3.34 \(\int (d+c d x)^4 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=107 \[ \frac {d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}+\frac {b d^4 (c x+1)^4}{20 c}+\frac {2 b d^4 (c x+1)^3}{15 c}+\frac {2 b d^4 (c x+1)^2}{5 c}+\frac {16 b d^4 \log (1-c x)}{5 c}+\frac {8}{5} b d^4 x \]

[Out]

8/5*b*d^4*x+2/5*b*d^4*(c*x+1)^2/c+2/15*b*d^4*(c*x+1)^3/c+1/20*b*d^4*(c*x+1)^4/c+1/5*d^4*(c*x+1)^5*(a+b*arctanh
(c*x))/c+16/5*b*d^4*ln(-c*x+1)/c

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Rubi [A]  time = 0.05, antiderivative size = 107, 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} \[ \frac {d^4 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}+\frac {b d^4 (c x+1)^4}{20 c}+\frac {2 b d^4 (c x+1)^3}{15 c}+\frac {2 b d^4 (c x+1)^2}{5 c}+\frac {16 b d^4 \log (1-c x)}{5 c}+\frac {8}{5} b d^4 x \]

Antiderivative was successfully verified.

[In]

Int[(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]

[Out]

(8*b*d^4*x)/5 + (2*b*d^4*(1 + c*x)^2)/(5*c) + (2*b*d^4*(1 + c*x)^3)/(15*c) + (b*d^4*(1 + c*x)^4)/(20*c) + (d^4
*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c) + (16*b*d^4*Log[1 - c*x])/(5*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)^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}-\frac {b \int \frac {(d+c d x)^5}{1-c^2 x^2} \, dx}{5 d}\\ &=\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}-\frac {b \int \frac {(d+c d x)^4}{\frac {1}{d}-\frac {c x}{d}} \, dx}{5 d}\\ &=\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}-\frac {b \int \left (-8 d^5+\frac {16 d^4}{\frac {1}{d}-\frac {c x}{d}}-4 d^4 (d+c d x)-2 d^3 (d+c d x)^2-d^2 (d+c d x)^3\right ) \, dx}{5 d}\\ &=\frac {8}{5} b d^4 x+\frac {2 b d^4 (1+c x)^2}{5 c}+\frac {2 b d^4 (1+c x)^3}{15 c}+\frac {b d^4 (1+c x)^4}{20 c}+\frac {d^4 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c}+\frac {16 b d^4 \log (1-c x)}{5 c}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 146, normalized size = 1.36 \[ \frac {d^4 \left (12 a c^5 x^5+60 a c^4 x^4+120 a c^3 x^3+120 a c^2 x^2+60 a c x+3 b c^4 x^4+20 b c^3 x^3+66 b c^2 x^2+6 b \log \left (1-c^2 x^2\right )+12 b c x \left (c^4 x^4+5 c^3 x^3+10 c^2 x^2+10 c x+5\right ) \tanh ^{-1}(c x)+180 b c x+180 b \log (1-c x)\right )}{60 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + c*d*x)^4*(a + b*ArcTanh[c*x]),x]

[Out]

(d^4*(60*a*c*x + 180*b*c*x + 120*a*c^2*x^2 + 66*b*c^2*x^2 + 120*a*c^3*x^3 + 20*b*c^3*x^3 + 60*a*c^4*x^4 + 3*b*
c^4*x^4 + 12*a*c^5*x^5 + 12*b*c*x*(5 + 10*c*x + 10*c^2*x^2 + 5*c^3*x^3 + c^4*x^4)*ArcTanh[c*x] + 180*b*Log[1 -
c*x] + 6*b*Log[1 - c^2*x^2]))/(60*c)

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fricas [A]  time = 0.43, size = 177, normalized size = 1.65 \[ \frac {12 \, a c^{5} d^{4} x^{5} + 3 \, {\left (20 \, a + b\right )} c^{4} d^{4} x^{4} + 20 \, {\left (6 \, a + b\right )} c^{3} d^{4} x^{3} + 6 \, {\left (20 \, a + 11 \, b\right )} c^{2} d^{4} x^{2} + 60 \, {\left (a + 3 \, b\right )} c d^{4} x + 6 \, b d^{4} \log \left (c x + 1\right ) + 186 \, b d^{4} \log \left (c x - 1\right ) + 6 \, {\left (b c^{5} d^{4} x^{5} + 5 \, b c^{4} d^{4} x^{4} + 10 \, b c^{3} d^{4} x^{3} + 10 \, b c^{2} d^{4} x^{2} + 5 \, b c d^{4} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*d^4*x^5 + 3*(20*a + b)*c^4*d^4*x^4 + 20*(6*a + b)*c^3*d^4*x^3 + 6*(20*a + 11*b)*c^2*d^4*x^2 + 6
0*(a + 3*b)*c*d^4*x + 6*b*d^4*log(c*x + 1) + 186*b*d^4*log(c*x - 1) + 6*(b*c^5*d^4*x^5 + 5*b*c^4*d^4*x^4 + 10*
b*c^3*d^4*x^3 + 10*b*c^2*d^4*x^2 + 5*b*c*d^4*x)*log(-(c*x + 1)/(c*x - 1)))/c

