3.10.100 \(\int \frac {1}{(b d+2 c d x)^2 (a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=140 \[ \frac {60 c^2}{d^2 \left (b^2-4 a c\right )^3 (b+2 c x)}-\frac {60 c^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{7/2}}+\frac {5 c}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \left (a+b x+c x^2\right )}-\frac {1}{2 d^2 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^2} \]

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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 = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {687, 693, 618, 206} \begin {gather*} \frac {60 c^2}{d^2 \left (b^2-4 a c\right )^3 (b+2 c x)}-\frac {60 c^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d^2 \left (b^2-4 a c\right )^{7/2}}+\frac {5 c}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \left (a+b x+c x^2\right )}-\frac {1}{2 d^2 \left (b^2-4 a c\right ) (b+2 c x) \left (a+b x+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^3),x]

[Out]

(60*c^2)/((b^2 - 4*a*c)^3*d^2*(b + 2*c*x)) - 1/(2*(b^2 - 4*a*c)*d^2*(b + 2*c*x)*(a + b*x + c*x^2)^2) + (5*c)/(
(b^2 - 4*a*c)^2*d^2*(b + 2*c*x)*(a + b*x + c*x^2)) - (60*c^2*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/((b^2 - 4
*a*c)^(7/2)*d^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 687

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

Rule 693

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

Rubi steps

\begin {align*} \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^3} \, dx &=-\frac {1}{2 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^2}-\frac {(5 c) \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^2} \, dx}{b^2-4 a c}\\ &=-\frac {1}{2 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^2}+\frac {5 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \left (a+b x+c x^2\right )}+\frac {\left (30 c^2\right ) \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right )^2}\\ &=\frac {60 c^2}{\left (b^2-4 a c\right )^3 d^2 (b+2 c x)}-\frac {1}{2 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^2}+\frac {5 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \left (a+b x+c x^2\right )}+\frac {\left (30 c^2\right ) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3 d^2}\\ &=\frac {60 c^2}{\left (b^2-4 a c\right )^3 d^2 (b+2 c x)}-\frac {1}{2 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^2}+\frac {5 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \left (a+b x+c x^2\right )}-\frac {\left (60 c^2\right ) \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 )^3 d^2}\\ &=\frac {60 c^2}{\left (b^2-4 a c\right )^3 d^2 (b+2 c x)}-\frac {1}{2 \left (b^2-4 a c\right ) d^2 (b+2 c x) \left (a+b x+c x^2\right )^2}+\frac {5 c}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \left (a+b x+c x^2\right )}-\frac {60 c^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2} d^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 119, normalized size = 0.85 \begin {gather*} \frac {\frac {120 c^2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {\left (b^2-4 a c\right ) (b+2 c x)}{(a+x (b+c x))^2}+\frac {14 c (b+2 c x)}{a+x (b+c x)}+\frac {64 c^2}{b+2 c x}}{2 d^2 \left (b^2-4 a c\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^3),x]

[Out]

((64*c^2)/(b + 2*c*x) - ((b^2 - 4*a*c)*(b + 2*c*x))/(a + x*(b + c*x))^2 + (14*c*(b + 2*c*x))/(a + x*(b + c*x))
+ (120*c^2*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c)^3*d^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^3),x]

[Out]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^3), x]

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fricas [B]  time = 0.42, size = 1170, normalized size = 8.36

