Integrand size = 16, antiderivative size = 119 \[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=-\frac {(c+d x)^3}{f}+\frac {(c+d x)^4}{4 d}+\frac {3 d (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}-\frac {3 d^3 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^4}-\frac {(c+d x)^3 \tanh (e+f x)}{f} \]
-(d*x+c)^3/f+1/4*(d*x+c)^4/d+3*d*(d*x+c)^2*ln(1+exp(2*f*x+2*e))/f^2+3*d^2* (d*x+c)*polylog(2,-exp(2*f*x+2*e))/f^3-3/2*d^3*polylog(3,-exp(2*f*x+2*e))/ f^4-(d*x+c)^3*tanh(f*x+e)/f
Time = 0.57 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.50 \[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=\frac {1}{4} \left (x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )+\frac {2 d e^{2 e} \left (\frac {4 e^{-2 e} (c+d x)^3}{d}+\frac {6 \left (1+e^{-2 e}\right ) (c+d x)^2 \log \left (1+e^{-2 (e+f x)}\right )}{f}-\frac {3 d \left (1+e^{-2 e}\right ) \left (2 f (c+d x) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )\right )}{f^3}\right )}{\left (1+e^{2 e}\right ) f}-\frac {4 (c+d x)^3 \text {sech}(e) \text {sech}(e+f x) \sinh (f x)}{f}\right ) \]
(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3) + (2*d*E^(2*e)*((4*(c + d*x )^3)/(d*E^(2*e)) + (6*(1 + E^(-2*e))*(c + d*x)^2*Log[1 + E^(-2*(e + f*x))] )/f - (3*d*(1 + E^(-2*e))*(2*f*(c + d*x)*PolyLog[2, -E^(-2*(e + f*x))] + d *PolyLog[3, -E^(-2*(e + f*x))]))/f^3))/((1 + E^(2*e))*f) - (4*(c + d*x)^3* Sech[e]*Sech[e + f*x]*Sinh[f*x])/f)/4
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.19, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 25, 4203, 17, 26, 3042, 26, 4201, 2620, 3011, 2720, 7143}
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.
\(\displaystyle \int (c+d x)^3 \tanh ^2(e+f x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -(c+d x)^3 \tan (i e+i f x)^2dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int (c+d x)^3 \tan (i e+i f x)^2dx\) |
\(\Big \downarrow \) 4203 |
\(\displaystyle -\frac {3 i d \int i (c+d x)^2 \tanh (e+f x)dx}{f}+\int (c+d x)^3dx-\frac {(c+d x)^3 \tanh (e+f x)}{f}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle -\frac {3 i d \int i (c+d x)^2 \tanh (e+f x)dx}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {3 d \int (c+d x)^2 \tanh (e+f x)dx}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 d \int -i (c+d x)^2 \tan (i e+i f x)dx}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {3 i d \int (c+d x)^2 \tan (i e+i f x)dx}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle -\frac {3 i d \left (2 i \int \frac {e^{2 (e+f x)} (c+d x)^2}{1+e^{2 (e+f x)}}dx-\frac {i (c+d x)^3}{3 d}\right )}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\frac {3 i d \left (2 i \left (\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {d \int (c+d x) \log \left (1+e^{2 (e+f x)}\right )dx}{f}\right )-\frac {i (c+d x)^3}{3 d}\right )}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -\frac {3 i d \left (2 i \left (\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {d \left (\frac {d \int \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )dx}{2 f}-\frac {(c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f}\right )}{f}\right )-\frac {i (c+d x)^3}{3 d}\right )}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\frac {3 i d \left (2 i \left (\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {d \left (\frac {d \int e^{-2 (e+f x)} \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )de^{2 (e+f x)}}{4 f^2}-\frac {(c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f}\right )}{f}\right )-\frac {i (c+d x)^3}{3 d}\right )}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -\frac {3 i d \left (2 i \left (\frac {(c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {d \left (\frac {d \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{4 f^2}-\frac {(c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f}\right )}{f}\right )-\frac {i (c+d x)^3}{3 d}\right )}{f}-\frac {(c+d x)^3 \tanh (e+f x)}{f}+\frac {(c+d x)^4}{4 d}\) |
(c + d*x)^4/(4*d) - ((3*I)*d*(((-1/3*I)*(c + d*x)^3)/d + (2*I)*(((c + d*x) ^2*Log[1 + E^(2*(e + f*x))])/(2*f) - (d*(-1/2*((c + d*x)*PolyLog[2, -E^(2* (e + f*x))])/f + (d*PolyLog[3, -E^(2*(e + f*x))])/(4*f^2)))/f)))/f - ((c + d*x)^3*Tanh[e + f*x])/f
3.1.6.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x _Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I Int[ (c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si mp[b*d*(m/(f*(n - 1))) Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] , x] - Simp[b^2 Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Leaf count of result is larger than twice the leaf count of optimal. \(335\) vs. \(2(115)=230\).
