3.9 \(\int (c e+d e x)^3 (a+b \tanh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=72 \[ \frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}+\frac {b e^3 (c+d x)^3}{12 d}-\frac {b e^3 \tanh ^{-1}(c+d x)}{4 d}+\frac {1}{4} b e^3 x \]

[Out]

1/4*b*e^3*x+1/12*b*e^3*(d*x+c)^3/d-1/4*b*e^3*arctanh(d*x+c)/d+1/4*e^3*(d*x+c)^4*(a+b*arctanh(d*x+c))/d

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Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6107, 12, 5916, 302, 206} \[ \frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}+\frac {b e^3 (c+d x)^3}{12 d}-\frac {b e^3 \tanh ^{-1}(c+d x)}{4 d}+\frac {1}{4} b e^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^3*x)/4 + (b*e^3*(c + d*x)^3)/(12*d) - (b*e^3*ArcTanh[c + d*x])/(4*d) + (e^3*(c + d*x)^4*(a + b*ArcTanh[c
+ d*x]))/(4*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int (c e+d e x)^3 \left (a+b \tanh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e^3 x^3 \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int x^3 \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,c+d x\right )}{4 d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,c+d x\right )}{4 d}\\ &=\frac {1}{4} b e^3 x+\frac {b e^3 (c+d x)^3}{12 d}+\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,c+d x\right )}{4 d}\\ &=\frac {1}{4} b e^3 x+\frac {b e^3 (c+d x)^3}{12 d}-\frac {b e^3 \tanh ^{-1}(c+d x)}{4 d}+\frac {e^3 (c+d x)^4 \left (a+b \tanh ^{-1}(c+d x)\right )}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 78, normalized size = 1.08 \[ \frac {e^3 \left (6 a (c+d x)^4+2 b (c+d x)^3+6 b (c+d x)+3 b \log (-c-d x+1)-3 b \log (c+d x+1)+6 b (c+d x)^4 \tanh ^{-1}(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(6*b*(c + d*x) + 2*b*(c + d*x)^3 + 6*a*(c + d*x)^4 + 6*b*(c + d*x)^4*ArcTanh[c + d*x] + 3*b*Log[1 - c - d
*x] - 3*b*Log[1 + c + d*x]))/(24*d)

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fricas [B]  time = 0.56, size = 159, normalized size = 2.21 \[ \frac {6 \, a d^{4} e^{3} x^{4} + 2 \, {\left (12 \, a c + b\right )} d^{3} e^{3} x^{3} + 6 \, {\left (6 \, a c^{2} + b c\right )} d^{2} e^{3} x^{2} + 6 \, {\left (4 \, a c^{3} + b c^{2} + b\right )} d e^{3} x + 3 \, {\left (b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 4 \, b c^{3} d e^{3} x + {\left (b c^{4} - b\right )} e^{3}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*a*d^4*e^3*x^4 + 2*(12*a*c + b)*d^3*e^3*x^3 + 6*(6*a*c^2 + b*c)*d^2*e^3*x^2 + 6*(4*a*c^3 + b*c^2 + b)*d
*e^3*x + 3*(b*d^4*e^3*x^4 + 4*b*c*d^3*e^3*x^3 + 6*b*c^2*d^2*e^3*x^2 + 4*b*c^3*d*e^3*x + (b*c^4 - b)*e^3)*log(-
(d*x + c + 1)/(d*x + c - 1)))/d

