3.10 \(\int (c e+d e x)^2 (a+b \tanh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=69 \[ \frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}+\frac {b e^2 (c+d x)^2}{6 d}+\frac {b e^2 \log \left (1-(c+d x)^2\right )}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 69, 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, 266, 43} \[ \frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}+\frac {b e^2 (c+d x)^2}{6 d}+\frac {b e^2 \log \left (1-(c+d x)^2\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2*(a + b*ArcTanh[c + d*x]),x]

[Out]

(b*e^2*(c + d*x)^2)/(6*d) + (e^2*(c + d*x)^3*(a + b*ArcTanh[c + d*x]))/(3*d) + (b*e^2*Log[1 - (c + d*x)^2])/(6
*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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)^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \tanh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{1-x^2} \, dx,x,c+d x\right )}{3 d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x}{1-x} \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \left (-1+\frac {1}{1-x}\right ) \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac {b e^2 (c+d x)^2}{6 d}+\frac {e^2 (c+d x)^3 \left (a+b \tanh ^{-1}(c+d x)\right )}{3 d}+\frac {b e^2 \log \left (1-(c+d x)^2\right )}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 59, normalized size = 0.86 \[ \frac {e^2 \left ((c+d x)^2 (2 a (c+d x)+b)+b \log \left (1-(c+d x)^2\right )+2 b (c+d x)^3 \tanh ^{-1}(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^2*(a + b*ArcTanh[c + d*x]),x]

[Out]

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

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fricas [B]  time = 0.50, size = 144, normalized size = 2.09 \[ \frac {2 \, a d^{3} e^{2} x^{3} + {\left (6 \, a c + b\right )} d^{2} e^{2} x^{2} + 2 \, {\left (3 \, a c^{2} + b c\right )} d e^{2} x + {\left (b c^{3} + b\right )} e^{2} \log \left (d x + c + 1\right ) - {\left (b c^{3} - b\right )} e^{2} \log \left (d x + c - 1\right ) + {\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(a+b*arctanh(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.34, size = 364, normalized size = 5.28 \[ -\frac {{\left (\frac {{\left (d x + c + 1\right )}^{3} b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{{\left (d x + c - 1\right )}^{3}} - \frac {3 \, {\left (d x + c + 1\right )}^{2} b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{{\left (d x + c - 1\right )}^{2}} + \frac {3 \, {\left (d x + c + 1\right )} b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{d x + c - 1} - b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right ) - \frac {{\left (d x + c + 1\right )}^{3} b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{{\left (d x + c - 1\right )}^{3}} - \frac {3 \, {\left (d x + c + 1\right )} b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} - \frac {6 \, {\left (d x + c + 1\right )}^{2} a e^{2}}{{\left (d x + c - 1\right )}^{2}} - 2 \, a e^{2} - \frac {2 \, {\left (d x + c + 1\right )}^{2} b e^{2}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} b e^{2}}{d x + c - 1}\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )}}{6 \, {\left (\frac {{\left (d x + c + 1\right )}^{3} d^{2}}{{\left (d x + c - 1\right )}^{3}} - \frac {3 \, {\left (d x + c + 1\right )}^{2} d^{2}}{{\left (d x + c - 1\right )}^{2}} + \frac {3 \, {\left (d x + c + 1\right )} d^{2}}{d x + c - 1} - d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(a+b*arctanh(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.03, size = 174, normalized size = 2.52 \[ \frac {d^{2} x^{3} a \,e^{2}}{3}+d \,x^{2} a c \,e^{2}+x a \,c^{2} e^{2}+\frac {a \,c^{3} e^{2}}{3 d}+\frac {d^{2} \arctanh \left (d x +c \right ) x^{3} b \,e^{2}}{3}+d \arctanh \left (d x +c \right ) x^{2} b c \,e^{2}+\arctanh \left (d x +c \right ) x b \,c^{2} e^{2}+\frac {\arctanh \left (d x +c \right ) b \,c^{3} e^{2}}{3 d}+\frac {d \,x^{2} b \,e^{2}}{6}+\frac {x b c \,e^{2}}{3}+\frac {b \,c^{2} e^{2}}{6 d}+\frac {e^{2} b \ln \left (d x +c -1\right )}{6 d}+\frac {e^{2} b \ln \left (d x +c +1\right )}{6 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2*(a+b*arctanh(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.32, size = 225, normalized size = 3.26 \[ \frac {1}{3} \, a d^{2} e^{2} x^{3} + a c d e^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b c d e^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} + \frac {{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac {{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} b d^{2} e^{2} + a c^{2} e^{2} x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b c^{2} e^{2}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(a+b*arctanh(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.62, size = 237, normalized size = 3.43 \[ \frac {a\,d^2\,e^2\,x^3}{3}+\frac {b\,c\,e^2\,x}{3}+\frac {b\,e^2\,\ln \left (c+d\,x-1\right )}{6\,d}+\frac {b\,e^2\,\ln \left (c+d\,x+1\right )}{6\,d}+a\,c^2\,e^2\,x+\frac {b\,d\,e^2\,x^2}{6}+a\,c\,d\,e^2\,x^2+\frac {b\,c^2\,e^2\,x\,\ln \left (c+d\,x+1\right )}{2}-\frac {b\,c^3\,e^2\,\ln \left (c+d\,x-1\right )}{6\,d}+\frac {b\,c^3\,e^2\,\ln \left (c+d\,x+1\right )}{6\,d}-\frac {b\,c^2\,e^2\,x\,\ln \left (1-d\,x-c\right )}{2}+\frac {b\,d^2\,e^2\,x^3\,\ln \left (c+d\,x+1\right )}{6}-\frac {b\,d^2\,e^2\,x^3\,\ln \left (1-d\,x-c\right )}{6}+\frac {b\,c\,d\,e^2\,x^2\,\ln \left (c+d\,x+1\right )}{2}-\frac {b\,c\,d\,e^2\,x^2\,\ln \left (1-d\,x-c\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^2*(a + b*atanh(c + d*x)),x)

[Out]

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

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sympy [A]  time = 3.03, size = 180, normalized size = 2.61 \[ \begin {cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac {a d^{2} e^{2} x^{3}}{3} + \frac {b c^{3} e^{2} \operatorname {atanh}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname {atanh}{\left (c + d x \right )} + b c d e^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )} + \frac {b c e^{2} x}{3} + \frac {b d^{2} e^{2} x^{3} \operatorname {atanh}{\left (c + d x \right )}}{3} + \frac {b d e^{2} x^{2}}{6} + \frac {b e^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{3 d} - \frac {b e^{2} \operatorname {atanh}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {atanh}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**2*(a+b*atanh(d*x+c)),x)

[Out]

Piecewise((a*c**2*e**2*x + a*c*d*e**2*x**2 + a*d**2*e**2*x**3/3 + b*c**3*e**2*atanh(c + d*x)/(3*d) + b*c**2*e*
*2*x*atanh(c + d*x) + b*c*d*e**2*x**2*atanh(c + d*x) + b*c*e**2*x/3 + b*d**2*e**2*x**3*atanh(c + d*x)/3 + b*d*
e**2*x**2/6 + b*e**2*log(c/d + x + 1/d)/(3*d) - b*e**2*atanh(c + d*x)/(3*d), Ne(d, 0)), (c**2*e**2*x*(a + b*at
anh(c)), True))

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