\(\int \frac {1}{(c+d x)^3 (a+b (c+d x)^3)} \, dx\) [2866]

Optimal result

Integrand size = 21, antiderivative size = 156 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=-\frac {1}{2 a d (c+d x)^2}+\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} d}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d} \]

[Out]

-1/2/a/d/(d*x+c)^2-1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(5/3)/d+1/6*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d
*x+c)+b^(2/3)*(d*x+c)^2)/a^(5/3)/d+1/3*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))/a^(1/3)*3^(1/2))/a^(5/3)
/d*3^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {379, 331, 206, 31, 648, 631, 210, 642} \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} d}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d}-\frac {1}{2 a d (c+d x)^2} \]

[In]

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

[Out]

-1/2*1/(a*d*(c + d*x)^2) + (b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3
)*d) - (b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*a^(5/3)*d) + (b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c +
d*x) + b^(2/3)*(c + d*x)^2])/(6*a^(5/3)*d)

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 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^3 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{d} \\ & = -\frac {1}{2 a d (c+d x)^2}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^3} \, dx,x,c+d x\right )}{a d} \\ & = -\frac {1}{2 a d (c+d x)^2}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 a^{5/3} d}-\frac {b \text {Subst}\left (\int \frac {2 \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 )}{3 a^{5/3} d} \\ & = -\frac {1}{2 a d (c+d x)^2}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d}+\frac {b^{2/3} \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 )}{6 a^{5/3} d}-\frac {b \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 )}{2 a^{4/3} d} \\ & = -\frac {1}{2 a d (c+d x)^2}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d}-\frac {b^{2/3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{a^{5/3} d} \\ & = -\frac {1}{2 a d (c+d x)^2}+\frac {b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{5/3} d}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {-\frac {3 a^{2/3}}{(c+d x)^2}-2 \sqrt {3} b^{2/3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )-2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )+b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d} \]

[In]

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

[Out]

((-3*a^(2/3))/(c + d*x)^2 - 2*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] - 2*b
^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] + b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^
2])/(6*a^(5/3)*d)

Maple [C] (verified)

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

Time = 3.96 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56

method	result	size
default	\(-\frac {1}{2 a d \left (d x +c \right )^{2}}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}}{3 d a}\)	\(87\)
risch	\(-\frac {1}{2 a d \left (d x +c \right )^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{5} d^{3} \textit {\_Z}^{3}+b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{5} d^{4} \textit {\_R}^{3}-3 b^{2} d \right ) x -4 a^{5} c \,d^{3} \textit {\_R}^{3}-a^{2} b d \textit {\_R} -3 b^{2} c \right )\right )}{3}\)	\(87\)

[In]

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

[Out]

-1/2/a/d/(d*x+c)^2-1/3/d*sum(1/(_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))/a

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {2 \, \sqrt {3} {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (a d x + a c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} + {\left (a b d x + a b c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b d x + b c - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 3}{6 \, {\left (a d^{3} x^{2} + 2 \, a c d^{2} x + a c^{2} d\right )}} \]

[In]

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

[Out]

1/6*(2*sqrt(3)*(d^2*x^2 + 2*c*d*x + c^2)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*(a*d*x + a*c)*(-b^2/a^2)^(2/3)
- sqrt(3)*b)/b) - (d^2*x^2 + 2*c*d*x + c^2)*(-b^2/a^2)^(1/3)*log(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + a^2*(-
b^2/a^2)^(2/3) + (a*b*d*x + a*b*c)*(-b^2/a^2)^(1/3)) + 2*(d^2*x^2 + 2*c*d*x + c^2)*(-b^2/a^2)^(1/3)*log(b*d*x
+ b*c - a*(-b^2/a^2)^(1/3)) - 3)/(a*d^3*x^2 + 2*a*c*d^2*x + a*c^2*d)

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.39 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=- \frac {1}{2 a c^{2} d + 4 a c d^{2} x + 2 a d^{3} x^{2}} + \frac {\operatorname {RootSum} {\left (27 t^{3} a^{5} + b^{2}, \left ( t \mapsto t \log {\left (x + \frac {- 3 t a^{2} + b c}{b d} \right )} \right )\right )}}{d} \]

[In]

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

[Out]

-1/(2*a*c**2*d + 4*a*c*d**2*x + 2*a*d**3*x**2) + RootSum(27*_t**3*a**5 + b**2, Lambda(_t, _t*log(x + (-3*_t*a*
*2 + b*c)/(b*d))))/d

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {2 \, \sqrt {3} \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | -b d x - b c + \left (-a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{6 \, a} - \frac {1}{2 \, {\left (d x + c\right )}^{2} a d} \]

[In]

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

[Out]

1/6*(2*sqrt(3)*(-b^2/(a^2*d^3))^(1/3)*arctan(-(b*d*x + b*c - (-a*b^2)^(1/3))/(sqrt(3)*b*d*x + sqrt(3)*b*c + sq
rt(3)*(-a*b^2)^(1/3))) - (-b^2/(a^2*d^3))^(1/3)*log(4*(sqrt(3)*b*d*x + sqrt(3)*b*c + sqrt(3)*(-a*b^2)^(1/3))^2
+ 4*(b*d*x + b*c - (-a*b^2)^(1/3))^2) + 2*(-b^2/(a^2*d^3))^(1/3)*log(abs(-b*d*x - b*c + (-a*b^2)^(1/3))))/a -
1/2/((d*x + c)^2*a*d)

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )} \, dx=\frac {b^{2/3}\,\ln \left (a^2\,b^{1/3}\,c-{\left (-a\right )}^{7/3}+a^2\,b^{1/3}\,d\,x\right )}{3\,{\left (-a\right )}^{5/3}\,d}-\frac {1}{2\,a\,d\,\left (c^2+2\,c\,d\,x+d^2\,x^2\right )}-\frac {b^{2/3}\,\ln \left (3\,a^2\,b^3\,c\,d^5+3\,a^2\,b^3\,d^6\,x+3\,{\left (-a\right )}^{7/3}\,b^{8/3}\,d^5\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,{\left (-a\right )}^{5/3}\,d}+\frac {b^{2/3}\,\ln \left (3\,a^2\,b^3\,c\,d^5+3\,a^2\,b^3\,d^6\,x-9\,{\left (-a\right )}^{7/3}\,b^{8/3}\,d^5\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{{\left (-a\right )}^{5/3}\,d} \]

[In]

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

[Out]

(b^(2/3)*log(a^2*b^(1/3)*c - (-a)^(7/3) + a^2*b^(1/3)*d*x))/(3*(-a)^(5/3)*d) - 1/(2*a*d*(c^2 + d^2*x^2 + 2*c*d
*x)) - (b^(2/3)*log(3*a^2*b^3*c*d^5 + 3*a^2*b^3*d^6*x + 3*(-a)^(7/3)*b^(8/3)*d^5*((3^(1/2)*1i)/2 + 1/2))*((3^(
1/2)*1i)/2 + 1/2))/(3*(-a)^(5/3)*d) + (b^(2/3)*log(3*a^2*b^3*c*d^5 + 3*a^2*b^3*d^6*x - 9*(-a)^(7/3)*b^(8/3)*d^
5*((3^(1/2)*1i)/6 - 1/6))*((3^(1/2)*1i)/6 - 1/6))/((-a)^(5/3)*d)