3.14.79 \(\int \frac {(b+2 c x) (a+b x+c x^2)^{5/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=464 \[ \frac {15 \left (16 c^2 e^2 \left (a^2 e^2-8 a b d e+10 b^2 d^2\right )-8 b^2 c e^3 (4 b d-3 a e)-128 c^3 d^2 e (2 b d-a e)+b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 \sqrt {c} e^7}-\frac {15 (2 c d-b e) \sqrt {a e^2-b d e+c d^2} \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{8 e^7}-\frac {5 \left (a+b x+c x^2\right )^{3/2} \left (-c e (7 b d-2 a e)+b^2 e^2+c e x (2 c d-b e)+8 c^2 d^2\right )}{4 e^4 (d+e x)}-\frac {15 \sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-2 a e)+4 b c e^2 (14 b d-5 a e)-7 b^3 e^3+64 c^3 d^3\right )}{32 e^6}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+3 c d+c e x)}{2 e^2 (d+e x)^2} \]

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Rubi [A]  time = 0.85, antiderivative size = 464, 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} \begin {gather*} \frac {15 \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 {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 \sqrt {c} e^7}-\frac {5 \left (a+b x+c x^2\right )^{3/2} \left (-c e (7 b d-2 a e)+b^2 e^2+c e x (2 c d-b e)+8 c^2 d^2\right )}{4 e^4 (d+e x)}-\frac {15 \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 (7 b d-2 a e)+4 b c e^2 (14 b d-5 a e)-7 b^3 e^3+64 c^3 d^3\right )}{32 e^6}-\frac {15 (2 c d-b e) \sqrt {a e^2-b d e+c d^2} \left (-4 c e (2 b d-a e)+b^2 e^2+8 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 )}{8 e^7}+\frac {\left (a+b x+c x^2\right )^{5/2} (-b e+3 c d+c e x)}{2 e^2 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^3,x]

[Out]

