\(\int \frac {1}{(c+d x)^2 (a+b (c+d x)^3)^2} \, dx\) [2874]

Optimal result

Integrand size = 21, antiderivative size = 189 \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}+\frac {4 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} d}+\frac {4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d}-\frac {2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d} \]

[Out]

-4/3/a^2/d/(d*x+c)+1/3/a/d/(d*x+c)/(a+b*(d*x+c)^3)+4/9*b^(1/3)*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(7/3)/d-2/9*b^(1/
3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(7/3)/d+4/9*b^(1/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*(
d*x+c))/a^(1/3)*3^(1/2))/a^(7/3)/d*3^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {379, 296, 331, 298, 31, 648, 631, 210, 642} \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {4 \sqrt [3]{b} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{7/3} d}-\frac {2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d}+\frac {4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d}-\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )} \]

[In]

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

[Out]

-4/(3*a^2*d*(c + d*x)) + 1/(3*a*d*(c + d*x)*(a + b*(c + d*x)^3)) + (4*b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c +
d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)*d) + (4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9*a^(7/3)*d)
- (2*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 379

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

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 648

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d} \\ & = \frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}+\frac {4 \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{3 a d} \\ & = -\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}-\frac {(4 b) \text {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{3 a^2 d} \\ & = -\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}+\frac {\left (4 b^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 a^{7/3} d}-\frac {\left (4 b^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{9 a^{7/3} d} \\ & = -\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}+\frac {4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d}-\frac {\left (2 \sqrt [3]{b}\right ) \text {Subst}\left (\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{9 a^{7/3} d}-\frac {\left (2 b^{2/3}\right ) \text {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{3 a^2 d} \\ & = -\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}+\frac {4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d}-\frac {2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d}-\frac {\left (4 \sqrt [3]{b}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{3 a^{7/3} d} \\ & = -\frac {4}{3 a^2 d (c+d x)}+\frac {1}{3 a d (c+d x) \left (a+b (c+d x)^3\right )}+\frac {4 \sqrt [3]{b} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{7/3} d}+\frac {4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d}-\frac {2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {-\frac {9 \sqrt [3]{a}}{c+d x}-\frac {3 \sqrt [3]{a} b (c+d x)^2}{a+b (c+d x)^3}-4 \sqrt {3} \sqrt [3]{b} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )+4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d} \]

[In]

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

[Out]

((-9*a^(1/3))/(c + d*x) - (3*a^(1/3)*b*(c + d*x)^2)/(a + b*(c + d*x)^3) - 4*Sqrt[3]*b^(1/3)*ArcTan[(-a^(1/3) +
2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] + 4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] - 2*b^(1/3)*Log[a^(2/3)
- a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 4.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.81

method	result	size
default	\(-\frac {1}{a^{2} d \left (d x +c \right )}-\frac {b \left (\frac {\frac {d \,x^{2}}{3}+\frac {2 c x}{3}+\frac {c^{2}}{3 d}}{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +c^{3} b +a}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{9 b d}\right )}{a^{2}}\)	\(154\)
risch	\(\frac {-\frac {4 b \,d^{2} x^{3}}{3 a^{2}}-\frac {4 b c d \,x^{2}}{a^{2}}-\frac {4 b x \,c^{2}}{a^{2}}-\frac {4 c^{3} b +3 a}{3 a^{2} d}}{\left (d x +c \right ) \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +c^{3} b +a \right )}+\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} d^{3} \textit {\_Z}^{3}-b \right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{7} d^{4} \textit {\_R}^{3}+3 b d \right ) x -4 a^{7} c \,d^{3} \textit {\_R}^{3}-a^{5} d^{2} \textit {\_R}^{2}+3 b c \right )\right )}{9}\)	\(166\)

[In]

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

[Out]

-1/a^2/d/(d*x+c)-b/a^2*((1/3*d*x^2+2/3*c*x+1/3*c^2/d)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)+4/9/b/d*su
m((_R*d+c)/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_R),_R=RootOf(_Z^3*b*d^3+3*_Z^2*b*c*d^2+3*_Z*b*c^2*d+b*c^3+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (148) = 296\).

Time = 0.25 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.03 \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=-\frac {12 \, b d^{3} x^{3} + 36 \, b c d^{2} x^{2} + 36 \, b c^{2} d x + 12 \, b c^{3} + 4 \, \sqrt {3} {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} + {\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (d x + c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + 2 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} + {\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} - {\left (a d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 4 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} + {\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b d x + b c + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 9 \, a}{9 \, {\left (a^{2} b d^{5} x^{4} + 4 \, a^{2} b c d^{4} x^{3} + 6 \, a^{2} b c^{2} d^{3} x^{2} + {\left (4 \, a^{2} b c^{3} + a^{3}\right )} d^{2} x + {\left (a^{2} b c^{4} + a^{3} c\right )} d\right )}} \]

[In]

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

[Out]

-1/9*(12*b*d^3*x^3 + 36*b*c*d^2*x^2 + 36*b*c^2*d*x + 12*b*c^3 + 4*sqrt(3)*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2
*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*arctan(2/3*sqrt(3)*(d*x + c)*(b/a)^(1/3) - 1/3*sqrt(3)
) + 2*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*log(b*d^2*x^
2 + 2*b*c*d*x + b*c^2 - (a*d*x + a*c)*(b/a)^(2/3) + a*(b/a)^(1/3)) - 4*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^
2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*log(b*d*x + b*c + a*(b/a)^(2/3)) + 9*a)/(a^2*b*d^5*x^4 +
4*a^2*b*c*d^4*x^3 + 6*a^2*b*c^2*d^3*x^2 + (4*a^2*b*c^3 + a^3)*d^2*x + (a^2*b*c^4 + a^3*c)*d)

