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

Optimal. Leaf size=196 \[ -\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-c (2 a d h-2 a e g+b d g+b e f)+b h (b d-a e)+2 c^2 d f\right )}{c \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac {\log \left (a+b x+c x^2\right ) (-a e h+b d h-c d g+c e f)}{2 c \left (a e^2-b d e+c d^2\right )}+\frac {\log (d+e x) \left (d^2 h-d e g+e^2 f\right )}{e \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1628, 634, 618, 206, 628} \[ -\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-c (2 a d h-2 a e g+b d g+b e f)+b h (b d-a e)+2 c^2 d f\right )}{c \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}-\frac {\log \left (a+b x+c x^2\right ) (-a e h+b d h-c d g+c e f)}{2 c \left (a e^2-b d e+c d^2\right )}+\frac {\log (d+e x) \left (d^2 h-d e g+e^2 f\right )}{e \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*c^2*d*f + b*(b*d - a*e)*h - c*(b*e*f + b*d*g - 2*a*e*g + 2*a*d*h))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]
])/(c*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2))) + ((e^2*f - d*e*g + d^2*h)*Log[d + e*x])/(e*(c*d^2 - b*d*e +
a*e^2)) - ((c*e*f - c*d*g + b*d*h - a*e*h)*Log[a + b*x + c*x^2])/(2*c*(c*d^2 - b*d*e + a*e^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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {f+g x+h x^2}{(d+e x) \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac {e^2 f-d e g+d^2 h}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {c d f-b e f+a e g-a d h-(c e f-c d g+b d h-a e h) x}{\left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {\left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}+\frac {\int \frac {c d f-b e f+a e g-a d h-(c e f-c d g+b d h-a e h) x}{a+b x+c x^2} \, dx}{c d^2-b d e+a e^2}\\ &=\frac {\left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}-\frac {(c e f-c d g+b d h-a e h) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c \left (c d^2-b d e+a e^2\right )}+\frac {\left (2 c^2 d f+b (b d-a e) h-c (b e f+b d g-2 a e g+2 a d h)\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c \left (c d^2-b d e+a e^2\right )}\\ &=\frac {\left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}-\frac {(c e f-c d g+b d h-a e h) \log \left (a+b x+c x^2\right )}{2 c \left (c d^2-b d e+a e^2\right )}-\frac {\left (2 c^2 d f+b (b d-a e) h-c (b e f+b d g-2 a e g+2 a d h)\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {\left (2 c^2 d f+b (b d-a e) h-c (b e f+b d g-2 a e g+2 a d h)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {\left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}-\frac {(c e f-c d g+b d h-a e h) \log \left (a+b x+c x^2\right )}{2 c \left (c d^2-b d e+a e^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 193, normalized size = 0.98 \[ \frac {-2 e \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right ) \left (c (2 a d h-2 a e g+b d g+b e f)+b h (a e-b d)-2 c^2 d f\right )+2 c \sqrt {4 a c-b^2} \log (d+e x) \left (d^2 h-d e g+e^2 f\right )-e \sqrt {4 a c-b^2} \log (a+x (b+c x)) (-a e h+b d h-c d g+c e f)}{2 c e \sqrt {4 a c-b^2} \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 83.61, size = 625, normalized size = 3.19 \[ \left [-\frac {\sqrt {b^{2} - 4 \, a c} {\left ({\left (2 \, c^{2} d e - b c e^{2}\right )} f - {\left (b c d e - 2 \, a c e^{2}\right )} g - {\left (a b e^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} h\right )} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} f - {\left (b^{2} c - 4 \, a c^{2}\right )} d e g + {\left ({\left (b^{3} - 4 \, a b c\right )} d e - {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )} h\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} f - {\left (b^{2} c - 4 \, a c^{2}\right )} d e g + {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} h\right )} \log \left (e x + d\right )}{2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}, -\frac {2 \, \sqrt {-b^{2} + 4 \, a c} {\left ({\left (2 \, c^{2} d e - b c e^{2}\right )} f - {\left (b c d e - 2 \, a c e^{2}\right )} g - {\left (a b e^{2} - {\left (b^{2} - 2 \, a c\right )} d e\right )} h\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} f - {\left (b^{2} c - 4 \, a c^{2}\right )} d e g + {\left ({\left (b^{3} - 4 \, a b c\right )} d e - {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\right )} h\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} f - {\left (b^{2} c - 4 \, a c^{2}\right )} d e g + {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2} h\right )} \log \left (e x + d\right )}{2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(b^2 - 4*a*c)*((2*c^2*d*e - b*c*e^2)*f - (b*c*d*e - 2*a*c*e^2)*g - (a*b*e^2 - (b^2 - 2*a*c)*d*e)*h)
*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + ((b^2*c - 4*a*c^
2)*e^2*f - (b^2*c - 4*a*c^2)*d*e*g + ((b^3 - 4*a*b*c)*d*e - (a*b^2 - 4*a^2*c)*e^2)*h)*log(c*x^2 + b*x + a) - 2
*((b^2*c - 4*a*c^2)*e^2*f - (b^2*c - 4*a*c^2)*d*e*g + (b^2*c - 4*a*c^2)*d^2*h)*log(e*x + d))/((b^2*c^2 - 4*a*c
^3)*d^2*e - (b^3*c - 4*a*b*c^2)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3), -1/2*(2*sqrt(-b^2 + 4*a*c)*((2*c^2*d*e - b
*c*e^2)*f - (b*c*d*e - 2*a*c*e^2)*g - (a*b*e^2 - (b^2 - 2*a*c)*d*e)*h)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/
(b^2 - 4*a*c)) + ((b^2*c - 4*a*c^2)*e^2*f - (b^2*c - 4*a*c^2)*d*e*g + ((b^3 - 4*a*b*c)*d*e - (a*b^2 - 4*a^2*c)
*e^2)*h)*log(c*x^2 + b*x + a) - 2*((b^2*c - 4*a*c^2)*e^2*f - (b^2*c - 4*a*c^2)*d*e*g + (b^2*c - 4*a*c^2)*d^2*h
)*log(e*x + d))/((b^2*c^2 - 4*a*c^3)*d^2*e - (b^3*c - 4*a*b*c^2)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3)]

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giac [A]  time = 0.16, size = 204, normalized size = 1.04 \[ \frac {{\left (c d g - b d h - c f e + a h e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )}} + \frac {{\left (d^{2} h - d g e + f e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} + \frac {{\left (2 \, c^{2} d f - b c d g + b^{2} d h - 2 \, a c d h - b c f e + 2 \, a c g e - a b h e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*d*g - b*d*h - c*f*e + a*h*e)*log(c*x^2 + b*x + a)/(c^2*d^2 - b*c*d*e + a*c*e^2) + (d^2*h - d*g*e + f*e^
2)*log(abs(x*e + d))/(c*d^2*e - b*d*e^2 + a*e^3) + (2*c^2*d*f - b*c*d*g + b^2*d*h - 2*a*c*d*h - b*c*f*e + 2*a*
c*g*e - a*b*h*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^2*d^2 - b*c*d*e + a*c*e^2)*sqrt(-b^2 + 4*a*c))

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maple [B]  time = 0.01, size = 622, normalized size = 3.17 \[ -\frac {a b e h \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}\, c}-\frac {2 a d h \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {2 a e g \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {b^{2} d h \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}\, c}-\frac {b d g \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}-\frac {b e f \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {2 c d f \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {a e h \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) c}-\frac {b d h \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) c}+\frac {d^{2} h \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) e}-\frac {d g \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}+\frac {d g \ln \left (c \,x^{2}+b x +a \right )}{2 a \,e^{2}-2 b d e +2 c \,d^{2}}+\frac {e f \ln \left (e x +d \right )}{a \,e^{2}-b d e +c \,d^{2}}-\frac {e f \ln \left (c \,x^{2}+b x +a \right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2+g*x+f)/(e*x+d)/(c*x^2+b*x+a),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 positive or negative?

