Integrand size = 18, antiderivative size = 148 \[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=\frac {1}{8} e^{-2 d+\frac {b^2 \log ^2(f)}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {4 f x-b \log (f)}{2 \sqrt {2} \sqrt {f}}\right )+\frac {1}{8} e^{2 d-\frac {b^2 \log ^2(f)}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {4 f x+b \log (f)}{2 \sqrt {2} \sqrt {f}}\right )-\frac {f^{a+b x}}{2 b \log (f)} \]
-1/2*f^(b*x+a)/b/ln(f)+1/16*exp(-2*d+1/8*b^2*ln(f)^2/f)*f^(-1/2+a)*erf(1/4 *(4*f*x-b*ln(f))*2^(1/2)/f^(1/2))*2^(1/2)*Pi^(1/2)+1/16*exp(2*d-1/8*b^2*ln (f)^2/f)*f^(-1/2+a)*erfi(1/4*(4*f*x+b*ln(f))*2^(1/2)/f^(1/2))*2^(1/2)*Pi^( 1/2)
Time = 0.56 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.01 \[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=\frac {1}{16} f^a \left (-\frac {8 f^{b x}}{b \log (f)}+\frac {e^{\frac {b^2 \log ^2(f)}{8 f}} \sqrt {2 \pi } \text {erf}\left (\frac {4 f x-b \log (f)}{2 \sqrt {2} \sqrt {f}}\right ) (\cosh (2 d)-\sinh (2 d))}{\sqrt {f}}+\frac {e^{-\frac {b^2 \log ^2(f)}{8 f}} \sqrt {2 \pi } \text {erfi}\left (\frac {4 f x+b \log (f)}{2 \sqrt {2} \sqrt {f}}\right ) (\cosh (2 d)+\sinh (2 d))}{\sqrt {f}}\right ) \]
(f^a*((-8*f^(b*x))/(b*Log[f]) + (E^((b^2*Log[f]^2)/(8*f))*Sqrt[2*Pi]*Erf[( 4*f*x - b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*(Cosh[2*d] - Sinh[2*d]))/Sqrt[f] + (Sqrt[2*Pi]*Erfi[(4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[f])]*(Cosh[2*d] + Sinh [2*d]))/(E^((b^2*Log[f]^2)/(8*f))*Sqrt[f])))/16
Time = 0.40 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6038, 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.
| \(\displaystyle \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx\) |
| \(\Big \downarrow \) 6038 |
| \(\displaystyle \int \left (\frac {1}{4} e^{-2 d-2 f x^2} f^{a+b x}+\frac {1}{4} e^{2 d+2 f x^2} f^{a+b x}-\frac {1}{2} f^{a+b x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{8} \sqrt {\frac {\pi }{2}} f^{a-\frac {1}{2}} e^{\frac {b^2 \log ^2(f)}{8 f}-2 d} \text {erf}\left (\frac {4 f x-b \log (f)}{2 \sqrt {2} \sqrt {f}}\right )+\frac {1}{8} \sqrt {\frac {\pi }{2}} f^{a-\frac {1}{2}} e^{2 d-\frac {b^2 \log ^2(f)}{8 f}} \text {erfi}\left (\frac {b \log (f)+4 f x}{2 \sqrt {2} \sqrt {f}}\right )-\frac {f^{a+b x}}{2 b \log (f)}\) |
(E^(-2*d + (b^2*Log[f]^2)/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erf[(4*f*x - b*Lo g[f])/(2*Sqrt[2]*Sqrt[f])])/8 + (E^(2*d - (b^2*Log[f]^2)/(8*f))*f^(-1/2 + a)*Sqrt[Pi/2]*Erfi[(4*f*x + b*Log[f])/(2*Sqrt[2]*Sqrt[f])])/8 - f^(a + b*x )/(2*b*Log[f])
3.4.43.3.1 Defintions of rubi rules used
Int[(F_)^(u_)*Sinh[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sinh[v] ^n, x], x] /; FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[ v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]
Time = 0.77 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85
| method | result | size |
| risch | \(-\frac {\operatorname {erf}\left (-\sqrt {2}\, \sqrt {f}\, x +\frac {\ln \left (f \right ) b \sqrt {2}}{4 \sqrt {f}}\right ) \sqrt {2}\, \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {b^{2} \ln \left (f \right )^{2}-16 d f}{8 f}}}{16 \sqrt {f}}-\frac {\operatorname {erf}\left (-\sqrt {-2 f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-2 f}}\right ) \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {b^{2} \ln \left (f \right )^{2}-16 d f}{8 f}}}{8 \sqrt {-2 f}}-\frac {f^{a} f^{b x}}{2 b \ln \left (f \right )}\) | \(126\) |
-1/16*erf(-2^(1/2)*f^(1/2)*x+1/4*ln(f)*b*2^(1/2)/f^(1/2))/f^(1/2)*2^(1/2)* Pi^(1/2)*f^a*exp(1/8*(b^2*ln(f)^2-16*d*f)/f)-1/8*erf(-(-2*f)^(1/2)*x+1/2*l n(f)*b/(-2*f)^(1/2))/(-2*f)^(1/2)*Pi^(1/2)*f^a*exp(-1/8*(b^2*ln(f)^2-16*d* f)/f)-1/2*f^a*f^(b*x)/b/ln(f)
Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (114) = 228\).
