3.102 \(\int (e+f x)^3 (a+b \coth ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=168 \[ \frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}+\frac {b f x \left (\left (6 c^2+1\right ) f^2-12 c d e f+6 d^2 e^2\right )}{4 d^3}+\frac {b f^2 (c+d x)^2 (d e-c f)}{2 d^4}-\frac {b (-c f+d e-f)^4 \log (c+d x+1)}{8 d^4 f}+\frac {b (-c f+d e+f)^4 \log (-c-d x+1)}{8 d^4 f}+\frac {b f^3 (c+d x)^3}{12 d^4} \]

[Out]

1/4*b*f*(6*d^2*e^2-12*c*d*e*f+(6*c^2+1)*f^2)*x/d^3+1/2*b*f^2*(-c*f+d*e)*(d*x+c)^2/d^4+1/12*b*f^3*(d*x+c)^3/d^4
+1/4*(f*x+e)^4*(a+b*arccoth(d*x+c))/f+1/8*b*(-c*f+d*e+f)^4*ln(-d*x-c+1)/d^4/f-1/8*b*(-c*f+d*e-f)^4*ln(d*x+c+1)
/d^4/f

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Rubi [A]  time = 0.34, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6112, 5927, 702, 633, 31} \[ \frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}+\frac {b f x \left (\left (6 c^2+1\right ) f^2-12 c d e f+6 d^2 e^2\right )}{4 d^3}+\frac {b f^2 (c+d x)^2 (d e-c f)}{2 d^4}-\frac {b (-c f+d e-f)^4 \log (c+d x+1)}{8 d^4 f}+\frac {b (-c f+d e+f)^4 \log (-c-d x+1)}{8 d^4 f}+\frac {b f^3 (c+d x)^3}{12 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f*(6*d^2*e^2 - 12*c*d*e*f + (1 + 6*c^2)*f^2)*x)/(4*d^3) + (b*f^2*(d*e - c*f)*(c + d*x)^2)/(2*d^4) + (b*f^3*
(c + d*x)^3)/(12*d^4) + ((e + f*x)^4*(a + b*ArcCoth[c + d*x]))/(4*f) + (b*(d*e + f - c*f)^4*Log[1 - c - d*x])/
(8*d^4*f) - (b*(d*e - f - c*f)^4*Log[1 + c + d*x])/(8*d^4*f)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 5927

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcCoth[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]

Rule 6112

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

Rubi steps

\begin {align*} \int (e+f x)^3 \left (a+b \coth ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^3 \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}-\frac {b \operatorname {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^4}{1-x^2} \, dx,x,c+d x\right )}{4 f}\\ &=\frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {f^2 \left (6 d^2 e^2-12 c d e f+\left (1+6 c^2\right ) f^2\right )}{d^4}-\frac {4 f^3 (d e-c f) x}{d^4}-\frac {f^4 x^2}{d^4}+\frac {d^4 e^4-4 c d^3 e^3 f+6 \left (1+c^2\right ) d^2 e^2 f^2-4 c \left (3+c^2\right ) d e f^3+\left (1+6 c^2+c^4\right ) f^4+4 f (d e-c f) \left (d^2 e^2-2 c d e f+f^2+c^2 f^2\right ) x}{d^4 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{4 f}\\ &=\frac {b f \left (6 d^2 e^2-12 c d e f+\left (1+6 c^2\right ) f^2\right ) x}{4 d^3}+\frac {b f^2 (d e-c f) (c+d x)^2}{2 d^4}+\frac {b f^3 (c+d x)^3}{12 d^4}+\frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}-\frac {b \operatorname {Subst}\left (\int \frac {d^4 e^4-4 c d^3 e^3 f+6 \left (1+c^2\right ) d^2 e^2 f^2-4 c \left (3+c^2\right ) d e f^3+\left (1+6 c^2+c^4\right ) f^4+4 f (d e-c f) \left (d^2 e^2-2 c d e f+f^2+c^2 f^2\right ) x}{1-x^2} \, dx,x,c+d x\right )}{4 d^4 f}\\ &=\frac {b f \left (6 d^2 e^2-12 c d e f+\left (1+6 c^2\right ) f^2\right ) x}{4 d^3}+\frac {b f^2 (d e-c f) (c+d x)^2}{2 d^4}+\frac {b f^3 (c+d x)^3}{12 d^4}+\frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}+\frac {\left (b (d e-f-c f)^4\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,c+d x\right )}{8 d^4 f}-\frac {\left (b (d e+f-c f)^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,c+d x\right )}{8 d^4 f}\\ &=\frac {b f \left (6 d^2 e^2-12 c d e f+\left (1+6 c^2\right ) f^2\right ) x}{4 d^3}+\frac {b f^2 (d e-c f) (c+d x)^2}{2 d^4}+\frac {b f^3 (c+d x)^3}{12 d^4}+\frac {(e+f x)^4 \left (a+b \coth ^{-1}(c+d x)\right )}{4 f}+\frac {b (d e+f-c f)^4 \log (1-c-d x)}{8 d^4 f}-\frac {b (d e-f-c f)^4 \log (1+c+d x)}{8 d^4 f}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 270, normalized size = 1.61 \[ \frac {6 d x \left (4 a d^3 e^3+b f \left (\left (3 c^2+1\right ) f^2-8 c d e f+6 d^2 e^2\right )\right )+6 d^2 f x^2 \left (6 a d^2 e^2+b f (2 d e-c f)\right )+2 d^3 f^2 x^3 (12 a d e+b f)+6 a d^4 f^3 x^4+6 b d^4 x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right ) \coth ^{-1}(c+d x)-3 b (c-1) \left (-6 (c-1) d^2 e^2 f+4 (c-1)^2 d e f^2-(c-1)^3 f^3+4 d^3 e^3\right ) \log (-c-d x+1)-3 b (c+1) \left (6 (c+1) d^2 e^2 f-4 (c+1)^2 d e f^2+(c+1)^3 f^3-4 d^3 e^3\right ) \log (c+d x+1)}{24 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*d*(4*a*d^3*e^3 + b*f*(6*d^2*e^2 - 8*c*d*e*f + (1 + 3*c^2)*f^2))*x + 6*d^2*f*(6*a*d^2*e^2 + b*f*(2*d*e - c*f
))*x^2 + 2*d^3*f^2*(12*a*d*e + b*f)*x^3 + 6*a*d^4*f^3*x^4 + 6*b*d^4*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x
^3)*ArcCoth[c + d*x] - 3*b*(-1 + c)*(4*d^3*e^3 - 6*(-1 + c)*d^2*e^2*f + 4*(-1 + c)^2*d*e*f^2 - (-1 + c)^3*f^3)
*Log[1 - c - d*x] - 3*b*(1 + c)*(-4*d^3*e^3 + 6*(1 + c)*d^2*e^2*f - 4*(1 + c)^2*d*e*f^2 + (1 + c)^3*f^3)*Log[1
+ c + d*x])/(24*d^4)