Time = 0.22 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.82
method | result | size |
risch | \(\frac {d^{3} x^{4}}{4}+d^{2} c \,x^{3}+\frac {3 d \,c^{2} x^{2}}{2}+c^{3} x +\frac {c^{4}}{4 d}+\frac {2 d^{3} x^{3}+6 c \,d^{2} x^{2}+6 c^{2} d x +2 c^{3}}{f \left (1+{\mathrm e}^{2 f x +2 e}\right )}-\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{4}}-\frac {2 d^{3} x^{3}}{f}+\frac {4 d^{3} e^{3}}{f^{4}}+\frac {3 d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f^{2}}-\frac {6 d^{2} c \,x^{2}}{f}-\frac {6 d^{2} c \,e^{2}}{f^{3}}+\frac {6 d^{2} c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {3 d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}-\frac {6 d^{3} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {6 d^{3} e^{2} x}{f^{3}}+\frac {3 d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{3}}-\frac {12 d^{2} c e x}{f^{2}}+\frac {12 d^{2} c e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {3 d \,c^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}\) | \(336\) |
1/4*d^3*x^4+d^2*c*x^3+3/2*d*c^2*x^2+c^3*x+1/4/d*c^4+2*(d^3*x^3+3*c*d^2*x^2 +3*c^2*d*x+c^3)/f/(1+exp(2*f*x+2*e))-3/2*d^3*polylog(3,-exp(2*f*x+2*e))/f^ 4-2/f*d^3*x^3+4/f^4*d^3*e^3+3/f^2*d^3*ln(1+exp(2*f*x+2*e))*x^2-6/f*d^2*c*x ^2-6/f^3*d^2*c*e^2+6/f^2*d^2*c*ln(1+exp(2*f*x+2*e))*x+3/f^3*d^2*c*polylog( 2,-exp(2*f*x+2*e))-6/f^4*d^3*e^2*ln(exp(f*x+e))+6/f^3*d^3*e^2*x+3/f^3*d^3* polylog(2,-exp(2*f*x+2*e))*x-12/f^2*d^2*c*e*x+12/f^3*d^2*c*e*ln(exp(f*x+e) )+3/f^2*d*c^2*ln(1+exp(2*f*x+2*e))-6/f^2*d*c^2*ln(exp(f*x+e))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 1505, normalized size of antiderivative = 12.65 \[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=\text {Too large to display} \]
1/4*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*c^2*d*f^4*x^2 + 4*c^3*f^4*x - 8*d^3 *e^3 + 24*c*d^2*e^2*f - 24*c^2*d*e*f^2 + 8*c^3*f^3 + (d^3*f^4*x^4 - 8*d^3* e^3 + 24*c*d^2*e^2*f - 24*c^2*d*e*f^2 + 4*(c*d^2*f^4 - 2*d^3*f^3)*x^3 + 6* (c^2*d*f^4 - 4*c*d^2*f^3)*x^2 + 4*(c^3*f^4 - 6*c^2*d*f^3)*x)*cosh(f*x + e) ^2 + 2*(d^3*f^4*x^4 - 8*d^3*e^3 + 24*c*d^2*e^2*f - 24*c^2*d*e*f^2 + 4*(c*d ^2*f^4 - 2*d^3*f^3)*x^3 + 6*(c^2*d*f^4 - 4*c*d^2*f^3)*x^2 + 4*(c^3*f^4 - 6 *c^2*d*f^3)*x)*cosh(f*x + e)*sinh(f*x + e) + (d^3*f^4*x^4 - 8*d^3*e^3 + 24 *c*d^2*e^2*f - 24*c^2*d*e*f^2 + 4*(c*d^2*f^4 - 2*d^3*f^3)*x^3 + 6*(c^2*d*f ^4 - 4*c*d^2*f^3)*x^2 + 4*(c^3*f^4 - 6*c^2*d*f^3)*x)*sinh(f*x + e)^2 + 24* (d^3*f*x + c*d^2*f + (d^3*f*x + c*d^2*f)*cosh(f*x + e)^2 + 2*(d^3*f*x + c* d^2*f)*cosh(f*x + e)*sinh(f*x + e) + (d^3*f*x + c*d^2*f)*sinh(f*x + e)^2)* dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) + 24*(d^3*f*x + c*d^2*f + (d^3*f* x + c*d^2*f)*cosh(f*x + e)^2 + 2*(d^3*f*x + c*d^2*f)*cosh(f*x + e)*sinh(f* x + e) + (d^3*f*x + c*d^2*f)*sinh(f*x + e)^2)*dilog(-I*cosh(f*x + e) - I*s