Integrand size = 20, antiderivative size = 546 \[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx=\frac {a b^2 f^2 x}{d^2}+\frac {b^3 f^2 (c+d x) \text {arctanh}(c+d x)}{d^3}-\frac {b f^2 (a+b \text {arctanh}(c+d x))^2}{2 d^3}+\frac {3 b f (d e-c f) (a+b \text {arctanh}(c+d x))^2}{d^3}+\frac {3 b f (d e-c f) (c+d x) (a+b \text {arctanh}(c+d x))^2}{d^3}+\frac {b f^2 (c+d x)^2 (a+b \text {arctanh}(c+d x))^2}{2 d^3}-\frac {(d e-c f) \left (d^2 e^2-2 c d e f+\left (3+c^2\right ) f^2\right ) (a+b \text {arctanh}(c+d x))^3}{3 d^3 f}+\frac {\left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) (a+b \text {arctanh}(c+d x))^3}{3 d^3}+\frac {(e+f x)^3 (a+b \text {arctanh}(c+d x))^3}{3 f}-\frac {6 b^2 f (d e-c f) (a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1-c-d x}\right )}{d^3}-\frac {b \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) (a+b \text {arctanh}(c+d x))^2 \log \left (\frac {2}{1-c-d x}\right )}{d^3}+\frac {b^3 f^2 \log \left (1-(c+d x)^2\right )}{2 d^3}-\frac {3 b^3 f (d e-c f) \operatorname {PolyLog}\left (2,-\frac {1+c+d x}{1-c-d x}\right )}{d^3}-\frac {b^2 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) (a+b \text {arctanh}(c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-c-d x}\right )}{d^3}+\frac {b^3 \left (3 d^2 e^2-6 c d e f+\left (1+3 c^2\right ) f^2\right ) \operatorname {PolyLog}\left (3,1-\frac {2}{1-c-d x}\right )}{2 d^3} \]
a*b^2*f^2*x/d^2+b^3*f^2*(d*x+c)*arctanh(d*x+c)/d^3-1/2*b*f^2*(a+b*arctanh( d*x+c))^2/d^3+3*b*f*(-c*f+d*e)*(a+b*arctanh(d*x+c))^2/d^3+3*b*f*(-c*f+d*e) *(d*x+c)*(a+b*arctanh(d*x+c))^2/d^3+1/2*b*f^2*(d*x+c)^2*(a+b*arctanh(d*x+c ))^2/d^3-1/3*(-c*f+d*e)*(d^2*e^2-2*c*d*e*f+(c^2+3)*f^2)*(a+b*arctanh(d*x+c ))^3/d^3/f+1/3*(3*d^2*e^2-6*c*d*e*f+(3*c^2+1)*f^2)*(a+b*arctanh(d*x+c))^3/ d^3+1/3*(f*x+e)^3*(a+b*arctanh(d*x+c))^3/f-6*b^2*f*(-c*f+d*e)*(a+b*arctanh (d*x+c))*ln(2/(-d*x-c+1))/d^3-b*(3*d^2*e^2-6*c*d*e*f+(3*c^2+1)*f^2)*(a+b*a rctanh(d*x+c))^2*ln(2/(-d*x-c+1))/d^3+1/2*b^3*f^2*ln(1-(d*x+c)^2)/d^3-3*b^ 3*f*(-c*f+d*e)*polylog(2,(-d*x-c-1)/(-d*x-c+1))/d^3-b^2*(3*d^2*e^2-6*c*d*e *f+(3*c^2+1)*f^2)*(a+b*arctanh(d*x+c))*polylog(2,1-2/(-d*x-c+1))/d^3+1/2*b ^3*(3*d^2*e^2-6*c*d*e*f+(3*c^2+1)*f^2)*polylog(3,1-2/(-d*x-c+1))/d^3
Leaf count is larger than twice the leaf count of optimal. \(1646\) vs. \(2(546)=1092\).