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

[-1/2*(b^6 - 22*a*b^4*c + 8*a^2*b^2*c^2 + 256*a^3*c^3 - 120*(b^2*c^4 - 4*a*c^5)*x^4 - 240*(b^3*c^3 - 4*a*b*c^4
)*x^3 - 10*(13*b^4*c^2 - 32*a*b^2*c^3 - 80*a^2*c^4)*x^2 + 60*(2*c^5*x^5 + 5*b*c^4*x^4 + a^2*b*c^2 + 4*(b^2*c^3
+ a*c^4)*x^3 + (b^3*c^2 + 6*a*b*c^3)*x^2 + 2*(a*b^2*c^2 + a^2*c^3)*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)) - 10*(b^5*c + 16*a*b^3*c^2 - 80*a^2*b*c^
3)*x)/(2*(b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6 + 256*a^4*c^7)*d^2*x^5 + 5*(b^9*c^2 - 16*a
*b^7*c^3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*d^2*x^4 + 4*(b^10*c - 15*a*b^8*c^2 + 80*a^2*b^6*c
^3 - 160*a^3*b^4*c^4 + 256*a^5*c^6)*d^2*x^3 + (b^11 - 10*a*b^9*c + 320*a^3*b^5*c^3 - 1280*a^4*b^3*c^4 + 1536*a
^5*b*c^5)*d^2*x^2 + 2*(a*b^10 - 15*a^2*b^8*c + 80*a^3*b^6*c^2 - 160*a^4*b^4*c^3 + 256*a^6*c^5)*d^2*x + (a^2*b^
9 - 16*a^3*b^7*c + 96*a^4*b^5*c^2 - 256*a^5*b^3*c^3 + 256*a^6*b*c^4)*d^2), -1/2*(b^6 - 22*a*b^4*c + 8*a^2*b^2*
c^2 + 256*a^3*c^3 - 120*(b^2*c^4 - 4*a*c^5)*x^4 - 240*(b^3*c^3 - 4*a*b*c^4)*x^3 - 10*(13*b^4*c^2 - 32*a*b^2*c^
3 - 80*a^2*c^4)*x^2 + 120*(2*c^5*x^5 + 5*b*c^4*x^4 + a^2*b*c^2 + 4*(b^2*c^3 + a*c^4)*x^3 + (b^3*c^2 + 6*a*b*c^
3)*x^2 + 2*(a*b^2*c^2 + a^2*c^3)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) -
10*(b^5*c + 16*a*b^3*c^2 - 80*a^2*b*c^3)*x)/(2*(b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6 + 2
56*a^4*c^7)*d^2*x^5 + 5*(b^9*c^2 - 16*a*b^7*c^3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*d^2*x^4 +
4*(b^10*c - 15*a*b^8*c^2 + 80*a^2*b^6*c^3 - 160*a^3*b^4*c^4 + 256*a^5*c^6)*d^2*x^3 + (b^11 - 10*a*b^9*c + 320*
a^3*b^5*c^3 - 1280*a^4*b^3*c^4 + 1536*a^5*b*c^5)*d^2*x^2 + 2*(a*b^10 - 15*a^2*b^8*c + 80*a^3*b^6*c^2 - 160*a^4
*b^4*c^3 + 256*a^6*c^5)*d^2*x + (a^2*b^9 - 16*a^3*b^7*c + 96*a^4*b^5*c^2 - 256*a^5*b^3*c^3 + 256*a^6*b*c^4)*d^
2)]

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giac [B]  time = 0.17, size = 302, normalized size = 2.16 \begin {gather*} \frac {32 \, c^{8} d^{11}}{{\left (b^{6} c^{6} d^{12} - 12 \, a b^{4} c^{7} d^{12} + 48 \, a^{2} b^{2} c^{8} d^{12} - 64 \, a^{3} c^{9} d^{12}\right )} {\left (2 \, c d x + b d\right )}} - \frac {60 \, c^{2} \arctan \left (-\frac {\frac {b^{2} d}{2 \, c d x + b d} - \frac {4 \, a c d}{2 \, c d x + b d}}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-b^{2} + 4 \, a c} d^{2}} - \frac {4 \, {\left (\frac {9 \, b^{2} c^{2} d}{{\left (2 \, c d x + b d\right )}^{3}} - \frac {36 \, a c^{3} d}{{\left (2 \, c d x + b d\right )}^{3}} - \frac {7 \, c^{2}}{{\left (2 \, c d x + b d\right )} d}\right )}}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} {\left (\frac {b^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {4 \, a c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

32*c^8*d^11/((b^6*c^6*d^12 - 12*a*b^4*c^7*d^12 + 48*a^2*b^2*c^8*d^12 - 64*a^3*c^9*d^12)*(2*c*d*x + b*d)) - 60*
c^2*arctan(-(b^2*d/(2*c*d*x + b*d) - 4*a*c*d/(2*c*d*x + b*d))/sqrt(-b^2 + 4*a*c))/((b^6 - 12*a*b^4*c + 48*a^2*
b^2*c^2 - 64*a^3*c^3)*sqrt(-b^2 + 4*a*c)*d^2) - 4*(9*b^2*c^2*d/(2*c*d*x + b*d)^3 - 36*a*c^3*d/(2*c*d*x + b*d)^
3 - 7*c^2/((2*c*d*x + b*d)*d))/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*(b^2*d^2/(2*c*d*x + b*d)^2 -
4*a*c*d^2/(2*c*d*x + b*d)^2 - 1)^2)

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maple [B]  time = 0.06, size = 273, normalized size = 1.95 \begin {gather*} -\frac {14 c^{3} x^{3}}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{2} d^{2}}-\frac {21 b \,c^{2} x^{2}}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{2} d^{2}}-\frac {18 a \,c^{2} x}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{2} d^{2}}-\frac {6 b^{2} c x}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{2} d^{2}}-\frac {9 a b c}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{2} d^{2}}+\frac {b^{3}}{2 \left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )^{2} d^{2}}-\frac {60 c^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {7}{2}} d^{2}}-\frac {32 c^{2}}{\left (4 a c -b^{2}\right )^{3} \left (2 c x +b \right ) d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^3,x)

