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

Optimal. Leaf size=208 \[ \frac {8 e^2 \sqrt {a+b x+c x^2} \left (-2 c e (4 a e+3 b d)+3 b^2 e^2+2 c e x (2 c d-b e)+8 c^2 d^2\right )}{3 c^2 \left (b^2-4 a c\right )}-\frac {16 e (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 e^3 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.23, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {768, 738, 779, 621, 206} \begin {gather*} \frac {8 e^2 \sqrt {a+b x+c x^2} \left (-2 c e (4 a e+3 b d)+3 b^2 e^2+2 c e x (2 c d-b e)+8 c^2 d^2\right )}{3 c^2 \left (b^2-4 a c\right )}-\frac {16 e (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 e^3 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^4)/(3*(a + b*x + c*x^2)^(3/2)) - (16*e*(d + e*x)^2*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*
a*c)*Sqrt[a + b*x + c*x^2]) + (8*e^2*(8*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(3*b*d + 4*a*e) + 2*c*e*(2*c*d - b*e)*x)*S
qrt[a + b*x + c*x^2])/(3*c^2*(b^2 - 4*a*c)) + (4*e^3*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x
+ c*x^2])])/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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 768

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

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) (d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac {1}{3} (8 e) \int \frac {(d+e x)^3}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 e (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {(16 e) \int \frac {(d+e x) (-2 e (b d-2 a e)-2 e (2 c d-b e) x)}{\sqrt {a+b x+c x^2}} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 e (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {8 e^2 \left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left (4 e^3 (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c^2}\\ &=-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 e (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {8 e^2 \left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {\left (8 e^3 (2 c d-b e)\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 )}{c^2}\\ &=-\frac {2 (d+e x)^4}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 e (d+e x)^2 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {8 e^2 \left (8 c^2 d^2+3 b^2 e^2-2 c e (3 b d+4 a e)+2 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )}+\frac {4 e^3 (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 354, normalized size = 1.70 \begin {gather*} \frac {2 \left (\frac {\sqrt {c} \left (b^2 \left (-12 a^2 e^4+24 a c e^3 x (2 d+e x)+c^2 \left (d^4+12 d^3 e x-18 d^2 e^2 x^2+28 d e^3 x^3-3 e^4 x^4\right )\right )+8 b c e \left (3 a^2 e^2 (d+3 e x)+a c \left (d^3-9 d^2 e x-3 d e^2 x^2+7 e^3 x^3\right )+3 c^2 d^2 x^2 (d-e x)\right )+4 c \left (8 a^3 e^4-12 a^2 c e^2 \left (d^2+d e x-e^2 x^2\right )-a c^2 \left (d^4+18 d^2 e^2 x^2+16 d e^3 x^3-3 e^4 x^4\right )+4 c^3 d^3 e x^3\right )-8 b^3 e^3 x (3 a e+c x (2 e x-3 d))-12 b^4 e^4 x^2\right )}{(a+x (b+c x))^{3/2}}+6 e^3 \left (b^2-4 a c\right ) (b e-2 c d) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{3 c^{5/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((Sqrt[c]*(-12*b^4*e^4*x^2 - 8*b^3*e^3*x*(3*a*e + c*x*(-3*d + 2*e*x)) + 8*b*c*e*(3*c^2*d^2*x^2*(d - e*x) +
