3.235 \(\int \frac {(g+h x) (d+e x+f x^2)}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=186 \[ \frac {2 (g+h x) \left (c \left (2 a e-b \left (\frac {a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {h \sqrt {a+b x+c x^2} \left (-8 a c f+3 b^2 f-2 b c e+4 c^2 d\right )}{c^2 \left (b^2-4 a c\right )}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (3 b f h-2 c (e h+f g))}{2 c^{5/2}} \]

[Out]

-1/2*(3*b*f*h-2*c*(e*h+f*g))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)+2*(c*(2*a*e-b*(d+a*f/c
))-(-2*a*c*f+b^2*f-b*c*e+2*c^2*d)*x)*(h*x+g)/c/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/2)+(4*c^2*d+3*b^2*f-2*c*(4*a*f+b*
e))*h*(c*x^2+b*x+a)^(1/2)/c^2/(-4*a*c+b^2)

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Rubi [A]  time = 0.23, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1644, 640, 621, 206} \[ \frac {2 (g+h x) \left (c \left (2 a e-b \left (\frac {a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {h \sqrt {a+b x+c x^2} \left (-8 a c f+3 b^2 f-2 b c e+4 c^2 d\right )}{c^2 \left (b^2-4 a c\right )}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (3 b f h-2 c (e h+f g))}{2 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x)*(g + h*x))/(c*(b^2 - 4*a*c)*Sqrt[a +
b*x + c*x^2]) + ((4*c^2*d - 2*b*c*e + 3*b^2*f - 8*a*c*f)*h*Sqrt[a + b*x + c*x^2])/(c^2*(b^2 - 4*a*c)) - ((3*b*
f*h - 2*c*(f*g + e*h))*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 1644

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[Po
lynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g +
(2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x
+ c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Q + g*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c
*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[p] ||  !IntegerQ[m
] ||  !RationalQ[a, b, c, d, e]) &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2,
0]))

