Integrand size = 29, antiderivative size = 130 \[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i (e+f x)^2}{a d}-\frac {i (e+f x)^3}{3 a f}-\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {4 i f^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \]
I*(f*x+e)^2/a/d-1/3*I*(f*x+e)^3/a/f-4*I*f*(f*x+e)*ln(1+I*exp(d*x+c))/a/d^2 -4*I*f^2*polylog(2,-I*exp(d*x+c))/a/d^3+I*(f*x+e)^2*tanh(1/2*c+1/4*I*Pi+1/ 2*d*x)/a/d
Time = 1.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.44 \[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {-i x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {-6 d (e+f x) \left (d (e+f x)+2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right )\right )+12 \left (1+i e^c\right ) f^2 \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )}{d^3 \left (-i+e^c\right )}+\frac {6 i (e+f x)^2 \sinh \left (\frac {d x}{2}\right )}{d \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}}{3 a} \]
((-I)*x*(3*e^2 + 3*e*f*x + f^2*x^2) + (-6*d*(e + f*x)*(d*(e + f*x) + 2*(1 + I*E^c)*f*Log[1 - I*E^(-c - d*x)]) + 12*(1 + I*E^c)*f^2*PolyLog[2, I*E^(- c - d*x)])/(d^3*(-I + E^c)) + ((6*I)*(e + f*x)^2*Sinh[(d*x)/2])/(d*(Cosh[c /2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))/(3*a)
Time = 0.72 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {6091, 17, 3042, 3799, 25, 25, 3042, 4672, 26, 3042, 26, 4199, 26, 2620, 2715, 2838}
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.
| \(\displaystyle \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 6091 |
| \(\displaystyle i \int \frac {(e+f x)^2}{i \sinh (c+d x) a+a}dx-\frac {i \int (e+f x)^2dx}{a}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle i \int \frac {(e+f x)^2}{i \sinh (c+d x) a+a}dx-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \int \frac {(e+f x)^2}{\sin (i c+i d x) a+a}dx-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 3799 |
| \(\displaystyle \frac {i \int -(e+f x)^2 \text {csch}^2\left (\frac {c}{2}+\frac {d x}{2}-\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i \int -(e+f x)^2 \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {i \int (e+f x)^2 \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {i \int (e+f x)^2 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 4672 |
| \(\displaystyle \frac {i \left (\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {4 i f \int -i (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {i \left (\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {4 f \int (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {i \left (\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {4 f \int -i (e+f x) \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {i \left (\frac {4 i f \int (e+f x) \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}+\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 4199 |
| \(\displaystyle \frac {i \left (\frac {4 i f \left (2 i \int \frac {i e^{c+d x} (e+f x)}{1+i e^{c+d x}}dx-\frac {i (e+f x)^2}{2 f}\right )}{d}+\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {i \left (\frac {4 i f \left (-2 \int \frac {e^{c+d x} (e+f x)}{1+i e^{c+d x}}dx-\frac {i (e+f x)^2}{2 f}\right )}{d}+\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {i \left (\frac {4 i f \left (-2 \left (\frac {i f \int \log \left (1+i e^{c+d x}\right )dx}{d}-\frac {i (e+f x) \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}+\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {i \left (\frac {4 i f \left (-2 \left (\frac {i f \int e^{-c-d x} \log \left (1+i e^{c+d x}\right )de^{c+d x}}{d^2}-\frac {i (e+f x) \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}+\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {i \left (\frac {4 i f \left (-2 \left (-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}-\frac {i (e+f x) \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}+\frac {2 (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^3}{3 a f}\) |
((-1/3*I)*(e + f*x)^3)/(a*f) + ((I/2)*(((4*I)*f*(((-1/2*I)*(e + f*x)^2)/f - 2*(((-I)*(e + f*x)*Log[1 + I*E^(c + d*x)])/d - (I*f*PolyLog[2, (-I)*E^(c + d*x)])/d^2)))/d + (2*(e + f*x)^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/d))/a
3.2.88.3.1 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[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Simp[(2*a)^n Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^ 2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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 ))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && In tegerQ[4*k] && IGtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_ .)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/b Int[(e + f*x)^m*Sinh[ c + d*x]^(n - 1), x], x] - Simp[a/b Int[(e + f*x)^m*(Sinh[c + d*x]^(n - 1 )/(a + b*Sinh[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (112 ) = 224\).