Time = 0.31 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.88 \[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } b \sqrt {-f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} - 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right ) \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \left (f\right )\right )} \sqrt {-f}}{4 \, f}\right ) \log \left (f\right ) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \cosh \left (\frac {b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right ) \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \left (f\right )\right )}}{4 \, \sqrt {f}}\right ) \log \left (f\right ) + \sqrt {2} \sqrt {\pi } b \sqrt {f} \operatorname {erf}\left (-\frac {\sqrt {2} {\left (4 \, f x - b \log \left (f\right )\right )}}{4 \, \sqrt {f}}\right ) \log \left (f\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right ) - \sqrt {2} \sqrt {\pi } b \sqrt {-f} \operatorname {erf}\left (\frac {\sqrt {2} {\left (4 \, f x + b \log \left (f\right )\right )} \sqrt {-f}}{4 \, f}\right ) \log \left (f\right ) \sinh \left (\frac {b^{2} \log \left (f\right )^{2} - 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right ) + 8 \, f \cosh \left ({\left (b x + a\right )} \log \left (f\right )\right ) + 8 \, f \sinh \left ({\left (b x + a\right )} \log \left (f\right )\right )}{16 \, b f \log \left (f\right )} \]
-1/16*(sqrt(2)*sqrt(pi)*b*sqrt(-f)*cosh(1/8*(b^2*log(f)^2 - 8*a*f*log(f) - 16*d*f)/f)*erf(1/4*sqrt(2)*(4*f*x + b*log(f))*sqrt(-f)/f)*log(f) + sqrt(2 )*sqrt(pi)*b*sqrt(f)*cosh(1/8*(b^2*log(f)^2 + 8*a*f*log(f) - 16*d*f)/f)*er f(-1/4*sqrt(2)*(4*f*x - b*log(f))/sqrt(f))*log(f) + sqrt(2)*sqrt(pi)*b*sqr t(f)*erf(-1/4*sqrt(2)*(4*f*x - b*log(f))/sqrt(f))*log(f)*sinh(1/8*(b^2*log (f)^2 + 8*a*f*log(f) - 16*d*f)/f) - sqrt(2)*sqrt(pi)*b*sqrt(-f)*erf(1/4*sq rt(2)*(4*f*x + b*log(f))*sqrt(-f)/f)*log(f)*sinh(1/8*(b^2*log(f)^2 - 8*a*f *log(f) - 16*d*f)/f) + 8*f*cosh((b*x + a)*log(f)) + 8*f*sinh((b*x + a)*log (f)))/(b*f*log(f))
\[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=\int f^{a + b x} \sinh ^{2}{\left (d + f x^{2} \right )}\, dx \]
Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.86 \[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=\frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {f} x - \frac {\sqrt {2} b \log \left (f\right )}{4 \, \sqrt {f}}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2}}{8 \, f} - 2 \, d\right )}}{16 \, \sqrt {f}} + \frac {\sqrt {2} \sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {2} \sqrt {-f} x - \frac {\sqrt {2} b \log \left (f\right )}{4 \, \sqrt {-f}}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2}}{8 \, f} + 2 \, d\right )}}{16 \, \sqrt {-f}} - \frac {f^{b x + a}}{2 \, b \log \left (f\right )} \]
1/16*sqrt(2)*sqrt(pi)*f^a*erf(sqrt(2)*sqrt(f)*x - 1/4*sqrt(2)*b*log(f)/sqr t(f))*e^(1/8*b^2*log(f)^2/f - 2*d)/sqrt(f) + 1/16*sqrt(2)*sqrt(pi)*f^a*erf (sqrt(2)*sqrt(-f)*x - 1/4*sqrt(2)*b*log(f)/sqrt(-f))*e^(-1/8*b^2*log(f)^2/ f + 2*d)/sqrt(-f) - 1/2*f^(b*x + a)/(b*log(f))
