3.2875 \(\int \frac {1}{(c+d x)^3 (a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=189 \[ -\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d}+\frac {5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} d}-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )} \]

[Out]

-5/6/a^2/d/(d*x+c)^2+1/3/a/d/(d*x+c)^2/(a+b*(d*x+c)^3)-5/9*b^(2/3)*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(8/3)/d+5/18*
b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(8/3)/d+5/9*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1
/3)*(d*x+c))/a^(1/3)*3^(1/2))/a^(8/3)/d*3^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {372, 290, 325, 200, 31, 634, 617, 204, 628} \[ -\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d}+\frac {5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3} d}-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-5/(6*a^2*d*(c + d*x)^2) + 1/(3*a*d*(c + d*x)^2*(a + b*(c + d*x)^3)) + (5*b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*
(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)*d) - (5*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9*a^(8/3)
*d) + (5*b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d)

Rule 31

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

Rule 200

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 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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {1}{(c+d x)^3 \left (a+b (c+d x)^3\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 \left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^3 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{3 a d}\\ &=-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{a+b x^3} \, dx,x,c+d x\right )}{3 a^2 d}\\ &=-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 a^{8/3} d}-\frac {(5 b) \operatorname {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 )}{9 a^{8/3} d}\\ &=-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d}+\frac {\left (5 b^{2/3}\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 )}{18 a^{8/3} d}-\frac {(5 b) \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 )}{6 a^{7/3} d}\\ &=-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d}+\frac {5 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 )}{18 a^{8/3} d}-\frac {\left (5 b^{2/3}\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 )}{3 a^{8/3} d}\\ &=-\frac {5}{6 a^2 d (c+d x)^2}+\frac {1}{3 a d (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac {5 b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{8/3} d}-\frac {5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d}+\frac {5 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 )}{18 a^{8/3} d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 166, normalized size = 0.88 \[ \frac {5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac {6 a^{2/3} b (c+d x)}{a+b (c+d x)^3}-\frac {9 a^{2/3}}{(c+d x)^2}-10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-10 \sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{18 a^{8/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^(2/3))/(c + d*x)^2 - (6*a^(2/3)*b*(c + d*x))/(a + b*(c + d*x)^3) - 10*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3)
+ 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] - 10*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] + 5*b^(2/3)*Log[a^(2/3
) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d)

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fricas [B]  time = 0.88, size = 509, normalized size = 2.69 \[ -\frac {15 \, b d^{3} x^{3} + 45 \, b c d^{2} x^{2} + 45 \, b c^{2} d x + 15 \, b c^{3} - 10 \, \sqrt {3} {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (a d x + a c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) + 5 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} + {\left (a b d x + a b c\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} + {\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} + {\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b d x + b c - a \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 9 \, a}{18 \, {\left (a^{2} b d^{6} x^{5} + 5 \, a^{2} b c d^{5} x^{4} + 10 \, a^{2} b c^{2} d^{4} x^{3} + {\left (10 \, a^{2} b c^{3} + a^{3}\right )} d^{3} x^{2} + {\left (5 \, a^{2} b c^{4} + 2 \, a^{3} c\right )} d^{2} x + {\left (a^{2} b c^{5} + a^{3} c^{2}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(15*b*d^3*x^3 + 45*b*c*d^2*x^2 + 45*b*c^2*d*x + 15*b*c^3 - 10*sqrt(3)*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*
c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*s
qrt(3)*(a*d*x + a*c)*(-b^2/a^2)^(2/3) - sqrt(3)*b)/b) + 5*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^
5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-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)) - 10*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2
*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*log(b*d*x + b*c -
a*(-b^2/a^2)^(1/3)) + 9*a)/(a^2*b*d^6*x^5 + 5*a^2*b*c*d^5*x^4 + 10*a^2*b*c^2*d^4*x^3 + (10*a^2*b*c^3 + a^3)*d^
3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*x + (a^2*b*c^5 + a^3*c^2)*d)

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giac [A]  time = 0.22, size = 244, normalized size = 1.29 \[ \frac {5 \, {\left (2 \, \sqrt {3} \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}}\right ) - \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c + \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (-\frac {b^{2}}{a^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | -b d x - b c + \left (-a b^{2}\right )^{\frac {1}{3}} \right |}\right )\right )}}{18 \, a^{2}} - \frac {b d x + b c}{3 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} a^{2} d} - \frac {1}{2 \, {\left (d x + c\right )}^{2} a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/18*(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 + s
qrt(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^
2 - 1/3*(b*d*x + b*c)/((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*a^2*d) - 1/2/((d*x + c)^2*a^2*d)

