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

Optimal. Leaf size=140 \[ \frac {3 (b+2 c x) (2 c g-b h)}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {-2 a h+x (2 c g-b h)+b g}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {6 c (2 c g-b h) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )^{5/2}} \]

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Rubi [A]  time = 0.10, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {998, 638, 614, 618, 206} \begin {gather*} \frac {3 (b+2 c x) (2 c g-b h)}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {-2 a h+x (2 c g-b h)+b g}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {6 c (2 c g-b h) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*g - 2*a*h + (2*c*g - b*h)*x)/(2*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^2) + (3*(2*c*g - b*h)*(b + 2*c*x))/(2*(b
^2 - 4*a*c)^2*d*(a + b*x + c*x^2)) - (6*c*(2*c*g - b*h)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)
^(5/2)*d)

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 614

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

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 638

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

Rule 998

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_)
, x_Symbol] :> Dist[(c/f)^p, Int[(g + h*x)^m*(d + e*x + f*x^2)^(p + q), x], x] /; FreeQ[{a, b, c, d, e, f, g,
h, p, q}, x] && EqQ[c*d - a*f, 0] && EqQ[b*d - a*e, 0] && (IntegerQ[p] || GtQ[c/f, 0]) && ( !IntegerQ[q] || Le
afCount[d + e*x + f*x^2] <= LeafCount[a + b*x + c*x^2])

Rubi steps

\begin {align*} \int \frac {g+h x}{\left (a+b x+c x^2\right )^2 \left (a d+b d x+c d x^2\right )} \, dx &=\frac {\int \frac {g+h x}{\left (a+b x+c x^2\right )^3} \, dx}{d}\\ &=-\frac {b g-2 a h+(2 c g-b h) x}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac {(3 (2 c g-b h)) \int \frac {1}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right ) d}\\ &=-\frac {b g-2 a h+(2 c g-b h) x}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {3 (2 c g-b h) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(3 c (2 c g-b h)) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d}\\ &=-\frac {b g-2 a h+(2 c g-b h) x}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {3 (2 c g-b h) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {(6 c (2 c g-b h)) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right )^2 d}\\ &=-\frac {b g-2 a h+(2 c g-b h) x}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {3 (2 c g-b h) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {6 c (2 c g-b h) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 131, normalized size = 0.94 \begin {gather*} \frac {\frac {\left (b^2-4 a c\right ) (2 a h-b g+b h x-2 c g x)}{(a+x (b+c x))^2}-\frac {12 c (b h-2 c g) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {3 (b+2 c x) (2 c g-b h)}{a+x (b+c x)}}{2 d \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((b^2 - 4*a*c)*(-(b*g) + 2*a*h - 2*c*g*x + b*h*x))/(a + x*(b + c*x))^2 + (3*(2*c*g - b*h)*(b + 2*c*x))/(a + x
*(b + c*x)) - (12*c*(-2*c*g + b*h)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c
)^2*d)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.61, size = 1130, normalized size = 8.07

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*g - (b^3*c^2 - 4*a*b*c^3)*h)*x^3 + 9*(2*(b^3*c^2 - 4*a*b*c^3)*g - (b^4*c - 4*a*
b^2*c^2)*h)*x^2 - 6*(2*a^2*c^2*g - a^2*b*c*h + (2*c^4*g - b*c^3*h)*x^4 + 2*(2*b*c^3*g - b^2*c^2*h)*x^3 + (2*(b
^2*c^2 + 2*a*c^3)*g - (b^3*c + 2*a*b*c^2)*h)*x^2 + 2*(2*a*b*c^2*g - a*b^2*c*h)*x)*sqrt(b^2 - 4*a*c)*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^5 - 14*a*b^3*c + 40*a^2*
b*c^2)*g - (a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*h + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*g - (b^5 + a*b^3*c - 2
0*a^2*b*c^2)*h)*x)/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 4
8*a^2*b^3*c^3 - 64*a^3*b*c^4)*d*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*d*x^2
+ 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*d*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 6
4*a^5*c^3)*d), 1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*g - (b^3*c^2 - 4*a*b*c^3)*h)*x^3 + 9*(2*(b^3*c^2 - 4*a*b*c^3)*g -
(b^4*c - 4*a*b^2*c^2)*h)*x^2 - 12*(2*a^2*c^2*g - a^2*b*c*h + (2*c^4*g - b*c^3*h)*x^4 + 2*(2*b*c^3*g - b^2*c^2
*h)*x^3 + (2*(b^2*c^2 + 2*a*c^3)*g - (b^3*c + 2*a*b*c^2)*h)*x^2 + 2*(2*a*b*c^2*g - a*b^2*c*h)*x)*sqrt(-b^2 + 4
*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - (b^5 - 14*a*b^3*c + 40*a^2*b*c^2)*g - (a*b^4 + 4
*a^2*b^2*c - 32*a^3*c^2)*h + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*g - (b^5 + a*b^3*c - 20*a^2*b*c^2)*h)*x)/((
b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^
3*b*c^4)*d*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*d*x^2 + 2*(a*b^7 - 12*a^2*
b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*d*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*d)]

