Integrand size = 18, antiderivative size = 162 \[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\frac {(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}+\frac {b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e-f-c f}\right )}{2 f (d e-(1+c) f) (1+m) (2+m)}-\frac {b d (e+f x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {d (e+f x)}{d e+f-c f}\right )}{2 f (d e+f-c f) (1+m) (2+m)} \]
(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)
\[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx \]
Time = 0.47 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6662, 6479, 485, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 6662 |
\(\displaystyle \frac {\int \left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^m \left (a+b \coth ^{-1}(c+d x)\right )d(c+d x)}{d}\) |
\(\Big \downarrow \) 6479 |
\(\displaystyle \frac {\frac {d \left (a+b \coth ^{-1}(c+d x)\right ) \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+1}}{f (m+1)}-\frac {b d \int \frac {\left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^{m+1}}{1-(c+d x)^2}d(c+d x)}{f (m+1)}}{d}\) |
\(\Big \downarrow \) 485 |
\(\displaystyle \frac {\frac {d \left (a+b \coth ^{-1}(c+d x)\right ) \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+1}}{f (m+1)}-\frac {b d \int \left (\frac {\left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^{m+1}}{2 (-c-d x+1)}+\frac {\left (e-\frac {c f}{d}+\frac {f (c+d x)}{d}\right )^{m+1}}{2 (c+d x+1)}\right )d(c+d x)}{f (m+1)}}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {d \left (a+b \coth ^{-1}(c+d x)\right ) \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+1}}{f (m+1)}-\frac {b d \left (\frac {d \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {d e-c f+f (c+d x)}{d e-c f+f}\right )}{2 (m+2) (-c f+d e+f)}-\frac {d \left (\frac {f (c+d x)}{d}-\frac {c f}{d}+e\right )^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {d e-c f+f (c+d x)}{d e-c f-f}\right )}{2 (m+2) (d e-(c+1) f)}\right )}{f (m+1)}}{d}\) |
((d*(e - (c*f)/d + (f*(c + d*x))/d)^(1 + m)*(a + b*ArcCoth[c + d*x]))/(f*( 1 + m)) - (b*d*(-1/2*(d*(e - (c*f)/d + (f*(c + d*x))/d)^(2 + m)*Hypergeome tric2F1[1, 2 + m, 3 + m, (d*e - c*f + f*(c + d*x))/(d*e - f - c*f)])/((d*e - (1 + c)*f)*(2 + m)) + (d*(e - (c*f)/d + (f*(c + d*x))/d)^(2 + m)*Hyperg eometric2F1[1, 2 + m, 3 + m, (d*e - c*f + f*(c + d*x))/(d*e + f - c*f)])/( 2*(d*e + f - c*f)*(2 + m))))/(f*(1 + m)))/d
3.2.19.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & & !IntegerQ[2*n]
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] - Simp[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]
Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[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] && IG tQ[p, 0]
\[\int \left (f x +e \right )^{m} \left (a +b \,\operatorname {arccoth}\left (d x +c \right )\right )d x\]
\[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\int { {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
\[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\int \left (a + b \operatorname {acoth}{\left (c + d x \right )}\right ) \left (e + f x\right )^{m}\, dx \]
\[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\int { {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
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))
\[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\int { {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )} {\left (f x + e\right )}^{m} \,d x } \]
Timed out. \[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx=\int {\left (e+f\,x\right )}^m\,\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right ) \,d x \]