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

Optimal. Leaf size=84 \[ \frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}+\frac {b d^3 (c x+1)^3}{12 c}+\frac {b d^3 (c x+1)^2}{4 c}+\frac {2 b d^3 \log (1-c x)}{c}+b d^3 x \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5926, 627, 43} \[ \frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}+\frac {b d^3 (c x+1)^3}{12 c}+\frac {b d^3 (c x+1)^2}{4 c}+\frac {2 b d^3 \log (1-c x)}{c}+b d^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*d^3*x + (b*d^3*(1 + c*x)^2)/(4*c) + (b*d^3*(1 + c*x)^3)/(12*c) + (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*c
) + (2*b*d^3*Log[1 - c*x])/c

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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rubi steps

\begin {align*} \int (d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}-\frac {b \int \frac {(d+c d x)^4}{1-c^2 x^2} \, dx}{4 d}\\ &=\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}-\frac {b \int \frac {(d+c d x)^3}{\frac {1}{d}-\frac {c x}{d}} \, dx}{4 d}\\ &=\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}-\frac {b \int \left (-4 d^4+\frac {8 d^3}{\frac {1}{d}-\frac {c x}{d}}-2 d^3 (d+c d x)-d^2 (d+c d x)^2\right ) \, dx}{4 d}\\ &=b d^3 x+\frac {b d^3 (1+c x)^2}{4 c}+\frac {b d^3 (1+c x)^3}{12 c}+\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c}+\frac {2 b d^3 \log (1-c x)}{c}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 115, normalized size = 1.37 \[ \frac {d^3 \left (6 a c^4 x^4+24 a c^3 x^3+36 a c^2 x^2+24 a c x+2 b c^3 x^3+12 b c^2 x^2+6 b c x \left (c^3 x^3+4 c^2 x^2+6 c x+4\right ) \tanh ^{-1}(c x)+42 b c x+45 b \log (1-c x)+3 b \log (c x+1)\right )}{24 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(24*a*c*x + 42*b*c*x + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 24*a*c^3*x^3 + 2*b*c^3*x^3 + 6*a*c^4*x^4 + 6*b*c*x*(
4 + 6*c*x + 4*c^2*x^2 + c^3*x^3)*ArcTanh[c*x] + 45*b*Log[1 - c*x] + 3*b*Log[1 + c*x]))/(24*c)

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fricas [A]  time = 0.49, size = 149, normalized size = 1.77 \[ \frac {6 \, a c^{4} d^{3} x^{4} + 2 \, {\left (12 \, a + b\right )} c^{3} d^{3} x^{3} + 12 \, {\left (3 \, a + b\right )} c^{2} d^{3} x^{2} + 6 \, {\left (4 \, a + 7 \, b\right )} c d^{3} x + 3 \, b d^{3} \log \left (c x + 1\right ) + 45 \, b d^{3} \log \left (c x - 1\right ) + 3 \, {\left (b c^{4} d^{3} x^{4} + 4 \, b c^{3} d^{3} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c d^{3} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*a*c^4*d^3*x^4 + 2*(12*a + b)*c^3*d^3*x^3 + 12*(3*a + b)*c^2*d^3*x^2 + 6*(4*a + 7*b)*c*d^3*x + 3*b*d^3*
log(c*x + 1) + 45*b*d^3*log(c*x - 1) + 3*(b*c^4*d^3*x^4 + 4*b*c^3*d^3*x^3 + 6*b*c^2*d^3*x^2 + 4*b*c*d^3*x)*log
(-(c*x + 1)/(c*x - 1)))/c

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giac [B]  time = 0.22, size = 425, normalized size = 5.06 \[ -\frac {1}{3} \, {\left (\frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {6 \, {\left (\frac {4 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} - b d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}} - \frac {\frac {48 \, {\left (c x + 1\right )}^{3} a d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {72 \, {\left (c x + 1\right )}^{2} a d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {48 \, {\left (c x + 1\right )} a d^{3}}{c x - 1} - 12 \, a d^{3} + \frac {18 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} - \frac {45 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} + \frac {38 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} - 11 \, b d^{3}}{\frac {{\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*b*d^3*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - 6*b*d^3*log(-(c*x + 1)/(c*x - 1))/c^2 - 6*(4*(c*x + 1)^3*b*d
^3/(c*x - 1)^3 - 6*(c*x + 1)^2*b*d^3/(c*x - 1)^2 + 4*(c*x + 1)*b*d^3/(c*x - 1) - b*d^3)*log(-(c*x + 1)/(c*x -
1))/((c*x + 1)^4*c^2/(c*x - 1)^4 - 4*(c*x + 1)^3*c^2/(c*x - 1)^3 + 6*(c*x + 1)^2*c^2/(c*x - 1)^2 - 4*(c*x + 1)
*c^2/(c*x - 1) + c^2) - (48*(c*x + 1)^3*a*d^3/(c*x - 1)^3 - 72*(c*x + 1)^2*a*d^3/(c*x - 1)^2 + 48*(c*x + 1)*a*
d^3/(c*x - 1) - 12*a*d^3 + 18*(c*x + 1)^3*b*d^3/(c*x - 1)^3 - 45*(c*x + 1)^2*b*d^3/(c*x - 1)^2 + 38*(c*x + 1)*
b*d^3/(c*x - 1) - 11*b*d^3)/((c*x + 1)^4*c^2/(c*x - 1)^4 - 4*(c*x + 1)^3*c^2/(c*x - 1)^3 + 6*(c*x + 1)^2*c^2/(
c*x - 1)^2 - 4*(c*x + 1)*c^2/(c*x - 1) + c^2))*c