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giac [B]  time = 0.24, size = 526, normalized size = 4.92 \[ -\frac {4}{15} \, {\left (\frac {12 \, b d^{4} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {12 \, b d^{4} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {12 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} b d^{4}}{{\left (c x - 1\right )}^{4}} - \frac {10 \, {\left (c x + 1\right )}^{3} b d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2} b d^{4}}{{\left (c x - 1\right )}^{2}} - \frac {5 \, {\left (c x + 1\right )} b d^{4}}{c x - 1} + b d^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{2}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}} - \frac {\frac {120 \, {\left (c x + 1\right )}^{4} a d^{4}}{{\left (c x - 1\right )}^{4}} - \frac {240 \, {\left (c x + 1\right )}^{3} a d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {240 \, {\left (c x + 1\right )}^{2} a d^{4}}{{\left (c x - 1\right )}^{2}} - \frac {120 \, {\left (c x + 1\right )} a d^{4}}{c x - 1} + 24 \, a d^{4} + \frac {48 \, {\left (c x + 1\right )}^{4} b d^{4}}{{\left (c x - 1\right )}^{4}} - \frac {156 \, {\left (c x + 1\right )}^{3} b d^{4}}{{\left (c x - 1\right )}^{3}} + \frac {196 \, {\left (c x + 1\right )}^{2} b d^{4}}{{\left (c x - 1\right )}^{2}} - \frac {113 \, {\left (c x + 1\right )} b d^{4}}{c x - 1} + 25 \, b d^{4}}{\frac {{\left (c x + 1\right )}^{5} c^{2}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{2}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\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)^4*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

-4/15*(12*b*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - 12*b*d^4*log(-(c*x + 1)/(c*x - 1))/c^2 - 12*(5*(c*x + 1)^4
*b*d^4/(c*x - 1)^4 - 10*(c*x + 1)^3*b*d^4/(c*x - 1)^3 + 10*(c*x + 1)^2*b*d^4/(c*x - 1)^2 - 5*(c*x + 1)*b*d^4/(
c*x - 1) + b*d^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^2/(c*x - 1)^5 - 5*(c*x + 1)^4*c^2/(c*x - 1)^4 + 10*
(c*x + 1)^3*c^2/(c*x - 1)^3 - 10*(c*x + 1)^2*c^2/(c*x - 1)^2 + 5*(c*x + 1)*c^2/(c*x - 1) - c^2) - (120*(c*x +
1)^4*a*d^4/(c*x - 1)^4 - 240*(c*x + 1)^3*a*d^4/(c*x - 1)^3 + 240*(c*x + 1)^2*a*d^4/(c*x - 1)^2 - 120*(c*x + 1)
*a*d^4/(c*x - 1) + 24*a*d^4 + 48*(c*x + 1)^4*b*d^4/(c*x - 1)^4 - 156*(c*x + 1)^3*b*d^4/(c*x - 1)^3 + 196*(c*x
+ 1)^2*b*d^4/(c*x - 1)^2 - 113*(c*x + 1)*b*d^4/(c*x - 1) + 25*b*d^4)/((c*x + 1)^5*c^2/(c*x - 1)^5 - 5*(c*x + 1
)^4*c^2/(c*x - 1)^4 + 10*(c*x + 1)^3*c^2/(c*x - 1)^3 - 10*(c*x + 1)^2*c^2/(c*x - 1)^2 + 5*(c*x + 1)*c^2/(c*x -
1) - c^2))*c

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maple [B]  time = 0.03, size = 202, normalized size = 1.89 \[ \frac {c^{4} d^{4} a \,x^{5}}{5}+c^{3} d^{4} a \,x^{4}+2 c^{2} d^{4} a \,x^{3}+2 c \,d^{4} a \,x^{2}+x a \,d^{4}+\frac {d^{4} a}{5 c}+\frac {c^{4} d^{4} b \arctanh \left (c x \right ) x^{5}}{5}+c^{3} d^{4} b \arctanh \left (c x \right ) x^{4}+2 c^{2} d^{4} b \arctanh \left (c x \right ) x^{3}+2 c \,d^{4} b \arctanh \left (c x \right ) x^{2}+d^{4} b \arctanh \left (c x \right ) x +\frac {d^{4} b \arctanh \left (c x \right )}{5 c}+\frac {c^{3} d^{4} b \,x^{4}}{20}+\frac {c^{2} d^{4} b \,x^{3}}{3}+\frac {11 c \,d^{4} b \,x^{2}}{10}+3 b \,d^{4} x +\frac {16 b \ln \left (c x -1\right ) d^{4}}{5 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^4*(a+b*arctanh(c*x)),x)