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.41 \[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} \sqrt {f} {\left (4 \, x - \frac {b \log \left (f\right )}{f}\right )}\right ) e^{\left (\frac {b^{2} \log \left (f\right )^{2} + 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right )}}{16 \, \sqrt {f}} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} \sqrt {-f} {\left (4 \, x + \frac {b \log \left (f\right )}{f}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 8 \, a f \log \left (f\right ) - 16 \, d f}{8 \, f}\right )}}{16 \, \sqrt {-f}} - {\left (\frac {2 \, b \cos \left (-\frac {1}{2} \, \pi b x \mathrm {sgn}\left (f\right ) + \frac {1}{2} \, \pi b x - \frac {1}{2} \, \pi a \mathrm {sgn}\left (f\right ) + \frac {1}{2} \, \pi a\right ) \log \left ({\left | f \right |}\right )}{4 \, b^{2} \log \left ({\left | f \right |}\right )^{2} + {\left (\pi b \mathrm {sgn}\left (f\right ) - \pi b\right )}^{2}} - \frac {{\left (\pi b \mathrm {sgn}\left (f\right ) - \pi b\right )} \sin \left (-\frac {1}{2} \, \pi b x \mathrm {sgn}\left (f\right ) + \frac {1}{2} \, \pi b x - \frac {1}{2} \, \pi a \mathrm {sgn}\left (f\right ) + \frac {1}{2} \, \pi a\right )}{4 \, b^{2} \log \left ({\left | f \right |}\right )^{2} + {\left (\pi b \mathrm {sgn}\left (f\right ) - \pi b\right )}^{2}}\right )} e^{\left (b x \log \left ({\left | f \right |}\right ) + a \log \left ({\left | f \right |}\right )\right )} + i \, {\left (-\frac {i \, e^{\left (\frac {1}{2} i \, \pi b x \mathrm {sgn}\left (f\right ) - \frac {1}{2} i \, \pi b x + \frac {1}{2} i \, \pi a \mathrm {sgn}\left (f\right ) - \frac {1}{2} i \, \pi a\right )}}{2 i \, \pi b \mathrm {sgn}\left (f\right ) - 2 i \, \pi b + 4 \, b \log \left ({\left | f \right |}\right )} + \frac {i \, e^{\left (-\frac {1}{2} i \, \pi b x \mathrm {sgn}\left (f\right ) + \frac {1}{2} i \, \pi b x - \frac {1}{2} i \, \pi a \mathrm {sgn}\left (f\right ) + \frac {1}{2} i \, \pi a\right )}}{-2 i \, \pi b \mathrm {sgn}\left (f\right ) + 2 i \, \pi b + 4 \, b \log \left ({\left | f \right |}\right )}\right )} e^{\left (b x \log \left ({\left | f \right |}\right ) + a \log \left ({\left | f \right |}\right )\right )} \]
-1/16*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*sqrt(f)*(4*x - b*log(f)/f))*e^(1/8 *(b^2*log(f)^2 + 8*a*f*log(f) - 16*d*f)/f)/sqrt(f) - 1/16*sqrt(2)*sqrt(pi) *erf(-1/4*sqrt(2)*sqrt(-f)*(4*x + b*log(f)/f))*e^(-1/8*(b^2*log(f)^2 - 8*a *f*log(f) - 16*d*f)/f)/sqrt(-f) - (2*b*cos(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)*log(abs(f))/(4*b^2*log(abs(f))^2 + (pi*b*sg n(f) - pi*b)^2) - (pi*b*sgn(f) - pi*b)*sin(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)/(4*b^2*log(abs(f))^2 + (pi*b*sgn(f) - pi*b) ^2))*e^(b*x*log(abs(f)) + a*log(abs(f))) + I*(-I*e^(1/2*I*pi*b*x*sgn(f) - 1/2*I*pi*b*x + 1/2*I*pi*a*sgn(f) - 1/2*I*pi*a)/(2*I*pi*b*sgn(f) - 2*I*pi*b + 4*b*log(abs(f))) + I*e^(-1/2*I*pi*b*x*sgn(f) + 1/2*I*pi*b*x - 1/2*I*pi* a*sgn(f) + 1/2*I*pi*a)/(-2*I*pi*b*sgn(f) + 2*I*pi*b + 4*b*log(abs(f))))*e^ (b*x*log(abs(f)) + a*log(abs(f)))
Timed out. \[ \int f^{a+b x} \sinh ^2\left (d+f x^2\right ) \, dx=\int f^{a+b\,x}\,{\mathrm {sinh}\left (f\,x^2+d\right )}^2 \,d x \]