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {- 3 a - 4 b c^{3} - 12 b c^{2} d x - 12 b c d^{2} x^{2} - 4 b d^{3} x^{3}}{3 a^{3} c d + 3 a^{2} b c^{4} d + 18 a^{2} b c^{2} d^{3} x^{2} + 12 a^{2} b c d^{4} x^{3} + 3 a^{2} b d^{5} x^{4} + x \left (3 a^{3} d^{2} + 12 a^{2} b c^{3} d^{2}\right )} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{7} - 64 b, \left ( t \mapsto t \log {\left (x + \frac {81 t^{2} a^{5} + 16 b c}{16 b d} \right )} \right )\right )}}{d} \]

[In]

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

[Out]

(-3*a - 4*b*c**3 - 12*b*c**2*d*x - 12*b*c*d**2*x**2 - 4*b*d**3*x**3)/(3*a**3*c*d + 3*a**2*b*c**4*d + 18*a**2*b
*c**2*d**3*x**2 + 12*a**2*b*c*d**4*x**3 + 3*a**2*b*d**5*x**4 + x*(3*a**3*d**2 + 12*a**2*b*c**3*d**2)) + RootSu
m(729*_t**3*a**7 - 64*b, Lambda(_t, _t*log(x + (81*_t**2*a**5 + 16*b*c)/(16*b*d))))/d

Maxima [F]

\[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\int { \frac {1}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{2} {\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

-1/3*(4*b*d^3*x^3 + 12*b*c*d^2*x^2 + 12*b*c^2*d*x + 4*b*c^3 + 3*a)/(a^2*b*d^5*x^4 + 4*a^2*b*c*d^4*x^3 + 6*a^2*
b*c^2*d^3*x^2 + (4*a^2*b*c^3 + a^3)*d^2*x + (a^2*b*c^4 + a^3*c)*d) - 4/3*b*integrate((d*x + c)/(b*d^3*x^3 + 3*
b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a), x)/a^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {4 \, \left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | -\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} - \frac {1}{{\left (d x + c\right )} d} \right |}\right )}{9 \, a^{2}} - \frac {4 \, \sqrt {3} \left (a^{2} b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} - \frac {2}{{\left (d x + c\right )} d}\right )}}{3 \, \left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}}}\right )}{9 \, a^{3} d} - \frac {2 \, \left (a^{2} b\right )^{\frac {1}{3}} \log \left (\left (\frac {b}{a d^{3}}\right )^{\frac {2}{3}} - \frac {\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}}}{{\left (d x + c\right )} d} + \frac {1}{{\left (d x + c\right )}^{2} d^{2}}\right )}{9 \, a^{3} d} - \frac {1}{{\left (d x + c\right )} a^{2} d} - \frac {b}{3 \, {\left (d x + c\right )} a^{2} {\left (b + \frac {a}{{\left (d x + c\right )}^{3}}\right )} d} \]

[In]

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

[Out]

4/9*(b/(a*d^3))^(1/3)*log(abs(-(b/(a*d^3))^(1/3) - 1/((d*x + c)*d)))/a^2 - 4/9*sqrt(3)*(a^2*b)^(1/3)*arctan(1/
3*sqrt(3)*((b/(a*d^3))^(1/3) - 2/((d*x + c)*d))/(b/(a*d^3))^(1/3))/(a^3*d) - 2/9*(a^2*b)^(1/3)*log((b/(a*d^3))
^(2/3) - (b/(a*d^3))^(1/3)/((d*x + c)*d) + 1/((d*x + c)^2*d^2))/(a^3*d) - 1/((d*x + c)*a^2*d) - 1/3*b/((d*x +
c)*a^2*(b + a/(d*x + c)^3)*d)

Mupad [B] (verification not implemented)

Time = 6.13 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^2} \, dx=\frac {4\,b^{1/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{7/3}\,d}-\frac {\frac {4\,b\,c^3+3\,a}{3\,a^2\,d}+\frac {4\,b\,d^2\,x^3}{3\,a^2}+\frac {4\,b\,c^2\,x}{a^2}+\frac {4\,b\,c\,d\,x^2}{a^2}}{a\,c+x\,\left (4\,b\,d\,c^3+a\,d\right )+b\,c^4+b\,d^4\,x^4+6\,b\,c^2\,d^2\,x^2+4\,b\,c\,d^3\,x^3}-\frac {4\,b^{1/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{7/3}\,d}+\frac {b^{1/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {2}{9}+\frac {\sqrt {3}\,2{}\mathrm {i}}{9}\right )}{a^{7/3}\,d} \]

[In]

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

[Out]

(4*b^(1/3)*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(9*a^(7/3)*d) - ((3*a + 4*b*c^3)/(3*a^2*d) + (4*b*d^2*x^3)/
(3*a^2) + (4*b*c^2*x)/a^2 + (4*b*c*d*x^2)/a^2)/(a*c + x*(a*d + 4*b*c^3*d) + b*c^4 + b*d^4*x^4 + 6*b*c^2*d^2*x^
2 + 4*b*c*d^3*x^3) - (4*b^(1/3)*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*c - a^(1/3) + 2*b^(1/3)*d*x)*((3^(1/2)*1i)/
2 + 1/2))/(9*a^(7/3)*d) + (b^(1/3)*log(2*b^(1/3)*c - 3^(1/2)*a^(1/3)*1i - a^(1/3) + 2*b^(1/3)*d*x)*((3^(1/2)*2
i)/9 - 2/9))/(a^(7/3)*d)