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]

(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)

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Rubi [A]  time = 0.135692, antiderivative size = 172, normalized size of antiderivative = 1., 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 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 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 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}{\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*}

Mathematica [A]  time = 0.0763395, 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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Maple [C]  time = 0.01, size = 144, normalized size = 0.8 \begin{align*}{\frac{1}{b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a} \left ({\frac{d{x}^{2}}{3\,a}}+{\frac{2\,cx}{3\,a}}+{\frac{{c}^{2}}{3\,ad}} \right ) }+{\frac{1}{9\,abd}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}b{d}^{3}+3\,{{\it \_Z}}^{2}bc{d}^{2}+3\,{\it \_Z}\,b{c}^{2}d+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)/(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(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 1.63137, size = 1897, normalized size = 11.03 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [A]  time = 1.57706, size = 105, normalized size = 0.61 \begin{align*} \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} \end{align*}

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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Giac [A]  time = 1.17771, size = 282, normalized size = 1.64 \begin{align*} -\frac{1}{9} \, \sqrt{3} \left (-\frac{1}{a^{4} 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}{18} \, \left (-\frac{1}{a^{4} 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}{9} \, \left (-\frac{1}{a^{4} b^{2} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | 3 \, a^{2} b d x + 3 \, a^{2} b c + 3 \, \left (-a^{2} b\right )^{\frac{2}{3}} a \right |}\right ) + \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} \end{align*}

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]

-1/9*sqrt(3)*(-1/(a^4*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)
) - 1/18*(-1/(a^4*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)) + 1/9*(-1/(
a^4*b^2*d^3))^(1/3)*log(abs(3*a^2*b*d*x + 3*a^2*b*c + 3*(-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)