3.2894 \(\int \frac {(c e+d e x)^4}{(a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=184 \[ \frac {e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 \sqrt [3]{a} b^{5/3} d}-\frac {2 e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 \sqrt [3]{a} b^{5/3} d}-\frac {2 e^4 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{5/3} d}-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )} \]

[Out]

-1/3*e^4*(d*x+c)^2/b/d/(a+b*(d*x+c)^3)-2/9*e^4*ln(a^(1/3)+b^(1/3)*(d*x+c))/a^(1/3)/b^(5/3)/d+1/9*e^4*ln(a^(2/3
)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(1/3)/b^(5/3)/d-2/9*e^4*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))/
a^(1/3)*3^(1/2))/a^(1/3)/b^(5/3)/d*3^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {372, 288, 292, 31, 634, 617, 204, 628} \[ \frac {e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 \sqrt [3]{a} b^{5/3} d}-\frac {2 e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 \sqrt [3]{a} b^{5/3} d}-\frac {2 e^4 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{5/3} d}-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*(c + d*x)^3)^2,x]

[Out]

-(e^4*(c + d*x)^2)/(3*b*d*(a + b*(c + d*x)^3)) - (2*e^4*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3
))])/(3*Sqrt[3]*a^(1/3)*b^(5/3)*d) - (2*e^4*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9*a^(1/3)*b^(5/3)*d) + (e^4*Log
[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(1/3)*b^(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 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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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 e+d e x)^4}{\left (a+b (c+d x)^3\right )^2} \, dx &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )}-\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 \sqrt [3]{a} b^{4/3} d}+\frac {\left (2 e^4\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 )}{9 \sqrt [3]{a} b^{4/3} d}\\ &=-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )}-\frac {2 e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 \sqrt [3]{a} b^{5/3} d}+\frac {e^4 \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 )}{9 \sqrt [3]{a} b^{5/3} d}+\frac {e^4 \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 )}{3 b^{4/3} d}\\ &=-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )}-\frac {2 e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 \sqrt [3]{a} b^{5/3} d}+\frac {e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 \sqrt [3]{a} b^{5/3} d}+\frac {\left (2 e^4\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 \sqrt [3]{a} b^{5/3} d}\\ &=-\frac {e^4 (c+d x)^2}{3 b d \left (a+b (c+d x)^3\right )}-\frac {2 e^4 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{5/3} d}-\frac {2 e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 \sqrt [3]{a} b^{5/3} d}+\frac {e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 \sqrt [3]{a} b^{5/3} d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 155, normalized size = 0.84 \[ \frac {e^4 \left (\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{\sqrt [3]{a}}-\frac {3 b^{2/3} (c+d x)^2}{a+b (c+d x)^3}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{\sqrt [3]{a}}+\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{a}}\right )}{9 b^{5/3} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*(c + d*x)^3)^2,x]

[Out]

(e^4*((-3*b^(2/3)*(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))])/a^(1/3) - (2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/a^(1/3) + Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x)
+ b^(2/3)*(c + d*x)^2]/a^(1/3)))/(9*b^(5/3)*d)