[Out]

1/5*c^4*d^4*a*x^5+c^3*d^4*a*x^4+2*c^2*d^4*a*x^3+2*c*d^4*a*x^2+x*a*d^4+1/5/c*d^4*a+1/5*c^4*d^4*b*arctanh(c*x)*x
^5+c^3*d^4*b*arctanh(c*x)*x^4+2*c^2*d^4*b*arctanh(c*x)*x^3+2*c*d^4*b*arctanh(c*x)*x^2+d^4*b*arctanh(c*x)*x+1/5
/c*d^4*b*arctanh(c*x)+1/20*c^3*d^4*b*x^4+1/3*c^2*d^4*b*x^3+11/10*c*d^4*b*x^2+3*b*d^4*x+16/5/c*b*ln(c*x-1)*d^4

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maxima [B]  time = 0.33, size = 283, normalized size = 2.64 \[ \frac {1}{5} \, a c^{4} d^{4} x^{5} + a c^{3} d^{4} x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{4} d^{4} + 2 \, a c^{2} d^{4} x^{3} + \frac {1}{6} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{3} d^{4} + {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{4} + 2 \, a c d^{4} x^{2} + {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d^{4} + a d^{4} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/5*a*c^4*d^4*x^5 + a*c^3*d^4*x^4 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c
^6))*b*c^4*d^4 + 2*a*c^2*d^4*x^3 + 1/6*(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^4 + (2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c^2*d^4 + 2*a*c*d^
4*x^2 + (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^4 + a*d^4*x + 1/2*(2*c*
x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*d^4/c

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mupad [B]  time = 1.05, size = 168, normalized size = 1.57 \[ \frac {d^4\,\left (60\,a\,x+180\,b\,x+60\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^4\,d^4\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{60}-\frac {d^4\,\left (180\,b\,\mathrm {atanh}\left (c\,x\right )-96\,b\,\ln \left (c^2\,x^2-1\right )\right )}{60\,c}+\frac {c\,d^4\,\left (120\,a\,x^2+66\,b\,x^2+120\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^3\,d^4\,\left (60\,a\,x^4+3\,b\,x^4+60\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c^2\,d^4\,\left (120\,a\,x^3+20\,b\,x^3+120\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))*(d + c*d*x)^4,x)

[Out]

(d^4*(60*a*x + 180*b*x + 60*b*x*atanh(c*x)))/60 + (c^4*d^4*(12*a*x^5 + 12*b*x^5*atanh(c*x)))/60 - (d^4*(180*b*
atanh(c*x) - 96*b*log(c^2*x^2 - 1)))/(60*c) + (c*d^4*(120*a*x^2 + 66*b*x^2 + 120*b*x^2*atanh(c*x)))/60 + (c^3*
d^4*(60*a*x^4 + 3*b*x^4 + 60*b*x^4*atanh(c*x)))/60 + (c^2*d^4*(120*a*x^3 + 20*b*x^3 + 120*b*x^3*atanh(c*x)))/6
0

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sympy [A]  time = 1.85, size = 226, normalized size = 2.11 \[ \begin {cases} \frac {a c^{4} d^{4} x^{5}}{5} + a c^{3} d^{4} x^{4} + 2 a c^{2} d^{4} x^{3} + 2 a c d^{4} x^{2} + a d^{4} x + \frac {b c^{4} d^{4} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + b c^{3} d^{4} x^{4} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{3} d^{4} x^{4}}{20} + 2 b c^{2} d^{4} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c^{2} d^{4} x^{3}}{3} + 2 b c d^{4} x^{2} \operatorname {atanh}{\left (c x \right )} + \frac {11 b c d^{4} x^{2}}{10} + b d^{4} x \operatorname {atanh}{\left (c x \right )} + 3 b d^{4} x + \frac {16 b d^{4} \log {\left (x - \frac {1}{c} \right )}}{5 c} + \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{5 c} & \text {for}\: c \neq 0 \\a d^{4} x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**4*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*c**4*d**4*x**5/5 + a*c**3*d**4*x**4 + 2*a*c**2*d**4*x**3 + 2*a*c*d**4*x**2 + a*d**4*x + b*c**4*d*
*4*x**5*atanh(c*x)/5 + b*c**3*d**4*x**4*atanh(c*x) + b*c**3*d**4*x**4/20 + 2*b*c**2*d**4*x**3*atanh(c*x) + b*c
**2*d**4*x**3/3 + 2*b*c*d**4*x**2*atanh(c*x) + 11*b*c*d**4*x**2/10 + b*d**4*x*atanh(c*x) + 3*b*d**4*x + 16*b*d
**4*log(x - 1/c)/(5*c) + b*d**4*atanh(c*x)/(5*c), Ne(c, 0)), (a*d**4*x, True))

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