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

Optimal. Leaf size=162 \[ \frac {(e+f x)^{m+1} \left (a+b \coth ^{-1}(c+d x)\right )}{f (m+1)}+\frac {b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {d (e+f x)}{d e-c f-f}\right )}{2 f (m+1) (m+2) (d e-(c+1) f)}-\frac {b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {d (e+f x)}{d e-c f+f}\right )}{2 f (m+1) (m+2) (-c f+d e+f)} \]

[Out]

(f*x+e)^(1+m)*(a+b*arccoth(d*x+c))/f/(1+m)+1/2*b*d*(f*x+e)^(2+m)*hypergeom([1, 2+m],[3+m],d*(f*x+e)/(-c*f+d*e-
f))/f/(d*e-(1+c)*f)/(1+m)/(2+m)-1/2*b*d*(f*x+e)^(2+m)*hypergeom([1, 2+m],[3+m],d*(f*x+e)/(-c*f+d*e+f))/f/(-c*f
+d*e+f)/(1+m)/(2+m)

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Rubi [A]  time = 0.25, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6112, 5927, 712, 68} \[ \frac {(e+f x)^{m+1} \left (a+b \coth ^{-1}(c+d x)\right )}{f (m+1)}+\frac {b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {d (e+f x)}{d e-c f-f}\right )}{2 f (m+1) (m+2) (d e-(c+1) f)}-\frac {b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac {d (e+f x)}{d e-c f+f}\right )}{2 f (m+1) (m+2) (-c f+d e+f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)^(1 + m)*(a + b*ArcCoth[c + d*x]))/(f*(1 + m)) + (b*d*(e + f*x)^(2 + m)*Hypergeometric2F1[1, 2 + m,
3 + m, (d*(e + f*x))/(d*e - f - c*f)])/(2*f*(d*e - (1 + c)*f)*(1 + m)*(2 + m)) - (b*d*(e + f*x)^(2 + m)*Hyperg
eometric2F1[1, 2 + m, 3 + m, (d*(e + f*x))/(d*e + f - c*f)])/(2*f*(d*e + f - c*f)*(1 + m)*(2 + m))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 712

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

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)^m \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 )^m \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}-\frac {b \operatorname {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{1-x^2} \, dx,x,c+d x\right )}{f (1+m)}\\ &=\frac {(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}-\frac {b \operatorname {Subst}\left (\int \left (\frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{2 (1-x)}+\frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{2 (1+x)}\right ) \, dx,x,c+d x\right )}{f (1+m)}\\ &=\frac {(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}-\frac {b \operatorname {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{1-x} \, dx,x,c+d x\right )}{2 f (1+m)}-\frac {b \operatorname {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^{1+m}}{1+x} \, dx,x,c+d x\right )}{2 f (1+m)}\\ &=\frac {(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}+\frac {b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e-f-c f}\right )}{2 f (d e-(1+c) f) (1+m) (2+m)}-\frac {b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac {d (e+f x)}{d e+f-c f}\right )}{2 f (d e+f-c f) (1+m) (2+m)}\\ \end {align*}

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Mathematica [F]  time = 2.53, size = 0, normalized size = 0.00 \[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arccoth(d*x + c) + a)*(f*x + e)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.93, size = 0, normalized size = 0.00 \[ \int \left (f x +e \right )^{m} \left (a +b \,\mathrm {arccoth}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, b {\left (\frac {{\left (f x + e\right )} {\left (f x + e\right )}^{m} \log \left (d x + c + 1\right )}{f {\left (m + 1\right )}} - \int \frac {{\left (d f x + d e + {\left (d f {\left (m + 1\right )} x + c f {\left (m + 1\right )} + f {\left (m + 1\right )}\right )} \log \left (d x + c - 1\right )\right )} {\left (f x + e\right )}^{m}}{d f {\left (m + 1\right )} x + c f {\left (m + 1\right )} + f {\left (m + 1\right )}}\,{d x}\right )} + \frac {{\left (f x + e\right )}^{m + 1} a}{f {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*((f*x + e)*(f*x + e)^m*log(d*x + c + 1)/(f*(m + 1)) - integrate((d*f*x + d*e + (d*f*(m + 1)*x + c*f*(m +
1) + f*(m + 1))*log(d*x + c - 1))*(f*x + e)^m/(d*f*(m + 1)*x + c*f*(m + 1) + f*(m + 1)), x)) + (f*x + e)^(m +
1)*a/(f*(m + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e+f\,x\right )}^m\,\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acoth(c + d*x))*(e + f*x)**m, x)

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