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

Optimal. Leaf size=168 \[ \frac {a^{2/3} e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}+\frac {a^{2/3} e^4 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {e^4 (c+d x)^2}{2 b d} \]

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Rubi [A]  time = 0.13, antiderivative size = 168, 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, 321, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {a^{2/3} e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}+\frac {a^{2/3} e^4 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {e^4 (c+d x)^2}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*(c + d*x)^2)/(2*b*d) + (a^(2/3)*e^4*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b
^(5/3)*d) + (a^(2/3)*e^4*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*b^(5/3)*d) - (a^(2/3)*e^4*Log[a^(2/3) - a^(1/3)*
b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(6*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 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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 {(c e+d e x)^4}{a+b (c+d x)^3} \, dx &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{a+b x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 (c+d x)^2}{2 b d}-\frac {\left (a e^4\right ) \operatorname {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{b d}\\ &=\frac {e^4 (c+d x)^2}{2 b d}+\frac {\left (a^{2/3} e^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 b^{4/3} d}-\frac {\left (a^{2/3} 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 )}{3 b^{4/3} d}\\ &=\frac {e^4 (c+d x)^2}{2 b d}+\frac {a^{2/3} e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {\left (a^{2/3} e^4\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 )}{6 b^{5/3} d}-\frac {\left (a e^4\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 )}{2 b^{4/3} d}\\ &=\frac {e^4 (c+d x)^2}{2 b d}+\frac {a^{2/3} e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}-\frac {\left (a^{2/3} 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 )}{b^{5/3} d}\\ &=\frac {e^4 (c+d x)^2}{2 b d}+\frac {a^{2/3} e^4 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{2/3} e^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} e^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 163, normalized size = 0.97 \begin {gather*} e^4 \left (\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{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 b^{5/3} d}-\frac {a^{2/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {(c+d x)^2}{2 b d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

e^4*((c + d*x)^2/(2*b*d) - (a^(2/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(5/
3)*d) + (a^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*b^(5/3)*d) - (a^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c +
d*x) + b^(2/3)*(c + d*x)^2])/(6*b^(5/3)*d))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(c e+d e x)^4}{a+b (c+d x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 179, normalized size = 1.07 \begin {gather*} \frac {3 \, d^{2} e^{4} x^{2} + 6 \, c d e^{4} x - 2 \, \sqrt {3} e^{4} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b d x + b c\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - e^{4} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a d^{2} x^{2} + 2 \, a c d x + a c^{2} - {\left (b d x + b c\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, e^{4} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a d x + a c + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{6 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 170, normalized size = 1.01 \begin {gather*} \frac {b d^{7} x^{2} e^{4} + 2 \, b c d^{6} x e^{4}}{2 \, b^{2} d^{6}} - \frac {2 \, \sqrt {3} \left (a^{2} b d^{15}\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 ) e^{4} + \left (a^{2} b d^{15}\right )^{\frac {1}{3}} e^{4} \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 (a^{2} b d^{15}\right )^{\frac {1}{3}} e^{4} \log \left ({\left | a b d x + a b c + \left (a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{6 \, b^{2} d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*d^7*x^2*e^4 + 2*b*c*d^6*x*e^4)/(b^2*d^6) - 1/6*(2*sqrt(3)*(a^2*b*d^15)^(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^4 + (a^2*b*d^15)^(1/3)*e^4*log((2*a*b*d*x + 2*a*b*c - (a^2*b)^
(2/3))^2 + 3*(a^2*b)^(4/3)) - 2*(a^2*b*d^15)^(1/3)*e^4*log(abs(a*b*d*x + a*b*c + (a^2*b)^(2/3))))/(b^2*d^6)

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maple [C]  time = 0.00, size = 102, normalized size = 0.61 \begin {gather*} \frac {d \,e^{4} x^{2}}{2 b}+\frac {c \,e^{4} x}{b}-\frac {a \,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 )}{3 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {-\frac {1}{6} \, {\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 )} a e^{4}}{b} + \frac {d e^{4} x^{2} + 2 \, c e^{4} x}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*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 + 1/2*(d*e^4*x^2 + 2*c*
e^4*x)/b

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mupad [B]  time = 1.70, size = 151, normalized size = 0.90 \begin {gather*} \frac {d\,e^4\,x^2}{2\,b}+\frac {c\,e^4\,x}{b}+\frac {a^{2/3}\,e^4\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{3\,b^{5/3}\,d}+\frac {a^{2/3}\,e^4\,\ln \left (a^{1/3}-2\,b^{1/3}\,c-2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^{5/3}\,d}-\frac {a^{2/3}\,e^4\,\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 )}{3\,b^{5/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.54, size = 66, normalized size = 0.39 \begin {gather*} \frac {e^{4} \operatorname {RootSum} {\left (27 t^{3} b^{5} - a^{2}, \left (t \mapsto t \log {\left (x + \frac {9 t^{2} b^{3} e^{8} + a c e^{8}}{a d e^{8}} \right )} \right )\right )}}{d} + \frac {c e^{4} x}{b} + \frac {d e^{4} x^{2}}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**4*RootSum(27*_t**3*b**5 - a**2, Lambda(_t, _t*log(x + (9*_t**2*b**3*e**8 + a*c*e**8)/(a*d*e**8))))/d + c*e*
*4*x/b + d*e**4*x**2/(2*b)

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