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fricas [B]  time = 0.55, size = 385, normalized size = 2.29 \[ \frac {6 \, a d^{4} f^{3} x^{4} + 2 \, {\left (12 \, a d^{4} e f^{2} + b d^{3} f^{3}\right )} x^{3} + 6 \, {\left (6 \, a d^{4} e^{2} f + 2 \, b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x^{2} + 6 \, {\left (4 \, a d^{4} e^{3} + 6 \, b d^{3} e^{2} f - 8 \, b c d^{2} e f^{2} + {\left (3 \, b c^{2} + b\right )} d f^{3}\right )} x + 3 \, {\left (4 \, {\left (b c + b\right )} d^{3} e^{3} - 6 \, {\left (b c^{2} + 2 \, b c + b\right )} d^{2} e^{2} f + 4 \, {\left (b c^{3} + 3 \, b c^{2} + 3 \, b c + b\right )} d e f^{2} - {\left (b c^{4} + 4 \, b c^{3} + 6 \, b c^{2} + 4 \, b c + b\right )} f^{3}\right )} \log \left (d x + c + 1\right ) - 3 \, {\left (4 \, {\left (b c - b\right )} d^{3} e^{3} - 6 \, {\left (b c^{2} - 2 \, b c + b\right )} d^{2} e^{2} f + 4 \, {\left (b c^{3} - 3 \, b c^{2} + 3 \, b c - b\right )} d e f^{2} - {\left (b c^{4} - 4 \, b c^{3} + 6 \, b c^{2} - 4 \, b c + b\right )} f^{3}\right )} \log \left (d x + c - 1\right ) + 3 \, {\left (b d^{4} f^{3} x^{4} + 4 \, b d^{4} e f^{2} x^{3} + 6 \, b d^{4} e^{2} f x^{2} + 4 \, b d^{4} e^{3} x\right )} \log \left (\frac {d x + c + 1}{d x + c - 1}\right )}{24 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*a*d^4*f^3*x^4 + 2*(12*a*d^4*e*f^2 + b*d^3*f^3)*x^3 + 6*(6*a*d^4*e^2*f + 2*b*d^3*e*f^2 - b*c*d^2*f^3)*x
^2 + 6*(4*a*d^4*e^3 + 6*b*d^3*e^2*f - 8*b*c*d^2*e*f^2 + (3*b*c^2 + b)*d*f^3)*x + 3*(4*(b*c + b)*d^3*e^3 - 6*(b
*c^2 + 2*b*c + b)*d^2*e^2*f + 4*(b*c^3 + 3*b*c^2 + 3*b*c + b)*d*e*f^2 - (b*c^4 + 4*b*c^3 + 6*b*c^2 + 4*b*c + b
)*f^3)*log(d*x + c + 1) - 3*(4*(b*c - b)*d^3*e^3 - 6*(b*c^2 - 2*b*c + b)*d^2*e^2*f + 4*(b*c^3 - 3*b*c^2 + 3*b*
c - b)*d*e*f^2 - (b*c^4 - 4*b*c^3 + 6*b*c^2 - 4*b*c + b)*f^3)*log(d*x + c - 1) + 3*(b*d^4*f^3*x^4 + 4*b*d^4*e*
f^2*x^3 + 6*b*d^4*e^2*f*x^2 + 4*b*d^4*e^3*x)*log((d*x + c + 1)/(d*x + c - 1)))/d^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )}^{3} {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^3*(b*arccoth(d*x + c) + a), x)

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maple [B]  time = 0.04, size = 786, normalized size = 4.68 \[ -\frac {3 b f \ln \left (d x +c +1\right ) c \,e^{2}}{2 d^{2}}-\frac {3 b f \ln \left (d x +c +1\right ) c^{2} e^{2}}{4 d^{2}}-\frac {3 b f \ln \left (d x +c -1\right ) c \,e^{2}}{2 d^{2}}-\frac {b \,f^{2} \ln \left (d x +c -1\right ) c^{3} e}{2 d^{3}}+\frac {3 b \,f^{2} \ln \left (d x +c -1\right ) c^{2} e}{2 d^{3}}+\frac {b \,f^{2} \ln \left (d x +c +1\right ) c^{3} e}{2 d^{3}}+\frac {3 b \,f^{2} \ln \left (d x +c +1\right ) c e}{2 d^{3}}-\frac {3 b \,f^{2} \ln \left (d x +c -1\right ) c e}{2 d^{3}}-\frac {2 b \,f^{2} c e x}{d^{2}}+\frac {3 b \,f^{2} \ln \left (d x +c +1\right ) c^{2} e}{2 d^{3}}+\frac {3 b f \ln \left (d x +c -1\right ) c^{2} e^{2}}{4 d^{2}}+\frac {b \ln \left (d x +c -1\right ) e^{3}}{2 d}+\frac {3 a f \,x^{2} e^{2}}{2}+a \,f^{2} x^{3} e +\mathrm {arccoth}\left (d x +c \right ) x b \,e^{3}+\frac {b \,\mathrm {arccoth}\left (d x +c \right ) e^{4}}{4 f}-\frac {b \ln \left (d x +c +1\right ) e^{4}}{8 f}+\frac {b \ln \left (d x +c -1\right ) e^{4}}{8 f}+\frac {b \,f^{3} x^{3}}{12 d}+\frac {b \,f^{3} \ln \left (d x +c -1\right )}{8 d^{4}}-\frac {b \,f^{3} \ln \left (d x +c +1\right )}{8 d^{4}}+\frac {b \,f^{3} x}{4 d^{3}}+\frac {b \ln \left (d x +c +1\right ) e^{3}}{2 d}+\frac {b \,f^{3} \mathrm {arccoth}\left (d x +c \right ) x^{4}}{4}+\frac {13 b \,f^{3} c^{3}}{12 d^{4}}+\frac {b \,f^{3} c}{4 d^{4}}+\frac {a \,e^{4}}{4 f}+\frac {3 b f c \,e^{2}}{2 d^{2}}-\frac {b \,f^{3} \ln \left (d x +c -1\right ) c^{3}}{2 d^{4}}+\frac {b \ln \left (d x +c +1\right ) c \,e^{3}}{2 d}-\frac {b \ln \left (d x +c -1\right ) c \,e^{3}}{2 d}+\frac {b \,f^{2} \ln \left (d x +c +1\right ) e}{2 d^{3}}-\frac {3 b f \ln \left (d x +c +1\right ) e^{2}}{4 d^{2}}+\frac {3 b \,f^{3} \ln \left (d x +c -1\right ) c^{2}}{4 d^{4}}-\frac {b \,f^{3} \ln \left (d x +c -1\right ) c}{2 d^{4}}-\frac {b \,f^{3} x^{2} c}{4 d^{2}}+\frac {b \,f^{2} e \,x^{2}}{2 d}+b \,f^{2} \mathrm {arccoth}\left (d x +c \right ) e \,x^{3}-\frac {b \,f^{3} \ln \left (d x +c +1\right ) c^{3}}{2 d^{4}}+\frac {b \,f^{2} \ln \left (d x +c -1\right ) e}{2 d^{3}}-\frac {b \,f^{3} \ln \left (d x +c +1\right ) c^{4}}{8 d^{4}}-\frac {b \,f^{3} \ln \left (d x +c +1\right ) c}{2 d^{4}}+\frac {3 b f \ln \left (d x +c -1\right ) e^{2}}{4 d^{2}}+\frac {3 b f \,\mathrm {arccoth}\left (d x +c \right ) e^{2} x^{2}}{2}+\frac {3 b f \,e^{2} x}{2 d}+\frac {3 b \,f^{3} c^{2} x}{4 d^{3}}-\frac {3 b \,f^{3} \ln \left (d x +c +1\right ) c^{2}}{4 d^{4}}+\frac {b \,f^{3} \ln \left (d x +c -1\right ) c^{4}}{8 d^{4}}+a x \,e^{3}+\frac {a \,f^{3} x^{4}}{4}-\frac {5 b \,f^{2} c^{2} e}{2 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/d^2*b*f*ln(d*x+c+1)*c*e^2-3/4/d^2*b*f*ln(d*x+c+1)*c^2*e^2-3/2/d^2*b*f*ln(d*x+c-1)*c*e^2-1/2/d^3*b*f^2*ln(
d*x+c-1)*c^3*e+3/2/d^3*b*f^2*ln(d*x+c-1)*c^2*e+1/2/d^3*b*f^2*ln(d*x+c+1)*c^3*e+3/2/d^3*b*f^2*ln(d*x+c+1)*c*e-3
/2/d^3*b*f^2*ln(d*x+c-1)*c*e-2*b/d^2*f^2*c*e*x+3/2/d^3*b*f^2*ln(d*x+c+1)*c^2*e+3/4/d^2*b*f*ln(d*x+c-1)*c^2*e^2
+1/2/d*b*ln(d*x+c-1)*e^3+3/2*a*f*x^2*e^2+a*f^2*x^3*e+arccoth(d*x+c)*x*b*e^3+1/4*b/f*arccoth(d*x+c)*e^4-1/8*b/f
*ln(d*x+c+1)*e^4+1/8*b/f*ln(d*x+c-1)*e^4+1/12/d*b*f^3*x^3+1/8/d^4*b*f^3*ln(d*x+c-1)-1/8/d^4*b*f^3*ln(d*x+c+1)+
1/4*b/d^3*f^3*x+1/2/d*b*ln(d*x+c+1)*e^3+1/4*b*f^3*arccoth(d*x+c)*x^4+13/12/d^4*b*f^3*c^3+1/4/d^4*b*f^3*c+1/4*a