Time = 1.66 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.47
| method | result | size |
| risch | \(\frac {2 i f^{2} x^{2}}{a d}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 i f^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {2 i c \,f^{2} \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a \,d^{3}}-\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {4 i e f \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {4 e f \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {2 i e f \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a \,d^{2}}+\frac {4 i f^{2} c x}{a \,d^{2}}-\frac {i f^{2} x^{3}}{3 a}-\frac {i f e \,x^{2}}{a}-\frac {4 i c \,f^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {i e^{3}}{3 a f}-\frac {4 c \,f^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {i e^{2} x}{a}\) | \(321\) |
2*I/a/d*f^2*x^2+2*I/a/d^3*f^2*c^2-4*I*f^2*polylog(2,-I*exp(d*x+c))/a/d^3+2 *I/a/d^3*c*f^2*ln(1+exp(2*d*x+2*c))-2*(f^2*x^2+2*e*f*x+e^2)/d/a/(exp(d*x+c )-I)+4*I/a/d^2*e*f*ln(exp(d*x+c))+4/a/d^2*e*f*arctan(exp(d*x+c))-4*I/a/d^3 *f^2*ln(1+I*exp(d*x+c))*c-2*I/a/d^2*e*f*ln(1+exp(2*d*x+2*c))+4*I/a/d^2*f^2 *c*x-1/3*I/a*f^2*x^3-I/a*f*e*x^2-4*I/a/d^3*c*f^2*ln(exp(d*x+c))-4*I/a/d^2* f^2*ln(1+I*exp(d*x+c))*x-1/3*I/a/f*e^3-4/a/d^3*c*f^2*arctan(exp(d*x+c))-I/ a*e^2*x
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (103) = 206\).
Time = 0.26 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.01 \[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 6 \, d^{2} e^{2} - 12 \, c d e f + 6 \, c^{2} f^{2} + 12 \, {\left (i \, f^{2} e^{\left (d x + c\right )} + f^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - {\left (-i \, d^{3} f^{2} x^{3} + 12 i \, c d e f - 6 i \, c^{2} f^{2} - 3 \, {\left (i \, d^{3} e f - 2 i \, d^{2} f^{2}\right )} x^{2} - 3 \, {\left (i \, d^{3} e^{2} - 4 i \, d^{2} e f\right )} x\right )} e^{\left (d x + c\right )} + 12 \, {\left (d e f - c f^{2} + {\left (i \, d e f - i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 12 \, {\left (d f^{2} x + c f^{2} + {\left (i \, d f^{2} x + i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, {\left (a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}\right )}} \]
-1/3*(d^3*f^2*x^3 + 3*d^3*e*f*x^2 + 3*d^3*e^2*x + 6*d^2*e^2 - 12*c*d*e*f + 6*c^2*f^2 + 12*(I*f^2*e^(d*x + c) + f^2)*dilog(-I*e^(d*x + c)) - (-I*d^3* f^2*x^3 + 12*I*c*d*e*f - 6*I*c^2*f^2 - 3*(I*d^3*e*f - 2*I*d^2*f^2)*x^2 - 3 *(I*d^3*e^2 - 4*I*d^2*e*f)*x)*e^(d*x + c) + 12*(d*e*f - c*f^2 + (I*d*e*f - I*c*f^2)*e^(d*x + c))*log(e^(d*x + c) - I) + 12*(d*f^2*x + c*f^2 + (I*d*f ^2*x + I*c*f^2)*e^(d*x + c))*log(I*e^(d*x + c) + 1))/(a*d^3*e^(d*x + c) - I*a*d^3)
\[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {- 2 e^{2} - 4 e f x - 2 f^{2} x^{2}}{a d e^{c} e^{d x} - i a d} - \frac {i \left (\int \left (- \frac {i d e^{2}}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {4 i e f}{e^{c} e^{d x} - i}\, dx + \int \frac {4 i f^{2} x}{e^{c} e^{d x} - i}\, dx + \int \left (- \frac {i d f^{2} x^{2}}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {d e^{2} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \left (- \frac {2 i d e f x}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {d f^{2} x^{2} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \frac {2 d e f x e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx\right )}{a d} \]
(-2*e**2 - 4*e*f*x - 2*f**2*x**2)/(a*d*exp(c)*exp(d*x) - I*a*d) - I*(Integ ral(-I*d*e**2/(exp(c)*exp(d*x) - I), x) + Integral(4*I*e*f/(exp(c)*exp(d*x ) - I), x) + Integral(4*I*f**2*x/(exp(c)*exp(d*x) - I), x) + Integral(-I*d *f**2*x**2/(exp(c)*exp(d*x) - I), x) + Integral(d*e**2*exp(c)*exp(d*x)/(ex p(c)*exp(d*x) - I), x) + Integral(-2*I*d*e*f*x/(exp(c)*exp(d*x) - I), x) + Integral(d*f**2*x**2*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x) + Integral (2*d*e*f*x*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x))/(a*d)
\[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sinh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \]
1/3*f^2*((-I*d*x^3*e^(d*x + c) - d*x^3 - 6*x^2)/(a*d*e^(d*x + c) - I*a*d) + 12*integrate(x/(a*d*e^(d*x + c) - I*a*d), x)) + e*f*((-I*d*x^2 + (d*x^2* e^c - 4*x*e^c)*e^(d*x))/(I*a*d*e^(d*x + c) + a*d) - 4*I*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) - e^2*(I*(d*x + c)/(a*d) + 2/((a*e^(-d*x - c) + I*a)* d))
\[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sinh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {\mathrm {sinh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]