a*e) + 16*c^2*e^2*(10*b^2*d^2 - 8*a*b*d*e + a^2*e^2))*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])])/(16*e^7*Sqrt[c*d^2 - b*d*e + a
*e^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 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)^4} \, dx &=\frac {(4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{3 e^2 (d+e x)^3}-\frac {5 \int \frac {\left (3 \left (4 b c d-b^2 e-4 a c e\right )+12 c (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^3} \, dx}{18 e^2}\\ &=-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4 (d+e x)^2}+\frac {(4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{3 e^2 (d+e x)^3}+\frac {5 \int \frac {\left (-6 \left (12 b^2 c d e+16 a c^2 d e-b^3 e^2-4 b c \left (4 c d^2+3 a e^2\right )\right )+12 c \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx}{48 e^4}\\ &=\frac {5 \left (64 c^3 d^3-b^3 e^3-16 c^2 d e (5 b d-2 a e)+12 b c e^2 (2 b d-a e)+2 c e \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{8 e^6 (d+e x)}-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4 (d+e x)^2}+\frac {(4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{3 e^2 (d+e x)^3}-\frac {5 \int \frac {6 \left (24 b^3 c d e^2-b^4 e^3-16 a c^2 e \left (4 c d^2+a e^2\right )+32 b c^2 d \left (2 c d^2+3 a e^2\right )-8 b^2 c e \left (10 c d^2+3 a e^2\right )\right )+48 c (2 c d-b e) \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{96 e^6}\\ &=\frac {5 \left (64 c^3 d^3-b^3 e^3-16 c^2 d e (5 b d-2 a e)+12 b c e^2 (2 b d-a e)+2 c e \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{8 e^6 (d+e x)}-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4 (d+e x)^2}+\frac {(4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{3 e^2 (d+e x)^3}-\frac {\left (5 c (2 c d-b e) \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 e^7}+\frac {\left (5 \left (128 c^4 d^4+b^4 e^4-8 b^2 c e^3 (4 b d-3 a e)-128 c^3 d^2 e (2 b d-a e)+16 c^2 e^2 \left (10 b^2 d^2-8 a b d e+a^2 e^2\right )\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{16 e^7}\\ &=\frac {5 \left (64 c^3 d^3-b^3 e^3-16 c^2 d e (5 b d-2 a e)+12 b c e^2 (2 b d-a e)+2 c e \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{8 e^6 (d+e x)}-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4 (d+e x)^2}+\frac {(4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{3 e^2 (d+e x)^3}-\frac {\left (5 c (2 c d-b e) \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\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 )}{e^7}-\frac {\left (5 \left (128 c^4 d^4+b^4 e^4-8 b^2 c e^3 (4 b d-3 a e)-128 c^3 d^2 e (2 b d-a e)+16 c^2 e^2 \left (10 b^2 d^2-8 a b d e+a^2 e^2\right )\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 )}{8 e^7}\\ &=\frac {5 \left (64 c^3 d^3-b^3 e^3-16 c^2 d e (5 b d-2 a e)+12 b c e^2 (2 b d-a e)+2 c e \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{8 e^6 (d+e x)}-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^4 (d+e x)^2}+\frac {(4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{5/2}}{3 e^2 (d+e x)^3}-\frac {5 \sqrt {c} (2 c d-b e) \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 e^7}+\frac {5 \left (128 c^4 d^4+b^4 e^4-8 b^2 c e^3 (4 b d-3 a e)-128 c^3 d^2 e (2 b d-a e)+16 c^2 e^2 \left (10 b^2 d^2-8 a b d e+a^2 e^2\right )\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 )}{16 e^7 \sqrt {c d^2-b d e+a e^2}}\\ \end {align*}

