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

Optimal. Leaf size=210 \[ \frac {e \sqrt {a+b x+c x^2} \left (-2 c (4 a B e+A b e+b B d)+4 A c^2 d+3 b^2 B e\right )}{c^2 \left (b^2-4 a c\right )}+\frac {2 (d+e x) \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (2 A c e-3 b B e+4 B c d)}{2 c^{5/2}} \]

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Rubi [A]  time = 0.19, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {818, 640, 621, 206} \begin {gather*} \frac {e \sqrt {a+b x+c x^2} \left (-2 c (4 a B e+A b e+b B d)+4 A c^2 d+3 b^2 B e\right )}{c^2 \left (b^2-4 a c\right )}+\frac {2 (d+e x) \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (2 A c e-3 b B e+4 B c d)}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x)*(2*a*c*(B*d + A*e) - b*(A*c*d + a*B*e) - (b^2*B*e - b*c*(B*d + A*e) + 2*c*(A*c*d - a*B*e))*x))/(c
*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (e*(4*A*c^2*d + 3*b^2*B*e - 2*c*(b*B*d + A*b*e + 4*a*B*e))*Sqrt[a + b*
x + c*x^2])/(c^2*(b^2 - 4*a*c)) + (e*(4*B*c*d - 3*b*B*e + 2*A*c*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x
+ c*x^2])])/(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 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 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.45, size = 238, normalized size = 1.13 \begin {gather*} \frac {\frac {2 \sqrt {c} \left (B \left (8 a^2 c e^2+a \left (-3 b^2 e^2+2 b c e (2 d+5 e x)-4 c^2 \left (d^2+2 d e x-e^2 x^2\right )\right )-b x \left (3 b^2 e^2+b c e (e x-4 d)+2 c^2 d^2\right )\right )+2 A c \left (a b e^2-2 a c e (2 d+e x)+b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )\right )}{\sqrt {a+x (b+c x)}}+e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) (-2 A c e+3 b B e-4 B c d)}{2 c^{5/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[c]*(2*A*c*(a*b*e^2 + 2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x) - 2*a*c*e*(2*d + e*x)) + B*(8*a^2*c*
e^2 - b*x*(2*c^2*d^2 + 3*b^2*e^2 + b*c*e*(-4*d + e*x)) + a*(-3*b^2*e^2 + 2*b*c*e*(2*d + 5*e*x) - 4*c^2*(d^2 +
2*d*e*x - e^2*x^2)))))/Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*e*(-4*B*c*d + 3*b*B*e - 2*A*c*e)*ArcTanh[(b + 2*c
*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(2*c^(5/2)*(-b^2 + 4*a*c))

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IntegrateAlgebraic [A]  time = 1.35, size = 281, normalized size = 1.34 \begin {gather*} \frac {\log \left (-2 c^{5/2} \sqrt {a+b x+c x^2}+b c^2+2 c^3 x\right ) \left (-2 A c e^2+3 b B e^2-4 B c d e\right )}{2 c^{5/2}}-\frac {-8 a^2 B c e^2-2 a A b c e^2+8 a A c^2 d e+4 a A c^2 e^2 x+3 a b^2 B e^2-4 a b B c d e-10 a b B c e^2 x+4 a B c^2 d^2+8 a B c^2 d e x-4 a B c^2 e^2 x^2-2 A b^2 c e^2 x-2 A b c^2 d^2+4 A b c^2 d e x-4 A c^3 d^2 x+3 b^3 B e^2 x-4 b^2 B c d e x+b^2 B c e^2 x^2+2 b B c^2 d^2 x}{c^2 \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((-2*A*b*c^2*d^2 + 4*a*B*c^2*d^2 - 4*a*b*B*c*d*e + 8*a*A*c^2*d*e + 3*a*b^2*B*e^2 - 2*a*A*b*c*e^2 - 8*a^2*B*c*
e^2 + 2*b*B*c^2*d^2*x - 4*A*c^3*d^2*x - 4*b^2*B*c*d*e*x + 4*A*b*c^2*d*e*x + 8*a*B*c^2*d*e*x + 3*b^3*B*e^2*x -
2*A*b^2*c*e^2*x - 10*a*b*B*c*e^2*x + 4*a*A*c^2*e^2*x + b^2*B*c*e^2*x^2 - 4*a*B*c^2*e^2*x^2)/(c^2*(-b^2 + 4*a*c
)*Sqrt[a + b*x + c*x^2])) + ((-4*B*c*d*e + 3*b*B*e^2 - 2*A*c*e^2)*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[a + b*x
+ c*x^2]])/(2*c^(5/2))