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fricas [B]  time = 0.98, size = 940, normalized size = 5.11 \[ \left [-\frac {3 \, a b^{2} d^{2} e^{4} x^{2} + 6 \, a b^{2} c d e^{4} x + 3 \, a b^{2} c^{2} e^{4} - 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} d^{3} e^{4} x^{3} + 3 \, a b^{2} c d^{2} e^{4} x^{2} + 3 \, a b^{2} c^{2} d e^{4} x + {\left (a b^{2} c^{3} + a^{2} b\right )} e^{4}\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} e^{4} x^{3} + 3 \, b c d^{2} e^{4} x^{2} + 3 \, b c^{2} d e^{4} x + {\left (b c^{3} + a\right )} e^{4}\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} e^{4} x^{3} + 3 \, b c d^{2} e^{4} x^{2} + 3 \, b c^{2} d e^{4} x + {\left (b c^{3} + a\right )} e^{4}\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 )}{9 \, {\left (a b^{4} d^{4} x^{3} + 3 \, a b^{4} c d^{3} x^{2} + 3 \, a b^{4} c^{2} d^{2} x + {\left (a b^{4} c^{3} + a^{2} b^{3}\right )} d\right )}}, -\frac {3 \, a b^{2} d^{2} e^{4} x^{2} + 6 \, a b^{2} c d e^{4} x + 3 \, a b^{2} c^{2} e^{4} - 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} d^{3} e^{4} x^{3} + 3 \, a b^{2} c d^{2} e^{4} x^{2} + 3 \, a b^{2} c^{2} d e^{4} x + {\left (a b^{2} c^{3} + a^{2} b\right )} e^{4}\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} e^{4} x^{3} + 3 \, b c d^{2} e^{4} x^{2} + 3 \, b c^{2} d e^{4} x + {\left (b c^{3} + a\right )} e^{4}\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} e^{4} x^{3} + 3 \, b c d^{2} e^{4} x^{2} + 3 \, b c^{2} d e^{4} x + {\left (b c^{3} + a\right )} e^{4}\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 )}{9 \, {\left (a b^{4} d^{4} x^{3} + 3 \, a b^{4} c d^{3} x^{2} + 3 \, a b^{4} c^{2} d^{2} x + {\left (a b^{4} c^{3} + a^{2} b^{3}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

[-1/9*(3*a*b^2*d^2*e^4*x^2 + 6*a*b^2*c*d*e^4*x + 3*a*b^2*c^2*e^4 - 3*sqrt(1/3)*(a*b^2*d^3*e^4*x^3 + 3*a*b^2*c*
d^2*e^4*x^2 + 3*a*b^2*c^2*d*e^4*x + (a*b^2*c^3 + a^2*b)*e^4)*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*e^4*x^3 + 3*b*c*d^2*e^4*x^2 + 3*b*c^2*d*e^4*x + (b*c^3 + a)*e^4)*(-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*e^
4*x^3 + 3*b*c*d^2*e^4*x^2 + 3*b*c^2*d*e^4*x + (b*c^3 + a)*e^4)*(-a*b^2)^(2/3)*log(b*d*x + b*c - (-a*b^2)^(1/3)
))/(a*b^4*d^4*x^3 + 3*a*b^4*c*d^3*x^2 + 3*a*b^4*c^2*d^2*x + (a*b^4*c^3 + a^2*b^3)*d), -1/9*(3*a*b^2*d^2*e^4*x^
2 + 6*a*b^2*c*d*e^4*x + 3*a*b^2*c^2*e^4 - 6*sqrt(1/3)*(a*b^2*d^3*e^4*x^3 + 3*a*b^2*c*d^2*e^4*x^2 + 3*a*b^2*c^2
*d*e^4*x + (a*b^2*c^3 + a^2*b)*e^4)*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*e^4*x^3 + 3*b*c*d^2*e^4*x^2 + 3*b*c^2*d*e^4*x + (b*c^3 + a)*e^4)*(-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*e^
4*x^3 + 3*b*c*d^2*e^4*x^2 + 3*b*c^2*d*e^4*x + (b*c^3 + a)*e^4)*(-a*b^2)^(2/3)*log(b*d*x + b*c - (-a*b^2)^(1/3)
))/(a*b^4*d^4*x^3 + 3*a*b^4*c*d^3*x^2 + 3*a*b^4*c^2*d^2*x + (a*b^4*c^3 + a^2*b^3)*d)]

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giac [A]  time = 0.25, size = 218, normalized size = 1.18 \[ -\frac {2 \, \sqrt {3} \left (-\frac {e^{12}}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (-\frac {e^{12}}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (-\frac {e^{12}}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{9 \, b} - \frac {d^{2} x^{2} e^{4} + 2 \, c d x e^{4} + c^{2} e^{4}}{3 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*(d*x+c)^3)^2,x, algorithm="giac")

[Out]