3*a^2*e^2*(d + 3*e*x) + a*c*(d^3 - 9*d^2*e*x - 3*d*e^2*x^2 + 7*e^3*x^3)) + 4*c*(8*a^3*e^4 + 4*c^3*d^3*e*x^3 -
12*a^2*c*e^2*(d^2 + d*e*x - e^2*x^2) - a*c^2*(d^4 + 18*d^2*e^2*x^2 + 16*d*e^3*x^3 - 3*e^4*x^4)) + b^2*(-12*a^2
*e^4 + 24*a*c*e^3*x*(2*d + e*x) + c^2*(d^4 + 12*d^3*e*x - 18*d^2*e^2*x^2 + 28*d*e^3*x^3 - 3*e^4*x^4))))/(a + x
*(b + c*x))^(3/2) + 6*(b^2 - 4*a*c)*e^3*(-2*c*d + b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])
)/(3*c^(5/2)*(-b^2 + 4*a*c))

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IntegrateAlgebraic [B]  time = 3.97, size = 455, normalized size = 2.19 \begin {gather*} -\frac {2 \left (-32 a^3 c e^4+12 a^2 b^2 e^4-24 a^2 b c d e^3-72 a^2 b c e^4 x+48 a^2 c^2 d^2 e^2+48 a^2 c^2 d e^3 x-48 a^2 c^2 e^4 x^2+24 a b^3 e^4 x-48 a b^2 c d e^3 x-24 a b^2 c e^4 x^2-8 a b c^2 d^3 e+72 a b c^2 d^2 e^2 x+24 a b c^2 d e^3 x^2-56 a b c^2 e^4 x^3+4 a c^3 d^4+72 a c^3 d^2 e^2 x^2+64 a c^3 d e^3 x^3-12 a c^3 e^4 x^4+12 b^4 e^4 x^2-24 b^3 c d e^3 x^2+16 b^3 c e^4 x^3-b^2 c^2 d^4-12 b^2 c^2 d^3 e x+18 b^2 c^2 d^2 e^2 x^2-28 b^2 c^2 d e^3 x^3+3 b^2 c^2 e^4 x^4-24 b c^3 d^3 e x^2+24 b c^3 d^2 e^2 x^3-16 c^4 d^3 e x^3\right )}{3 c^2 \left (4 a c-b^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {4 \left (2 c d e^3-b e^4\right ) \log \left (-2 c^{5/2} \sqrt {a+b x+c x^2}+b c^2+2 c^3 x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-(b^2*c^2*d^4) + 4*a*c^3*d^4 - 8*a*b*c^2*d^3*e + 48*a^2*c^2*d^2*e^2 - 24*a^2*b*c*d*e^3 + 12*a^2*b^2*e^4 -
32*a^3*c*e^4 - 12*b^2*c^2*d^3*e*x + 72*a*b*c^2*d^2*e^2*x - 48*a*b^2*c*d*e^3*x + 48*a^2*c^2*d*e^3*x + 24*a*b^3
*e^4*x - 72*a^2*b*c*e^4*x - 24*b*c^3*d^3*e*x^2 + 18*b^2*c^2*d^2*e^2*x^2 + 72*a*c^3*d^2*e^2*x^2 - 24*b^3*c*d*e^
3*x^2 + 24*a*b*c^2*d*e^3*x^2 + 12*b^4*e^4*x^2 - 24*a*b^2*c*e^4*x^2 - 48*a^2*c^2*e^4*x^2 - 16*c^4*d^3*e*x^3 + 2
4*b*c^3*d^2*e^2*x^3 - 28*b^2*c^2*d*e^3*x^3 + 64*a*c^3*d*e^3*x^3 + 16*b^3*c*e^4*x^3 - 56*a*b*c^2*e^4*x^3 + 3*b^
2*c^2*e^4*x^4 - 12*a*c^3*e^4*x^4))/(3*c^2*(-b^2 + 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (4*(2*c*d*e^3 - b*e^4)*Log
[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[a + b*x + c*x^2]])/c^(5/2)

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fricas [B]  time = 1.64, size = 1493, normalized size = 7.18

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-2/3*(3*(2*(a^2*b^2*c - 4*a^3*c^2)*d*e^3 - (a^2*b^3 - 4*a^3*b*c)*e^4 + (2*(b^2*c^3 - 4*a*c^4)*d*e^3 - (b^3*c^
