\(\int f^{a+b x+c x^2} \sinh (d+e x) \, dx\) [357]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 153 \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\frac {e^{-d-\frac {(e-b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{d-\frac {(e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

-1/4*exp(-d-1/4*(e-b*ln(f))^2/c/ln(f))*f^a*erfi(1/2*(-e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(
1/2)/ln(f)^(1/2)+1/4*exp(d-1/4*(e+b*ln(f))^2/c/ln(f))*f^a*erfi(1/2*(e+b*ln(f)+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2)
)*Pi^(1/2)/c^(1/2)/ln(f)^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5623, 2325, 2266, 2235} \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\frac {\sqrt {\pi } f^a e^{-\frac {(e-b \log (f))^2}{4 c \log (f)}-d} \text {erfi}\left (\frac {-b \log (f)-2 c x \log (f)+e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a e^{d-\frac {(b \log (f)+e)^2}{4 c \log (f)}} \text {erfi}\left (\frac {b \log (f)+2 c x \log (f)+e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(a + b*x + c*x^2)*Sinh[d + e*x],x]

[Out]

(E^(-d - (e - b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]
])])/(4*Sqrt[c]*Sqrt[Log[f]]) + (E^(d - (e + b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(e + b*Log[f] + 2*c*x
*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(4*Sqrt[c]*Sqrt[Log[f]])

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2325

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 5623

Int[(F_)^(u_)*Sinh[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sinh[v]^n, x], x] /; FreeQ[F, x] && (Linea
rQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2} e^{-d-e x} f^{a+b x+c x^2}+\frac {1}{2} e^{d+e x} f^{a+b x+c x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int e^{-d-e x} f^{a+b x+c x^2} \, dx\right )+\frac {1}{2} \int e^{d+e x} f^{a+b x+c x^2} \, dx \\ & = -\left (\frac {1}{2} \int \exp \left (-d+a \log (f)+c x^2 \log (f)-x (e-b \log (f))\right ) \, dx\right )+\frac {1}{2} \int \exp \left (d+a \log (f)+c x^2 \log (f)+x (e+b \log (f))\right ) \, dx \\ & = -\left (\frac {1}{2} \left (e^{-d-\frac {(e-b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(-e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx\right )+\frac {1}{2} \left (e^{d-\frac {(e+b \log (f))^2}{4 c \log (f)}} f^a\right ) \int e^{\frac {(e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}} \, dx \\ & = \frac {e^{-d-\frac {(e-b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{d-\frac {(e+b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.88 \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\frac {e^{-\frac {e (e+2 b \log (f))}{4 c \log (f)}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (-e^{\frac {b e}{c}} \text {erfi}\left (\frac {-e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (d)-\sinh (d))+\text {erfi}\left (\frac {e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cosh (d)+\sinh (d))\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Integrate[f^(a + b*x + c*x^2)*Sinh[d + e*x],x]

[Out]

(f^(a - b^2/(4*c))*Sqrt[Pi]*(-(E^((b*e)/c)*Erfi[(-e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[d] -
 Sinh[d])) + Erfi[(e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cosh[d] + Sinh[d])))/(4*Sqrt[c]*E^((e*(e
 + 2*b*Log[f]))/(4*c*Log[f]))*Sqrt[Log[f]])

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.06

method result size
risch \(-\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {2 \ln \left (f \right ) b e -4 d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}}}{4 \sqrt {-c \ln \left (f \right )}}+\frac {\operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-e}{2 \sqrt {-c \ln \left (f \right )}}\right ) \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {2 \ln \left (f \right ) b e -4 d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}}}{4 \sqrt {-c \ln \left (f \right )}}\) \(162\)

[In]

int(f^(c*x^2+b*x+a)*sinh(e*x+d),x,method=_RETURNVERBOSE)

[Out]

-1/4*erf(-(-c*ln(f))^(1/2)*x+1/2*(e+b*ln(f))/(-c*ln(f))^(1/2))/(-c*ln(f))^(1/2)*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*ex
p(-1/4*(2*ln(f)*b*e-4*d*ln(f)*c+e^2)/ln(f)/c)+1/4*erf(-(-c*ln(f))^(1/2)*x+1/2*(b*ln(f)-e)/(-c*ln(f))^(1/2))/(-
c*ln(f))^(1/2)*Pi^(1/2)*f^a*f^(-1/4*b^2/c)*exp(1/4*(2*ln(f)*b*e-4*d*ln(f)*c-e^2)/ln(f)/c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (121) = 242\).

