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]

(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)

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Rubi [A]  time = 0.341981, antiderivative size = 168, normalized size of antiderivative = 1., 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 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]

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 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 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 31

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

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*}

Mathematica [A]  time = 0.281329, size = 270, normalized size = 1.61 \[ \frac{6 d x \left (4 a d^3 e^3+b f \left (\left (3 c^2+1\right ) f^2-8 c d e f+6 d^2 e^2\right )\right )+6 d^2 f x^2 \left (6 a d^2 e^2+b f (2 d e-c f)\right )+2 d^3 f^2 x^3 (12 a d e+b f)+6 a d^4 f^3 x^4+6 b d^4 x \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right ) \coth ^{-1}(c+d x)-3 b (c-1) \left (-6 (c-1) d^2 e^2 f+4 (c-1)^2 d e f^2-(c-1)^3 f^3+4 d^3 e^3\right ) \log (-c-d x+1)-3 b (c+1) \left (6 (c+1) d^2 e^2 f-4 (c+1)^2 d e f^2+(c+1)^3 f^3-4 d^3 e^3\right ) \log (c+d x+1)}{24 d^4} \]

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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Maple [B]  time = 0.046, size = 786, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.992158, size = 450, normalized size = 2.68 \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 2.11405, size = 856, normalized size = 5.1 \begin{align*} \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}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}^{3}{\left (b \operatorname{arcoth}\left (d x + c\right ) + a\right )}\,{d x} \end{align*}

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)