(-15*(64*c^3*d^3 - 7*b^3*e^3 + 4*b*c*e^2*(14*b*d - 5*a*e) - 16*c^2*d*e*(7*b*d - 2*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])/(32*e^6) - (5*(8*c^2*d^2 + b^2*e^2 - c*e*(7*b*d - 2*
a*e) + c*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(3/2))/(4*e^4*(d + e*x)) + ((3*c*d - b*e + c*e*x)*(a + b*x + c*x
^2)^(5/2))/(2*e^2*(d + e*x)^2) + (15*(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 + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2
])])/(64*Sqrt[c]*e^7) - (15*(2*c*d - b*e)*Sqrt[c*d^2 - b*d*e + a*e^2]*(8*c^2*d^2 + b^2*e^2 - 4*c*e*(2*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])])/(8*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)^3} \, dx &=\frac {(3 c d-b e+c e x) \left (a+b x+c x^2\right )^{5/2}}{2 e^2 (d+e x)^2}-\frac {5 \int \frac {\left (4 \left (3 b c d-b^2 e-2 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)^2} \, dx}{16 e^2}\\ &=-\frac {5 \left (8 c^2 d^2+b^2 e^2-c e (7 b d-2 a e)+c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{4 e^4 (d+e x)}+\frac {(3 c d-b e+c e x) \left (a+b x+c x^2\right )^{5/2}}{2 e^2 (d+e x)^2}+\frac {5 \int \frac {\left (-12 \left (7 b^2 c d e+4 a c^2 d e-b^3 e^2-4 b c \left (2 c d^2+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} \, dx}{32 e^4}\\ &=-\frac {15 \left (64 c^3 d^3-7 b^3 e^3+4 b c e^2 (14 b d-5 a e)-16 c^2 d e (7 b d-2 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}}{32 e^6}-\frac {5 \left (8 c^2 d^2+b^2 e^2-c e (7 b d-2 a e)+c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{4 e^4 (d+e x)}+\frac {(3 c d-b e+c e x) \left (a+b x+c x^2\right )^{5/2}}{2 e^2 (d+e x)^2}-\frac {5 \int \frac {6 c \left (7 b^4 d e^3+16 a c^2 d e \left (4 c d^2+3 a e^2\right )+8 b^2 c d e \left (14 c d^2+11 a e^2\right )-8 b^3 \left (7 c d^2 e^2+a e^4\right )-32 b c \left (2 c^2 d^4+5 a c d^2 e^2+a^2 e^4\right )\right )-6 c \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 ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{128 c e^6}\\ &=-\frac {15 \left (64 c^3 d^3-7 b^3 e^3+4 b c e^2 (14 b d-5 a e)-16 c^2 d e (7 b d-2 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}}{32 e^6}-\frac {5 \left (8 c^2 d^2+b^2 e^2-c e (7 b d-2 a e)+c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{4 e^4 (d+e x)}+\frac {(3 c d-b e+c e x) \left (a+b x+c x^2\right )^{5/2}}{2 e^2 (d+e x)^2}+\frac {\left (15 \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}{\sqrt {a+b x+c x^2}} \, dx}{64 e^7}-\frac {\left (5 \left (6 c d \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 )+6 c e \left (7 b^4 d e^3+16 a c^2 d e \left (4 c d^2+3 a e^2\right )+8 b^2 c d e \left (14 c d^2+11 a e^2\right )-8 b^3 \left (7 c d^2 e^2+a e^4\right )-32 b c \left (2 c^2 d^4+5 a c d^2 e^2+a^2 e^4\right )\right )\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{128 c e^7}\\ &=-\frac {15 \left (64 c^3 d^3-7 b^3 e^3+4 b c e^2 (14 b d-5 a e)-16 c^2 d e (7 b d-2 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}}{32 e^6}-\frac {5 \left (8 c^2 d^2+b^2 e^2-c e (7 b d-2 a e)+c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{4 e^4 (d+e x)}+\frac {(3 c d-b e+c e x) \left (a+b x+c x^2\right )^{5/2}}{2 e^2 (d+e x)^2}+\frac {\left (15 \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-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{32 e^7}+\frac {\left (5 \left (6 c d \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 )+6 c e \left (7 b^4 d e^3+16 a c^2 d e \left (4 c d^2+3 a e^2\right )+8 b^2 c d e \left (14 c d^2+11 a e^2\right )-8 b^3 \left (7 c d^2 e^2+a e^4\right )-32 b c \left (2 c^2 d^4+5 a c d^2 e^2+a^2 e^4\right )\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 )}{64 c e^7}\\ &=-\frac {15 \left (64 c^3 d^3-7 b^3 e^3+4 b c e^2 (14 b d-5 a e)-16 c^2 d e (7 b d-2 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}}{32 e^6}-\frac {5 \left (8 c^2 d^2+b^2 e^2-c e (7 b d-2 a e)+c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{4 e^4 (d+e x)}+\frac {(3 c d-b e+c e x) \left (a+b x+c x^2\right )^{5/2}}{2 e^2 (d+e x)^2}+\frac {15 \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+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 \sqrt {c} e^7}-\frac {15 (2 c d-b e) \sqrt {c d^2-b d e+a e^2} \left (8 c^2 d^2-8 b c d e+b^2 e^2+4 a c e^2\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 )}{8 e^7}\\ \end {align*}

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Mathematica [A]  time = 2.13, size = 535, normalized size = 1.15 \begin {gather*} \frac {-\frac {2 e \sqrt {a+x (b+c x)} \left (2 c e^2 \left (16 a^2 e^2 (d+2 e x)-2 a b e \left (145 d^2+234 d e x+65 e^2 x^2\right )+b^2 \left (420 d^3+655 d^2 e x+166 d e^2 x^2-37 e^3 x^3\right )\right )+b e^3 \left (16 a^2 e^2+8 a b e (5 d+9 e x)-\left (b^2 \left (105 d^2+170 d e x+49 e^2 x^2\right )\right )\right )-8 c^2 e \left (a e \left (-100 d^3-155 d^2 e x-38 d e^2 x^2+9 e^3 x^3\right )+b \left (210 d^4+320 d^3 e x+75 d^2 e^2 x^2-18 d e^3 x^3+7 e^4 x^4\right )\right )+16 c^3 \left (60 d^5+90 d^4 e x+20 d^3 e^2 x^2-5 d^2 e^3 x^3+2 d e^4 x^4-e^5 x^5\right )\right )}{(d+e x)^2}+\frac {15 \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 {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+120 (2 c d-b e) \left (4 c e (a e-2 b d)+b^2 e^2+8 c^2 d^2\right ) \sqrt {e (a e-b d)+c d^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 e^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^3,x]

[Out]

