3.2871 \(\int \frac {c+d x}{(a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=172 \[ -\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{4/3} b^{2/3} d}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{4/3} b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3} d}+\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )} \]

[Out]

1/3*(d*x+c)^2/a/d/(a+b*(d*x+c)^3)-1/9*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(4/3)/b^(2/3)/d+1/18*ln(a^(2/3)-a^(1/3)*b^
(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(4/3)/b^(2/3)/d-1/9*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))/a^(1/3)*3^(1/2))
/a^(4/3)/b^(2/3)/d*3^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {372, 290, 292, 31, 634, 617, 204, 628} \[ -\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{4/3} b^{2/3} d}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{4/3} b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3} d}+\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^2/(3*a*d*(a + b*(c + d*x)^3)) - ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))]/(3*Sqrt[3]
*a^(4/3)*b^(2/3)*d) - Log[a^(1/3) + b^(1/3)*(c + d*x)]/(9*a^(4/3)*b^(2/3)*d) + Log[a^(2/3) - a^(1/3)*b^(1/3)*(
c + d*x) + b^(2/3)*(c + d*x)^2]/(18*a^(4/3)*b^(2/3)*d)

Rule 31

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

Rule 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 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 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 {c+d x}{\left (a+b (c+d x)^3\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{\left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )}+\frac {\operatorname {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{3 a d}\\ &=\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 a^{4/3} \sqrt [3]{b} d}+\frac {\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 )}{9 a^{4/3} \sqrt [3]{b} d}\\ &=\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{4/3} b^{2/3} d}+\frac {\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^{4/3} b^{2/3} d}+\frac {\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 \sqrt [3]{b} d}\\ &=\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{4/3} b^{2/3} d}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{4/3} b^{2/3} d}+\frac {\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^{4/3} b^{2/3} d}\\ &=\frac {(c+d x)^2}{3 a d \left (a+b (c+d x)^3\right )}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3} d}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{4/3} b^{2/3} d}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{4/3} b^{2/3} d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 152, normalized size = 0.88 \[ \frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{b^{2/3}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{b^{2/3}}+\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{b^{2/3}}+\frac {6 \sqrt [3]{a} (c+d x)^2}{a+b (c+d x)^3}}{18 a^{4/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^(1/3)*(c + d*x)^2)/(a + b*(c + d*x)^3) + (2*Sqrt[3]*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(
1/3))])/b^(2/3) - (2*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]/b^(2/3))/(18*a^(4/3)*d)

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fricas [B]  time = 0.84, size = 852, normalized size = 4.95 \[ \left [\frac {6 \, a b^{2} d^{2} x^{2} + 12 \, a b^{2} c d x + 6 \, a b^{2} c^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} d^{3} x^{3} + 3 \, a b^{2} c d^{2} x^{2} + 3 \, a b^{2} c^{2} d x + a b^{2} c^{3} + a^{2} b\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} d^{3} x^{3} + 6 \, b^{2} c d^{2} x^{2} + 6 \, b^{2} c^{2} d x + 2 \, b^{2} c^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b d x + a b c + 2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )}}{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}\right ) + {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} {\left (b d x + b c\right )} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{18 \, {\left (a^{2} b^{3} d^{4} x^{3} + 3 \, a^{2} b^{3} c d^{3} x^{2} + 3 \, a^{2} b^{3} c^{2} d^{2} x + {\left (a^{2} b^{3} c^{3} + a^{3} b^{2}\right )} d\right )}}, \frac {6 \, a b^{2} d^{2} x^{2} + 12 \, a b^{2} c d x + 6 \, a b^{2} c^{2} + 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} d^{3} x^{3} + 3 \, a b^{2} c d^{2} x^{2} + 3 \, a b^{2} c^{2} d x + a b^{2} c^{3} + a^{2} b\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b d x + 2 \, b c + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} {\left (b d x + b c\right )} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{18 \, {\left (a^{2} b^{3} d^{4} x^{3} + 3 \, a^{2} b^{3} c d^{3} x^{2} + 3 \, a^{2} b^{3} c^{2} d^{2} x + {\left (a^{2} b^{3} c^{3} + a^{3} b^{2}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(6*a*b^2*d^2*x^2 + 12*a*b^2*c*d*x + 6*a*b^2*c^2 + 3*sqrt(1/3)*(a*b^2*d^3*x^3 + 3*a*b^2*c*d^2*x^2 + 3*a*b
^2*c^2*d*x + a*b^2*c^3 + a^2*b)*sqrt((-a*b^2)^(1/3)/a)*log((2*b^2*d^3*x^3 + 6*b^2*c*d^2*x^2 + 6*b^2*c^2*d*x +
2*b^2*c^3 - a*b + 3*sqrt(1/3)*(a*b*d*x + a*b*c + 2*(d^2*x^2 + 2*c*d*x + c^2)*(-a*b^2)^(2/3) + (-a*b^2)^(1/3)*a
)*sqrt((-a*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*(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))
+ (b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(-a*b^2)^(2/3)*log(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2
+ (-a*b^2)^(1/3)*(b*d*x + b*c) + (-a*b^2)^(2/3)) - 2*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(-
a*b^2)^(2/3)*log(b*d*x + b*c - (-a*b^2)^(1/3)))/(a^2*b^3*d^4*x^3 + 3*a^2*b^3*c*d^3*x^2 + 3*a^2*b^3*c^2*d^2*x +
(a^2*b^3*c^3 + a^3*b^2)*d), 1/18*(6*a*b^2*d^2*x^2 + 12*a*b^2*c*d*x + 6*a*b^2*c^2 + 6*sqrt(1/3)*(a*b^2*d^3*x^3
+ 3*a*b^2*c*d^2*x^2 + 3*a*b^2*c^2*d*x + a*b^2*c^3 + a^2*b)*sqrt(-(-a*b^2)^(1/3)/a)*arctan(sqrt(1/3)*(2*b*d*x
+ 2*b*c + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3)/a)/b) + (b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(
-a*b^2)^(2/3)*log(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + (-a*b^2)^(1/3)*(b*d*x + b*c) + (-a*b^2)^(2/3)) - 2*(b*
d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(-a*b^2)^(2/3)*log(b*d*x + b*c - (-a*b^2)^(1/3)))/(a^2*b^3*
d^4*x^3 + 3*a^2*b^3*c*d^3*x^2 + 3*a^2*b^3*c^2*d^2*x + (a^2*b^3*c^3 + a^3*b^2)*d)]

