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

Optimal. Leaf size=158 \[ \frac {4 e^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac {4 e (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 (d+e x)^3}{3 \left (a+b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 158, 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, 640, 621, 206} \begin {gather*} \frac {4 e^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac {4 e (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e^3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 (d+e x)^3}{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)^3)/(a + b*x + c*x^2)^(5/2),x]

[Out]

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

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 640

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.49, size = 285, normalized size = 1.80 \begin {gather*} \frac {6 e^3 \sqrt {a+x (b+c x)} \left (4 a^2 c+a \left (-b^2+4 b c x+4 c^2 x^2\right )-b^2 x (b+c x)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+2 \sqrt {c} \left (6 b e \left (a^2 e^2+a c \left (d^2-6 d e x-e^2 x^2\right )+c^2 d x^2 (3 d-2 e x)\right )-4 c \left (3 a^2 e^2 (2 d+e x)+a c \left (d^3+9 d e^2 x^2+4 e^3 x^3\right )-3 c^2 d^2 e x^3\right )+b^2 \left (12 a e^3 x+c \left (d^3+9 d^2 e x-9 d e^2 x^2+7 e^3 x^3\right )\right )+6 b^3 e^3 x^2\right )}{3 c^{3/2} \left (4 a c-b^2\right ) (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.67, size = 273, normalized size = 1.73 \begin {gather*} \frac {2 \left (6 a^2 b e^3-24 a^2 c d e^2-12 a^2 c e^3 x+12 a b^2 e^3 x+6 a b c d^2 e-36 a b c d e^2 x-6 a b c e^3 x^2-4 a c^2 d^3-36 a c^2 d e^2 x^2-16 a c^2 e^3 x^3+6 b^3 e^3 x^2+b^2 c d^3+9 b^2 c d^2 e x-9 b^2 c d e^2 x^2+7 b^2 c e^3 x^3+18 b c^2 d^2 e x^2-12 b c^2 d e^2 x^3+12 c^3 d^2 e x^3\right )}{3 c \left (4 a c-b^2\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 e^3 \log \left (-2 c^{3/2} \sqrt {a+b x+c x^2}+b c+2 c^2 x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2*c*d^3 - 4*a*c^2*d^3 + 6*a*b*c*d^2*e - 24*a^2*c*d*e^2 + 6*a^2*b*e^3 + 9*b^2*c*d^2*e*x - 36*a*b*c*d*e^2*
x + 12*a*b^2*e^3*x - 12*a^2*c*e^3*x + 18*b*c^2*d^2*e*x^2 - 9*b^2*c*d*e^2*x^2 - 36*a*c^2*d*e^2*x^2 + 6*b^3*e^3*
x^2 - 6*a*b*c*e^3*x^2 + 12*c^3*d^2*e*x^3 - 12*b*c^2*d*e^2*x^3 + 7*b^2*c*e^3*x^3 - 16*a*c^2*e^3*x^3))/(3*c*(-b^
2 + 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (2*e^3*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[a + b*x + c*x^2]])/c^(3/2)