Rubi steps

\begin {align*} \int \frac {(g+h x) \left (d+e x+f x^2\right )}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac {2 \left (c \left (2 a e-b \left (d+\frac {a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {-\frac {b^2 f g+2 b (c d+a f) h-4 a c (f g+e h)}{2 c}-\frac {1}{2} \left (4 c d-2 b e-8 a f+\frac {3 b^2 f}{c}\right ) h x}{\sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=\frac {2 \left (c \left (2 a e-b \left (d+\frac {a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (4 c^2 d+3 b^2 f-2 c (b e+4 a f)\right ) h \sqrt {a+b x+c x^2}}{c^2 \left (b^2-4 a c\right )}-\frac {(3 b f h-2 c (f g+e h)) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c^2}\\ &=\frac {2 \left (c \left (2 a e-b \left (d+\frac {a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (4 c^2 d+3 b^2 f-2 c (b e+4 a f)\right ) h \sqrt {a+b x+c x^2}}{c^2 \left (b^2-4 a c\right )}-\frac {(3 b f h-2 c (f g+e h)) \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 \left (c \left (2 a e-b \left (d+\frac {a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {\left (4 c^2 d+3 b^2 f-2 c (b e+4 a f)\right ) h \sqrt {a+b x+c x^2}}{c^2 \left (b^2-4 a c\right )}-\frac {(3 b f h-2 c (f g+e h)) \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 = 0.73, size = 205, normalized size = 1.10 \[ \frac {\frac {2 \sqrt {c} \left (4 c \left (2 a^2 f h-a c (d h+e (g+h x)+f x (g-h x))+c^2 d g x\right )+b^2 (c x (2 e h+2 f g-f h x)-3 a f h)+2 b c (a e h+a f (g+5 h x)+c d (g-h x)-c e g x)-3 b^3 f h x\right )}{\sqrt {a+x (b+c x)}}+\left (b^2-4 a c\right ) \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right ) (3 b f h-2 c (e h+f g))}{2 c^{5/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[c]*(-3*b^3*f*h*x + 2*b*c*(a*e*h - c*e*g*x + c*d*(g - h*x) + a*f*(g + 5*h*x)) + b^2*(-3*a*f*h + c*x*(2
*f*g + 2*e*h - f*h*x)) + 4*c*(2*a^2*f*h + c^2*d*g*x - a*c*(d*h + f*x*(g - h*x) + e*(g + h*x)))))/Sqrt[a + x*(b
+ c*x)] + (b^2 - 4*a*c)*(3*b*f*h - 2*c*(f*g + e*h))*Log[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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fricas [B]  time = 14.45, size = 905, normalized size = 4.87 \[ \left [-\frac {{\left (2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} f g + {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f g + {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e - 3 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} f\right )} h\right )} x^{2} + {\left (2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e - 3 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} f\right )} h + {\left (2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} f g + {\left (2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e - 3 \, {\left (b^{4} - 4 \, a b^{2} c\right )} f\right )} h\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 ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f h x^{2} - 2 \, {\left (b c^{3} d - 2 \, a c^{3} e + a b c^{2} f\right )} g + {\left (4 \, a c^{3} d - 2 \, a b c^{2} e + {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} f\right )} h - {\left (2 \, {\left (2 \, c^{4} d - b c^{3} e + {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} f\right )} g - {\left (2 \, b c^{3} d - 2 \, {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} e + {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} f\right )} h\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 (2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} f g + {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f g + {\left (2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e - 3 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} f\right )} h\right )} x^{2} + {\left (2 \, {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e - 3 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} f\right )} h + {\left (2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} f g + {\left (2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e - 3 \, {\left (b^{4} - 4 \, a b^{2} c\right )} f\right )} h\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 ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f h x^{2} - 2 \, {\left (b c^{3} d - 2 \, a c^{3} e + a b c^{2} f\right )} g + {\left (4 \, a c^{3} d - 2 \, a b c^{2} e + {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} f\right )} h - {\left (2 \, {\left (2 \, c^{4} d - b c^{3} e + {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} f\right )} g - {\left (2 \, b c^{3} d - 2 \, {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} e + {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} f\right )} h\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 ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*(a*b^2*c - 4*a^2*c^2)*f*g + (2*(b^2*c^2 - 4*a*c^3)*f*g + (2*(b^2*c^2 - 4*a*c^3)*e - 3*(b^3*c - 4*a*b
*c^2)*f)*h)*x^2 + (2*(a*b^2*c - 4*a^2*c^2)*e - 3*(a*b^3 - 4*a^2*b*c)*f)*h + (2*(b^3*c - 4*a*b*c^2)*f*g + (2*(b
^3*c - 4*a*b*c^2)*e - 3*(b^4 - 4*a*b^2*c)*f)*h)*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) - 4*((b^2*c^2 - 4*a*c^3)*f*h*x^2 - 2*(b*c^3*d - 2*a*c^3*e + a*b*c^2*f)*g +
(4*a*c^3*d - 2*a*b*c^2*e + (3*a*b^2*c - 8*a^2*c^2)*f)*h - (2*(2*c^4*d - b*c^3*e + (b^2*c^2 - 2*a*c^3)*f)*g - (
2*b*c^3*d - 2*(b^2*c^2 - 2*a*c^3)*e + (3*b^3*c - 10*a*b*c^2)*f)*h)*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*((2*(a*b^2*c - 4*a^2*c^2)*f*g + (2*(b^2*c^2 -
