Integrand size = 23, antiderivative size = 149 \[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=-\frac {a^2 \cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d}+\frac {3 a^2 \log (c+d x)}{2 d}+\frac {2 i a^2 \text {Chi}\left (\frac {c f}{d}+f x\right ) \sinh \left (e-\frac {c f}{d}\right )}{d}+\frac {2 i a^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{d}-\frac {a^2 \sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 c f}{d}+2 f x\right )}{2 d} \]
-1/2*a^2*Chi(2*c*f/d+2*f*x)*cosh(-2*e+2*c*f/d)/d+3/2*a^2*ln(d*x+c)/d+2*I*a ^2*cosh(-e+c*f/d)*Shi(c*f/d+f*x)/d+1/2*a^2*Shi(2*c*f/d+2*f*x)*sinh(-2*e+2* c*f/d)/d-2*I*a^2*Chi(c*f/d+f*x)*sinh(-e+c*f/d)/d
Time = 0.77 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=-\frac {a^2 \left (\cosh \left (2 e-\frac {2 c f}{d}\right ) \text {Chi}\left (\frac {2 f (c+d x)}{d}\right )-3 \log (c+d x)-4 i \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right ) \sinh \left (e-\frac {c f}{d}\right )-4 i \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )+\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (\frac {2 f (c+d x)}{d}\right )\right )}{2 d} \]
-1/2*(a^2*(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*f*(c + d*x))/d] - 3*Log[c + d*x] - (4*I)*CoshIntegral[f*(c/d + x)]*Sinh[e - (c*f)/d] - (4*I)*Cosh[e - (c*f)/d]*SinhIntegral[f*(c/d + x)] + Sinh[2*e - (2*c*f)/d]*SinhIntegral [(2*f*(c + d*x))/d]))/d
Time = 0.58 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3799, 3042, 3793, 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.
\(\displaystyle \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+a \sin (i e+i f x))^2}{c+d x}dx\) |
\(\Big \downarrow \) 3799 |
\(\displaystyle 4 a^2 \int \frac {\sinh ^4\left (\frac {e}{2}+\frac {f x}{2}-\frac {i \pi }{4}\right )}{c+d x}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 4 a^2 \int \frac {\sin \left (\frac {i e}{2}+\frac {i f x}{2}+\frac {\pi }{4}\right )^4}{c+d x}dx\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle 4 a^2 \int \left (-\frac {\cosh (2 e+2 f x)}{8 (c+d x)}+\frac {i \sinh (e+f x)}{2 (c+d x)}+\frac {3}{8 (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 a^2 \left (\frac {i \text {Chi}\left (x f+\frac {c f}{d}\right ) \sinh \left (e-\frac {c f}{d}\right )}{2 d}-\frac {\text {Chi}\left (2 x f+\frac {2 c f}{d}\right ) \cosh \left (2 e-\frac {2 c f}{d}\right )}{8 d}-\frac {\sinh \left (2 e-\frac {2 c f}{d}\right ) \text {Shi}\left (2 x f+\frac {2 c f}{d}\right )}{8 d}+\frac {i \cosh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{2 d}+\frac {3 \log (c+d x)}{8 d}\right )\) |
4*a^2*(-1/8*(Cosh[2*e - (2*c*f)/d]*CoshIntegral[(2*c*f)/d + 2*f*x])/d + (3 *Log[c + d*x])/(8*d) + ((I/2)*CoshIntegral[(c*f)/d + f*x]*Sinh[e - (c*f)/d ])/d + ((I/2)*Cosh[e - (c*f)/d]*SinhIntegral[(c*f)/d + f*x])/d - (Sinh[2*e - (2*c*f)/d]*SinhIntegral[(2*c*f)/d + 2*f*x])/(8*d))
3.2.5.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 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])
Time = 2.36 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.30
method | result | size |