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fricas [B]  time = 1.41, size = 958, normalized size = 6.06 \begin {gather*} \left [\frac {3 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{3} x^{4} + 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3} x^{3} + {\left (b^{4} - 2 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} e^{3} x^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} e^{3} x + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 2 \, {\left (6 \, a b c^{2} d^{2} e - 24 \, a^{2} c^{2} d e^{2} + 6 \, a^{2} b c e^{3} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{3} + {\left (12 \, c^{4} d^{2} e - 12 \, b c^{3} d e^{2} + {\left (7 \, b^{2} c^{2} - 16 \, a c^{3}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, b c^{3} d^{2} e - 3 \, {\left (b^{2} c^{2} + 4 \, a c^{3}\right )} d e^{2} + 2 \, {\left (b^{3} c - a b c^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (3 \, b^{2} c^{2} d^{2} e - 12 \, a b c^{2} d e^{2} + 4 \, {\left (a b^{2} c - a^{2} c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{2} c^{2} - 4 \, a^{3} c^{3} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} + 2 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{3} + {\left (b^{4} c^{2} - 2 \, a b^{2} c^{3} - 8 \, a^{2} c^{4}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b c^{3}\right )} x\right )}}, -\frac {2 \, {\left (3 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{3} x^{4} + 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3} x^{3} + {\left (b^{4} - 2 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} e^{3} x^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} e^{3} x + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (6 \, a b c^{2} d^{2} e - 24 \, a^{2} c^{2} d e^{2} + 6 \, a^{2} b c e^{3} + {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{3} + {\left (12 \, c^{4} d^{2} e - 12 \, b c^{3} d e^{2} + {\left (7 \, b^{2} c^{2} - 16 \, a c^{3}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, b c^{3} d^{2} e - 3 \, {\left (b^{2} c^{2} + 4 \, a c^{3}\right )} d e^{2} + 2 \, {\left (b^{3} c - a b c^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (3 \, b^{2} c^{2} d^{2} e - 12 \, a b c^{2} d e^{2} + 4 \, {\left (a b^{2} c - a^{2} c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (a^{2} b^{2} c^{2} - 4 \, a^{3} c^{3} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} + 2 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{3} + {\left (b^{4} c^{2} - 2 \, a b^{2} c^{3} - 8 \, a^{2} c^{4}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b c^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.35, size = 492, normalized size = 3.11 \begin {gather*} -\frac {2 \, e^{3} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {3}{2}}} - \frac {2 \, {\left ({\left ({\left (\frac {{\left (12 \, b^{2} c^{3} d^{2} e - 48 \, a c^{4} d^{2} e - 12 \, b^{3} c^{2} d e^{2} + 48 \, a b c^{3} d e^{2} + 7 \, b^{4} c e^{3} - 44 \, a b^{2} c^{2} e^{3} + 64 \, a^{2} c^{3} e^{3}\right )} x}{b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}} + \frac {3 \, {\left (6 \, b^{3} c^{2} d^{2} e - 24 \, a b c^{3} d^{2} e - 3 \, b^{4} c d e^{2} + 48 \, a^{2} c^{3} d e^{2} + 2 \, b^{5} e^{3} - 10 \, a b^{3} c e^{3} + 8 \, a^{2} b c^{2} e^{3}\right )}}{b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}}\right )} x + \frac {3 \, {\left (3 \, b^{4} c d^{2} e - 12 \, a b^{2} c^{2} d^{2} e - 12 \, a b^{3} c d e^{2} + 48 \, a^{2} b c^{2} d e^{2} + 4 \, a b^{4} e^{3} - 20 \, a^{2} b^{2} c e^{3} + 16 \, a^{3} c^{2} e^{3}\right )}}{b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}}\right )} x + \frac {b^{4} c d^{3} - 8 \, a b^{2} c^{2} d^{3} + 16 \, a^{2} c^{3} d^{3} + 6 \, a b^{3} c d^{2} e - 24 \, a^{2} b c^{2} d^{2} e - 24 \, a^{2} b^{2} c d e^{2} + 96 \, a^{3} c^{2} d e^{2} + 6 \, a^{2} b^{3} e^{3} - 24 \, a^{3} b c e^{3}}{b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2*e^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2) - 2/3*((((12*b^2*c^3*d^2*e - 48*a*
c^4*d^2*e - 12*b^3*c^2*d*e^2 + 48*a*b*c^3*d*e^2 + 7*b^4*c*e^3 - 44*a*b^2*c^2*e^3 + 64*a^2*c^3*e^3)*x/(b^4*c -
8*a*b^2*c^2 + 16*a^2*c^3) + 3*(6*b^3*c^2*d^2*e - 24*a*b*c^3*d^2*e - 3*b^4*c*d*e^2 + 48*a^2*c^3*d*e^2 + 2*b^5*e
^3 - 10*a*b^3*c*e^3 + 8*a^2*b*c^2*e^3)/(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3))*x + 3*(3*b^4*c*d^2*e - 12*a*b^2*c^2
*d^2*e - 12*a*b^3*c*d*e^2 + 48*a^2*b*c^2*d*e^2 + 4*a*b^4*e^3 - 20*a^2*b^2*c*e^3 + 16*a^3*c^2*e^3)/(b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3))*x + (b^4*c*d^3 - 8*a*b^2*c^2*d^3 + 16*a^2*c^3*d^3 + 6*a*b^3*c*d^2*e - 24*a^2*b*c^2*d^2
*e - 24*a^2*b^2*c*d*e^2 + 96*a^3*c^2*d*e^2 + 6*a^2*b^3*e^3 - 24*a^3*b*c*e^3)/(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3
))/(c*x^2 + b*x + a)^(3/2)

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maple [B]  time = 0.06, size = 865, normalized size = 5.47 \begin {gather*} -\frac {32 a b c d \,e^{2} x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {32 a \,c^{2} d^{2} e x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {8 b^{3} d \,e^{2} x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {8 b^{2} c \,d^{2} e x}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {16 a \,b^{2} d \,e^{2}}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {16 a b c \,d^{2} e}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {4 a b d \,e^{2} x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {4 a c \,d^{2} e x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {4 b^{4} d \,e^{2}}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, c}+\frac {b^{3} d \,e^{2} x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}-\frac {4 b^{3} d^{2} e}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {b^{2} d^{2} e x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {2 e^{3} x^{3}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {2 a \,b^{2} d \,e^{2}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}+\frac {2 a b \,d^{2} e}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b^{4} d \,e^{2}}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}-\frac {b^{3} d^{2} e}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}+\frac {2 b^{2} e^{3} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {6 d \,e^{2} x^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b^{3} e^{3}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 b d \,e^{2} x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}-\frac {3 d^{2} e x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {4 a d \,e^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}+\frac {b^{2} d \,e^{2}}{2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{2}}-\frac {b \,d^{2} e}{2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c}-\frac {2 e^{3} x}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {2 d^{3}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {2 e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {b \,e^{3}}{\sqrt {c \,x^{2}+b x +a}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^3/(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 = -0.01 \begin {gather*} \int \frac {\left (b+2\,c\,x\right )\,{\left (d+e\,x\right )}^3}{{\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)^3)/(a + b*x + c*x^2)^(5/2),x)

[Out]

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

[Out]

Timed out

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