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mupad [B]  time = 10.45, size = 2467, normalized size = 12.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a^2*b*e^4*g - 2*a*b^2*e^4*f - 2*a^3*e^4*h + 6*a^2*c*e^4*f - 4*a*c^2*d^4*h + b^2*c*d^4*h + b^3*d^3*e*h - 2
*b^3*e^4*f*x + a^2*e^4*g*(b^2 - 4*a*c)^(1/2) + a*b^2*d*e^3*g + 6*a*c^2*d^3*e*g + b*c^2*d^3*e*f + 3*a^2*b*d*e^3
*h - 10*a^2*c*d*e^3*g - 2*b^2*c*d^3*e*g + a*b^2*e^4*g*x - a^2*b*e^4*h*x - 2*a^2*c*e^4*g*x + b^3*d*e^3*g*x + 2*
c^3*d^3*e*f*x - 3*a^2*d*e^3*h*(b^2 - 4*a*c)^(1/2) - c^2*d^3*e*f*(b^2 - 4*a*c)^(1/2) - b^2*d^3*e*h*(b^2 - 4*a*c
)^(1/2) - 2*b^2*e^4*f*x*(b^2 - 4*a*c)^(1/2) - a^2*e^4*h*x*(b^2 - 4*a*c)^(1/2) - 2*c^2*d^4*h*x*(b^2 - 4*a*c)^(1
/2) - 10*a*c^2*d^2*e^2*f - 4*a*b^2*d^2*e^2*h + b^2*c*d^2*e^2*f + 10*a^2*c*d^2*e^2*h - b^3*d^2*e^2*h*x - 2*a*b*
e^4*f*(b^2 - 4*a*c)^(1/2) - b*c*d^4*h*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*d*e^3*f - 3*a*b*c*d^3*e*h + 7*a*b*c*e^4*f*
x - 5*c^2*d^2*e^2*f*x*(b^2 - 4*a*c)^(1/2) - b^2*d^2*e^2*h*x*(b^2 - 4*a*c)^(1/2) + a*b*d*e^3*g*(b^2 - 4*a*c)^(1
/2) + 7*a*c*d*e^3*f*(b^2 - 4*a*c)^(1/2) + 5*a*c*d^3*e*h*(b^2 - 4*a*c)^(1/2) + 2*b*c*d^3*e*g*(b^2 - 4*a*c)^(1/2
) + a*b*e^4*g*x*(b^2 - 4*a*c)^(1/2) + 3*a*c*e^4*f*x*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*d^2*e^2*g - 14*a*c^2*d*e^3*f
*x + 5*b^2*c*d*e^3*f*x - 10*a*c^2*d^3*e*h*x - b*c^2*d^3*e*g*x + 6*a^2*c*d*e^3*h*x + 3*b^2*c*d^3*e*h*x + 2*a*b*
d^2*e^2*h*(b^2 - 4*a*c)^(1/2) - 7*a*c*d^2*e^2*g*(b^2 - 4*a*c)^(1/2) - b*c*d^2*e^2*f*(b^2 - 4*a*c)^(1/2) + b^2*
d*e^3*g*x*(b^2 - 4*a*c)^(1/2) + 3*c^2*d^3*e*g*x*(b^2 - 4*a*c)^(1/2) + 14*a*c^2*d^2*e^2*g*x - 3*b*c^2*d^2*e^2*f
*x - 2*b^2*c*d^2*e^2*g*x + 5*a*c*d^2*e^2*h*x*(b^2 - 4*a*c)^(1/2) - 2*b*c*d^2*e^2*g*x*(b^2 - 4*a*c)^(1/2) - 7*a
*b*c*d*e^3*g*x - 5*a*c*d*e^3*g*x*(b^2 - 4*a*c)^(1/2) + 5*b*c*d*e^3*f*x*(b^2 - 4*a*c)^(1/2) + b*c*d^3*e*h*x*(b^
2 - 4*a*c)^(1/2) + a*b*c*d^2*e^2*h*x)*(b^3*d*h + 4*a*c^2*d*g - 4*a*c^2*e*f - a*b^2*e*h - b^2*c*d*g + b^2*c*e*f
+ 4*a^2*c*e*h - 2*c^2*d*f*(b^2 - 4*a*c)^(1/2) - b^2*d*h*(b^2 - 4*a*c)^(1/2) - 4*a*b*c*d*h + a*b*e*h*(b^2 - 4*
a*c)^(1/2) + 2*a*c*d*h*(b^2 - 4*a*c)^(1/2) - 2*a*c*e*g*(b^2 - 4*a*c)^(1/2) + b*c*d*g*(b^2 - 4*a*c)^(1/2) + b*c
*e*f*(b^2 - 4*a*c)^(1/2)))/(2*(4*a*c^3*d^2 + 4*a^2*c^2*e^2 - b^2*c^2*d^2 + b^3*c*d*e - a*b^2*c*e^2 - 4*a*b*c^2