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giac [A]  time = 0.17, size = 207, normalized size = 1.48 \begin {gather*} \frac {6 \, {\left (2 \, c^{2} g - b c h\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {12 \, c^{3} g x^{3} - 6 \, b c^{2} h x^{3} + 18 \, b c^{2} g x^{2} - 9 \, b^{2} c h x^{2} + 4 \, b^{2} c g x + 20 \, a c^{2} g x - 2 \, b^{3} h x - 10 \, a b c h x - b^{3} g + 10 \, a b c g - a b^{2} h - 8 \, a^{2} c h}{2 \, {\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )} {\left (c x^{2} + b x + a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*(2*c^2*g - b*c*h)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*sqrt(-b^2 + 4
*a*c)) + 1/2*(12*c^3*g*x^3 - 6*b*c^2*h*x^3 + 18*b*c^2*g*x^2 - 9*b^2*c*h*x^2 + 4*b^2*c*g*x + 20*a*c^2*g*x - 2*b
^3*h*x - 10*a*b*c*h*x - b^3*g + 10*a*b*c*g - a*b^2*h - 8*a^2*c*h)/((b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)*(c*x^2
+ b*x + a)^2)

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maple [B]  time = 0.01, size = 340, normalized size = 2.43 \begin {gather*} -\frac {3 b c h x}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right ) d}-\frac {6 b c h \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}} d}+\frac {6 c^{2} g x}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right ) d}+\frac {12 c^{2} g \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}} d}-\frac {3 b^{2} h}{2 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right ) d}+\frac {3 b c g}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right ) d}-\frac {b h x}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2} d}+\frac {c g x}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2} d}-\frac {a h}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2} d}+\frac {b g}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/(4*a*c-b^2)/(c*x^2+b*x+a)^2*x*b*h+1/d/(4*a*c-b^2)/(c*x^2+b*x+a)^2*x*c*g-1/d/(4*a*c-b^2)/(c*x^2+b*x+a)^2
*a*h+1/2/d/(4*a*c-b^2)/(c*x^2+b*x+a)^2*b*g-3/d/(4*a*c-b^2)^2/(c*x^2+b*x+a)*c*x*b*h+6/d/(4*a*c-b^2)^2/(c*x^2+b*
x+a)*c^2*x*g-3/2/d/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b^2*h+3/d/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b*c*g-6/d/(4*a*c-b^2)^(5/
2)*c*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*h+12/d/(4*a*c-b^2)^(5/2)*c^2*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*g