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maple [B]  time = 0.03, size = 162, normalized size = 1.93 \[ \frac {c^{3} d^{3} a \,x^{4}}{4}+c^{2} d^{3} a \,x^{3}+\frac {3 c \,d^{3} a \,x^{2}}{2}+a x \,d^{3}+\frac {d^{3} a}{4 c}+\frac {c^{3} d^{3} b \arctanh \left (c x \right ) x^{4}}{4}+c^{2} d^{3} b \arctanh \left (c x \right ) x^{3}+\frac {3 c \,d^{3} b \arctanh \left (c x \right ) x^{2}}{2}+d^{3} b \arctanh \left (c x \right ) x +\frac {d^{3} b \arctanh \left (c x \right )}{4 c}+\frac {c^{2} d^{3} b \,x^{3}}{12}+\frac {c \,d^{3} b \,x^{2}}{2}+\frac {7 b \,d^{3} x}{4}+\frac {2 d^{3} b \ln \left (c x -1\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^3*d^3*a*x^4+c^2*d^3*a*x^3+3/2*c*d^3*a*x^2+a*x*d^3+1/4/c*d^3*a+1/4*c^3*d^3*b*arctanh(c*x)*x^4+c^2*d^3*b*a
rctanh(c*x)*x^3+3/2*c*d^3*b*arctanh(c*x)*x^2+d^3*b*arctanh(c*x)*x+1/4/c*d^3*b*arctanh(c*x)+1/12*c^2*d^3*b*x^3+
1/2*c*d^3*b*x^2+7/4*b*d^3*x+2/c*d^3*b*ln(c*x-1)

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maxima [B]  time = 0.32, size = 219, normalized size = 2.61 \[ \frac {1}{4} \, a c^{3} d^{3} x^{4} + a c^{2} d^{3} x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{3} d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{3} + \frac {3}{2} \, a c d^{3} x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d^{3} + a d^{3} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.96, size = 136, normalized size = 1.62 \[ \frac {d^3\,\left (12\,a\,x+21\,b\,x+12\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c^3\,d^3\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12}-\frac {d^3\,\left (21\,b\,\mathrm {atanh}\left (c\,x\right )-12\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12\,c}+\frac {c\,d^3\,\left (18\,a\,x^2+6\,b\,x^2+18\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c^2\,d^3\,\left (12\,a\,x^3+b\,x^3+12\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^3*(12*a*x + 21*b*x + 12*b*x*atanh(c*x)))/12 + (c^3*d^3*(3*a*x^4 + 3*b*x^4*atanh(c*x)))/12 - (d^3*(21*b*atan
h(c*x) - 12*b*log(c^2*x^2 - 1)))/(12*c) + (c*d^3*(18*a*x^2 + 6*b*x^2 + 18*b*x^2*atanh(c*x)))/12 + (c^2*d^3*(12
*a*x^3 + b*x^3 + 12*b*x^3*atanh(c*x)))/12

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sympy [A]  time = 1.31, size = 182, normalized size = 2.17 \[ \begin {cases} \frac {a c^{3} d^{3} x^{4}}{4} + a c^{2} d^{3} x^{3} + \frac {3 a c d^{3} x^{2}}{2} + a d^{3} x + \frac {b c^{3} d^{3} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + b c^{2} d^{3} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{2} d^{3} x^{3}}{12} + \frac {3 b c d^{3} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b c d^{3} x^{2}}{2} + b d^{3} x \operatorname {atanh}{\left (c x \right )} + \frac {7 b d^{3} x}{4} + \frac {2 b d^{3} \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{4 c} & \text {for}\: c \neq 0 \\a d^{3} x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*d**3*x**4/4 + a*c**2*d**3*x**3 + 3*a*c*d**3*x**2/2 + a*d**3*x + b*c**3*d**3*x**4*atanh(c*x)/
4 + b*c**2*d**3*x**3*atanh(c*x) + b*c**2*d**3*x**3/12 + 3*b*c*d**3*x**2*atanh(c*x)/2 + b*c*d**3*x**2/2 + b*d**
3*x*atanh(c*x) + 7*b*d**3*x/4 + 2*b*d**3*log(x - 1/c)/c + b*d**3*atanh(c*x)/(4*c), Ne(c, 0)), (a*d**3*x, True)
)

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