2 - 4*a*b*c^3)*e^4)*x^4 + 2*(2*(b^3*c^2 - 4*a*b*c^3)*d*e^3 - (b^4*c - 4*a*b^2*c^2)*e^4)*x^3 + (2*(b^4*c - 2*a*
b^2*c^2 - 8*a^2*c^3)*d*e^3 - (b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*e^4)*x^2 + 2*(2*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 - (
a*b^4 - 4*a^2*b^2*c)*e^4)*x)*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) + (8*a*b*c^3*d^3*e - 48*a^2*c^3*d^2*e^2 + 24*a^2*b*c^2*d*e^3 - 3*(b^2*c^3 - 4*a*c^4)*e^4*x^4 + (b
^2*c^3 - 4*a*c^4)*d^4 - 4*(3*a^2*b^2*c - 8*a^3*c^2)*e^4 + 4*(4*c^5*d^3*e - 6*b*c^4*d^2*e^2 + (7*b^2*c^3 - 16*a
*c^4)*d*e^3 - 2*(2*b^3*c^2 - 7*a*b*c^3)*e^4)*x^3 + 6*(4*b*c^4*d^3*e - 3*(b^2*c^3 + 4*a*c^4)*d^2*e^2 + 4*(b^3*c
^2 - a*b*c^3)*d*e^3 - 2*(b^4*c - 2*a*b^2*c^2 - 4*a^2*c^3)*e^4)*x^2 + 12*(b^2*c^3*d^3*e - 6*a*b*c^3*d^2*e^2 + 4
*(a*b^2*c^2 - a^2*c^3)*d*e^3 - 2*(a*b^3*c - 3*a^2*b*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^2*c^3 - 4*a^3*c
^4 + (b^2*c^5 - 4*a*c^6)*x^4 + 2*(b^3*c^4 - 4*a*b*c^5)*x^3 + (b^4*c^3 - 2*a*b^2*c^4 - 8*a^2*c^5)*x^2 + 2*(a*b^
3*c^3 - 4*a^2*b*c^4)*x), -2/3*(6*(2*(a^2*b^2*c - 4*a^3*c^2)*d*e^3 - (a^2*b^3 - 4*a^3*b*c)*e^4 + (2*(b^2*c^3 -
4*a*c^4)*d*e^3 - (b^3*c^2 - 4*a*b*c^3)*e^4)*x^4 + 2*(2*(b^3*c^2 - 4*a*b*c^3)*d*e^3 - (b^4*c - 4*a*b^2*c^2)*e^4
)*x^3 + (2*(b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d*e^3 - (b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*e^4)*x^2 + 2*(2*(a*b^3*c
- 4*a^2*b*c^2)*d*e^3 - (a*b^4 - 4*a^2*b^2*c)*e^4)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqr
t(-c)/(c^2*x^2 + b*c*x + a*c)) + (8*a*b*c^3*d^3*e - 48*a^2*c^3*d^2*e^2 + 24*a^2*b*c^2*d*e^3 - 3*(b^2*c^3 - 4*a
*c^4)*e^4*x^4 + (b^2*c^3 - 4*a*c^4)*d^4 - 4*(3*a^2*b^2*c - 8*a^3*c^2)*e^4 + 4*(4*c^5*d^3*e - 6*b*c^4*d^2*e^2 +
(7*b^2*c^3 - 16*a*c^4)*d*e^3 - 2*(2*b^3*c^2 - 7*a*b*c^3)*e^4)*x^3 + 6*(4*b*c^4*d^3*e - 3*(b^2*c^3 + 4*a*c^4)*
d^2*e^2 + 4*(b^3*c^2 - a*b*c^3)*d*e^3 - 2*(b^4*c - 2*a*b^2*c^2 - 4*a^2*c^3)*e^4)*x^2 + 12*(b^2*c^3*d^3*e - 6*a
*b*c^3*d^2*e^2 + 4*(a*b^2*c^2 - a^2*c^3)*d*e^3 - 2*(a*b^3*c - 3*a^2*b*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(a^2
*b^2*c^3 - 4*a^3*c^4 + (b^2*c^5 - 4*a*c^6)*x^4 + 2*(b^3*c^4 - 4*a*b*c^5)*x^3 + (b^4*c^3 - 2*a*b^2*c^4 - 8*a^2*
c^5)*x^2 + 2*(a*b^3*c^3 - 4*a^2*b*c^4)*x)]

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giac [B]  time = 0.38, size = 754, normalized size = 3.62 \begin {gather*} \frac {2 \, {\left ({\left ({\left ({\left (\frac {3 \, {\left (b^{4} c^{2} e^{4} - 8 \, a b^{2} c^{3} e^{4} + 16 \, a^{2} c^{4} e^{4}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} - \frac {4 \, {\left (4 \, b^{2} c^{4} d^{3} e - 16 \, a c^{5} d^{3} e - 6 \, b^{3} c^{3} d^{2} e^{2} + 24 \, a b c^{4} d^{2} e^{2} + 7 \, b^{4} c^{2} d e^{3} - 44 \, a b^{2} c^{3} d e^{3} + 64 \, a^{2} c^{4} d e^{3} - 4 \, b^{5} c e^{4} + 30 \, a b^{3} c^{2} e^{4} - 56 \, a^{2} b