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Mathematica [A]  time = 2.62, size = 823, normalized size = 1.74 \begin {gather*} \frac {-120 \sqrt {c} (2 c d-b e) \left (8 c^3 d^4-4 c^2 e (4 b d-3 a e) d^2+c e^2 (3 b d-2 a e)^2+b^2 e^3 (a e-b d)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) (d+e x)^3-15 \sqrt {c d^2+e (a e-b d)} \left (128 c^4 d^4-128 c^3 e (2 b d-a e) d^2+b^4 e^4-8 b^2 c e^3 (4 b d-3 a e)+16 c^2 e^2 \left (10 b^2 d^2-8 a b e d+a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {-b d-2 c x d+2 a e+b e x}{2 \sqrt {c d^2+e (a e-b d)} \sqrt {a+x (b+c x)}}\right ) (d+e x)^3+2 e \sqrt {a+x (b+c x)} \left (16 d^2 \left (60 d^5+150 e x d^4+110 e^2 x^2 d^3+15 e^3 x^3 d^2-3 e^4 x^4 d+e^5 x^5\right ) c^4+8 e \left (a e \left (160 d^5+405 e x d^4+303 e^2 x^2 d^3+44 e^3 x^3 d^2-6 e^4 x^4 d+2 e^5 x^5\right )-b d \left (270 d^5+680 e x d^4+505 e^2 x^2 d^3+72 e^3 x^3 d^2-14 e^4 x^4 d+2 e^5 x^5\right )\right ) c^3+2 e^2 \left (d \left (780 d^4+1985 e x d^3+1501 e^2 x^2 d^2+228 e^3 x^3 d-32 e^4 x^4\right ) b^2-2 a e \left (395 d^4+1014 e x d^3+777 e^2 x^2 d^2+112 e^3 x^3 d-16 e^4 x^4\right ) b+4 a^2 e^2 \left (39 d^3+102 e x d^2+83 e^2 x^2 d+14 e^3 x^3\right )\right ) c^2-e^3 \left (5 d \left (75 d^3+194 e x d^2+151 e^2 x^2 d+24 e^3 x^3\right ) b^3-2 a e \left (205 d^3+540 e x d^2+443 e^2 x^2 d+60 e^3 x^3\right ) b^2+4 a^2 e^2 \left (15 d^2+38 e x d+41 e^2 x^2\right ) b+8 a^3 e^3 (d+3 e x)\right ) c+b e^4 (b d-a e) \left (\left (15 d^2+40 e x d+33 e^2 x^2\right ) b^2+2 a e (5 d+13 e x) b+8 a^2 e^2\right )\right )}{48 e^7 \left (c d^2+e (a e-b d)\right ) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

(2*e*Sqrt[a + x*(b + c*x)]*(16*c^4*d^2*(60*d^5 + 150*d^4*e*x + 110*d^3*e^2*x^2 + 15*d^2*e^3*x^3 - 3*d*e^4*x^4
+ e^5*x^5) + b*e^4*(b*d - a*e)*(8*a^2*e^2 + 2*a*b*e*(5*d + 13*e*x) + b^2*(15*d^2 + 40*d*e*x + 33*e^2*x^2)) - c
*e^3*(8*a^3*e^3*(d + 3*e*x) + 4*a^2*b*e^2*(15*d^2 + 38*d*e*x + 41*e^2*x^2) + 5*b^3*d*(75*d^3 + 194*d^2*e*x + 1
51*d*e^2*x^2 + 24*e^3*x^3) - 2*a*b^2*e*(205*d^3 + 540*d^2*e*x + 443*d*e^2*x^2 + 60*e^3*x^3)) + 2*c^2*e^2*(4*a^
2*e^2*(39*d^3 + 102*d^2*e*x + 83*d*e^2*x^2 + 14*e^3*x^3) + b^2*d*(780*d^4 + 1985*d^3*e*x + 1501*d^2*e^2*x^2 +
228*d*e^3*x^3 - 32*e^4*x^4) - 2*a*b*e*(395*d^4 + 1014*d^3*e*x + 777*d^2*e^2*x^2 + 112*d*e^3*x^3 - 16*e^4*x^4))
+ 8*c^3*e*(-(b*d*(270*d^5 + 680*d^4*e*x + 505*d^3*e^2*x^2 + 72*d^2*e^3*x^3 - 14*d*e^4*x^4 + 2*e^5*x^5)) + a*e
*(160*d^5 + 405*d^4*e*x + 303*d^3*e^2*x^2 + 44*d^2*e^3*x^3 - 6*d*e^4*x^4 + 2*e^5*x^5))) - 120*Sqrt[c]*(2*c*d -
b*e)*(8*c^3*d^4 - 4*c^2*d^2*e*(4*b*d - 3*a*e) + c*e^2*(3*b*d - 2*a*e)^2 + b^2*e^3*(-(b*d) + a*e))*(d + e*x)^3
*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - 15*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*(128*c^4*d^4 + b^4
*e^4 - 8*b^2*c*e^3*(4*b*d - 3*a*e) - 128*c^3*d^2*e*(2*b*d - a*e) + 16*c^2*e^2*(10*b^2*d^2 - 8*a*b*d*e + a^2*e^
2))*(d + e*x)^3*ArcTanh[(-(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)])])/(48*e^7*(c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^3)