risch | \(-\frac {i a^{2} {\mathrm e}^{-\frac {c f -d e}{d}} \operatorname {Ei}_{1}\left (-f x -e -\frac {c f -d e}{d}\right )}{d}+\frac {3 a^{2} \ln \left (d x +c \right )}{2 d}+\frac {a^{2} {\mathrm e}^{-\frac {2 \left (c f -d e \right )}{d}} \operatorname {Ei}_{1}\left (-2 f x -2 e -\frac {2 \left (c f -d e \right )}{d}\right )}{4 d}+\frac {a^{2} {\mathrm e}^{\frac {2 c f -2 d e}{d}} \operatorname {Ei}_{1}\left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right )}{4 d}+\frac {i a^{2} {\mathrm e}^{\frac {c f -d e}{d}} \operatorname {Ei}_{1}\left (f x +e +\frac {c f -d e}{d}\right )}{d}\) | \(193\) |
-I*a^2/d*exp(-(c*f-d*e)/d)*Ei(1,-f*x-e-(c*f-d*e)/d)+3/2*a^2*ln(d*x+c)/d+1/ 4*a^2/d*exp(-2*(c*f-d*e)/d)*Ei(1,-2*f*x-2*e-2*(c*f-d*e)/d)+1/4*a^2/d*exp(2 *(c*f-d*e)/d)*Ei(1,2*f*x+2*e+2*(c*f-d*e)/d)+I*a^2/d*exp((c*f-d*e)/d)*Ei(1, f*x+e+(c*f-d*e)/d)
Time = 0.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=-\frac {a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (\frac {2 \, {\left (d e - c f\right )}}{d}\right )} - 4 i \, a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (\frac {d e - c f}{d}\right )} + 4 i \, a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-\frac {d e - c f}{d}\right )} + a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right )} - 6 \, a^{2} \log \left (\frac {d x + c}{d}\right )}{4 \, d} \]
-1/4*(a^2*Ei(2*(d*f*x + c*f)/d)*e^(2*(d*e - c*f)/d) - 4*I*a^2*Ei((d*f*x + c*f)/d)*e^((d*e - c*f)/d) + 4*I*a^2*Ei(-(d*f*x + c*f)/d)*e^(-(d*e - c*f)/d ) + a^2*Ei(-2*(d*f*x + c*f)/d)*e^(-2*(d*e - c*f)/d) - 6*a^2*log((d*x + c)/ d))/d
\[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=- a^{2} \left (\int \frac {\sinh ^{2}{\left (e + f x \right )}}{c + d x}\, dx + \int \left (- \frac {2 i \sinh {\left (e + f x \right )}}{c + d x}\right )\, dx + \int \left (- \frac {1}{c + d x}\right )\, dx\right ) \]
-a**2*(Integral(sinh(e + f*x)**2/(c + d*x), x) + Integral(-2*I*sinh(e + f* x)/(c + d*x), x) + Integral(-1/(c + d*x), x))
Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=\frac {1}{4} \, a^{2} {\left (\frac {e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{1}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{1}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {2 \, \log \left (d x + c\right )}{d}\right )} + i \, a^{2} {\left (\frac {e^{\left (-e + \frac {c f}{d}\right )} E_{1}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {e^{\left (e - \frac {c f}{d}\right )} E_{1}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} + \frac {a^{2} \log \left (d x + c\right )}{d} \]
1/4*a^2*(e^(-2*e + 2*c*f/d)*exp_integral_e(1, 2*(d*x + c)*f/d)/d + e^(2*e - 2*c*f/d)*exp_integral_e(1, -2*(d*x + c)*f/d)/d + 2*log(d*x + c)/d) + I*a ^2*(e^(-e + c*f/d)*exp_integral_e(1, (d*x + c)*f/d)/d - e^(e - c*f/d)*exp_ integral_e(1, -(d*x + c)*f/d)/d) + a^2*log(d*x + c)/d
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91 \[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=-\frac {a^{2} {\rm Ei}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} - 4 i \, a^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} + 4 i \, a^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + a^{2} {\rm Ei}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} - 6 \, a^{2} \log \left (d x + c\right )}{4 \, d} \]
-1/4*(a^2*Ei(2*(d*f*x + c*f)/d)*e^(2*e - 2*c*f/d) - 4*I*a^2*Ei((d*f*x + c* f)/d)*e^(e - c*f/d) + 4*I*a^2*Ei(-(d*f*x + c*f)/d)*e^(-e + c*f/d) + a^2*Ei (-2*(d*f*x + c*f)/d)*e^(-2*e + 2*c*f/d) - 6*a^2*log(d*x + c))/d
Timed out. \[ \int \frac {(a+i a \sinh (e+f x))^2}{c+d x} \, dx=\int \frac {{\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{c+d\,x} \,d x \]