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giac [B]  time = 3.08, size = 282, normalized size = 3.92 \[ \frac {{\left (\frac {3 \, {\left (d x + c + 1\right )}^{3} b e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{3}} + \frac {3 \, {\left (d x + c + 1\right )} b e^{3} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {6 \, {\left (d x + c + 1\right )}^{3} a e^{3}}{{\left (d x + c - 1\right )}^{3}} + \frac {6 \, {\left (d x + c + 1\right )} a e^{3}}{d x + c - 1} + \frac {3 \, {\left (d x + c + 1\right )}^{3} b e^{3}}{{\left (d x + c - 1\right )}^{3}} - \frac {6 \, {\left (d x + c + 1\right )}^{2} b e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {5 \, {\left (d x + c + 1\right )} b e^{3}}{d x + c - 1} - 2 \, b e^{3}\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )}}{6 \, {\left (\frac {{\left (d x + c + 1\right )}^{4} d^{2}}{{\left (d x + c - 1\right )}^{4}} - \frac {4 \, {\left (d x + c + 1\right )}^{3} d^{2}}{{\left (d x + c - 1\right )}^{3}} + \frac {6 \, {\left (d x + c + 1\right )}^{2} d^{2}}{{\left (d x + c - 1\right )}^{2}} - \frac {4 \, {\left (d x + c + 1\right )} d^{2}}{d x + c - 1} + d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 242, normalized size = 3.36 \[ \frac {d^{3} x^{4} a \,e^{3}}{4}+d^{2} x^{3} a c \,e^{3}+\frac {3 d \,x^{2} a \,c^{2} e^{3}}{2}+x a \,c^{3} e^{3}+\frac {a \,c^{4} e^{3}}{4 d}+\frac {d^{3} \arctanh \left (d x +c \right ) x^{4} b \,e^{3}}{4}+d^{2} \arctanh \left (d x +c \right ) x^{3} b c \,e^{3}+\frac {3 d \arctanh \left (d x +c \right ) x^{2} b \,c^{2} e^{3}}{2}+\arctanh \left (d x +c \right ) x b \,c^{3} e^{3}+\frac {\arctanh \left (d x +c \right ) b \,c^{4} e^{3}}{4 d}+\frac {d^{2} x^{3} b \,e^{3}}{12}+\frac {d \,x^{2} b c \,e^{3}}{4}+\frac {x b \,c^{2} e^{3}}{4}+\frac {b \,c^{3} e^{3}}{12 d}+\frac {b \,e^{3} x}{4}+\frac {b c \,e^{3}}{4 d}+\frac {e^{3} b \ln \left (d x +c -1\right )}{8 d}-\frac {e^{3} b \ln \left (d x +c +1\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^3*x^4*a*e^3+d^2*x^3*a*c*e^3+3/2*d*x^2*a*c^2*e^3+x*a*c^3*e^3+1/4/d*a*c^4*e^3+1/4*d^3*arctanh(d*x+c)*x^4*b
*e^3+d^2*arctanh(d*x+c)*x^3*b*c*e^3+3/2*d*arctanh(d*x+c)*x^2*b*c^2*e^3+arctanh(d*x+c)*x*b*c^3*e^3+1/4/d*arctan
h(d*x+c)*b*c^4*e^3+1/12*d^2*x^3*b*e^3+1/4*d*x^2*b*c*e^3+1/4*x*b*c^2*e^3+1/12/d*b*c^3*e^3+1/4*b*e^3*x+1/4/d*b*c
*e^3+1/8/d*e^3*b*ln(d*x+c-1)-1/8/d*e^3*b*ln(d*x+c+1)

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maxima [B]  time = 0.33, size = 357, normalized size = 4.96 \[ \frac {1}{4} \, a d^{3} e^{3} x^{4} + a c d^{2} e^{3} x^{3} + \frac {3}{2} \, a c^{2} d e^{3} x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b c^{2} d e^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} + \frac {{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac {{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} b c d^{2} e^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, {\left (d^{2} x^{3} - 3 \, c d x^{2} + 3 \, {\left (3 \, c^{2} + 1\right )} x\right )}}{d^{4}} - \frac {3 \, {\left (c^{4} + 4 \, c^{3} + 6 \, c^{2} + 4 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{5}} + \frac {3 \, {\left (c^{4} - 4 \, c^{3} + 6 \, c^{2} - 4 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{5}}\right )}\right )} b d^{3} e^{3} + a c^{3} e^{3} x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b c^{3} e^{3}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*d^3*e^3*x^4 + a*c*d^2*e^3*x^3 + 3/2*a*c^2*d*e^3*x^2 + 3/4*(2*x^2*arctanh(d*x + c) + d*(2*x/d^2 - (c^2 +
2*c + 1)*log(d*x + c + 1)/d^3 + (c^2 - 2*c + 1)*log(d*x + c - 1)/d^3))*b*c^2*d*e^3 + 1/2*(2*x^3*arctanh(d*x +
c) + d*((d*x^2 - 4*c*x)/d^3 + (c^3 + 3*c^2 + 3*c + 1)*log(d*x + c + 1)/d^4 - (c^3 - 3*c^2 + 3*c - 1)*log(d*x +
c - 1)/d^4))*b*c*d^2*e^3 + 1/24*(6*x^4*arctanh(d*x + c) + d*(2*(d^2*x^3 - 3*c*d*x^2 + 3*(3*c^2 + 1)*x)/d^4 -
3*(c^4 + 4*c^3 + 6*c^2 + 4*c + 1)*log(d*x + c + 1)/d^5 + 3*(c^4 - 4*c^3 + 6*c^2 - 4*c + 1)*log(d*x + c - 1)/d^
5))*b*d^3*e^3 + a*c^3*e^3*x + 1/2*(2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1))*b*c^3*e^3/d