[Out]

-32/d^2/(4*a*c-b^2)^3*c^2/(2*c*x+b)-14/d^2/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*c^3*x^3-21/d^2/(4*a*c-b^2)^3/(c*x^2+b
*x+a)^2*b*c^2*x^2-18/d^2/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*a*c^2*x-6/d^2/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*b^2*c*x-9/d
^2/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*a*b*c+1/2/d^2/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*b^3-60/d^2/(4*a*c-b^2)^(7/2)*c^2*
arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))

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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(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^3,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 = 0.89, size = 494, normalized size = 3.53 \begin {gather*} \frac {60\,c^2\,\mathrm {atan}\left (\frac {\frac {30\,c^2\,\left (-64\,a^3\,b\,c^3\,d^2+48\,a^2\,b^3\,c^2\,d^2-12\,a\,b^5\,c\,d^2+b^7\,d^2\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {60\,c^3\,x\,\left (-64\,a^3\,c^3\,d^2+48\,a^2\,b^2\,c^2\,d^2-12\,a\,b^4\,c\,d^2+b^6\,d^2\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{7/2}}}{30\,c^2}\right )}{d^2\,{\left (4\,a\,c-b^2\right )}^{7/2}}-\frac {\frac {64\,a^2\,c^2+18\,a\,b^2\,c-b^4}{2\,\left (4\,a\,c-b^2\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {60\,c^4\,x^4}{\left (4\,a\,c-b^2\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {5\,c\,x\,\left (b^3+20\,a\,c\,b\right )}{\left (4\,a\,c-b^2\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {5\,c\,x^2\,\left (13\,b^2\,c+20\,a\,c^2\right )}{\left (4\,a\,c-b^2\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {120\,b\,c^3\,x^3}{\left (4\,a\,c-b^2\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^3\,\left (4\,b^2\,c\,d^2+4\,a\,c^2\,d^2\right )+x\,\left (2\,c\,a^2\,d^2+2\,a\,b^2\,d^2\right )+x^2\,\left (b^3\,d^2+6\,a\,c\,b\,d^2\right )+a^2\,b\,d^2+2\,c^3\,d^2\,x^5+5\,b\,c^2\,d^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^3),x)

[Out]

(60*c^2*atan(((30*c^2*(b^7*d^2 - 64*a^3*b*c^3*d^2 + 48*a^2*b^3*c^2*d^2 - 12*a*b^5*c*d^2))/(d^2*(4*a*c - b^2)^(
7/2)) + (60*c^3*x*(b^6*d^2 - 64*a^3*c^3*d^2 + 48*a^2*b^2*c^2*d^2 - 12*a*b^4*c*d^2))/(d^2*(4*a*c - b^2)^(7/2)))
/(30*c^2)))/(d^2*(4*a*c - b^2)^(7/2)) - ((64*a^2*c^2 - b^4 + 18*a*b^2*c)/(2*(4*a*c - b^2)*(b^4 + 16*a^2*c^2 -
8*a*b^2*c)) + (60*c^4*x^4)/((4*a*c - b^2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (5*c*x*(b^3 + 20*a*b*c))/((4*a*c -
b^2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (5*c*x^2*(20*a*c^2 + 13*b^2*c))/((4*a*c - b^2)*(b^4 + 16*a^2*c^2 - 8*a
*b^2*c)) + (120*b*c^3*x^3)/((4*a*c - b^2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^3*(4*a*c^2*d^2 + 4*b^2*c*d^2) +
x*(2*a*b^2*d^2 + 2*a^2*c*d^2) + x^2*(b^3*d^2 + 6*a*b*c*d^2) + a^2*b*d^2 + 2*c^3*d^2*x^5 + 5*b*c^2*d^2*x^4)