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fricas [B]  time = 2.51, size = 945, normalized size = 4.50 \begin {gather*} \left [\frac {{\left (4 \, {\left (B a b^{2} c - 4 \, B a^{2} c^{2}\right )} d e - {\left (3 \, B a b^{3} + 8 \, A a^{2} c^{2} - 2 \, {\left (6 \, B a^{2} b + A a b^{2}\right )} c\right )} e^{2} + {\left (4 \, {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d e - {\left (3 \, B b^{3} c + 8 \, A a c^{3} - 2 \, {\left (6 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x^{2} + {\left (4 \, {\left (B b^{3} c - 4 \, B a b c^{2}\right )} d e - {\left (3 \, B b^{4} + 8 \, A a b c^{2} - 2 \, {\left (6 \, B a b^{2} + A b^{3}\right )} c\right )} e^{2}\right )} x\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 ) + 4 \, {\left (2 \, {\left (2 \, B a - A b\right )} c^{3} d^{2} + {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} e^{2} x^{2} - 4 \, {\left (B a b c^{2} - 2 \, A a c^{3}\right )} d e + {\left (3 \, B a b^{2} c - 2 \, {\left (4 \, B a^{2} + A a b\right )} c^{2}\right )} e^{2} + {\left (2 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} - 4 \, {\left (B b^{2} c^{2} - {\left (2 \, B a + A b\right )} c^{3}\right )} d e + {\left (3 \, B b^{3} c + 4 \, A a c^{3} - 2 \, {\left (5 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x\right )}}, -\frac {{\left (4 \, {\left (B a b^{2} c - 4 \, B a^{2} c^{2}\right )} d e - {\left (3 \, B a b^{3} + 8 \, A a^{2} c^{2} - 2 \, {\left (6 \, B a^{2} b + A a b^{2}\right )} c\right )} e^{2} + {\left (4 \, {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d e - {\left (3 \, B b^{3} c + 8 \, A a c^{3} - 2 \, {\left (6 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x^{2} + {\left (4 \, {\left (B b^{3} c - 4 \, B a b c^{2}\right )} d e - {\left (3 \, B b^{4} + 8 \, A a b c^{2} - 2 \, {\left (6 \, B a b^{2} + A b^{3}\right )} c\right )} e^{2}\right )} x\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 ) - 2 \, {\left (2 \, {\left (2 \, B a - A b\right )} c^{3} d^{2} + {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} e^{2} x^{2} - 4 \, {\left (B a b c^{2} - 2 \, A a c^{3}\right )} d e + {\left (3 \, B a b^{2} c - 2 \, {\left (4 \, B a^{2} + A a b\right )} c^{2}\right )} e^{2} + {\left (2 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} - 4 \, {\left (B b^{2} c^{2} - {\left (2 \, B a + A b\right )} c^{3}\right )} d e + {\left (3 \, B b^{3} c + 4 \, A a c^{3} - 2 \, {\left (5 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((4*(B*a*b^2*c - 4*B*a^2*c^2)*d*e - (3*B*a*b^3 + 8*A*a^2*c^2 - 2*(6*B*a^2*b + A*a*b^2)*c)*e^2 + (4*(B*b^2
*c^2 - 4*B*a*c^3)*d*e - (3*B*b^3*c + 8*A*a*c^3 - 2*(6*B*a*b + A*b^2)*c^2)*e^2)*x^2 + (4*(B*b^3*c - 4*B*a*b*c^2
)*d*e - (3*B*b^4 + 8*A*a*b*c^2 - 2*(6*B*a*b^2 + A*b^3)*c)*e^2)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*s
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(2*(2*B*a - A*b)*c^3*d^2 + (B*b^2*c^2 - 4*B*a*c^3)*e^2*x
^2 - 4*(B*a*b*c^2 - 2*A*a*c^3)*d*e + (3*B*a*b^2*c - 2*(4*B*a^2 + A*a*b)*c^2)*e^2 + (2*(B*b*c^3 - 2*A*c^4)*d^2
- 4*(B*b^2*c^2 - (2*B*a + A*b)*c^3)*d*e + (3*B*b^3*c + 4*A*a*c^3 - 2*(5*B*a*b + A*b^2)*c^2)*e^2)*x)*sqrt(c*x^2
+ b*x + a))/(a*b^2*c^3 - 4*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x), -1/2*((4*(B*a*b^2*c
- 4*B*a^2*c^2)*d*e - (3*B*a*b^3 + 8*A*a^2*c^2 - 2*(6*B*a^2*b + A*a*b^2)*c)*e^2 + (4*(B*b^2*c^2 - 4*B*a*c^3)*d*
e - (3*B*b^3*c + 8*A*a*c^3 - 2*(6*B*a*b + A*b^2)*c^2)*e^2)*x^2 + (4*(B*b^3*c - 4*B*a*b*c^2)*d*e - (3*B*b^4 + 8
*A*a*b*c^2 - 2*(6*B*a*b^2 + A*b^3)*c)*e^2)*x)*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)) - 2*(2*(2*B*a - A*b)*c^3*d^2 + (B*b^2*c^2 - 4*B*a*c^3)*e^2*x^2 - 4*(B*a*b*c^2 - 2*A*a*
c^3)*d*e + (3*B*a*b^2*c - 2*(4*B*a^2 + A*a*b)*c^2)*e^2 + (2*(B*b*c^3 - 2*A*c^4)*d^2 - 4*(B*b^2*c^2 - (2*B*a +
A*b)*c^3)*d*e + (3*B*b^3*c + 4*A*a*c^3 - 2*(5*B*a*b + A*b^2)*c^2)*e^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^3 -
4*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x)]