Time = 6.96 (sec) , antiderivative size = 1646, normalized size of antiderivative = 3.01 \[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx =\text {Too large to display} \]
a^2*(a*e^2 + (b*f*(3*d*e - 2*c*f))/d^2)*x + (a^2*f*(2*a*d*e + b*f)*x^2)/(2 *d) + (a^3*f^2*x^3)/3 + a^2*b*x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcTanh[c + d* x] - (a^2*b*(-1 + c)*(3*d^2*e^2 - 3*(-1 + c)*d*e*f + (-1 + c)^2*f^2)*Log[1 - c - d*x])/(2*d^3) + (a^2*b*(1 + c)*(3*d^2*e^2 - 3*(1 + c)*d*e*f + (1 + c)^2*f^2)*Log[1 + c + d*x])/(2*d^3) + (3*a*b^2*e^2*(ArcTanh[c + d*x]*((-1 + c + d*x)*ArcTanh[c + d*x] - 2*Log[1 + E^(-2*ArcTanh[c + d*x])]) + PolyLo g[2, -E^(-2*ArcTanh[c + d*x])]))/d - (3*a*b^2*e*f*((1 - 2*c + c^2 - d^2*x^ 2)*ArcTanh[c + d*x]^2 - 2*ArcTanh[c + d*x]*(c + d*x + 2*c*Log[1 + E^(-2*Ar cTanh[c + d*x])]) + 2*Log[1/Sqrt[1 - (c + d*x)^2]] + 2*c*PolyLog[2, -E^(-2 *ArcTanh[c + d*x])]))/d^2 + (b^3*e^2*(2*ArcTanh[c + d*x]^2*((-1 + c + d*x) *ArcTanh[c + d*x] - 3*Log[1 + E^(-2*ArcTanh[c + d*x])]) + 6*ArcTanh[c + d* x]*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] + 3*PolyLog[3, -E^(-2*ArcTanh[c + d*x])]))/(2*d) - (b^3*e*f*(ArcTanh[c + d*x]*((1 - 2*c + c^2 - d^2*x^2)*Arc Tanh[c + d*x]^2 + 6*Log[1 + E^(-2*ArcTanh[c + d*x])] - 3*ArcTanh[c + d*x]* (-1 + c + d*x + 2*c*Log[1 + E^(-2*ArcTanh[c + d*x])])) + (-3 + 6*c*ArcTanh [c + d*x])*PolyLog[2, -E^(-2*ArcTanh[c + d*x])] + 3*c*PolyLog[3, -E^(-2*Ar cTanh[c + d*x])]))/d^2 - (a*b^2*f^2*(1 - (c + d*x)^2)^(3/2)*(-((c + d*x)/S qrt[1 - (c + d*x)^2]) + (6*c*(c + d*x)*ArcTanh[c + d*x])/Sqrt[1 - (c + d*x )^2] + (3*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c + d*x)^2] - (3*c^2*(c + d*x)*ArcTanh[c + d*x]^2)/Sqrt[1 - (c + d*x)^2] + ArcTanh[c + d*x]^2*C...
Time = 1.15 (sec) , antiderivative size = 533, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6661, 27, 6480, 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.