c^{3} e^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {6 \, {\left (4 \, b^{3} c^{3} d^{3} e - 16 \, a b c^{4} d^{3} e - 3 \, b^{4} c^{2} d^{2} e^{2} + 48 \, a^{2} c^{4} d^{2} e^{2} + 4 \, b^{5} c d e^{3} - 20 \, a b^{3} c^{2} d e^{3} + 16 \, a^{2} b c^{3} d e^{3} - 2 \, b^{6} e^{4} + 12 \, a b^{4} c e^{4} - 8 \, a^{2} b^{2} c^{2} e^{4} - 32 \, a^{3} c^{3} e^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {12 \, {\left (b^{4} c^{2} d^{3} e - 4 \, a b^{2} c^{3} d^{3} e - 6 \, a b^{3} c^{2} d^{2} e^{2} + 24 \, a^{2} b c^{3} d^{2} e^{2} + 4 \, a b^{4} c d e^{3} - 20 \, a^{2} b^{2} c^{2} d e^{3} + 16 \, a^{3} c^{3} d e^{3} - 2 \, a b^{5} e^{4} + 14 \, a^{2} b^{3} c e^{4} - 24 \, a^{3} b c^{2} e^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {b^{4} c^{2} d^{4} - 8 \, a b^{2} c^{3} d^{4} + 16 \, a^{2} c^{4} d^{4} + 8 \, a b^{3} c^{2} d^{3} e - 32 \, a^{2} b c^{3} d^{3} e - 48 \, a^{2} b^{2} c^{2} d^{2} e^{2} + 192 \, a^{3} c^{3} d^{2} e^{2} + 24 \, a^{2} b^{3} c d e^{3} - 96 \, a^{3} b c^{2} d e^{3} - 12 \, a^{2} b^{4} e^{4} + 80 \, a^{3} b^{2} c e^{4} - 128 \, a^{4} c^{2} e^{4}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} - \frac {4 \, {\left (2 \, c d e^{3} - b e^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((((3*(b^4*c^2*e^4 - 8*a*b^2*c^3*e^4 + 16*a^2*c^4*e^4)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) - 4*(4*b^2*c
^4*d^3*e - 16*a*c^5*d^3*e - 6*b^3*c^3*d^2*e^2 + 24*a*b*c^4*d^2*e^2 + 7*b^4*c^2*d*e^3 - 44*a*b^2*c^3*d*e^3 + 64
*a^2*c^4*d*e^3 - 4*b^5*c*e^4 + 30*a*b^3*c^2*e^4 - 56*a^2*b*c^3*e^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x -
6*(4*b^3*c^3*d^3*e - 16*a*b*c^4*d^3*e - 3*b^4*c^2*d^2*e^2 + 48*a^2*c^4*d^2*e^2 + 4*b^5*c*d*e^3 - 20*a*b^3*c^2*
d*e^3 + 16*a^2*b*c^3*d*e^3 - 2*b^6*e^4 + 12*a*b^4*c*e^4 - 8*a^2*b^2*c^2*e^4 - 32*a^3*c^3*e^4)/(b^4*c^2 - 8*a*b
^2*c^3 + 16*a^2*c^4))*x - 12*(b^4*c^2*d^3*e - 4*a*b^2*c^3*d^3*e - 6*a*b^3*c^2*d^2*e^2 + 24*a^2*b*c^3*d^2*e^2 +
4*a*b^4*c*d*e^3 - 20*a^2*b^2*c^2*d*e^3 + 16*a^3*c^3*d*e^3 - 2*a*b^5*e^4 + 14*a^2*b^3*c*e^4 - 24*a^3*b*c^2*e^4
)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x - (b^4*c^2*d^4 - 8*a*b^2*c^3*d^4 + 16*a^2*c^4*d^4 + 8*a*b^3*c^2*d^3*
e - 32*a^2*b*c^3*d^3*e - 48*a^2*b^2*c^2*d^2*e^2 + 192*a^3*c^3*d^2*e^2 + 24*a^2*b^3*c*d*e^3 - 96*a^3*b*c^2*d*e^
3 - 12*a^2*b^4*e^4 + 80*a^3*b^2*c*e^4 - 128*a^4*c^2*e^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x +
a)^(3/2) - 4*(2*c*d*e^3 - b*e^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.07, size = 1692, normalized size = 8.13

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*x^2/(c*x^2+b*x+a)^(3/2)*d^2*e^2-4*x/(c*x^2+b*x+a)^(3/2)*d^3*e+16/3/c^2*e^4*a^2/(c*x^2+b*x+a)^(3/2)+1/12/c^
4*e^4*b^4/(c*x^2+b*x+a)^(3/2)-2/c^3*e^4*b^2/(c*x^2+b*x+a)^(1/2)-8/3*x^3/(c*x^2+b*x+a)^(3/2)*d*e^3-4/c^(5/2)*e^