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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((h*x+g)/(c*x^2+b*x+a)^2/(c*d*x^2+b*d*x+a*d),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 = 4.02, size = 375, normalized size = 2.68 \begin {gather*} \frac {6\,c\,\mathrm {atan}\left (\frac {d\,\left (\frac {6\,c^2\,x\,\left (b\,h-2\,c\,g\right )}{d\,{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {3\,c\,\left (b\,h-2\,c\,g\right )\,\left (16\,d\,a^2\,b\,c^2-8\,d\,a\,b^3\,c+d\,b^5\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{6\,c^2\,g-3\,b\,c\,h}\right )\,\left (b\,h-2\,c\,g\right )}{d\,{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {\frac {8\,c\,h\,a^2+h\,a\,b^2-10\,c\,g\,a\,b+g\,b^3}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x\,\left (b^2+5\,a\,c\right )\,\left (b\,h-2\,c\,g\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {3\,c^2\,x^3\,\left (b\,h-2\,c\,g\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {9\,b\,c\,x^2\,\left (b\,h-2\,c\,g\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{a^2\,d+x^2\,\left (d\,b^2+2\,a\,c\,d\right )+c^2\,d\,x^4+2\,b\,c\,d\,x^3+2\,a\,b\,d\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*c*atan((d*((6*c^2*x*(b*h - 2*c*g))/(d*(4*a*c - b^2)^(5/2)) + (3*c*(b*h - 2*c*g)*(b^5*d - 8*a*b^3*c*d + 16*a
^2*b*c^2*d))/(d^2*(4*a*c - b^2)^(5/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(6*c^2*
g - 3*b*c*h))*(b*h - 2*c*g))/(d*(4*a*c - b^2)^(5/2)) - ((b^3*g + a*b^2*h + 8*a^2*c*h - 10*a*b*c*g)/(2*(b^4 + 1
6*a^2*c^2 - 8*a*b^2*c)) + (x*(5*a*c + b^2)*(b*h - 2*c*g))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (3*c^2*x^3*(b*h - 2
*c*g))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (9*b*c*x^2*(b*h - 2*c*g))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(a^2*d +
x^2*(b^2*d + 2*a*c*d) + c^2*d*x^4 + 2*b*c*d*x^3 + 2*a*b*d*x)

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sympy [B]  time = 2.26, size = 680, normalized size = 4.86 \begin {gather*} \frac {3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) \log {\left (x + \frac {- 192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{2} c h - 6 b c^{2} g}{6 b c^{2} h - 12 c^{3} g} \right )}}{d} - \frac {3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) \log {\left (x + \frac {192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) - 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b h - 2 c g\right ) + 3 b^{2} c h - 6 b c^{2} g}{6 b c^{2} h - 12 c^{3} g} \right )}}{d} + \frac {- 8 a^{2} c h - a b^{2} h + 10 a b c g - b^{3} g + x^{3} \left (- 6 b c^{2} h + 12 c^{3} g\right ) + x^{2} \left (- 9 b^{2} c h + 18 b c^{2} g\right ) + x \left (- 10 a b c h + 20 a c^{2} g - 2 b^{3} h + 4 b^{2} c g\right )}{32 a^{4} c^{2} d - 16 a^{3} b^{2} c d + 2 a^{2} b^{4} d + x^{4} \left (32 a^{2} c^{4} d - 16 a b^{2} c^{3} d + 2 b^{4} c^{2} d\right ) + x^{3} \left (64 a^{2} b c^{3} d - 32 a b^{3} c^{2} d + 4 b^{5} c d\right ) + x^{2} \left (64 a^{3} c^{3} d - 12 a b^{4} c d + 2 b^{6} d\right ) + x \left (64 a^{3} b c^{2} d - 32 a^{2} b^{3} c d + 4 a b^{5} d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)*log(x + (-192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)
+ 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) - 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5)*(b*h
- 2*c*g) + 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**2*c*h - 6*b*c**2*g)/(6*b*c**2*h - 12*c**3
*g))/d - 3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g)*log(x + (192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*h -
2*c*g) - 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)*
*5)*(b*h - 2*c*g) - 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*h - 2*c*g) + 3*b**2*c*h - 6*b*c**2*g)/(6*b*c**2*h -
12*c**3*g))/d + (-8*a**2*c*h - a*b**2*h + 10*a*b*c*g - b**3*g + x**3*(-6*b*c**2*h + 12*c**3*g) + x**2*(-9*b**
2*c*h + 18*b*c**2*g) + x*(-10*a*b*c*h + 20*a*c**2*g - 2*b**3*h + 4*b**2*c*g))/(32*a**4*c**2*d - 16*a**3*b**2*c
*d + 2*a**2*b**4*d + x**4*(32*a**2*c**4*d - 16*a*b**2*c**3*d + 2*b**4*c**2*d) + x**3*(64*a**2*b*c**3*d - 32*a*
b**3*c**2*d + 4*b**5*c*d) + x**2*(64*a**3*c**3*d - 12*a*b**4*c*d + 2*b**6*d) + x*(64*a**3*b*c**2*d - 32*a**2*b
**3*c*d + 4*a*b**5*d))

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