3.2859 \(\int \frac{(c+d x)^4}{a+b (c+d x)^3} \, dx\)

Optimal. Leaf size=156 \[ \frac{a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac{a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}+\frac{a^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{5/3} d}+\frac{(c+d x)^2}{2 b d} \]

[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)

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Rubi [A]  time = 0.166688, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {372, 321, 292, 31, 634, 617, 204, 628} \[ \frac{a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac{a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}+\frac{a^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{5/3} d}+\frac{(c+d x)^2}{2 b d} \]

Antiderivative was successfully verified.

[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 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 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 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 31

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^4}{a+b (c+d x)^3} \, dx &=\frac{\operatorname{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 \operatorname{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} \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{a^{2/3} \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{(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} \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{a \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{(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} \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{(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]  time = 0.0485128, size = 159, normalized size = 1.02 \[ \frac{a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{5/3} d}-\frac{a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{5/3} d}-\frac{a^{2/3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{5/3} d}+\frac{(c+d x)^2}{2 b d} \]

Antiderivative was successfully verified.

[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)

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Maple [C]  time = 0.005, size = 93, normalized size = 0.6 \begin{align*}{\frac{d{x}^{2}}{2\,b}}+{\frac{cx}{b}}-{\frac{a}{3\,{b}^{2}d}\sum _{{\it \_R}={\it RootOf} \left ( b{d}^{3}{{\it \_Z}}^{3}+3\,bc{d}^{2}{{\it \_Z}}^{2}+3\,b{c}^{2}d{\it \_Z}+b{c}^{3}+a \right ) }{\frac{ \left ({\it \_R}\,d+c \right ) \ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*x^2*d+1/b*x*c-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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\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}{b} + \frac{d x^{2} + 2 \, c x}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.54433, size = 393, normalized size = 2.52 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.517771, size = 46, normalized size = 0.29 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.14198, size = 286, normalized size = 1.83 \begin{align*} -\frac{1}{3} \, \sqrt{3} \left (\frac{a^{2}}{b^{5} d^{3}}\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, a b^{2} d^{7} x + 2 \, a b^{2} c d^{6} - \left (a^{2} b\right )^{\frac{2}{3}} b d^{6}\right )}}{3 \, \left (a^{2} b\right )^{\frac{2}{3}} b d^{6}}\right ) - \frac{1}{6} \, \left (\frac{a^{2}}{b^{5} d^{3}}\right )^{\frac{1}{3}} \log \left (3 \, \left (a^{2} b\right )^{\frac{4}{3}} b^{2} d^{12} +{\left (2 \, a b^{2} d^{7} x + 2 \, a b^{2} c d^{6} - \left (a^{2} b\right )^{\frac{2}{3}} b d^{6}\right )}^{2}\right ) + \frac{1}{3} \, \left (\frac{a^{2}}{b^{5} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | a b^{3} d^{7} x + a b^{3} c d^{6} + \left (a^{2} b\right )^{\frac{2}{3}} b^{2} d^{6} \right |}\right ) + \frac{b d^{7} x^{2} + 2 \, b c d^{6} x}{2 \, b^{2} d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*(a^2/(b^5*d^3))^(1/3)*arctan(1/3*sqrt(3)*(2*a*b^2*d^7*x + 2*a*b^2*c*d^6 - (a^2*b)^(2/3)*b*d^6)/((
a^2*b)^(2/3)*b*d^6)) - 1/6*(a^2/(b^5*d^3))^(1/3)*log(3*(a^2*b)^(4/3)*b^2*d^12 + (2*a*b^2*d^7*x + 2*a*b^2*c*d^6
- (a^2*b)^(2/3)*b*d^6)^2) + 1/3*(a^2/(b^5*d^3))^(1/3)*log(abs(a*b^3*d^7*x + a*b^3*c*d^6 + (a^2*b)^(2/3)*b^2*d
^6)) + 1/2*(b*d^7*x^2 + 2*b*c*d^6*x)/(b^2*d^6)