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giac [A]  time = 0.29, size = 205, normalized size = 1.19 \[ -\frac {2 \, \sqrt {3} \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{18 \, a} + \frac {d^{2} x^{2} + 2 \, c d x + c^{2}}{3 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*sqrt(3)*(-1/(a*b^2*d^3))^(1/3)*arctan(1/3*sqrt(3)*(2*a*b*d*x + 2*a*b*c - (-a^2*b)^(2/3))/(-a^2*b)^(2/
3)) + (-1/(a*b^2*d^3))^(1/3)*log((2*a*b*d*x + 2*a*b*c - (-a^2*b)^(2/3))^2 + 3*(-a^2*b)^(4/3)) - 2*(-1/(a*b^2*d
^3))^(1/3)*log(abs(a*b*d*x + a*b*c + (-a^2*b)^(2/3))))/a + 1/3*(d^2*x^2 + 2*c*d*x + c^2)/((b*d^3*x^3 + 3*b*c*d
^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*a*d)

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maple [C]  time = 0.01, size = 144, normalized size = 0.84 \[ \frac {\left (\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right ) d +c \right ) \ln \left (-\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+x \right )}{9 a b d \left (d^{2} \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )^{2}+2 c d \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+c^{2}\right )}+\frac {\frac {d \,x^{2}}{3 a}+\frac {2 c x}{3 a}+\frac {c^{2}}{3 a d}}{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/3*d/a*x^2+2/3*c/a*x+1/3*c^2/d/a)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)+1/9/a/b/d*sum((_R*d+c)/(_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 \[ \frac {d^{2} x^{2} + 2 \, c d x + c^{2}}{3 \, {\left (a b d^{4} x^{3} + 3 \, a b c d^{3} x^{2} + 3 \, a b c^{2} d^{2} x + {\left (a b c^{3} + a^{2}\right )} d\right )}} + \frac {-\frac {1}{3} \, \sqrt {3} \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) - \frac {1}{6} \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) + \frac {1}{3} \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d^2*x^2 + 2*c*d*x + c^2)/(a*b*d^4*x^3 + 3*a*b*c*d^3*x^2 + 3*a*b*c^2*d^2*x + (a*b*c^3 + a^2)*d) + 1/3*inte
grate((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

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mupad [B]  time = 1.42, size = 216, normalized size = 1.26 \[ \frac {\frac {d\,x^2}{3\,a}+\frac {c^2}{3\,a\,d}+\frac {2\,c\,x}{3\,a}}{b\,c^3+3\,b\,c^2\,d\,x+3\,b\,c\,d^2\,x^2+b\,d^3\,x^3+a}-\frac {\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{4/3}\,b^{2/3}\,d}-\frac {\ln \left (\frac {b^{2/3}\,d^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{36\,a^{5/3}}+\frac {b\,c\,d^4}{9\,a^2}+\frac {b\,d^5\,x}{9\,a^2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{4/3}\,b^{2/3}\,d}+\frac {\ln \left (\frac {b^{2/3}\,d^4\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{36\,a^{5/3}}+\frac {b\,c\,d^4}{9\,a^2}+\frac {b\,d^5\,x}{9\,a^2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{18\,a^{4/3}\,b^{2/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x^2)/(3*a) + c^2/(3*a*d) + (2*c*x)/(3*a))/(a + b*c^3 + b*d^3*x^3 + 3*b*c^2*d*x + 3*b*c*d^2*x^2) - log(b^(1
/3)*c + a^(1/3) + b^(1/3)*d*x)/(9*a^(4/3)*b^(2/3)*d) - (log((b^(2/3)*d^4*(3^(1/2)*1i - 1)^2)/(36*a^(5/3)) + (b
*c*d^4)/(9*a^2) + (b*d^5*x)/(9*a^2))*(3^(1/2)*1i - 1))/(18*a^(4/3)*b^(2/3)*d) + (log((b^(2/3)*d^4*(3^(1/2)*1i
+ 1)^2)/(36*a^(5/3)) + (b*c*d^4)/(9*a^2) + (b*d^5*x)/(9*a^2))*(3^(1/2)*1i + 1))/(18*a^(4/3)*b^(2/3)*d)

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sympy [A]  time = 1.00, size = 105, normalized size = 0.61 \[ \frac {c^{2} + 2 c d x + d^{2} x^{2}}{3 a^{2} d + 3 a b c^{3} d + 9 a b c^{2} d^{2} x + 9 a b c d^{3} x^{2} + 3 a b d^{4} x^{3}} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{4} b^{2} + 1, \left (t \mapsto t \log {\left (x + \frac {81 t^{2} a^{3} b + c}{d} \right )} \right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c**2 + 2*c*d*x + d**2*x**2)/(3*a**2*d + 3*a*b*c**3*d + 9*a*b*c**2*d**2*x + 9*a*b*c*d**3*x**2 + 3*a*b*d**4*x**
3) + RootSum(729*_t**3*a**4*b**2 + 1, Lambda(_t, _t*log(x + (81*_t**2*a**3*b + c)/d)))/d

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