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IntegrateAlgebraic [F]  time = 180.05, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

$Aborted

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

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)^4,x, algorithm="fricas")

[Out]

Timed out

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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((2*c*x+b)*(c*x^2+b*x+a)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 42.98Unable to divide, perhaps due to rounding error%%%{%%{[-1,0]:[1,0,%%%{-1,[1]%%%}]%%},[8,0,0,
11,0]%%%}+%%%{%%%{8,[1]%%%},[7,0,0,10,1]%%%}+%%%{%%{[4,0]:[1,0,%%%{-1,[1]%%%}]%%},[6,1,0,11,0]%%%}+%%%{%%{[-4,
0]:[1,0,%%%{-1,[1]%%%}]%%},[6,0,1,10,1]%%%}+%%%{%%{[%%%{-24,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[6,0,0,9,2]%%%}
+%%%{%%%{-24,[1]%%%},[5,1,0,10,1]%%%}+%%%{%%%{24,[1]%%%},[5,0,1,9,2]%%%}+%%%{%%%{32,[2]%%%},[5,0,0,8,3]%%%}+%%
%{%%{[-6,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,2,0,11,0]%%%}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,1,1,10,1]%%%}+%%
%{%%{[%%%{48,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,1,0,9,2]%%%}+%%%{%%{[-6,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,2,9
,2]%%%}+%%%{%%{[%%%{-48,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,1,8,3]%%%}+%%%{%%{[%%%{-16,[2]%%%},0]:[1,0,%%%
{-1,[1]%%%}]%%},[4,0,0,7,4]%%%}+%%%{%%%{24,[1]%%%},[3,2,0,10,1]%%%}+%%%{%%%{-48,[1]%%%},[3,1,1,9,2]%%%}+%%%{%%
%{-32,[2]%%%},[3,1,0,8,3]%%%}+%%%{%%%{24,[1]%%%},[3,0,2,8,3]%%%}+%%%{%%%{32,[2]%%%},[3,0,1,7,4]%%%}+%%%{%%{[4,
0]:[1,0,%%%{-1,[1]%%%}]%%},[2,3,0,11,0]%%%}+%%%{%%{[-12,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,2,1,10,1]%%%}+%%%{%%{[%%
%{-24,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,2,0,9,2]%%%}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,1,2,9,2]%%%}
+%%%{%%{[%%%{48,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,1,1,8,3]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,
3,8,3]%%%}+%%%{%%{[%%%{-24,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,2,7,4]%%%}+%%%{%%%{-8,[1]%%%},[1,3,0,10,1]%
%%}+%%%{%%%{24,[1]%%%},[1,2,1,9,2]%%%}+%%%{%%%{-24,[1]%%%},[1,1,2,8,3]%%%}+%%%{%%%{8,[1]%%%},[1,0,3,7,4]%%%}+%
%%{%%{[-1,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,4,0,11,0]%%%}+%%%{%%{[4,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,3,1,10,1]%%%}+%%
%{%%{[-6,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,2,2,9,2]%%%}+%%%{%%{[4,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,1,3,8,3]%%%}+%%%{%
%{[-1,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,4,7,4]%%%} / %%%{%%%{1,[2]%%%},[8,0,0,4,0]%%%}+%%%{%%{poly1[%%%{-8,[2]%%