-1/9*(2*sqrt(3)*(-e^12/(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)) + (-e^12/(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*(-e^12/
(a*b^2*d^3))^(1/3)*log(abs(a*b*d*x + a*b*c + (-a^2*b)^(2/3))))/b - 1/3*(d^2*x^2*e^4 + 2*c*d*x*e^4 + c^2*e^4)/(
(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*b*d)

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maple [C]  time = 0.01, size = 221, normalized size = 1.20 \[ -\frac {d \,e^{4} x^{2}}{3 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right ) b}-\frac {2 c \,e^{4} x}{3 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right ) b}-\frac {c^{2} e^{4}}{3 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right ) b d}+\frac {2 e^{4} \left (\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right ) d +c \right ) \ln \left (-\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+x \right )}{9 b^{2} d \left (d^{2} \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )^{2}+2 c d \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+c^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*(d*x+c)^3)^2,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

2/3*e^4*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)/b - 1/3*(d^2*e^4*x^2 + 2
*c*d*e^4*x + c^2*e^4)/(b^2*d^4*x^3 + 3*b^2*c*d^3*x^2 + 3*b^2*c^2*d^2*x + (b^2*c^3 + a*b)*d)

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mupad [B]  time = 1.56, size = 266, normalized size = 1.45 \[ \frac {2\,e^4\,\ln \left (b^{1/3}\,c-{\left (-a\right )}^{1/3}+b^{1/3}\,d\,x\right )}{9\,{\left (-a\right )}^{1/3}\,b^{5/3}\,d}-\frac {\frac {d\,e^4\,x^2}{3\,b}+\frac {c^2\,e^4}{3\,b\,d}+\frac {2\,c\,e^4\,x}{3\,b}}{b\,c^3+3\,b\,c^2\,d\,x+3\,b\,c\,d^2\,x^2+b\,d^3\,x^3+a}-\frac {\ln \left (\frac {4\,c\,d^4\,e^8}{9\,b}-\frac {{\left (-a\right )}^{1/3}\,d^4\,{\left (e^4+\sqrt {3}\,e^4\,1{}\mathrm {i}\right )}^2}{9\,b^{4/3}}+\frac {4\,d^5\,e^8\,x}{9\,b}\right )\,\left (e^4+\sqrt {3}\,e^4\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{1/3}\,b^{5/3}\,d}+\frac {e^4\,\ln \left (\frac {4\,c\,d^4\,e^8}{9\,b}+\frac {4\,d^5\,e^8\,x}{9\,b}-\frac {9\,{\left (-a\right )}^{1/3}\,d^4\,e^8\,{\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2}{b^{4/3}}\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}{{\left (-a\right )}^{1/3}\,b^{5/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*(c + d*x)^3)^2,x)

[Out]

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

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sympy [A]  time = 1.67, size = 133, normalized size = 0.72 \[ \frac {- c^{2} e^{4} - 2 c d e^{4} x - d^{2} e^{4} x^{2}}{3 a b d + 3 b^{2} c^{3} d + 9 b^{2} c^{2} d^{2} x + 9 b^{2} c d^{3} x^{2} + 3 b^{2} d^{4} x^{3}} + \frac {e^{4} \operatorname {RootSum} {\left (729 t^{3} a b^{5} + 8, \left (t \mapsto t \log {\left (x + \frac {81 t^{2} a b^{3} e^{8} + 4 c e^{8}}{4 d e^{8}} \right )} \right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*(d*x+c)**3)**2,x)

[Out]

(-c**2*e**4 - 2*c*d*e**4*x - d**2*e**4*x**2)/(3*a*b*d + 3*b**2*c**3*d + 9*b**2*c**2*d**2*x + 9*b**2*c*d**3*x**
2 + 3*b**2*d**4*x**3) + e**4*RootSum(729*_t**3*a*b**5 + 8, Lambda(_t, _t*log(x + (81*_t**2*a*b**3*e**8 + 4*c*e
**8)/(4*d*e**8))))/d

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