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maple [C]  time = 0.02, size = 174, normalized size = 0.92 \[ -\frac {b x}{3 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right ) a^{2}}-\frac {b c}{3 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right ) a^{2} d}-\frac {5 \ln \left (-\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+x \right )}{9 a^{2} d \left (d^{2} \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )^{2}+2 c d \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+c^{2}\right )}-\frac {1}{2 \left (d x +c \right )^{2} a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a^2/d/(d*x+c)^2-1/3/a^2*b/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)*x-1/3/a^2*b/(b*d^3*x^3+3*b*c*d^2*
x^2+3*b*c^2*d*x+b*c^3+a)*c/d-5/9/a^2/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(-_R+x),_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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {5 \, b d^{3} x^{3} + 15 \, b c d^{2} x^{2} + 15 \, b c^{2} d x + 5 \, b c^{3} + 3 \, a}{6 \, {\left (a^{2} b d^{6} x^{5} + 5 \, a^{2} b c d^{5} x^{4} + 10 \, a^{2} b c^{2} d^{4} x^{3} + {\left (10 \, a^{2} b c^{3} + a^{3}\right )} d^{3} x^{2} + {\left (5 \, a^{2} b c^{4} + 2 \, a^{3} c\right )} d^{2} x + {\left (a^{2} b c^{5} + a^{3} c^{2}\right )} d\right )}} - \frac {\frac {5}{6} \, b {\left (\frac {2 \, \sqrt {3} \left (\frac {1}{a^{2} b}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}}\right )}{d} - \frac {\left (\frac {1}{a^{2} b}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right )}{d} + \frac {2 \, \left (\frac {1}{a^{2} b}\right )^{\frac {1}{3}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{d}\right )}}{3 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(5*b*d^3*x^3 + 15*b*c*d^2*x^2 + 15*b*c^2*d*x + 5*b*c^3 + 3*a)/(a^2*b*d^6*x^5 + 5*a^2*b*c*d^5*x^4 + 10*a^2
*b*c^2*d^4*x^3 + (10*a^2*b*c^3 + a^3)*d^3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*x + (a^2*b*c^5 + a^3*c^2)*d) - 5/3
*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^2

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mupad [B]  time = 1.98, size = 255, normalized size = 1.35 \[ -\frac {\frac {5\,b\,c^3+3\,a}{6\,a^2\,d}+\frac {5\,b\,d^2\,x^3}{6\,a^2}+\frac {5\,b\,c^2\,x}{2\,a^2}+\frac {5\,b\,c\,d\,x^2}{2\,a^2}}{x^2\,\left (10\,b\,c^3\,d^2+a\,d^2\right )+a\,c^2+b\,c^5+x\,\left (5\,b\,d\,c^4+2\,a\,d\,c\right )+b\,d^5\,x^5+10\,b\,c^2\,d^3\,x^3+5\,b\,c\,d^4\,x^4}-\frac {5\,b^{2/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{8/3}\,d}+\frac {5\,b^{2/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}\,d}-\frac {5\,b^{2/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*b^(2/3)*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)*1i)/2 + 1/2))/(9*a^(8/3)*
d) - (5*b^(2/3)*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(9*a^(8/3)*d) - ((3*a + 5*b*c^3)/(6*a^2*d) + (5*b*d^2*
x^3)/(6*a^2) + (5*b*c^2*x)/(2*a^2) + (5*b*c*d*x^2)/(2*a^2))/(x^2*(a*d^2 + 10*b*c^3*d^2) + a*c^2 + b*c^5 + x*(2
*a*c*d + 5*b*c^4*d) + b*d^5*x^5 + 10*b*c^2*d^3*x^3 + 5*b*c*d^4*x^4) - (5*b^(2/3)*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^(8/3)*d)

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sympy [A]  time = 2.50, size = 199, normalized size = 1.05 \[ \frac {- 3 a - 5 b c^{3} - 15 b c^{2} d x - 15 b c d^{2} x^{2} - 5 b d^{3} x^{3}}{6 a^{3} c^{2} d + 6 a^{2} b c^{5} d + 60 a^{2} b c^{2} d^{4} x^{3} + 30 a^{2} b c d^{5} x^{4} + 6 a^{2} b d^{6} x^{5} + x^{2} \left (6 a^{3} d^{3} + 60 a^{2} b c^{3} d^{3}\right ) + x \left (12 a^{3} c d^{2} + 30 a^{2} b c^{4} d^{2}\right )} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{8} + 125 b^{2}, \left (t \mapsto t \log {\left (x + \frac {- 9 t a^{3} + 5 b c}{5 b d} \right )} \right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a - 5*b*c**3 - 15*b*c**2*d*x - 15*b*c*d**2*x**2 - 5*b*d**3*x**3)/(6*a**3*c**2*d + 6*a**2*b*c**5*d + 60*a**
2*b*c**2*d**4*x**3 + 30*a**2*b*c*d**5*x**4 + 6*a**2*b*d**6*x**5 + x**2*(6*a**3*d**3 + 60*a**2*b*c**3*d**3) + x
*(12*a**3*c*d**2 + 30*a**2*b*c**4*d**2)) + RootSum(729*_t**3*a**8 + 125*b**2, Lambda(_t, _t*log(x + (-9*_t*a**
3 + 5*b*c)/(5*b*d))))/d

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