| \(\displaystyle \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 6661 |
| \(\displaystyle \frac {\int \frac {\left (d \left (e-\frac {c f}{d}\right )+f (c+d x)\right )^2 (a+b \text {arctanh}(c+d x))^3}{d^2}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (d e-c f+f (c+d x))^2 (a+b \text {arctanh}(c+d x))^3d(c+d x)}{d^3}\) |
| \(\Big \downarrow \) 6480 |
| \(\displaystyle \frac {\frac {(f (c+d x)-c f+d e)^3 (a+b \text {arctanh}(c+d x))^3}{3 f}-\frac {b \int \left (-\left ((c+d x) (a+b \text {arctanh}(c+d x))^2 f^3\right )-3 (d e-c f) (a+b \text {arctanh}(c+d x))^2 f^2+\frac {\left ((d e-c f) \left (d^2 e^2-2 c d f e+\left (c^2+3\right ) f^2\right )+f \left (3 d^2 e^2-6 c d f e+\left (3 c^2+1\right ) f^2\right ) (c+d x)\right ) (a+b \text {arctanh}(c+d x))^2}{1-(c+d x)^2}\right )d(c+d x)}{f}}{d^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {(f (c+d x)-c f+d e)^3 (a+b \text {arctanh}(c+d x))^3}{3 f}-\frac {b \left (b f \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))-\frac {f \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) (a+b \text {arctanh}(c+d x))^3}{3 b}+\frac {(d e-c f) \left (\left (c^2+3\right ) f^2-2 c d e f+d^2 e^2\right ) (a+b \text {arctanh}(c+d x))^3}{3 b}+f \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \log \left (\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))^2-3 f^2 (d e-c f) (a+b \text {arctanh}(c+d x))^2-3 f^2 (c+d x) (d e-c f) (a+b \text {arctanh}(c+d x))^2+6 b f^2 (d e-c f) \log \left (\frac {2}{-c-d x+1}\right ) (a+b \text {arctanh}(c+d x))-\frac {1}{2} f^3 (c+d x)^2 (a+b \text {arctanh}(c+d x))^2+\frac {1}{2} f^3 (a+b \text {arctanh}(c+d x))^2-a b f^3 (c+d x)-b^2 f^3 (c+d x) \text {arctanh}(c+d x)-\frac {1}{2} b^2 f \left (\left (3 c^2+1\right ) f^2-6 c d e f+3 d^2 e^2\right ) \operatorname {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right )+3 b^2 f^2 (d e-c f) \operatorname {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )-\frac {1}{2} b^2 f^3 \log \left (1-(c+d x)^2\right )\right )}{f}}{d^3}\) |
(((d*e - c*f + f*(c + d*x))^3*(a + b*ArcTanh[c + d*x])^3)/(3*f) - (b*(-(a* b*f^3*(c + d*x)) - b^2*f^3*(c + d*x)*ArcTanh[c + d*x] + (f^3*(a + b*ArcTan h[c + d*x])^2)/2 - 3*f^2*(d*e - c*f)*(a + b*ArcTanh[c + d*x])^2 - 3*f^2*(d *e - c*f)*(c + d*x)*(a + b*ArcTanh[c + d*x])^2 - (f^3*(c + d*x)^2*(a + b*A rcTanh[c + d*x])^2)/2 + ((d*e - c*f)*(d^2*e^2 - 2*c*d*e*f + (3 + c^2)*f^2) *(a + b*ArcTanh[c + d*x])^3)/(3*b) - (f*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^ 2)*f^2)*(a + b*ArcTanh[c + d*x])^3)/(3*b) + 6*b*f^2*(d*e - c*f)*(a + b*Arc Tanh[c + d*x])*Log[2/(1 - c - d*x)] + f*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^ 2)*f^2)*(a + b*ArcTanh[c + d*x])^2*Log[2/(1 - c - d*x)] - (b^2*f^3*Log[1 - (c + d*x)^2])/2 + 3*b^2*f^2*(d*e - c*f)*PolyLog[2, -((1 + c + d*x)/(1 - c - d*x))] + b*f*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^2)*f^2)*(a + b*ArcTanh[c + d*x])*PolyLog[2, 1 - 2/(1 - c - d*x)] - (b^2*f*(3*d^2*e^2 - 6*c*d*e*f + (1 + 3*c^2)*f^2)*PolyLog[3, 1 - 2/(1 - c - d*x)])/2))/f)/d^3