inh(f*x + e)) + 12*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 + (d^3*e^2 - 2*c*d^2 *e*f + c^2*d*f^2)*cosh(f*x + e)^2 + 2*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)* cosh(f*x + e)*sinh(f*x + e) + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*sinh(f*x + e)^2)*log(cosh(f*x + e) + sinh(f*x + e) + I) + 12*(d^3*e^2 - 2*c*d^2*e* f + c^2*d*f^2 + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cosh(f*x + e)^2 + 2*(d ^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cosh(f*x + e)*sinh(f*x + e) + (d^3*e^...
\[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=\int \left (c + d x\right )^{3} \tanh ^{2}{\left (e + f x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (114) = 228\).
Time = 0.37 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.88 \[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=c^{3} {\left (x + \frac {e}{f} - \frac {2}{f {\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} - \frac {3}{2} \, c^{2} d {\left (\frac {2 \, x e^{\left (2 \, f x + 2 \, e\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac {f x^{2} + {\left (f x^{2} e^{\left (2 \, e\right )} - 2 \, x e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac {2 \, \log \left ({\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-2 \, e\right )}\right )}{f^{2}}\right )} + \frac {3 \, {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} c d^{2}}{f^{3}} + \frac {d^{3} f x^{4} + 24 \, c d^{2} x^{2} + 4 \, {\left (c d^{2} f + 2 \, d^{3}\right )} x^{3} + {\left (d^{3} f x^{4} e^{\left (2 \, e\right )} + 4 \, c d^{2} f x^{3} e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{4 \, {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )}} + \frac {3 \, {\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} d^{3}}{2 \, f^{4}} - \frac {2 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2}\right )}}{f^{4}} \]
c^3*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) - 3/2*c^2*d*(2*x*e^(2*f*x + 2 *e)/(f*e^(2*f*x + 2*e) + f) - (f*x^2 + (f*x^2*e^(2*e) - 2*x*e^(2*e))*e^(2* f*x))/(f*e^(2*f*x + 2*e) + f) - 2*log((e^(2*f*x + 2*e) + 1)*e^(-2*e))/f^2) + 3*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*c*d^2/f^3 + 1/4*(d^3*f*x^4 + 24*c*d^2*x^2 + 4*(c*d^2*f + 2*d^3)*x^3 + (d^3*f*x^4*e^( 2*e) + 4*c*d^2*f*x^3*e^(2*e))*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) + 3/2*(2* f^2*x^2*log(e^(2*f*x + 2*e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - polylog (3, -e^(2*f*x + 2*e)))*d^3/f^4 - 2*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2)/f^4
\[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=\int { {\left (d x + c\right )}^{3} \tanh \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int (c+d x)^3 \tanh ^2(e+f x) \, dx=\int {\mathrm {tanh}\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^3 \,d x \]