3.2883 \(\int \frac{1}{(c+d x)^2 (a+b (c+d x)^3)^3} \, dx\)

Optimal. Leaf size=219 \[ -\frac{7 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{27 a^{10/3} d}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{14 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{10/3} d}+\frac{14 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{10/3} d}-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2} \]

[Out]

-14/(9*a^3*d*(c + d*x)) + 1/(6*a*d*(c + d*x)*(a + b*(c + d*x)^3)^2) + 7/(18*a^2*d*(c + d*x)*(a + b*(c + d*x)^3
)) + (14*b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(10/3)*d) + (14*b^(1/
3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(27*a^(10/3)*d) - (7*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^
(2/3)*(c + d*x)^2])/(27*a^(10/3)*d)

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Rubi [A]  time = 0.187479, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {372, 290, 325, 292, 31, 634, 617, 204, 628} \[ -\frac{7 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{27 a^{10/3} d}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{14 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{10/3} d}+\frac{14 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{10/3} d}-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-14/(9*a^3*d*(c + d*x)) + 1/(6*a*d*(c + d*x)*(a + b*(c + d*x)^3)^2) + 7/(18*a^2*d*(c + d*x)*(a + b*(c + d*x)^3
)) + (14*b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(10/3)*d) + (14*b^(1/
3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(27*a^(10/3)*d) - (7*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^
(2/3)*(c + d*x)^2])/(27*a^(10/3)*d)

Rule 372

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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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 292

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 31

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

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 617

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 204

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

Rubi steps

\begin{align*} \int \frac{1}{(c+d x)^2 \left (a+b (c+d x)^3\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^3\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{6 a d}\\ &=\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{14 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{9 a^2 d}\\ &=-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}-\frac{(14 b) \operatorname{Subst}\left (\int \frac{x}{a+b x^3} \, dx,x,c+d x\right )}{9 a^3 d}\\ &=-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{\left (14 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{27 a^{10/3} d}-\frac{\left (14 b^{2/3}\right ) \operatorname{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 )}{27 a^{10/3} d}\\ &=-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{14 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{10/3} d}-\frac{\left (7 \sqrt [3]{b}\right ) \operatorname{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 )}{27 a^{10/3} d}-\frac{\left (7 b^{2/3}\right ) \operatorname{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 )}{9 a^3 d}\\ &=-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{14 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{10/3} d}-\frac{7 \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 )}{27 a^{10/3} d}-\frac{\left (14 \sqrt [3]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{9 a^{10/3} d}\\ &=-\frac{14}{9 a^3 d (c+d x)}+\frac{1}{6 a d (c+d x) \left (a+b (c+d x)^3\right )^2}+\frac{7}{18 a^2 d (c+d x) \left (a+b (c+d x)^3\right )}+\frac{14 \sqrt [3]{b} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{9 \sqrt{3} a^{10/3} d}+\frac{14 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{10/3} d}-\frac{7 \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 )}{27 a^{10/3} d}\\ \end{align*}

Mathematica [A]  time = 0.135873, size = 196, normalized size = 0.89 \[ \frac{-14 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac{9 a^{4/3} b (c+d x)^2}{\left (a+b (c+d x)^3\right )^2}-\frac{30 \sqrt [3]{a} b (c+d x)^2}{a+b (c+d x)^3}+28 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-28 \sqrt{3} \sqrt [3]{b} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )-\frac{54 \sqrt [3]{a}}{c+d x}}{54 a^{10/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-54*a^(1/3))/(c + d*x) - (9*a^(4/3)*b*(c + d*x)^2)/(a + b*(c + d*x)^3)^2 - (30*a^(1/3)*b*(c + d*x)^2)/(a + b
*(c + d*x)^3) - 28*Sqrt[3]*b^(1/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] + 28*b^(1/3)*Log
[a^(1/3) + b^(1/3)*(c + d*x)] - 14*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(54
*a^(10/3)*d)