Time = 0.31 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.72 \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=-\frac {\sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} - 2 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} - 2 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) - \sqrt {-c \log \left (f\right )} {\left (\sqrt {\pi } \cosh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} + 2 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right ) + \sqrt {\pi } \sinh \left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} + e^{2} + 2 \, {\left (2 \, c d - b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right )}{4 \, c \log \left (f\right )} \]

[In]

integrate(f^(c*x^2+b*x+a)*sinh(e*x+d),x, algorithm="fricas")

[Out]

-1/4*(sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 - 2*(2*c*d - b*e)*log(f))/(c*log(f)))
+ sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 - 2*(2*c*d - b*e)*log(f))/(c*log(f))))*erf(1/2*((2*c*x + b)
*log(f) + e)*sqrt(-c*log(f))/(c*log(f))) - sqrt(-c*log(f))*(sqrt(pi)*cosh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 +
 2*(2*c*d - b*e)*log(f))/(c*log(f))) + sqrt(pi)*sinh(-1/4*((b^2 - 4*a*c)*log(f)^2 + e^2 + 2*(2*c*d - b*e)*log(
f))/(c*log(f))))*erf(1/2*((2*c*x + b)*log(f) - e)*sqrt(-c*log(f))/(c*log(f))))/(c*log(f))

Sympy [F]

\[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\int f^{a + b x + c x^{2}} \sinh {\left (d + e x \right )}\, dx \]

[In]

integrate(f**(c*x**2+b*x+a)*sinh(e*x+d),x)

[Out]

Integral(f**(a + b*x + c*x**2)*sinh(d + e*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.84 \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right ) + e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (d - \frac {{\left (b \log \left (f\right ) + e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right ) - e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (-d - \frac {{\left (b \log \left (f\right ) - e\right )}^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} \]

[In]

integrate(f^(c*x^2+b*x+a)*sinh(e*x+d),x, algorithm="maxima")

[Out]

1/4*sqrt(pi)*f^a*erf(sqrt(-c*log(f))*x - 1/2*(b*log(f) + e)/sqrt(-c*log(f)))*e^(d - 1/4*(b*log(f) + e)^2/(c*lo
g(f)))/sqrt(-c*log(f)) - 1/4*sqrt(pi)*f^a*erf(sqrt(-c*log(f))*x - 1/2*(b*log(f) - e)/sqrt(-c*log(f)))*e^(-d -
1/4*(b*log(f) - e)^2/(c*log(f)))/sqrt(-c*log(f))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.09 \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b \log \left (f\right ) - e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} + 4 \, c d \log \left (f\right ) - 2 \, b e \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b \log \left (f\right ) + e}{c \log \left (f\right )}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right )^{2} - 4 \, a c \log \left (f\right )^{2} - 4 \, c d \log \left (f\right ) + 2 \, b e \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, \sqrt {-c \log \left (f\right )}} \]

[In]

integrate(f^(c*x^2+b*x+a)*sinh(e*x+d),x, algorithm="giac")

[Out]

1/4*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2*x + (b*log(f) - e)/(c*log(f))))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^
2 + 4*c*d*log(f) - 2*b*e*log(f) + e^2)/(c*log(f)))/sqrt(-c*log(f)) - 1/4*sqrt(pi)*erf(-1/2*sqrt(-c*log(f))*(2*
x + (b*log(f) + e)/(c*log(f))))*e^(-1/4*(b^2*log(f)^2 - 4*a*c*log(f)^2 - 4*c*d*log(f) + 2*b*e*log(f) + e^2)/(c
*log(f)))/sqrt(-c*log(f))

Mupad [F(-1)]

Timed out. \[ \int f^{a+b x+c x^2} \sinh (d+e x) \, dx=\int f^{c\,x^2+b\,x+a}\,\mathrm {sinh}\left (d+e\,x\right ) \,d x \]

[In]

int(f^(a + b*x + c*x^2)*sinh(d + e*x),x)

[Out]

int(f^(a + b*x + c*x^2)*sinh(d + e*x), x)