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

Optimal. Leaf size=48 \[ \frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d}-\frac {b e \tanh ^{-1}(c+d x)}{2 d}+\frac {b e x}{2} \]

[Out]

1/2*b*e*x-1/2*b*e*arctanh(d*x+c)/d+1/2*e*(d*x+c)^2*(a+b*arctanh(d*x+c))/d

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Rubi [A]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6107, 12, 5916, 321, 206} \[ \frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d}-\frac {b e \tanh ^{-1}(c+d x)}{2 d}+\frac {b e x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*x)/2 - (b*e*ArcTanh[c + d*x])/(2*d) + (e*(c + d*x)^2*(a + b*ArcTanh[c + d*x]))/(2*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 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 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) \left (a+b \tanh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e x \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e \operatorname {Subst}\left (\int x \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d}-\frac {(b e) \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {b e x}{2}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d}-\frac {(b e) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac {b e x}{2}-\frac {b e \tanh ^{-1}(c+d x)}{2 d}+\frac {e (c+d x)^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 77, normalized size = 1.60 \[ \frac {e \left (2 a c^2+4 a c d x+2 a d^2 x^2+b \log (-c-d x+1)-b \log (c+d x+1)+2 b (c+d x)^2 \tanh ^{-1}(c+d x)+2 b c+2 b d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.56, size = 73, normalized size = 1.52 \[ \frac {2 \, a d^{2} e x^{2} + 2 \, {\left (2 \, a c + b\right )} d e x + {\left (b d^{2} e x^{2} + 2 \, b c d e x + {\left (b c^{2} - b\right )} e\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.39, size = 137, normalized size = 2.85 \[ \frac {{\left (\frac {{\left (d x + c + 1\right )} b e \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {2 \, {\left (d x + c + 1\right )} a e}{d x + c - 1} + \frac {{\left (d x + c + 1\right )} b e}{d x + c - 1} - b e\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )}}{2 \, {\left (\frac {{\left (d x + c + 1\right )}^{2} d^{2}}{{\left (d x + c - 1\right )}^{2}} - \frac {2 \, {\left (d x + c + 1\right )} d^{2}}{d x + c - 1} + d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.03, size = 107, normalized size = 2.23 \[ \frac {a d e \,x^{2}}{2}+x a c e +\frac {a \,c^{2} e}{2 d}+\frac {d \arctanh \left (d x +c \right ) x^{2} b e}{2}+\arctanh \left (d x +c \right ) x b c e +\frac {\arctanh \left (d x +c \right ) b \,c^{2} e}{2 d}+\frac {b e x}{2}+\frac {b e c}{2 d}+\frac {b e \ln \left (d x +c -1\right )}{4 d}-\frac {b e \ln \left (d x +c +1\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*d*e*x^2+x*a*c*e+1/2/d*a*c^2*e+1/2*d*arctanh(d*x+c)*x^2*b*e+arctanh(d*x+c)*x*b*c*e+1/2/d*arctanh(d*x+c)*b
*c^2*e+1/2*b*e*x+1/2/d*b*e*c+1/4/d*b*e*ln(d*x+c-1)-1/4/d*b*e*ln(d*x+c+1)

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maxima [B]  time = 0.31, size = 113, normalized size = 2.35 \[ \frac {1}{2} \, a d e x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b d e + a c e x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b c e}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.84, size = 73, normalized size = 1.52 \[ \frac {b\,e\,x}{2}+a\,c\,e\,x-\frac {b\,e\,\mathrm {atanh}\left (c+d\,x\right )}{2\,d}+\frac {a\,d\,e\,x^2}{2}+\frac {b\,c^2\,e\,\mathrm {atanh}\left (c+d\,x\right )}{2\,d}+b\,c\,e\,x\,\mathrm {atanh}\left (c+d\,x\right )+\frac {b\,d\,e\,x^2\,\mathrm {atanh}\left (c+d\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e*x)/2 + a*c*e*x - (b*e*atanh(c + d*x))/(2*d) + (a*d*e*x^2)/2 + (b*c^2*e*atanh(c + d*x))/(2*d) + b*c*e*x*at
anh(c + d*x) + (b*d*e*x^2*atanh(c + d*x))/2

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sympy [A]  time = 1.66, size = 95, normalized size = 1.98 \[ \begin {cases} a c e x + \frac {a d e x^{2}}{2} + \frac {b c^{2} e \operatorname {atanh}{\left (c + d x \right )}}{2 d} + b c e x \operatorname {atanh}{\left (c + d x \right )} + \frac {b d e x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2} + \frac {b e x}{2} - \frac {b e \operatorname {atanh}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\c e x \left (a + b \operatorname {atanh}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c*e*x + a*d*e*x**2/2 + b*c**2*e*atanh(c + d*x)/(2*d) + b*c*e*x*atanh(c + d*x) + b*d*e*x**2*atanh(
c + d*x)/2 + b*e*x/2 - b*e*atanh(c + d*x)/(2*d), Ne(d, 0)), (c*e*x*(a + b*atanh(c)), True))

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