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Maple [C]  time = 0.023, size = 524, normalized size = 2.4 \begin{align*} -{\frac{1}{{a}^{3}d \left ( dx+c \right ) }}-{\frac{5\,{b}^{2}{d}^{4}{x}^{5}}{9\,{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{25\,{b}^{2}c{d}^{3}{x}^{4}}{9\,{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{50\,{b}^{2}{c}^{2}{d}^{2}{x}^{3}}{9\,{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{50\,{b}^{2}{x}^{2}{c}^{3}d}{9\,{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{13\,d{x}^{2}b}{18\,{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{25\,{b}^{2}x{c}^{4}}{9\,{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{13\,bcx}{9\,{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{5\,{b}^{2}{c}^{5}}{9\,{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}d}}-{\frac{13\,b{c}^{2}}{18\,{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}d}}-{\frac{14}{27\,{a}^{3}d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}b{d}^{3}+3\,{{\it \_Z}}^{2}bc{d}^{2}+3\,{\it \_Z}\,b{c}^{2}d+b{c}^{3}+a \right ) }{\frac{ \left ({\it \_R}\,d+c \right ) \ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^3/d/(d*x+c)-5/9*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*d^4*x^5-25/9*b^2/a^3/(b*d^3*x^3+3
*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c*d^3*x^4-50/9*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c^2
*d^2*x^3-50/9*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*x^2*c^3*d-13/18*b/a^2/(b*d^3*x^3+3*b*c*d
^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*x^2*d-25/9*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*x*c^4-13/9*b/
a^2/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c*x-5/9*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3
+a)^2*c^5/d-13/18*b/a^2/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c^2/d-14/27/a^3/d*sum((_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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{28 \, b^{2} d^{6} x^{6} + 168 \, b^{2} c d^{5} x^{5} + 420 \, b^{2} c^{2} d^{4} x^{4} + 28 \, b^{2} c^{6} + 7 \,{\left (80 \, b^{2} c^{3} + 7 \, a b\right )} d^{3} x^{3} + 49 \, a b c^{3} + 21 \,{\left (20 \, b^{2} c^{4} + 7 \, a b c\right )} d^{2} x^{2} + 21 \,{\left (8 \, b^{2} c^{5} + 7 \, a b c^{2}\right )} d x + 18 \, a^{2}}{18 \,{\left (a^{3} b^{2} d^{8} x^{7} + 7 \, a^{3} b^{2} c d^{7} x^{6} + 21 \, a^{3} b^{2} c^{2} d^{6} x^{5} +{\left (35 \, a^{3} b^{2} c^{3} + 2 \, a^{4} b\right )} d^{5} x^{4} +{\left (35 \, a^{3} b^{2} c^{4} + 8 \, a^{4} b c\right )} d^{4} x^{3} + 3 \,{\left (7 \, a^{3} b^{2} c^{5} + 4 \, a^{4} b c^{2}\right )} d^{3} x^{2} +{\left (7 \, a^{3} b^{2} c^{6} + 8 \, a^{4} b c^{3} + a^{5}\right )} d^{2} x +{\left (a^{3} b^{2} c^{7} + 2 \, a^{4} b c^{4} + a^{5} c\right )} d\right )}} - \frac{-\frac{7}{3} \,{\left (2 \, \sqrt{3} \left (-\frac{1}{a b^{2} d^{3}}\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac{2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac{2}{3}}}\right ) + \left (-\frac{1}{a b^{2} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac{2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac{4}{3}}\right ) - 2 \, \left (-\frac{1}{a b^{2} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac{2}{3}} \right |}\right )\right )} b}{9 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(28*b^2*d^6*x^6 + 168*b^2*c*d^5*x^5 + 420*b^2*c^2*d^4*x^4 + 28*b^2*c^6 + 7*(80*b^2*c^3 + 7*a*b)*d^3*x^3
+ 49*a*b*c^3 + 21*(20*b^2*c^4 + 7*a*b*c)*d^2*x^2 + 21*(8*b^2*c^5 + 7*a*b*c^2)*d*x + 18*a^2)/(a^3*b^2*d^8*x^7 +
7*a^3*b^2*c*d^7*x^6 + 21*a^3*b^2*c^2*d^6*x^5 + (35*a^3*b^2*c^3 + 2*a^4*b)*d^5*x^4 + (35*a^3*b^2*c^4 + 8*a^4*b
*c)*d^4*x^3 + 3*(7*a^3*b^2*c^5 + 4*a^4*b*c^2)*d^3*x^2 + (7*a^3*b^2*c^6 + 8*a^4*b*c^3 + a^5)*d^2*x + (a^3*b^2*c
^7 + 2*a^4*b*c^4 + a^5*c)*d) - 14/9*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^3