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sympy [B]  time = 3.74, size = 804, normalized size = 5.74 \begin {gather*} \frac {30 c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \log {\left (x + \frac {- 7680 a^{4} c^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 7680 a^{3} b^{2} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 2880 a^{2} b^{4} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 480 a b^{6} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 30 b^{8} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 30 b c^{2}}{60 c^{3}} \right )}}{d^{2}} - \frac {30 c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \log {\left (x + \frac {7680 a^{4} c^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 7680 a^{3} b^{2} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 2880 a^{2} b^{4} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 480 a b^{6} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 30 b^{8} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 30 b c^{2}}{60 c^{3}} \right )}}{d^{2}} + \frac {- 64 a^{2} c^{2} - 18 a b^{2} c + b^{4} - 240 b c^{3} x^{3} - 120 c^{4} x^{4} + x^{2} \left (- 200 a c^{3} - 130 b^{2} c^{2}\right ) + x \left (- 200 a b c^{2} - 10 b^{3} c\right )}{128 a^{5} b c^{3} d^{2} - 96 a^{4} b^{3} c^{2} d^{2} + 24 a^{3} b^{5} c d^{2} - 2 a^{2} b^{7} d^{2} + x^{5} \left (256 a^{3} c^{6} d^{2} - 192 a^{2} b^{2} c^{5} d^{2} + 48 a b^{4} c^{4} d^{2} - 4 b^{6} c^{3} d^{2}\right ) + x^{4} \left (640 a^{3} b c^{5} d^{2} - 480 a^{2} b^{3} c^{4} d^{2} + 120 a b^{5} c^{3} d^{2} - 10 b^{7} c^{2} d^{2}\right ) + x^{3} \left (512 a^{4} c^{5} d^{2} + 128 a^{3} b^{2} c^{4} d^{2} - 288 a^{2} b^{4} c^{3} d^{2} + 88 a b^{6} c^{2} d^{2} - 8 b^{8} c d^{2}\right ) + x^{2} \left (768 a^{4} b c^{4} d^{2} - 448 a^{3} b^{3} c^{3} d^{2} + 48 a^{2} b^{5} c^{2} d^{2} + 12 a b^{7} c d^{2} - 2 b^{9} d^{2}\right ) + x \left (256 a^{5} c^{4} d^{2} + 64 a^{4} b^{2} c^{3} d^{2} - 144 a^{3} b^{4} c^{2} d^{2} + 44 a^{2} b^{6} c d^{2} - 4 a b^{8} d^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**2/(c*x**2+b*x+a)**3,x)

[Out]

30*c**2*sqrt(-1/(4*a*c - b**2)**7)*log(x + (-7680*a**4*c**6*sqrt(-1/(4*a*c - b**2)**7) + 7680*a**3*b**2*c**5*s
qrt(-1/(4*a*c - b**2)**7) - 2880*a**2*b**4*c**4*sqrt(-1/(4*a*c - b**2)**7) + 480*a*b**6*c**3*sqrt(-1/(4*a*c -
b**2)**7) - 30*b**8*c**2*sqrt(-1/(4*a*c - b**2)**7) + 30*b*c**2)/(60*c**3))/d**2 - 30*c**2*sqrt(-1/(4*a*c - b*
*2)**7)*log(x + (7680*a**4*c**6*sqrt(-1/(4*a*c - b**2)**7) - 7680*a**3*b**2*c**5*sqrt(-1/(4*a*c - b**2)**7) +
2880*a**2*b**4*c**4*sqrt(-1/(4*a*c - b**2)**7) - 480*a*b**6*c**3*sqrt(-1/(4*a*c - b**2)**7) + 30*b**8*c**2*sqr
t(-1/(4*a*c - b**2)**7) + 30*b*c**2)/(60*c**3))/d**2 + (-64*a**2*c**2 - 18*a*b**2*c + b**4 - 240*b*c**3*x**3 -
120*c**4*x**4 + x**2*(-200*a*c**3 - 130*b**2*c**2) + x*(-200*a*b*c**2 - 10*b**3*c))/(128*a**5*b*c**3*d**2 - 9
6*a**4*b**3*c**2*d**2 + 24*a**3*b**5*c*d**2 - 2*a**2*b**7*d**2 + x**5*(256*a**3*c**6*d**2 - 192*a**2*b**2*c**5
*d**2 + 48*a*b**4*c**4*d**2 - 4*b**6*c**3*d**2) + x**4*(640*a**3*b*c**5*d**2 - 480*a**2*b**3*c**4*d**2 + 120*a
*b**5*c**3*d**2 - 10*b**7*c**2*d**2) + x**3*(512*a**4*c**5*d**2 + 128*a**3*b**2*c**4*d**2 - 288*a**2*b**4*c**3
*d**2 + 88*a*b**6*c**2*d**2 - 8*b**8*c*d**2) + x**2*(768*a**4*b*c**4*d**2 - 448*a**3*b**3*c**3*d**2 + 48*a**2*
b**5*c**2*d**2 + 12*a*b**7*c*d**2 - 2*b**9*d**2) + x*(256*a**5*c**4*d**2 + 64*a**4*b**2*c**3*d**2 - 144*a**3*b
**4*c**2*d**2 + 44*a**2*b**6*c*d**2 - 4*a*b**8*d**2))

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