*d*e)) - (log(a^2*b*e^4*g - 2*a*b^2*e^4*f - 2*a^3*e^4*h + 6*a^2*c*e^4*f - 4*a*c^2*d^4*h + b^2*c*d^4*h + b^3*d^
3*e*h - 2*b^3*e^4*f*x - a^2*e^4*g*(b^2 - 4*a*c)^(1/2) + a*b^2*d*e^3*g + 6*a*c^2*d^3*e*g + b*c^2*d^3*e*f + 3*a^
2*b*d*e^3*h - 10*a^2*c*d*e^3*g - 2*b^2*c*d^3*e*g + a*b^2*e^4*g*x - a^2*b*e^4*h*x - 2*a^2*c*e^4*g*x + b^3*d*e^3
*g*x + 2*c^3*d^3*e*f*x + 3*a^2*d*e^3*h*(b^2 - 4*a*c)^(1/2) + c^2*d^3*e*f*(b^2 - 4*a*c)^(1/2) + b^2*d^3*e*h*(b^
2 - 4*a*c)^(1/2) + 2*b^2*e^4*f*x*(b^2 - 4*a*c)^(1/2) + a^2*e^4*h*x*(b^2 - 4*a*c)^(1/2) + 2*c^2*d^4*h*x*(b^2 -
4*a*c)^(1/2) - 10*a*c^2*d^2*e^2*f - 4*a*b^2*d^2*e^2*h + b^2*c*d^2*e^2*f + 10*a^2*c*d^2*e^2*h - b^3*d^2*e^2*h*x
+ 2*a*b*e^4*f*(b^2 - 4*a*c)^(1/2) + b*c*d^4*h*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*d*e^3*f - 3*a*b*c*d^3*e*h + 7*a*b
*c*e^4*f*x + 5*c^2*d^2*e^2*f*x*(b^2 - 4*a*c)^(1/2) + b^2*d^2*e^2*h*x*(b^2 - 4*a*c)^(1/2) - a*b*d*e^3*g*(b^2 -
4*a*c)^(1/2) - 7*a*c*d*e^3*f*(b^2 - 4*a*c)^(1/2) - 5*a*c*d^3*e*h*(b^2 - 4*a*c)^(1/2) - 2*b*c*d^3*e*g*(b^2 - 4*
a*c)^(1/2) - a*b*e^4*g*x*(b^2 - 4*a*c)^(1/2) - 3*a*c*e^4*f*x*(b^2 - 4*a*c)^(1/2) + 3*a*b*c*d^2*e^2*g - 14*a*c^
2*d*e^3*f*x + 5*b^2*c*d*e^3*f*x - 10*a*c^2*d^3*e*h*x - b*c^2*d^3*e*g*x + 6*a^2*c*d*e^3*h*x + 3*b^2*c*d^3*e*h*x
- 2*a*b*d^2*e^2*h*(b^2 - 4*a*c)^(1/2) + 7*a*c*d^2*e^2*g*(b^2 - 4*a*c)^(1/2) + b*c*d^2*e^2*f*(b^2 - 4*a*c)^(1/
2) - b^2*d*e^3*g*x*(b^2 - 4*a*c)^(1/2) - 3*c^2*d^3*e*g*x*(b^2 - 4*a*c)^(1/2) + 14*a*c^2*d^2*e^2*g*x - 3*b*c^2*
d^2*e^2*f*x - 2*b^2*c*d^2*e^2*g*x - 5*a*c*d^2*e^2*h*x*(b^2 - 4*a*c)^(1/2) + 2*b*c*d^2*e^2*g*x*(b^2 - 4*a*c)^(1
/2) - 7*a*b*c*d*e^3*g*x + 5*a*c*d*e^3*g*x*(b^2 - 4*a*c)^(1/2) - 5*b*c*d*e^3*f*x*(b^2 - 4*a*c)^(1/2) - b*c*d^3*
e*h*x*(b^2 - 4*a*c)^(1/2) + a*b*c*d^2*e^2*h*x)*(4*a*c^2*e*f - 4*a*c^2*d*g - b^3*d*h + a*b^2*e*h + b^2*c*d*g -
b^2*c*e*f - 4*a^2*c*e*h - 2*c^2*d*f*(b^2 - 4*a*c)^(1/2) - b^2*d*h*(b^2 - 4*a*c)^(1/2) + 4*a*b*c*d*h + a*b*e*h*
(b^2 - 4*a*c)^(1/2) + 2*a*c*d*h*(b^2 - 4*a*c)^(1/2) - 2*a*c*e*g*(b^2 - 4*a*c)^(1/2) + b*c*d*g*(b^2 - 4*a*c)^(1
/2) + b*c*e*f*(b^2 - 4*a*c)^(1/2)))/(2*(4*a*c^3*d^2 + 4*a^2*c^2*e^2 - b^2*c^2*d^2 + b^3*c*d*e - a*b^2*c*e^2 -
4*a*b*c^2*d*e)) + (log(d + e*x)*(e^2*f + d^2*h - d*e*g))/(a*e^3 - b*d*e^2 + c*d^2*e)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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