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giac [A]  time = 0.31, size = 294, normalized size = 1.40 \begin {gather*} \frac {{\left (\frac {{\left (B b^{2} c e^{2} - 4 \, B a c^{2} e^{2}\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} + \frac {2 \, B b c^{2} d^{2} - 4 \, A c^{3} d^{2} - 4 \, B b^{2} c d e + 8 \, B a c^{2} d e + 4 \, A b c^{2} d e + 3 \, B b^{3} e^{2} - 10 \, B a b c e^{2} - 2 \, A b^{2} c e^{2} + 4 \, A a c^{2} e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x + \frac {4 \, B a c^{2} d^{2} - 2 \, A b c^{2} d^{2} - 4 \, B a b c d e + 8 \, A a c^{2} d e + 3 \, B a b^{2} e^{2} - 8 \, B a^{2} c e^{2} - 2 \, A a b c e^{2}}{b^{2} c^{2} - 4 \, a c^{3}}}{\sqrt {c x^{2} + b x + a}} - \frac {{\left (4 \, B c d e - 3 \, B b e^{2} + 2 \, A c e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((B*b^2*c*e^2 - 4*B*a*c^2*e^2)*x/(b^2*c^2 - 4*a*c^3) + (2*B*b*c^2*d^2 - 4*A*c^3*d^2 - 4*B*b^2*c*d*e + 8*B*a*c
^2*d*e + 4*A*b*c^2*d*e + 3*B*b^3*e^2 - 10*B*a*b*c*e^2 - 2*A*b^2*c*e^2 + 4*A*a*c^2*e^2)/(b^2*c^2 - 4*a*c^3))*x
+ (4*B*a*c^2*d^2 - 2*A*b*c^2*d^2 - 4*B*a*b*c*d*e + 8*A*a*c^2*d*e + 3*B*a*b^2*e^2 - 8*B*a^2*c*e^2 - 2*A*a*b*c*e
^2)/(b^2*c^2 - 4*a*c^3))/sqrt(c*x^2 + b*x + a) - 1/2*(4*B*c*d*e - 3*B*b*e^2 + 2*A*c*e^2)*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 = 0.05, size = 779, normalized size = 3.71 \begin {gather*} \frac {A \,b^{2} e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {4 A b d e x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {4 B a b \,e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {3 B \,b^{3} e^{2} x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {2 B \,b^{2} d e x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {2 B b \,d^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {A \,b^{3} e^{2}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 A \,b^{2} d e}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 B a \,b^{2} e^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 B \,b^{4} e^{2}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {B \,b^{3} d e}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {B \,b^{2} d^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {B \,e^{2} x^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {A \,e^{2} x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) A \,d^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {3 B b \,e^{2} x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 B d e x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {A \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}-\frac {3 B b \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}+\frac {2 B d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {A b \,e^{2}}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 A d e}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 B a \,e^{2}}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 B \,b^{2} e^{2}}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {B b d e}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {B \,d^{2}}{\sqrt {c \,x^{2}+b x +a}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*d^2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3/4*B*e^2*b^2/c^3/(c*x^2+b*x+a)^(1/2)-2*b/(4*a*c-b^2)/(c*x^2
+b*x+a)^(1/2)*x*B*d^2-2*x/c/(c*x^2+b*x+a)^(1/2)*B*d*e+b/c^2/(c*x^2+b*x+a)^(1/2)*B*d*e+1/2*b^3/c^2/(4*a*c-b^2)/
(c*x^2+b*x+a)^(1/2)*A*e^2+3/2*B*e^2*b/c^2*x/(c*x^2+b*x+a)^(1/2)-3/4*B*e^2*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1
/2)-b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d^2-1/c/(c*x^2+b*x+a)^(1/2)*B*d^2+1/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)
+(c*x^2+b*x+a)^(1/2))*A*e^2+2*B*e^2*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-2*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a
)^(1/2)*A*d*e-4*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*d*e+b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d*e+b^2/c/
(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*e^2-3/2*B*e^2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+4*B*e^2*a/c*b/(4*a
*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+2*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d*e+1/2*b/c^2/(c*x^2+b*x+a)^(1/2)*A*
e^2+B*e^2*x^2/c/(c*x^2+b*x+a)^(1/2)+2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d*e-2/c/(c*x^2+b*x
+a)^(1/2)*A*d*e-3/2*B*e^2*b/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2*B*e^2*a/c^2/(c*x^2+b*x+a)^(1
/2)-x/c/(c*x^2+b*x+a)^(1/2)*A*e^2

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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((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a)^(3/2),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 (A+B\,x\right )\,{\left (d+e\,x\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )^{2}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**2/(c*x**2+b*x+a)**(3/2),x)

[Out]

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

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