4*a*c^3)*f*g + (2*(b^2*c^2 - 4*a*c^3)*e - 3*(b^3*c - 4*a*b*c^2)*f)*h)*x^2 + (2*(a*b^2*c - 4*a^2*c^2)*e - 3*(a
*b^3 - 4*a^2*b*c)*f)*h + (2*(b^3*c - 4*a*b*c^2)*f*g + (2*(b^3*c - 4*a*b*c^2)*e - 3*(b^4 - 4*a*b^2*c)*f)*h)*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*((b^2*c^2 - 4*a*c^
3)*f*h*x^2 - 2*(b*c^3*d - 2*a*c^3*e + a*b*c^2*f)*g + (4*a*c^3*d - 2*a*b*c^2*e + (3*a*b^2*c - 8*a^2*c^2)*f)*h -
(2*(2*c^4*d - b*c^3*e + (b^2*c^2 - 2*a*c^3)*f)*g - (2*b*c^3*d - 2*(b^2*c^2 - 2*a*c^3)*e + (3*b^3*c - 10*a*b*c
^2)*f)*h)*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 = 0.28, size = 271, normalized size = 1.46 \[ \frac {{\left (\frac {{\left (b^{2} c f h - 4 \, a c^{2} f h\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} - \frac {4 \, c^{3} d g + 2 \, b^{2} c f g - 4 \, a c^{2} f g - 2 \, b c^{2} d h - 3 \, b^{3} f h + 10 \, a b c f h - 2 \, b c^{2} g e + 2 \, b^{2} c h e - 4 \, a c^{2} h e}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x - \frac {2 \, b c^{2} d g + 2 \, a b c f g - 4 \, a c^{2} d h - 3 \, a b^{2} f h + 8 \, a^{2} c f h - 4 \, a c^{2} g e + 2 \, a b c h e}{b^{2} c^{2} - 4 \, a c^{3}}}{\sqrt {c x^{2} + b x + a}} - \frac {{\left (2 \, c f g - 3 \, b f h + 2 \, c h e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((b^2*c*f*h - 4*a*c^2*f*h)*x/(b^2*c^2 - 4*a*c^3) - (4*c^3*d*g + 2*b^2*c*f*g - 4*a*c^2*f*g - 2*b*c^2*d*h - 3*b
^3*f*h + 10*a*b*c*f*h - 2*b*c^2*g*e + 2*b^2*c*h*e - 4*a*c^2*h*e)/(b^2*c^2 - 4*a*c^3))*x - (2*b*c^2*d*g + 2*a*b
*c*f*g - 4*a*c^2*d*h - 3*a*b^2*f*h + 8*a^2*c*f*h - 4*a*c^2*g*e + 2*a*b*c*h*e)/(b^2*c^2 - 4*a*c^3))/sqrt(c*x^2
+ b*x + a) - 1/2*(2*c*f*g - 3*b*f*h + 2*c*h*e)*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 = 0.01, size = 735, normalized size = 3.95 \[ \frac {4 a b f h x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {3 b^{3} f h x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {b^{2} e h x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {b^{2} f g x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {2 b d h x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {2 b e g x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {2 a \,b^{2} f h}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 b^{4} f h}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {b^{3} e h}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {b^{3} f g}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {b^{2} d h}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {b^{2} e g}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {f h \,x^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {3 b f h x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {e h x}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {f g x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) d g}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {3 b f h \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}+\frac {e h \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {f g \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {2 a f h}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 b^{2} f h}{4 \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {b e h}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {b f g}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {d h}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {e g}{\sqrt {c \,x^{2}+b x +a}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f*g-1/c/(c*x^2+b*x+a)^(1/2)*d*h-1/c/(c*x^2+b*x+a)^(1/2)*
e*g+1/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e*h-3/2*f*h/c^2*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*
x+2*f*h*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+4*f*h*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/c*b^2/(4*a*c
-b^2)/(c*x^2+b*x+a)^(1/2)*x*e*h+1/c*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*f*g-x/c/(c*x^2+b*x+a)^(1/2)*e*h+2*f*
h*a/c^2/(c*x^2+b*x+a)^(1/2)-3/2*f*h/c^(5/2)*b*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+f*h*x^2/c/(c*x^2+b*x
+a)^(1/2)-3/4*f*h/c^3*b^2/(c*x^2+b*x+a)^(1/2)+2*d*g*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+1/2/c^2*b/(c*x^2
+b*x+a)^(1/2)*f*g-2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*e*g-2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*h+1/2/c^2*
b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*f*g+1/2/c^2*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*e*h+3/2*f*h/c^2*b*x/(c*x^2
+b*x+a)^(1/2)-3/4*f*h/c^3*b^4/(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)*d*h-b^2/c/
(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*e*g-x/c/(c*x^2+b*x+a)^(1/2)*f*g+1/2/c^2*b/(c*x^2+b*x+a)^(1/2)*e*h

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(f*x^2+e*x+d)/(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 = -0.01 \[ \int \frac {\left (g+h\,x\right )\,\left (f\,x^2+e\,x+d\right )}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((g + h*x)*(d + e*x + f*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 \[ \int \frac {\left (g + h x\right ) \left (d + e x + f x^{2}\right )}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(f*x**2+e*x+d)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((g + h*x)*(d + e*x + f*x**2)/(a + b*x + c*x**2)**(3/2), x)

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