4*b*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+8/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e^3-2/
3/(c*x^2+b*x+a)^(3/2)*d^4+2*e^4*x^4/(c*x^2+b*x+a)^(3/2)-64*b*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*c*d^2*e^2+b
^4/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*d^2*e^2-64/3/c*e^4*b^3*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-8/3/c^2*e^
4*b^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+16*b^3/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*d^2*e^2+2/c^2*e^4*a*b*x/(
c*x^2+b*x+a)^(3/2)+4/c^2*e^4*a^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+32/c*e^4*a^2*b^2/(4*a*c-b^2)^2/(c*x^2+b*x
+a)^(1/2)-6*b/c*x/(c*x^2+b*x+a)^(3/2)*d^2*e^2+4/c^2*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d*e^3+64*e^4*a^2*b/(4*
a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-1/2/c^3*e^4*b^3*x/(c*x^2+b*x+a)^(3/2)-2/c^3*e^4*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)
^(1/2)-4/c^2*e^4*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-4/3*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x*d^3*e-2/3*b^3
/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*d^3*e+8/3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*b*d^3*e+1/12/c^4*e^4*b^6/(4*a*c
-b^2)/(c*x^2+b*x+a)^(3/2)+2/3/c^3*e^4*b^6/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)-5/3/c^3*e^4*b^2*a/(c*x^2+b*x+a)^(3
/2)+4/c^2*e^4*b*x/(c*x^2+b*x+a)^(1/2)+8/c*e^4*a*x^2/(c*x^2+b*x+a)^(3/2)+4/3/c*e^4*b*x^3/(c*x^2+b*x+a)^(3/2)-2/
c^2*e^4*b^2*x^2/(c*x^2+b*x+a)^(3/2)+b^2/c^2/(c*x^2+b*x+a)^(3/2)*d^2*e^2-8*a/c/(c*x^2+b*x+a)^(3/2)*d^2*e^2-8/c*
x/(c*x^2+b*x+a)^(1/2)*d*e^3+4/c^2*b/(c*x^2+b*x+a)^(1/2)*d*e^3+4/3/c^2*e^4*b^5/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2
)*x-4/3/c^3*e^4*b^4*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-32/3/c^2*e^4*b^4*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/6
/c^3*e^4*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+8*b^4/c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*d^2*e^2-32*b^2*a/(4*a
*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*d^2*e^2-2/3*b/c/(c*x^2+b*x+a)^(3/2)*d^3*e-16/3*b^3/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(
1/2)*d^3*e+8/c*e^4*a^2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-32/3*b^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*c*d^3*
e+16/3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x*c*d^3*e+128/3*a*c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x*d^3*e+64/3*
a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*b*c*d^3*e+2*b^3/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x*d^2*e^2-8*b*a/(4*a*c-b
^2)/(c*x^2+b*x+a)^(3/2)*x*d^2*e^2-4*b^2/c*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*d^2*e^2+8/c*b^2/(4*a*c-b^2)/(c*x^2
+b*x+a)^(1/2)*x*d*e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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