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Fricas [B]  time = 2.19139, size = 1832, normalized size = 8.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*(84*b^2*d^6*x^6 + 504*b^2*c*d^5*x^5 + 1260*b^2*c^2*d^4*x^4 + 84*b^2*c^6 + 21*(80*b^2*c^3 + 7*a*b)*d^3*x^
3 + 147*a*b*c^3 + 63*(20*b^2*c^4 + 7*a*b*c)*d^2*x^2 + 63*(8*b^2*c^5 + 7*a*b*c^2)*d*x + 28*sqrt(3)*(b^2*d^7*x^7
+ 7*b^2*c*d^6*x^6 + 21*b^2*c^2*d^5*x^5 + b^2*c^7 + (35*b^2*c^3 + 2*a*b)*d^4*x^4 + (35*b^2*c^4 + 8*a*b*c)*d^3*
x^3 + 2*a*b*c^4 + 3*(7*b^2*c^5 + 4*a*b*c^2)*d^2*x^2 + a^2*c + (7*b^2*c^6 + 8*a*b*c^3 + a^2)*d*x)*(b/a)^(1/3)*a
rctan(2/3*sqrt(3)*(d*x + c)*(b/a)^(1/3) - 1/3*sqrt(3)) + 14*(b^2*d^7*x^7 + 7*b^2*c*d^6*x^6 + 21*b^2*c^2*d^5*x^
5 + b^2*c^7 + (35*b^2*c^3 + 2*a*b)*d^4*x^4 + (35*b^2*c^4 + 8*a*b*c)*d^3*x^3 + 2*a*b*c^4 + 3*(7*b^2*c^5 + 4*a*b
*c^2)*d^2*x^2 + a^2*c + (7*b^2*c^6 + 8*a*b*c^3 + a^2)*d*x)*(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)) - 28*(b^2*d^7*x^7 + 7*b^2*c*d^6*x^6 + 21*b^2*c^2*d^5*x^5 + b^2*c^7 + (
35*b^2*c^3 + 2*a*b)*d^4*x^4 + (35*b^2*c^4 + 8*a*b*c)*d^3*x^3 + 2*a*b*c^4 + 3*(7*b^2*c^5 + 4*a*b*c^2)*d^2*x^2 +
a^2*c + (7*b^2*c^6 + 8*a*b*c^3 + a^2)*d*x)*(b/a)^(1/3)*log(b*d*x + b*c + a*(b/a)^(2/3)) + 54*a^2)/(a^3*b^2*d^
8*x^7 + 7*a^3*b^2*c*d^7*x^6 + 21*a^3*b^2*c^2*d^6*x^5 + (35*a^3*b^2*c^3 + 2*a^4*b)*d^5*x^4 + (35*a^3*b^2*c^4 +
8*a^4*b*c)*d^4*x^3 + 3*(7*a^3*b^2*c^5 + 4*a^4*b*c^2)*d^3*x^2 + (7*a^3*b^2*c^6 + 8*a^4*b*c^3 + a^5)*d^2*x + (a^
3*b^2*c^7 + 2*a^4*b*c^4 + a^5*c)*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.20566, size = 301, normalized size = 1.37 \begin{align*} \frac{14 \, \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 )}{27 \, a^{3}} - \frac{14 \, \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 )}{27 \, a^{4} d} - \frac{7 \, \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 )}{27 \, a^{4} d} - \frac{\frac{10 \, b^{2}}{{\left (d x + c\right )} d} + \frac{13 \, a b}{{\left (d x + c\right )}^{4} d}}{18 \, a^{3}{\left (b + \frac{a}{{\left (d x + c\right )}^{3}}\right )}^{2}} - \frac{1}{{\left (d x + c\right )} a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

14/27*(b/(a*d^3))^(1/3)*log(abs(-(b/(a*d^3))^(1/3) - 1/((d*x + c)*d)))/a^3 - 14/27*sqrt(3)*(a^2*b)^(1/3)*arcta
n(1/3*sqrt(3)*((b/(a*d^3))^(1/3) - 2/((d*x + c)*d))/(b/(a*d^3))^(1/3))/(a^4*d) - 7/27*(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^4*d) - 1/18*(10*b^2/((d*x + c)*d) + 13
*a*b/((d*x + c)^4*d))/(a^3*(b + a/(d*x + c)^3)^2) - 1/((d*x + c)*a^3*d)