/f*e^4+3/2/d^2*b*f*c*e^2-1/2/d^4*b*f^3*ln(d*x+c-1)*c^3+1/2/d*b*ln(d*x+c+1)*c*e^3-1/2/d*b*ln(d*x+c-1)*c*e^3+1/2
/d^3*b*f^2*ln(d*x+c+1)*e-3/4/d^2*b*f*ln(d*x+c+1)*e^2+3/4/d^4*b*f^3*ln(d*x+c-1)*c^2-1/2/d^4*b*f^3*ln(d*x+c-1)*c
-1/4/d^2*b*f^3*x^2*c+1/2/d*b*f^2*e*x^2+b*f^2*arccoth(d*x+c)*e*x^3-1/2/d^4*b*f^3*ln(d*x+c+1)*c^3+1/2/d^3*b*f^2*
ln(d*x+c-1)*e-1/8/d^4*b*f^3*ln(d*x+c+1)*c^4-1/2/d^4*b*f^3*ln(d*x+c+1)*c+3/4/d^2*b*f*ln(d*x+c-1)*e^2+3/2*b*f*ar
ccoth(d*x+c)*e^2*x^2+3/2*b/d*f*e^2*x+3/4*b/d^3*f^3*c^2*x-3/4/d^4*b*f^3*ln(d*x+c+1)*c^2+1/8/d^4*b*f^3*ln(d*x+c-
1)*c^4+a*x*e^3+1/4*a*f^3*x^4-5/2/d^3*b*f^2*c^2*e

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maxima [B]  time = 0.31, size = 333, normalized size = 1.98 \[ \frac {1}{4} \, a f^{3} x^{4} + a e f^{2} x^{3} + \frac {3}{2} \, a e^{2} f x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \operatorname {arcoth}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b e^{2} f + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {arcoth}\left (d x + c\right ) + d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} + \frac {{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac {{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} b e f^{2} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {arcoth}\left (d x + c\right ) + d {\left (\frac {2 \, {\left (d^{2} x^{3} - 3 \, c d x^{2} + 3 \, {\left (3 \, c^{2} + 1\right )} x\right )}}{d^{4}} - \frac {3 \, {\left (c^{4} + 4 \, c^{3} + 6 \, c^{2} + 4 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{5}} + \frac {3 \, {\left (c^{4} - 4 \, c^{3} + 6 \, c^{2} - 4 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{5}}\right )}\right )} b f^{3} + a e^{3} x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b e^{3}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.17, size = 742, normalized size = 4.42 \[ x\,\left (\frac {e\,\left (6\,a\,c^2\,f^2+12\,a\,c\,d\,e\,f+2\,a\,d^2\,e^2+3\,b\,d\,e\,f-6\,a\,f^2\right )}{2\,d^2}-\frac {\left (4\,c^2-4\right )\,\left (\frac {f^2\,\left (b\,f+8\,a\,c\,f+12\,a\,d\,e\right )}{4\,d}-\frac {2\,a\,c\,f^3}{d}\right )}{4\,d^2}+\frac {2\,c\,\left (\frac {2\,c\,\left (\frac {f^2\,\left (b\,f+8\,a\,c\,f+12\,a\,d\,e\right )}{4\,d}-\frac {2\,a\,c\,f^3}{d}\right )}{d}-\frac {4\,a\,c^2\,f^3+24\,a\,c\,d\,e\,f^2+12\,a\,d^2\,e^2\,f+4\,b\,d\,e\,f^2-4\,a\,f^3}{4\,d^2}+\frac {a\,f^3\,\left (4\,c^2-4\right )}{4\,d^2}\right )}{d}\right )-\ln \left (1-\frac {1}{c+d\,x}\right )\,\left (\frac {b\,e^3\,x}{2}+\frac {3\,b\,e^2\,f\,x^2}{4}+\frac {b\,e\,f^2\,x^3}{2}+\frac {b\,f^3\,x^4}{8}\right )-x^2\,\left (\frac {c\,\left (\frac {f^2\,\left (b\,f+8\,a\,c\,f+12\,a\,d\,e\right )}{4\,d}-\frac {2\,a\,c\,f^3}{d}\right )}{d}-\frac {4\,a\,c^2\,f^3+24\,a\,c\,d\,e\,f^2+12\,a\,d^2\,e^2\,f+4\,b\,d\,e\,f^2-4\,a\,f^3}{8\,d^2}+\frac {a\,f^3\,\left (4\,c^2-4\right )}{8\,d^2}\right )+x^3\,\left (\frac {f^2\,\left (b\,f+8\,a\,c\,f+12\,a\,d\,e\right )}{12\,d}-\frac {2\,a\,c\,f^3}{3\,d}\right )+\ln \left (\frac {1}{c+d\,x}+1\right )\,\left (\frac {b\,e^3\,x}{2}+\frac {3\,b\,e^2\,f\,x^2}{4}+\frac {b\,e\,f^2\,x^3}{2}+\frac {b\,f^3\,x^4}{8}\right )+\frac {a\,f^3\,x^4}{4}+\frac {\ln \left (c+d\,x-1\right )\,\left (b\,c^4\,f^3-4\,b\,c^3\,d\,e\,f^2-4\,b\,c^3\,f^3+6\,b\,c^2\,d^2\,e^2\,f+12\,b\,c^2\,d\,e\,f^2+6\,b\,c^2\,f^3-4\,b\,c\,d^3\,e^3-12\,b\,c\,d^2\,e^2\,f-12\,b\,c\,d\,e\,f^2-4\,b\,c\,f^3+4\,b\,d^3\,e^3+6\,b\,d^2\,e^2\,f+4\,b\,d\,e\,f^2+b\,f^3\right )}{8\,d^4}-\frac {\ln \left (c+d\,x+1\right )\,\left (b\,c^4\,f^3-4\,b\,c^3\,d\,e\,f^2+4\,b\,c^3\,f^3+6\,b\,c^2\,d^2\,e^2\,f-12\,b\,c^2\,d\,e\,f^2+6\,b\,c^2\,f^3-4\,b\,c\,d^3\,e^3+12\,b\,c\,d^2\,e^2\,f-12\,b\,c\,d\,e\,f^2+4\,b\,c\,f^3-4\,b\,d^3\,e^3+6\,b\,d^2\,e^2\,f-4\,b\,d\,e\,f^2+b\,f^3\right )}{8\,d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((e*(6*a*c^2*f^2 - 6*a*f^2 + 2*a*d^2*e^2 + 3*b*d*e*f + 12*a*c*d*e*f))/(2*d^2) - ((4*c^2 - 4)*((f^2*(b*f + 8*
a*c*f + 12*a*d*e))/(4*d) - (2*a*c*f^3)/d))/(4*d^2) + (2*c*((2*c*((f^2*(b*f + 8*a*c*f + 12*a*d*e))/(4*d) - (2*a
*c*f^3)/d))/d - (4*a*c^2*f^3 - 4*a*f^3 + 4*b*d*e*f^2 + 12*a*d^2*e^2*f + 24*a*c*d*e*f^2)/(4*d^2) + (a*f^3*(4*c^
2 - 4))/(4*d^2)))/d) - log(1 - 1/(c + d*x))*((b*f^3*x^4)/8 + (b*e^3*x)/2 + (3*b*e^2*f*x^2)/4 + (b*e*f^2*x^3)/2
) - x^2*((c*((f^2*(b*f + 8*a*c*f + 12*a*d*e))/(4*d) - (2*a*c*f^3)/d))/d - (4*a*c^2*f^3 - 4*a*f^3 + 4*b*d*e*f^2
+ 12*a*d^2*e^2*f + 24*a*c*d*e*f^2)/(8*d^2) + (a*f^3*(4*c^2 - 4))/(8*d^2)) + x^3*((f^2*(b*f + 8*a*c*f + 12*a*d
*e))/(12*d) - (2*a*c*f^3)/(3*d)) + log(1/(c + d*x) + 1)*((b*f^3*x^4)/8 + (b*e^3*x)/2 + (3*b*e^2*f*x^2)/4 + (b*
e*f^2*x^3)/2) + (a*f^3*x^4)/4 + (log(c + d*x - 1)*(b*f^3 + 6*b*c^2*f^3 - 4*b*c^3*f^3 + 4*b*d^3*e^3 + b*c^4*f^3
- 4*b*c*f^3 + 4*b*d*e*f^2 - 4*b*c*d^3*e^3 + 6*b*d^2*e^2*f - 12*b*c*d^2*e^2*f + 12*b*c^2*d*e*f^2 - 4*b*c^3*d*e
*f^2 + 6*b*c^2*d^2*e^2*f - 12*b*c*d*e*f^2))/(8*d^4) - (log(c + d*x + 1)*(b*f^3 + 6*b*c^2*f^3 + 4*b*c^3*f^3 - 4
*b*d^3*e^3 + b*c^4*f^3 + 4*b*c*f^3 - 4*b*d*e*f^2 - 4*b*c*d^3*e^3 + 6*b*d^2*e^2*f + 12*b*c*d^2*e^2*f - 12*b*c^2
*d*e*f^2 - 4*b*c^3*d*e*f^2 + 6*b*c^2*d^2*e^2*f - 12*b*c*d*e*f^2))/(8*d^4)