3.1.45.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_S ymbol] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTanh[c*x])^p/(e*(q + 1))), x] - Simp[b*c*(p/(e*(q + 1))) Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p - 1 ), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && NeQ[q, -1]
Int[((a_.) + ArcTanh[(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* ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IG tQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 24.71 (sec) , antiderivative size = 10013, normalized size of antiderivative = 18.34
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(10013\) |
| default | \(\text {Expression too large to display}\) | \(10013\) |
| parts | \(\text {Expression too large to display}\) | \(10025\) |
\[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx=\int { {\left (f x + e\right )}^{2} {\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3} \,d x } \]
integral(a^3*f^2*x^2 + 2*a^3*e*f*x + a^3*e^2 + (b^3*f^2*x^2 + 2*b^3*e*f*x + b^3*e^2)*arctanh(d*x + c)^3 + 3*(a*b^2*f^2*x^2 + 2*a*b^2*e*f*x + a*b^2*e ^2)*arctanh(d*x + c)^2 + 3*(a^2*b*f^2*x^2 + 2*a^2*b*e*f*x + a^2*b*e^2)*arc tanh(d*x + c), x)
\[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx=\int \left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{3} \left (e + f x\right )^{2}\, dx \]
\[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx=\int { {\left (f x + e\right )}^{2} {\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3} \,d x } \]
1/3*a^3*f^2*x^3 + a^3*e*f*x^2 + 3/2*(2*x^2*arctanh(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))*a^2*b*e*f + 1/2*(2*x^3*arctanh(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))*a^2*b*f^2 + a^3*e^2*x + 3/2*(2*(d*x + c)*arctanh(d*x + c ) + log(-(d*x + c)^2 + 1))*a^2*b*e^2/d - 1/24*((b^3*d^3*f^2*x^3 + 3*b^3*d^ 3*e*f*x^2 + 3*b^3*d^3*e^2*x + (c^3*f^2 - 3*d^2*e^2 - 3*(d*e*f + f^2)*c^2 - 3*d*e*f + 3*(d^2*e^2 + 2*d*e*f + f^2)*c - f^2)*b^3)*log(-d*x - c + 1)^3 - 3*(2*a*b^2*d^3*f^2*x^3 + (6*a*b^2*d^3*e*f + b^3*d^2*f^2)*x^2 + 2*(3*a*b^2 *d^3*e^2 + (3*d^2*e*f - 2*c*d*f^2)*b^3)*x + (b^3*d^3*f^2*x^3 + 3*b^3*d^3*e *f*x^2 + 3*b^3*d^3*e^2*x + (c^3*f^2 + 3*d^2*e^2 - 3*(d*e*f - f^2)*c^2 - 3* d*e*f + 3*(d^2*e^2 - 2*d*e*f + f^2)*c + f^2)*b^3)*log(d*x + c + 1))*log(-d *x - c + 1)^2)/d^3 - integrate(-1/8*((b^3*d^3*f^2*x^3 + (2*d^3*e*f + c*d^2 *f^2 - d^2*f^2)*b^3*x^2 + (d^3*e^2 + 2*c*d^2*e*f - 2*d^2*e*f)*b^3*x + (c*d ^2*e^2 - d^2*e^2)*b^3)*log(d*x + c + 1)^3 + 6*(a*b^2*d^3*f^2*x^3 + (2*d^3* e*f + c*d^2*f^2 - d^2*f^2)*a*b^2*x^2 + (d^3*e^2 + 2*c*d^2*e*f - 2*d^2*e*f) *a*b^2*x + (c*d^2*e^2 - d^2*e^2)*a*b^2)*log(d*x + c + 1)^2 - (4*a*b^2*d^3* f^2*x^3 + 2*(6*a*b^2*d^3*e*f + b^3*d^2*f^2)*x^2 + 3*(b^3*d^3*f^2*x^3 + (2* d^3*e*f + c*d^2*f^2 - d^2*f^2)*b^3*x^2 + (d^3*e^2 + 2*c*d^2*e*f - 2*d^2*e* f)*b^3*x + (c*d^2*e^2 - d^2*e^2)*b^3)*log(d*x + c + 1)^2 + 4*(3*a*b^2*d...
\[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx=\int { {\left (f x + e\right )}^{2} {\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3} \,d x } \]
Timed out. \[ \int (e+f x)^2 (a+b \text {arctanh}(c+d x))^3 \, dx=\int {\left (e+f\,x\right )}^2\,{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3 \,d x \]