%},0]:[1,0,%%%{-1,[1]%%%}]%%},[7,0,0,3,1]%%%}+%%%{%%%{-4,[2]%%%},[6,1,0,4,0]%%%}+%%%{%%%{4,[2]%%%},[6,0,1,3,1]
%%%}+%%%{%%%{24,[3]%%%},[6,0,0,2,2]%%%}+%%%{%%{[%%%{24,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[5,1,0,3,1]%%%}+%%%{
%%{poly1[%%%{-24,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[5,0,1,2,2]%%%}+%%%{%%{poly1[%%%{-32,[3]%%%},0]:[1,0,%%%{-
1,[1]%%%}]%%},[5,0,0,1,3]%%%}+%%%{%%%{6,[2]%%%},[4,2,0,4,0]%%%}+%%%{%%%{-12,[2]%%%},[4,1,1,3,1]%%%}+%%%{%%%{-4
8,[3]%%%},[4,1,0,2,2]%%%}+%%%{%%%{6,[2]%%%},[4,0,2,2,2]%%%}+%%%{%%%{48,[3]%%%},[4,0,1,1,3]%%%}+%%%{%%%{16,[4]%
%%},[4,0,0,0,4]%%%}+%%%{%%{poly1[%%%{-24,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,2,0,3,1]%%%}+%%%{%%{[%%%{48,[2]
%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1,1,2,2]%%%}+%%%{%%{[%%%{32,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1,0,1,3]%
%%}+%%%{%%{poly1[%%%{-24,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,2,1,3]%%%}+%%%{%%{poly1[%%%{-32,[3]%%%},0]:[1
,0,%%%{-1,[1]%%%}]%%},[3,0,1,0,4]%%%}+%%%{%%%{-4,[2]%%%},[2,3,0,4,0]%%%}+%%%{%%%{12,[2]%%%},[2,2,1,3,1]%%%}+%%
%{%%%{24,[3]%%%},[2,2,0,2,2]%%%}+%%%{%%%{-12,[2]%%%},[2,1,2,2,2]%%%}+%%%{%%%{-48,[3]%%%},[2,1,1,1,3]%%%}+%%%{%
%%{4,[2]%%%},[2,0,3,1,3]%%%}+%%%{%%%{24,[3]%%%},[2,0,2,0,4]%%%}+%%%{%%{[%%%{8,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%
%},[1,3,0,3,1]%%%}+%%%{%%{poly1[%%%{-24,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,2,1,2,2]%%%}+%%%{%%{[%%%{24,[2]%
%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1,2,1,3]%%%}+%%%{%%{poly1[%%%{-8,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,0,3,0
,4]%%%}+%%%{%%%{1,[2]%%%},[0,4,0,4,0]%%%}+%%%{%%%{-4,[2]%%%},[0,3,1,3,1]%%%}+%%%{%%%{6,[2]%%%},[0,2,2,2,2]%%%}
+%%%{%%%{-4,[2]%%%},[0,1,3,1,3]%%%}+%%%{%%%{1,[2]%%%},[0,0,4,0,4]%%%} Error: Bad Argument Value

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maple [B]  time = 0.32, size = 28593, normalized size = 60.45 \begin {gather*} \text {output too large to display} \end {gather*}

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)^4,x)

[Out]

result too large to display

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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((2*c*x+b)*(c*x^2+b*x+a)^(5/2)/(e*x+d)^4,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 \begin {gather*} \int \frac {\left (b+2\,c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

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)^4,x)

[Out]

int(((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b + 2 c x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

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)**4,x)

[Out]

Integral((b + 2*c*x)*(a + b*x + c*x**2)**(5/2)/(d + e*x)**4, x)

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