\(\int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx\) [2859]

Optimal result

Integrand size = 21, antiderivative size = 156 \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {(c+d x)^2}{2 b d}+\frac {a^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (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, 327, 298, 31, 648, 631, 210, 642} \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {a^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}+\frac {(c+d x)^2}{2 b d} \]

[In]

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

[Out]

(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)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 298

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 327

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 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 {x^4}{a+b x^3} \, dx,x,c+d x\right )}{d} \\ & = \frac {(c+d x)^2}{2 b d}-\frac {a \text {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{b d} \\ & = \frac {(c+d x)^2}{2 b d}+\frac {a^{2/3} \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 b^{4/3} d}-\frac {a^{2/3} \text {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 {(c+d x)^2}{2 b d}+\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} \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 b^{5/3} d}-\frac {a \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 b^{4/3} d} \\ & = \frac {(c+d x)^2}{2 b d}+\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} \text {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 {(c+d x)^2}{2 b d}+\frac {a^{2/3} \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} \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.02 \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {(c+d x)^2}{2 b d}-\frac {a^{2/3} \arctan \left (\frac {-\sqrt [3]{a}+2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d} \]

[In]

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

[Out]

(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)

Maple [C] (verified)

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

Time = 4.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.59

method	result	size
default	\(\frac {c x +\frac {1}{2} d \,x^{2}}{b}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (\textit {\_R} d +c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right ) a}{3 b^{2} d}\)	\(92\)
risch	\(\frac {c x}{b}+\frac {d \,x^{2}}{2 b}+\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,d^{3} \textit {\_Z}^{3}+3 b c \,d^{2} \textit {\_Z}^{2}+3 b \,c^{2} d \textit {\_Z} +c^{3} b +a \right )}{\sum }\frac {\left (-\textit {\_R} d -c \right ) \ln \left (x -\textit {\_R} \right )}{d^{2} \textit {\_R}^{2}+2 c d \textit {\_R} +c^{2}}\right )}{3 b^{2} d}\)	\(96\)

[In]

int((d*x+c)^4/(a+b*(d*x+c)^3),x,method=_RETURNVERBOSE)

[Out]

1/b*(c*x+1/2*d*x^2)-1/3/b^2/d*sum((_R*d+c)/(_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 = 0.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.05 \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {3 \, d^{2} x^{2} + 6 \, c d x - 2 \, \sqrt {3} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b d x + b c\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a d^{2} x^{2} + 2 \, a c d x + a c^{2} - {\left (b d x + b c\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a d x + a c + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{6 \, b d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.29 \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {\operatorname {RootSum} {\left (27 t^{3} b^{5} - a^{2}, \left ( t \mapsto t \log {\left (x + \frac {9 t^{2} b^{3} + a c}{a d} \right )} \right )\right )}}{d} + \frac {c x}{b} + \frac {d x^{2}}{2 b} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\int { \frac {{\left (d x + c\right )}^{4}}{{\left (d x + c\right )}^{3} b + a} \,d x } \]

[In]

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

[Out]

-a*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*x^2 + 2*c*x)/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {b d^{7} x^{2} + 2 \, b c d^{6} x}{2 \, b^{2} d^{6}} - \frac {2 \, \sqrt {3} \left (a^{2} b d^{15}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (a^{2} b d^{15}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (a^{2} b d^{15}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (a^{2} b\right )^{\frac {2}{3}} \right |}\right )}{6 \, b^{2} d^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x)^4}{a+b (c+d x)^3} \, dx=\frac {d\,x^2}{2\,b}+\frac {c\,x}{b}+\frac {a^{2/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{3\,b^{5/3}\,d}-\frac {a^{2/3}\,\ln \left (\frac {a^2\,c\,d^4}{b}+\frac {a^2\,d^5\,x}{b}+\frac {a^{7/3}\,d^4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{5/3}\,d}+\frac {a^{2/3}\,\ln \left (\frac {a^2\,c\,d^4}{b}+\frac {a^2\,d^5\,x}{b}+\frac {9\,a^{7/3}\,d^4\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2}{b^{4/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^{5/3}\,d} \]

[In]

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

[Out]

(d*x^2)/(2*b) + (c*x)/b + (a^(2/3)*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(3*b^(5/3)*d) - (a^(2/3)*log((a^2*c
*d^4)/b + (a^2*d^5*x)/b + (a^(7/3)*d^4*((3^(1/2)*1i)/2 + 1/2)^2)/b^(4/3))*((3^(1/2)*1i)/2 + 1/2))/(3*b^(5/3)*d
) + (a^(2/3)*log((a^2*c*d^4)/b + (a^2*d^5*x)/b + (9*a^(7/3)*d^4*((3^(1/2)*1i)/6 - 1/6)^2)/b^(4/3))*((3^(1/2)*1
i)/6 - 1/6))/(b^(5/3)*d)