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sympy [A]  time = 7.46, size = 644, normalized size = 3.83 \[ \begin {cases} a e^{3} x + \frac {3 a e^{2} f x^{2}}{2} + a e f^{2} x^{3} + \frac {a f^{3} x^{4}}{4} - \frac {b c^{4} f^{3} \operatorname {acoth}{\left (c + d x \right )}}{4 d^{4}} + \frac {b c^{3} e f^{2} \operatorname {acoth}{\left (c + d x \right )}}{d^{3}} - \frac {b c^{3} f^{3} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d^{4}} + \frac {b c^{3} f^{3} \operatorname {acoth}{\left (c + d x \right )}}{d^{4}} - \frac {3 b c^{2} e^{2} f \operatorname {acoth}{\left (c + d x \right )}}{2 d^{2}} + \frac {3 b c^{2} e f^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d^{3}} - \frac {3 b c^{2} e f^{2} \operatorname {acoth}{\left (c + d x \right )}}{d^{3}} + \frac {3 b c^{2} f^{3} x}{4 d^{3}} - \frac {3 b c^{2} f^{3} \operatorname {acoth}{\left (c + d x \right )}}{2 d^{4}} + \frac {b c e^{3} \operatorname {acoth}{\left (c + d x \right )}}{d} - \frac {3 b c e^{2} f \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d^{2}} + \frac {3 b c e^{2} f \operatorname {acoth}{\left (c + d x \right )}}{d^{2}} - \frac {2 b c e f^{2} x}{d^{2}} - \frac {b c f^{3} x^{2}}{4 d^{2}} + \frac {3 b c e f^{2} \operatorname {acoth}{\left (c + d x \right )}}{d^{3}} - \frac {b c f^{3} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d^{4}} + \frac {b c f^{3} \operatorname {acoth}{\left (c + d x \right )}}{d^{4}} + b e^{3} x \operatorname {acoth}{\left (c + d x \right )} + \frac {3 b e^{2} f x^{2} \operatorname {acoth}{\left (c + d x \right )}}{2} + b e f^{2} x^{3} \operatorname {acoth}{\left (c + d x \right )} + \frac {b f^{3} x^{4} \operatorname {acoth}{\left (c + d x \right )}}{4} + \frac {b e^{3} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d} - \frac {b e^{3} \operatorname {acoth}{\left (c + d x \right )}}{d} + \frac {3 b e^{2} f x}{2 d} + \frac {b e f^{2} x^{2}}{2 d} + \frac {b f^{3} x^{3}}{12 d} - \frac {3 b e^{2} f \operatorname {acoth}{\left (c + d x \right )}}{2 d^{2}} + \frac {b e f^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d^{3}} - \frac {b e f^{2} \operatorname {acoth}{\left (c + d x \right )}}{d^{3}} + \frac {b f^{3} x}{4 d^{3}} - \frac {b f^{3} \operatorname {acoth}{\left (c + d x \right )}}{4 d^{4}} & \text {for}\: d \neq 0 \\\left (a + b \operatorname {acoth}{\relax (c )}\right ) \left (e^{3} x + \frac {3 e^{2} f x^{2}}{2} + e f^{2} x^{3} + \frac {f^{3} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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