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mupad [B]  time = 1.42, size = 414, normalized size = 5.75 \[ x^3\,\left (\frac {d^2\,e^3\,\left (b+20\,a\,c\right )}{12}-\frac {2\,a\,c\,d^2\,e^3}{3}\right )+\ln \left (c+d\,x+1\right )\,\left (\frac {b\,c^3\,e^3\,x}{2}+\frac {3\,b\,c^2\,d\,e^3\,x^2}{4}+\frac {b\,c\,d^2\,e^3\,x^3}{2}+\frac {b\,d^3\,e^3\,x^4}{8}\right )-\ln \left (1-d\,x-c\right )\,\left (\frac {b\,c^3\,e^3\,x}{2}+\frac {3\,b\,c^2\,d\,e^3\,x^2}{4}+\frac {b\,c\,d^2\,e^3\,x^3}{2}+\frac {b\,d^3\,e^3\,x^4}{8}\right )-x^2\,\left (\frac {c\,\left (\frac {d^2\,e^3\,\left (b+20\,a\,c\right )}{4}-2\,a\,c\,d^2\,e^3\right )}{d}-\frac {d\,e^3\,\left (10\,a\,c^2+b\,c-a\right )}{2}+\frac {a\,d\,e^3\,\left (4\,c^2-4\right )}{8}\right )+x\,\left (\frac {c\,e^3\,\left (20\,a\,c^2+3\,b\,c-6\,a\right )}{2}-\frac {\left (4\,c^2-4\right )\,\left (\frac {d^2\,e^3\,\left (b+20\,a\,c\right )}{4}-2\,a\,c\,d^2\,e^3\right )}{4\,d^2}+\frac {2\,c\,\left (\frac {2\,c\,\left (\frac {d^2\,e^3\,\left (b+20\,a\,c\right )}{4}-2\,a\,c\,d^2\,e^3\right )}{d}-d\,e^3\,\left (10\,a\,c^2+b\,c-a\right )+\frac {a\,d\,e^3\,\left (4\,c^2-4\right )}{4}\right )}{d}\right )+\frac {\ln \left (c+d\,x-1\right )\,\left (b\,e^3-b\,c^4\,e^3\right )}{8\,d}+\frac {a\,d^3\,e^3\,x^4}{4}+\frac {b\,e^3\,\ln \left (c+d\,x+1\right )\,\left (c^2+1\right )\,\left (c-1\right )\,\left (c+1\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*((d^2*e^3*(b + 20*a*c))/12 - (2*a*c*d^2*e^3)/3) + log(c + d*x + 1)*((b*d^3*e^3*x^4)/8 + (b*c^3*e^3*x)/2 +
(3*b*c^2*d*e^3*x^2)/4 + (b*c*d^2*e^3*x^3)/2) - log(1 - d*x - c)*((b*d^3*e^3*x^4)/8 + (b*c^3*e^3*x)/2 + (3*b*c^
2*d*e^3*x^2)/4 + (b*c*d^2*e^3*x^3)/2) - x^2*((c*((d^2*e^3*(b + 20*a*c))/4 - 2*a*c*d^2*e^3))/d - (d*e^3*(b*c -
a + 10*a*c^2))/2 + (a*d*e^3*(4*c^2 - 4))/8) + x*((c*e^3*(3*b*c - 6*a + 20*a*c^2))/2 - ((4*c^2 - 4)*((d^2*e^3*(
b + 20*a*c))/4 - 2*a*c*d^2*e^3))/(4*d^2) + (2*c*((2*c*((d^2*e^3*(b + 20*a*c))/4 - 2*a*c*d^2*e^3))/d - d*e^3*(b
*c - a + 10*a*c^2) + (a*d*e^3*(4*c^2 - 4))/4))/d) + (log(c + d*x - 1)*(b*e^3 - b*c^4*e^3))/(8*d) + (a*d^3*e^3*
x^4)/4 + (b*e^3*log(c + d*x + 1)*(c^2 + 1)*(c - 1)*(c + 1))/(8*d)

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sympy [A]  time = 4.27, size = 231, normalized size = 3.21 \[ \begin {cases} a c^{3} e^{3} x + \frac {3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac {a d^{3} e^{3} x^{4}}{4} + \frac {b c^{4} e^{3} \operatorname {atanh}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname {atanh}{\left (c + d x \right )} + \frac {3 b c^{2} d e^{3} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2} + \frac {b c^{2} e^{3} x}{4} + b c d^{2} e^{3} x^{3} \operatorname {atanh}{\left (c + d x \right )} + \frac {b c d e^{3} x^{2}}{4} + \frac {b d^{3} e^{3} x^{4} \operatorname {atanh}{\left (c + d x \right )}}{4} + \frac {b d^{2} e^{3} x^{3}}{12} + \frac {b e^{3} x}{4} - \frac {b e^{3} \operatorname {atanh}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {atanh}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*e**3*x + 3*a*c**2*d*e**3*x**2/2 + a*c*d**2*e**3*x**3 + a*d**3*e**3*x**4/4 + b*c**4*e**3*atan
h(c + d*x)/(4*d) + b*c**3*e**3*x*atanh(c + d*x) + 3*b*c**2*d*e**3*x**2*atanh(c + d*x)/2 + b*c**2*e**3*x/4 + b*
c*d**2*e**3*x**3*atanh(c + d*x) + b*c*d*e**3*x**2/4 + b*d**3*e**3*x**4*atanh(c + d*x)/4 + b*d**2*e**3*x**3/12
+ b*e**3*x/4 - b*e**3*atanh(c + d*x)/(4*d), Ne(d, 0)), (c**3*e**3*x*(a + b*atanh(c)), True))

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