((-2*e*Sqrt[a + x*(b + c*x)]*(16*c^3*(60*d^5 + 90*d^4*e*x + 20*d^3*e^2*x^2 - 5*d^2*e^3*x^3 + 2*d*e^4*x^4 - e^5
*x^5) + b*e^3*(16*a^2*e^2 + 8*a*b*e*(5*d + 9*e*x) - b^2*(105*d^2 + 170*d*e*x + 49*e^2*x^2)) + 2*c*e^2*(16*a^2*
e^2*(d + 2*e*x) - 2*a*b*e*(145*d^2 + 234*d*e*x + 65*e^2*x^2) + b^2*(420*d^3 + 655*d^2*e*x + 166*d*e^2*x^2 - 37
*e^3*x^3)) - 8*c^2*e*(a*e*(-100*d^3 - 155*d^2*e*x - 38*d*e^2*x^2 + 9*e^3*x^3) + b*(210*d^4 + 320*d^3*e*x + 75*
d^2*e^2*x^2 - 18*d*e^3*x^3 + 7*e^4*x^4))))/(d + e*x)^2 + (15*(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 + 2*c*x)/(2*Sqrt
[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c] + 120*(2*c*d - b*e)*(8*c^2*d^2 + b^2*e^2 + 4*c*e*(-2*b*d + a*e))*Sqrt[c*d
^2 + e*(-(b*d) + a*e)]*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)])])/(64*e^7)

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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)^3,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)^3,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)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 1.07Unable to divide, perhaps due to rounding error%%%{1,[6,0,10,0,0]%%%}+%%%{%%{[-6,0]:[1,0,%%%{
-1,[1]%%%}]%%},[5,0,9,0,1]%%%}+%%%{-3,[4,1,10,0,0]%%%}+%%%{3,[4,0,9,1,1]%%%}+%%%{%%%{12,[1]%%%},[4,0,8,0,2]%%%
}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1,9,0,1]%%%}+%%%{%%{[-12,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,8,1,2]%%%}
+%%%{%%{[%%%{-8,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,7,0,3]%%%}+%%%{3,[2,2,10,0,0]%%%}+%%%{-6,[2,1,9,1,1]%%
%}+%%%{%%%{-12,[1]%%%},[2,1,8,0,2]%%%}+%%%{3,[2,0,8,2,2]%%%}+%%%{%%%{12,[1]%%%},[2,0,7,1,3]%%%}+%%%{%%{[-6,0]:
[1,0,%%%{-1,[1]%%%}]%%},[1,2,9,0,1]%%%}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1,8,1,2]%%%}+%%%{%%{[-6,0]:[1
,0,%%%{-1,[1]%%%}]%%},[1,0,7,2,3]%%%}+%%%{-1,[0,3,10,0,0]%%%}+%%%{3,[0,2,9,1,1]%%%}+%%%{-3,[0,1,8,2,2]%%%}+%%%
{1,[0,0,7,3,3]%%%} / %%%{%%{poly1[%%%{1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[6,0,3,0,0]%%%}+%%%{%%%{-6,[2]%%%},
[5,0,2,0,1]%%%}+%%%{%%{[%%%{-3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,1,3,0,0]%%%}+%%%{%%{poly1[%%%{3,[1]%%%},0
]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,2,1,1]%%%}+%%%{%%{poly1[%%%{12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,1,0,2]%%
%}+%%%{%%%{12,[2]%%%},[3,1,2,0,1]%%%}+%%%{%%%{-12,[2]%%%},[3,0,1,1,2]%%%}+%%%{%%%{-8,[3]%%%},[3,0,0,0,3]%%%}+%
%%{%%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,2,3,0,0]%%%}+%%%{%%{[%%%{-6,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]
%%},[2,1,2,1,1]%%%}+%%%{%%{[%%%{-12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,1,1,0,2]%%%}+%%%{%%{poly1[%%%{3,[1]%
%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,1,2,2]%%%}+%%%{%%{poly1[%%%{12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,0,1
,3]%%%}+%%%{%%%{-6,[2]%%%},[1,2,2,0,1]%%%}+%%%{%%%{12,[2]%%%},[1,1,1,1,2]%%%}+%%%{%%%{-6,[2]%%%},[1,0,0,2,3]%%
%}+%%%{%%{[%%%{-1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,3,3,0,0]%%%}+%%%{%%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%
%%}]%%},[0,2,2,1,1]%%%}+%%%{%%{[%%%{-3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,1,1,2,2]%%%}+%%%{%%{poly1[%%%{1,[
1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,0,3,3]%%%} Error: Bad Argument Value

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maple [B]  time = 0.08, size = 18705, normalized size = 40.31 \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)^3,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)^3,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 )}^3} \,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)^3,x)

[Out]

int(((b + 2*c*x)*(a + b*x + c*x^2)^(5/2))/(d + e*x)^3, 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 )^{3}}\, 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)**3,x)

[Out]

Integral((b + 2*c*x)*(a + b*x + c*x**2)**(5/2)/(d + e*x)**3, x)

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