3.2862 \(\int \frac {c+d x}{a+b (c+d x)^3} \, dx\)

Optimal. Leaf size=140 \[ \frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 \sqrt [3]{a} b^{2/3} d}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3} d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {372, 292, 31, 634, 617, 204, 628} \[ \frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 \sqrt [3]{a} b^{2/3} d}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3} d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*(c + d*x)^3),x]

[Out]

-(ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))]/(Sqrt[3]*a^(1/3)*b^(2/3)*d)) - Log[a^(1/3) + b^(1/
3)*(c + d*x)]/(3*a^(1/3)*b^(2/3)*d) + Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2]/(6*a^(1/3
)*b^(2/3)*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 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+d x}{a+b (c+d x)^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 \sqrt [3]{a} \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 )}{3 \sqrt [3]{a} \sqrt [3]{b} d}\\ &=-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} 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 )}{6 \sqrt [3]{a} 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 )}{2 \sqrt [3]{b} d}\\ &=-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} 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 )}{6 \sqrt [3]{a} 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 )}{\sqrt [3]{a} b^{2/3} d}\\ &=-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3} d}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 \sqrt [3]{a} 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 )}{6 \sqrt [3]{a} b^{2/3} d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 114, normalized size = 0.81 \[ \frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt [3]{a} b^{2/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 452, normalized size = 3.23 \[ \left [\frac {3 \, \sqrt {\frac {1}{3}} a b \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} d^{3} x^{3} + 6 \, b^{2} c d^{2} x^{2} + 6 \, b^{2} c^{2} d x + 2 \, b^{2} c^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b d x + a b c + 2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} {\left (d x + c\right )}}{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}\right ) + \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} {\left (b d x + b c\right )} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a b^{2} d}, \frac {6 \, \sqrt {\frac {1}{3}} a b \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b d x + 2 \, b c + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} {\left (b d x + b c\right )} + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b d x + b c - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a b^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 141, normalized size = 1.01 \[ -\frac {1}{3} \, \sqrt {3} \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) - \frac {1}{6} \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) + \frac {1}{3} \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.00, size = 76, normalized size = 0.54 \[ \frac {\left (\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right ) d +c \right ) \ln \left (-\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+x \right )}{3 b d \left (d^{2} \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )^{2}+2 c d \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+c^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/b/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 \[ \int \frac {d x + c}{{\left (d x + c\right )}^{3} b + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.44, size = 150, normalized size = 1.07 \[ \frac {\ln \left (b^{1/3}\,c-{\left (-a\right )}^{1/3}+b^{1/3}\,d\,x\right )}{3\,{\left (-a\right )}^{1/3}\,b^{2/3}\,d}+\frac {\ln \left (b\,c\,d^4+b\,d^5\,x-\frac {{\left (-a\right )}^{1/3}\,b^{2/3}\,d^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,{\left (-a\right )}^{1/3}\,b^{2/3}\,d}-\frac {\ln \left (b\,c\,d^4+b\,d^5\,x-\frac {{\left (-a\right )}^{1/3}\,b^{2/3}\,d^4\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,{\left (-a\right )}^{1/3}\,b^{2/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*(c + d*x)^3),x)

[Out]

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

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sympy [A]  time = 0.26, size = 29, normalized size = 0.21 \[ \frac {\operatorname {RootSum} {\left (27 t^{3} a b^{2} + 1, \left (t \mapsto t \log {\left (x + \frac {9 t^{2} a b + c}{d} \right )} \right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(27*_t**3*a*b**2 + 1, Lambda(_t, _t*log(x + (9*_t**2*a*b + c)/d)))/d

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